Search: a060006 -id:a060006
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A100283
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a(n) = floor(p*(n+1)) - floor(p*(n)) - 1 where p = Padovan plastic number = 1.324718... (cf. A060006).
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+20
1
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0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1
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OFFSET
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0,1
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COMMENTS
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A rabbit-like sequence generated by the Padovan plastic number.
The well-known rabbit sequence is generated by taking the difference between the nearest integer less than phi*(n+1) minus the nearest integer less than phi*(n). If this value is 2, then the n-th rabbit sequence value is one. If this value is 1, the n-th rabbit sequence is 0. The sequence given is calculated in a similar manner, but using the plastic constant = 1.324717957244... instead of phi = 1.618033... = (1+sqrt(5))/2. It is 0001 followed by 11 copies of 001 followed by 0001 followed by 12 copies of 001 followed by 11 copies of 001 followed by similar patterns of 0001 followed by n copies of 001 where n is 11 or 12.
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REFERENCES
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Midhat J. Gazale, Gnomon: From Pharaohs to Fractals, Princeton University Press, 1999
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LINKS
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PROG
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(PARI) p=(sqrt(23/108)+.5)^(1/3) + (abs( sqrt(23/108) -.5))^(1/3); for(n = 0, n = 200, r = floor(p*(n+1)) - floor(p*n) -1; print (r ))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A116397
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Decimal expansion of P^24 where P = plastic constant (A060006).
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+20
0
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8, 5, 3, 0, 2, 5, 7, 9, 1, 9, 1, 9, 1, 9, 6, 2, 4, 8, 8, 8, 9, 5, 4, 2, 6, 2, 5, 7, 0, 8, 4, 0, 1, 5, 3, 3, 6, 7, 3, 5, 3, 8, 3, 9, 8, 6, 3, 3, 5, 1, 7, 9, 7, 0, 3, 9, 9, 3, 7, 0, 7, 8, 2, 4, 5, 9, 4, 6, 5, 5, 1, 1, 6, 0, 5, 6, 8, 6, 3, 3, 0, 5, 7, 2, 1, 4, 0, 7, 4, 5, 7, 8, 3, 6, 2, 3, 6, 9, 0, 0, 1, 7, 1, 0, 1
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OFFSET
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3,1
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LINKS
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EXAMPLE
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853.0257919191962488...
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MATHEMATICA
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RealDigits[(x/.First[Solve[x^3-x-1==0, x]])^24, 10, 120][[1]] (* Harvey P. Dale, Jun 24 2011 *)
Root[ #^3 - 853#^2 - 22# - 1 & , 1] // RealDigits[#, 10, 105]& // First (* Jean-François Alcover, Mar 05 2013 *)
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CROSSREFS
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KEYWORD
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STATUS
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approved
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A167286
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Signature sequence of the smallest Pisot number (A060006).
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+20
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1, 2, 1, 3, 2, 1, 4, 3, 2, 1, 5, 4, 3, 2, 6, 1, 5, 4, 3, 7, 2, 6, 1, 5, 4, 8, 3, 7, 2, 6, 1, 5, 9, 4, 8, 3, 7, 2, 6, 10, 1, 5, 9, 4, 8, 3, 7, 11, 2, 6, 10, 1, 5, 9, 4, 8, 12, 3, 7, 11, 2, 6, 10, 1, 5, 9, 13, 4, 8, 12, 3, 7, 11, 2, 6, 10, 14, 1, 5, 9, 13, 4, 8, 12, 3, 7, 11, 15, 2, 6, 10, 14, 1, 5
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OFFSET
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1,2
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LINKS
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MATHEMATICA
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m = x /. Solve[x^3 - x - 1 == 0, x][[1]]
Take[Transpose[Sort[Flatten[Table[{i + j*m, i}, {i, 25}, {j, 17}], 1], #1[[1]] < #2[[1]] &]][[2]], 95]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A240982
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Decimal expansion of the limit of a recursive sequence connected to the Plastic constant (A060006).
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+20
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1, 8, 1, 6, 8, 8, 3, 4, 2, 4, 2, 4, 4, 7, 4, 0, 3, 1, 2, 4, 4, 8, 1, 8, 8, 2, 0, 2, 2, 2, 4, 8, 0, 7, 4, 5, 2, 9, 6, 5, 9, 2, 1, 7, 5, 7, 7, 5, 8, 7, 3, 4, 2, 3, 1, 5, 8, 1, 2, 5, 2, 9, 1, 6, 7, 0, 3, 9, 4, 7, 1, 7, 7, 1, 6, 0, 4, 1, 5, 3, 6, 7, 7, 5, 8, 0, 5, 7, 8, 6, 8, 7, 9, 6, 3, 9, 2, 3, 9
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OFFSET
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1,2
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REFERENCES
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Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.2.2 Cubic Variations of the Golden Mean, p. 9.
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LINKS
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FORMULA
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psi(1)=1, psi(n) = (1+psi(n-1))^(1/3),
lim_(n -> infinity) (psi0-psi(n))*(3*(1+1/psi0))^n, where psi0 = A060006 = the Plastic constant.
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EXAMPLE
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1.8168834242447403124481882022248074529659217577587342315812529167...
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MATHEMATICA
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digits = 99; n0 = 10; dn = 10; psi0 = A060006 = Root[x^3 - x - 1, x, 1] // N[#, 3*digits]&; Clear[psi, limPsi]; psi[1] = 1; psi[n_] := psi[n] = (1 + psi[n - 1])^(1/3) // N[#, 3*digits]&; limPsi[n_] := limPsi[n] = (psi0 - psi[n])*(3*(1 + 1/psi0))^n; limPsi[n = n0]; limPsi[n = n0 + dn]; While[RealDigits[limPsi[n], 10, digits] != RealDigits[limPsi[n - dn], 10, digits], Print["n = ", n ]; n = n + dn]; RealDigits[limPsi[n], 10, digits] // First
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CROSSREFS
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KEYWORD
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approved
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A000931
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Padovan sequence (or Padovan numbers): a(n) = a(n-2) + a(n-3) with a(0) = 1, a(1) = a(2) = 0.
(Formerly M0284 N0102)
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+10
246
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1, 0, 0, 1, 0, 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200, 265, 351, 465, 616, 816, 1081, 1432, 1897, 2513, 3329, 4410, 5842, 7739, 10252, 13581, 17991, 23833, 31572, 41824, 55405, 73396, 97229, 128801, 170625
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OFFSET
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0,9
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COMMENTS
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Number of compositions of n into parts congruent to 2 mod 3 (offset -1). - Vladeta Jovovic, Feb 09 2005
a(n) is the number of compositions of n into parts that are odd and >= 3. Example: a(10)=3 counts 3+7, 5+5, 7+3. - David Callan, Jul 14 2006
Referred to as N0102 in R. K. Guy's "Anyone for Twopins?" - Rainer Rosenthal, Dec 05 2006
Zagier conjectures that a(n+3) is the maximum number of multiple zeta values of weight n > 1 which are linearly independent over the rationals. - Jonathan Sondow and Sergey Zlobin (sirg_zlobin(AT)mail.ru), Dec 20 2006
Starting with offset 6: (1, 1, 2, 2, 3, 4, 5, ...) = INVERT transform of A106510: (1, 1, -1, 0, 1, -1, 0, 1, -1, ...). - Gary W. Adamson, Oct 10 2008
Starting with offset 7, the sequence 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, ... is called the Fibonacci quilt sequence by Catral et al., in Fib. Q. 2017. - N. J. A. Sloane, Dec 24 2021
Triangle A145462: right border = A000931 starting with offset 6. Row sums = Padovan sequence starting with offset 7. - Gary W. Adamson, Oct 10 2008
Starting with offset 3 = row sums of triangle A146973 and INVERT transform of [1, -1, 2, -2, 3, -3, ...]. - Gary W. Adamson, Nov 03 2008
a(n+5) corresponds to the diagonal sums of "triangle": 1; 1; 1,1; 1,1; 1,2,1; 1,2,1; 1,3,3,1; 1,3,3,1; 1,4,6,4,1; ..., rows of Pascal's triangle (A007318) repeated. - Philippe Deléham, Dec 12 2008
With offset 3: (1, 0, 1, 1, 1, 2, 2, ...) convolved with the tribonacci numbers prefaced with a "1": (1, 1, 1, 2, 4, 7, 13, ...) = the tribonacci numbers, A000073. (Cf. triangle A153462.) - Gary W. Adamson, Dec 27 2008
a(n) is also the number of strings of length (n-8) from an alphabet {A, B} with no more than one A or 2 B's consecutively. (E.g., n = 4: {ABAB,ABBA,BABA,BABB,BBAB} and a(4+8) = 5.) - Toby Gottfried, Mar 02 2010
p(n):=A000931(n+3), n >= 1, is the number of partitions of the numbers {1,2,3,...,n} into lists of length two or three containing neighboring numbers. The 'or' is inclusive. For n=0 one takes p(0)=1. For details see the W. Lang link. There the explicit formula for p(n) (analog of the Binet-de Moivre formula for Fibonacci numbers) is also given. Padovan sequences with different inputs are also considered there. - Wolfdieter Lang, Jun 15 2010
Equals the INVERTi transform of Fibonacci numbers prefaced with three 1's, i.e., (1 + x + x^2 + x^3 + x^4 + 2x^5 + 3x^6 + 5x^7 + 8x^8 + 13x^9 + ...). - Gary W. Adamson, Apr 01 2011
When run backwards gives (-1)^n*A050935(n).
a(n) is the top left entry of the n-th power of the 3 X 3 matrix [0, 0, 1; 1, 0, 1; 0, 1, 0] or of the 3 X 3 matrix [0, 1, 0; 0, 0, 1; 1, 1, 0]. - R. J. Mathar, Feb 03 2014
Figure 4 of Brauchart et al., 2014, shows a way to "visualize the Padovan sequence as cuboid spirals, where the dimensions of each cuboid made up by the previous ones are given by three consecutive numbers in the sequence". - N. J. A. Sloane, Mar 26 2014
a(n) is the number of closed walks from a vertex of a unidirectional triangle containing an opposing directed edge (arc) between the second and third vertices. Equivalently the (1,1) entry of A^n where the adjacency matrix of digraph is A=(0,1,0;0,0,1;1,1,0). - David Neil McGrath, Dec 19 2014
Number of compositions of n-3 (n >= 4) into 2's and 3's. Example: a(12)=5 because we have 333, 3222, 2322, 2232, and 2223. - Emeric Deutsch, Dec 28 2014
The Hoffman (2015) paper "offers significant evidence that the number of quantities needed to generate the weight-n multiple harmonic sums mod p is" a(n). - N. J. A. Sloane, Jun 24 2016
a(n) gives the number of compositions of n-5 into odd parts where the order of the 1's does not matter. For example, a(11)=4 counts the following compositions of 6: (5,1)=(1,5), (3,3), (3,1,1,1)=(1,3,1,1)=(1,1,3,1)=(1,1,1,3), (1,1,1,1,1,1). - Gregory L. Simay, Aug 04 2016
For n > 6, a(n) is the number of maximal matchings in the (n-5)-path graph, maximal independent vertex sets and minimal vertex covers in the (n-6)-path graph, and minimal edge covers in the (n-5)-pan graph and (n-3)-path graphs. - Eric W. Weisstein, Mar 30, Aug 03, and Aug 07 2017
a(2n + 5) + 2n - 4, n > 2, is the number of maximal subsemigroups of the monoid of order-preserving mappings on a set with n elements.
a(n + 6) + n - 3, n > 3, is the number of maximal subsemigroups of the monoid of order-preserving or reversing mappings on a set with n elements.
(End)
Has the property that the largest of any four consecutive terms equals the sum of the two smallest. - N. J. A. Sloane, Aug 29 2017 [David Nacin points out that there are many sequences with this property, such as 1,1,1,2,1,1,1,2,1,1,1,2,... or 2,3,4,5,2,3,4,5,2,3,4,5,... or 2,2,1,3,3, 4,1,4, 5,5,1,6,6, 7,1,7, 8,8,1,9,9, 10,1,10, ... (spaces added for clarity), and a conjecture I made here in 2017 was simply wrong. I have deleted it. - N. J. A. Sloane, Oct 23 2018]
a(n) is also the number of maximal cliques in the (n+6)-path complement graph. - Eric W. Weisstein, Apr 12 2018
a(n+8) is the number of solus bitstrings of length n with no runs of 3 zeros. - Steven Finch, Mar 25 2020
Named after the architect Richard Padovan (b. 1935). - Amiram Eldar, Jun 08 2021
Shannon et al. (2006) credit a French architecture student Gérard Cordonnier with the discovery of these numbers.
For n >= 3, a(n) is the number of sequences of 0s and 1s of length (n-2) that begin with a 0, end with a 0, contain no two consecutive 0s, and contain no three consecutive 1s. - Yifan Xie, Oct 20 2022
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REFERENCES
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A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 47, ex. 4.
Minerva Catral, Pari L. Ford, Pamela E. Harris, Steven J. Miller, Dawn Nelson, Zhao Pan, and Huanzhong Xu, Legal Decompositions Arising from Non-positive Linear Recurrences, Fib. Quart., 55:3 (2017), 252-275. [Note that there is an earlier version of this paper, with only five authors, on the arXiv in 2016. Note to editors: do not merge these two citations. - N. J. A. Sloane, Dec 24 2021]
Richard K. Guy, "Anyone for Twopins?" in D. A. Klarner, editor, The Mathematical Gardner. Prindle, Weber and Schmidt, Boston, 1981, pp. 10-11.
Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
A. G. Shannon, P. G. Anderson and A. F. Horadam, Properties of Cordonnier, Perrin and Van der Laan numbers, International Journal of Mathematical Education in Science and Technology, Volume 37:7 (2006), 825-831. See P_n.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Ian Stewart, L'univers des nombres, "La sculpture et les nombres", pp. 19-20, Belin-Pour La Science, Paris, 2000.
Steven J. Tedford, Combinatorial identities for the Padovan numbers, Fib. Q., Vol. 57, No. 4 (2019), pp. 291-298.
Hans van der Laan, Het plastische getal. XV lessen over de grondslagen van de architectonische ordonnantie. Leiden, E.J. Brill, 1967.
Don Zagier, Values of zeta functions and their applications, in First European Congress of Mathematics (Paris, 1992), Vol. II, A. Joseph et al. (eds.), Birkhäuser, Basel, 1994, pp. 497-512.
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LINKS
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David Applegate, Marc LeBrun and N. J. A. Sloane, Dismal Arithmetic, J. Int. Seq. 14 (2011) # 11.9.8.
Richard K. Guy, Anyone for Twopins?, in D. A. Klarner, editor, The Mathematical Gardner. Prindle, Weber and Schmidt, Boston, 1981, pp. 2-15. [Annotated scanned copy, with permission]
Dov Jarden, Recurring Sequences, Riveon Lematematika, Jerusalem, 1966. [Annotated scanned copy] See p. 90.
Steven J. Miller and Alexandra Newlon, The Fibonacci Quilt Game, arXiv preprint arXiv:1909.01938 [math.NT], 2019. Also Fib. Q., Vol. 58, No. 2 (2020), pp. 157-168. (See Fig. 2, The "Fibonacci Quilt" sequence.)
Richard Padovan, Dom Hans van der Laan and the Plastic Number, Chapter 74, pp. 407-419, Volume II of K. Williams and M. J. Ostwald (eds.), Architecture and Mathematics from Antiquity to the Future, DOI 10.1007/978-3-319-00143-2_27, Springer International Publishing Switzerland, 2015.
Eric Weisstein's World of Mathematics, Pan Graph.
Eric Weisstein's World of Mathematics, Path Graph.
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FORMULA
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G.f.: (1-x^2)/(1-x^2-x^3).
a(n) is asymptotic to r^n / (2*r+3) where r = 1.3247179572447... = A060006, the real root of x^3 = x + 1. - Philippe Deléham, Jan 13 2004
a(n)^2 + a(n+2)^2 + a(n+6)^2 = a(n+1)^2 + a(n+3)^2 + a(n+4)^2 + a(n+5)^2 (Barniville, Question 16884, Ed. Times 1911).
a(n+5) = a(0) + a(1) + ... + a(n).
a(n) = central and lower right terms in the (n-3)-th power of the 3 X 3 matrix M = [0 1 0 / 0 0 1 / 1 1 0]. E.g., a(13) = 7. M^10 = [3 5 4 / 4 7 5 / 5 9 7]. - Gary W. Adamson, Feb 01 2004
G.f.: 1/(1 - x^3 - x^5 - x^7 - x^9 - ...). - Jon Perry, Jul 04 2004
a(n+4) = Sum_{k=0..floor((n-1)/2)} binomial(floor((n+k-2)/3), k). - Paul Barry, Jul 06 2004
a(n+3) is diagonal sum of A026729 (as a number triangle), with formula a(n+3) = Sum_{k=0..floor(n/2)} Sum_{i=0..n-k} (-1)^(n-k+i)*binomial(n-k, i)*binomial(i+k, i-k). - Paul Barry, Sep 23 2004
a(n+3) = Sum_{k=0..floor(n/2)} binomial((n-k)/2, k)(1+(-1)^(n-k))/2. - Paul Barry, Sep 09 2005
The sequence 1/(1-x^2-x^3) (a(n+3)) is given by the diagonal sums of the Riordan array (1/(1-x^3), x/(1-x^3)). The row sums are A000930. - Paul Barry, Feb 25 2005
a(n+5) corresponds to the diagonal sums of A030528. The binomial transform of a(n+5) is A052921. a(n+5) = Sum_{k=0..floor(n/2)} Sum_{k=0..n} (-1)^(n-k+i)*binomial(n-k, i)binomial(i+k+1, 2k+1). - Paul Barry, Jun 21 2004
r^(n-1) = (1/r)*a(n) + r*(n+1) + a(n+2), where r = 1.32471... is the real root of x^3 - x - 1 = 0. Example: r^8 = (1/r)*a(9) + r*a(10) + a(11) = (1/r)*2 + r*3 + 4 = 9.483909... - Gary W. Adamson, Oct 22 2006
a(n) = (r^n)/(2r+3) + (s^n)/(2s+3) + (t^n)/(2t+3) where r, s, t are the three roots of x^3-x-1. - Keith Schneider (schneidk(AT)email.unc.edu), Sep 07 2007
a(n) = -k*a(n-1) + a(n-2) + (k+1)a(n-2) + k*a(n-4), n > 3, for any value of k. - Gary Detlefs, Sep 13 2010
a(0) + a(2) + a(4) + a(6) + ... + a(2*n) = a(2*n+3).
a(0) + a(3) + a(6) + a(9) + ... + a(3*n) = a(3*n+2)+1.
a(0) + a(5) + a(10) + a(15) + ... + a(5*n) = a(5*n+1)+1.
a(0) + a(7) + a(14) + a(21) + ... + a(7*n) = (a(7*n) + a(7*n+1) + 1)/2. (End)
a(n+3) = Sum_{k=0..floor((n+1)/2)} binomial((n+k)/3,k), where binomial((n+k)/3,k)=0 for noninteger (n+k)/3. - Nikita Gogin, Dec 07 2012
a(n) = the k-th difference of a(n+5k) - a(n+5k-1), k>=1. For example, a(10)=3 => a(15)-a(14) => 2nd difference of a(20)-a(19) => 3rd difference of a(25)-a(24)... - Bob Selcoe, Mar 18 2014
Construct the power matrix T(n,j) = [A^*j]*[S^*(j-1)] where A=(0,0,1,0,1,0,1,...) and S=(0,1,0,0,...) or A063524. [* is convolution operation] Define S^*0=I with I=(1,0,0,...). Then a(n) = Sum_{j=1...n} T(n,j). - David Neil McGrath, Dec 19 2014
If x=a(n), y=a(n+1), z=a(n+2), then x^3 + 2*y*x^2 - z^2*x - 3*y*z*x + y^2*x + y^3 - y^2*z + z^3 = 1. - Alexander Samokrutov, Jul 20 2015
For the sequence shifted by 6 terms, a(n) = Sum_{k=ceiling(n/3)..ceiling(n/2)} binomial(k+1,3*k-n) [Doslic-Zubac]. - N. J. A. Sloane, Apr 23 2017
a(2n) = 2*a(n-1)*a(n) + a(n)^2 + a(n+1)^2, for n > 8.
a(2n-1) = 2*a(n)*a(n+1) + a(n-1)^2, for n > 8.
a(2n+1) = 2*a(n+1)*a(n+2) + a(n)^2, for n > 7. (End)
0*a(0) + 1*a(1) + 2*a(2) + ... + n*a(n) = n*a(n+5) - a(n+9) + 2. - Greg Dresden and Zi Ye, Jul 02 2021
2*a(n) = a(n+2) + a(n-5) for n >= 5.
3*a(n) = a(n+4) - a(n-9) for n >= 9.
4*a(n) = a(n+5) - a(n-9) for n >= 9. (End)
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EXAMPLE
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G.f. = 1 + x^3 + x^5 + x^6 + x^7 + 2*x^8 + 2*x^9 + 3*x^10 + 4*x^11 + ...
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MAPLE
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A000931 := proc(n) option remember; if n = 0 then 1 elif n <= 2 then 0 else procname(n-2)+procname(n-3); fi; end;
A000931:=-(1+z)/(-1+z^2+z^3); # Simon Plouffe in his 1992 dissertation; gives sequence without five leading terms
a[0]:=1; a[1]:=0; a[2]:=0; for n from 3 to 50 do a[n]:=a[n-2]+a[n-3]; end do; # Francesco Daddi, Aug 04 2011
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MATHEMATICA
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CoefficientList[Series[(1-x^2)/(1-x^2-x^3), {x, 0, 50}], x]
a[0]=1; a[1]=a[2]=0; a[n_]:= a[n]= a[n-2] + a[n-3]; Table[a[n], {n, 0, 50}] (* Robert G. Wilson v, May 04 2006 *)
LinearRecurrence[{0, 1, 1}, {1, 0, 0}, 50] (* Harvey P. Dale, Jan 10 2012 *)
Table[RootSum[-1 -# +#^3 &, 5#^n -6#^(n+1) +4#^(n+2) &]/23, {n, 0, 50}] (* Eric W. Weisstein, Nov 09 2017 *)
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PROG
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(Haskell)
a000931 n = a000931_list !! n
a000931_list = 1 : 0 : 0 : zipWith (+) a000931_list (tail a000931_list)
(PARI) {a(n) = if( n<0, polcoeff(1/(1+x-x^3) + x * O(x^-n), -n), polcoeff( (1 - x^2)/(1-x^2-x^3) + x * O(x^n), n))}; /* Michael Somos, Sep 18 2012 */
(Magma) I:=[1, 0, 0]; [n le 3 select I[n] else Self(n-2) + Self(n-3): n in [1..60]]; // Vincenzo Librandi, Jul 21 2015
(Sage)
P.<x> = PowerSeriesRing(ZZ, prec)
return P( (1-x^2)/(1-x^2-x^3) ).list()
(GAP) a:=[1, 0, 0];; for n in [4..50] do a[n]:=a[n-2]+a[n-3]; od; a; # G. C. Greubel, Dec 30 2019
(Python)
def aupton(nn):
alst = [1, 0, 0]
for n in range(3, nn+1): alst.append(alst[n-2]+alst[n-3])
return alst
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021
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STATUS
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approved
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A001608
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Perrin sequence (or Ondrej Such sequence): a(n) = a(n-2) + a(n-3) with a(0) = 3, a(1) = 0, a(2) = 2.
(Formerly M0429 N0163)
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+10
73
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3, 0, 2, 3, 2, 5, 5, 7, 10, 12, 17, 22, 29, 39, 51, 68, 90, 119, 158, 209, 277, 367, 486, 644, 853, 1130, 1497, 1983, 2627, 3480, 4610, 6107, 8090, 10717, 14197, 18807, 24914, 33004, 43721, 57918, 76725, 101639, 134643, 178364, 236282, 313007
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OFFSET
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0,1
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COMMENTS
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Has been called the skiponacci sequence or skiponacci numbers. - N. J. A. Sloane, May 24 2013
For n >= 3, also the numbers of maximal independent vertex sets, maximal matchings, minimal edge covers, and minimal vertex covers in the n-cycle graph C_n. - Eric W. Weisstein, Mar 30 2017 and Aug 03 2017
With the terms indexed as shown, has property that p prime => p divides a(p). The smallest composite n such that n divides a(n) is 521^2. For quotients a(p)/p, where p is prime, see A014981.
Asymptotically, a(n) ~ r^n, with r=1.3247179572447... the inverse of the real root of 1-x^2-x^3=0 (see A060006). If n>9 then a(n)=round(r^n). - Ralf Stephan, Dec 13 2002
The recursion can be used to compute a(-n). The result is -A078712(n). - T. D. Noe, Oct 10 2006
For n>=3, a(n) is the number of maximal independent sets in a cycle of order n. - Vincent Vatter, Oct 24 2006
Let r1, r2 and r3 denote the roots of x^3 - x - 1. Then the following identity holds: a(k*n) + (a(k))^n - (a(k) - r1^k)^n - (a(k) - r2^k)^n - (a(k) - r3^k)^n
= 0 for n = 0, 1, 2,
= 6 for n = 3,
= 12*a(k) for n = 4,
= 10*[2*(a(k))^2 - a(-k)] for n = 5,
= 30*a(k)*[(a(k))^2 - a(-k)] for n = 6,
= 7*[6*(a(k))^4 - 9*a(-k)*(a(k))^2 + 2*(a(-k))^2 - a(k)] for n = 7,
= 56*a(k)*[((a(k))^2 - a(-k))^2 - a(k)/2] for n = 8,
where a(-k) = -A078712(k) and the formula (5.40) from the paper of Witula and Slota is used. (End)
The parity sequence of a(n) is periodic with period 7 and has the form (1,0,0,1,0,1,1). Hence we get that a(n) and a(2*n) are congruent modulo 2. Similarly we deduce that a(n) and a(3*n) are congruent modulo 3. Is it true that a(n) and a(p*n) are congruent modulo p for every prime p? - Roman Witula, Feb 09 2013
The trinomial x^3 - x - 1 divides the polynomial x^(3*n) - a(n)*x^(2*n) + ((a(n)^2 - a(2*n))/2)*x^n - 1 for every n>=1. For example, for n=3 we obtain the factorization x^9 - 3*x^6 + 2*x^3 - 1 = (x^3 - x - 1)*(x^6 + x^4 - 2*x^3 + x^2 - x + 1). Sketch of the proof: Let p,s,t be roots of the Perrin polynomial x^3 - x - 1. Then we have (a(n))^2 = (p^n + s^n + t^n)^2 = a(2*n) + 2*a(n)*x^n -2*x^n + 2/x^n for every x = p,s,t, i.e., x^(3*n) - a(n)*x^(2*n) + ((a(n)^2 - a(2*n))/2)*x^n - 1 = 0 for every x = p,s,t, which finishes the proof. By discussion of the power(a(n))^3 = (p^n + s^n + t^n)^3 it can be deduced that the trinomial x^3 - x - 1 divides the polynomial 2*x^(4*n) - a(n)*x^(3*n) - a(2*n)*x^(2*n) + ((a(n)^3 - a(3*n) - 3)/3)*x^n - a(n) = 0. Co-author of these divisibility relations is also my young student Szymon Gorczyca (13 years old as of 2013). - Roman Witula, Feb 09 2013
The sum of powers of the real root and complex roots of x^3-x-1=0 as expressed as powers of the plastic number r, (see A060006). Let r0=1, r1=r, r2=1+r^(-1) and c0=2, c1=-r and c3 = r^(-5) then a(n) = r(n-2)+r(n-3) + c(n-2)+c(n-3). Example: a(5) = 1 + r^(-1) + 1 + r + 2 - r + r^(-5) = 4 + r^(-1) + r^(-5) = 5. - Richard Turk, Jul 14 2016
Also the number of minimal total dominating sets in the n-sun graph. - Eric W. Weisstein, Apr 27 2018
Named after the French engineer François Olivier Raoul Perrin (1841-1910). - Amiram Eldar, Jun 05 2021
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REFERENCES
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Olivier Bordellès, Thèmes d'Arithmétique, Ellipses, 2006, Exercice 4.11, p. 127.0
Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.2.2.
Dmitry Fomin, On the properties of a certain recursive sequence, Mathematics and Informatics Quarterly, Vol. 3 (1993), pp. 50-53.
Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 70.
Manfred Schroeder, Number Theory in Science and Communication, 3rd ed., Springer, 1997.
A. G. Shannon, P. G. Anderson and A. F. Horadam, Properties of Cordonnier, Perrin and Van der Laan numbers, International Journal of Mathematical Education in Science and Technology, Volume 37:7 (2006), 825-831. See Q_n.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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J. Chick, Problem 81G, Math. Gazette, Vol. 81, No. 491 (1997), p. 304.
E. B. Escott, Problem 151, Amer. Math. Monthly, Vol. 15, No. 11 (1908), p. 209.
Christian Holzbaur, Perrin pseudoprimes [Original link broke many years ago. This is a cached copy from the WayBack machine, dated Apr 24 2006]
Dov Jarden, Recurring Sequences, Riveon Lematematika, Jerusalem, 1966. [Annotated scanned copy] See p. 90.
Mathilde Noual, Dynamics of Circuits and Intersecting Circuits, in Language and Automata Theory and Applications, Lecture Notes in Computer Science, 2012, Volume 7183/2012, 433-444, DOI; also on arXiv, arXiv 1011.3930 [cs.DM], 2010.
R. Perrin, Query 1484, L'Intermédiaire des Mathématiciens, Vol. 6 (1899), p. 76.
Razvan Tudoran, Problem 653, College Math. J., Vol. 31, No. 3 (2000), pp. 223-224.
Eric Weisstein's World of Mathematics, Sun Graph
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FORMULA
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G.f.: (3 - x^2)/(1 - x^2 - x^3). - Simon Plouffe in his 1992 dissertation
a(n) = r1^n + r2^n + r3^n where r1, r2, r3 are three roots of x^3-x-1=0.
a(n-1) + a(n) + a(n+1) = a(n+4), a(n) - a(n-1) = a(n-5). - Jon Perry, Jun 05 2003
a(n) = trace(M^n) where M is the 3 X 3 matrix [0 1 0 / 0 0 1 / 1 1 0], the companion matrix of the characteristic polynomial of this sequence, P = X^3 - X - 1.
M^n * [3, 0, 2] = [a(n), a(n+1), a(n+2)]; e.g., M^7 * [3, 0, 2] = [7, 10, 12].
a(0) + a(1) + a(2) + ... + a(n) = a(n+5) - 2.
a(0) + a(2) + a(4) + ... + a(2*n) = a(2*n+3).
a(1) + a(3) + a(5) + ... + a(2*n+1) = a(2*n+4) - 2. (End)
a(0) + a(3) + a(6) + a(9) + ... + a(3*n) = a(3*n+2) + 1.
a(0) + a(5) + a(10) + a(15) + ... + a(5*n) = a(5*n+1)+3.
a(0) + a(7) + a(14) + a(21) + ... + a(7*n) = (a(7*n) + a(7*n+1) + 3)/2. (End)
a(n) = n*Sum_{k=1..floor(n/2)} binomial(k,n-2*k)/k, n > 0, a(0)=3. - Vladimir Kruchinin, Oct 21 2011
(a(n)^3)/2 + a(3n) - 3*a(n)*a(2n)/2 - 3 = 0. - Richard Turk, Apr 26 2017
2*a(4n) - 2*a(n) - 2*a(n)*a(3n) - a(2n)^2 + a(n)^2*a(2n) = 0. - Richard Turk, May 02 2017
a(n)^4 + 6*a(4n) - 4*a(3n)*a(n) - 3*a(2n)^2 - 12a(n) = 0. - Richard Turk, May 02 2017
a(n+5)^2 + a(n+1)^2 - a(n)^2 = a(2*(n+5)) + a(2*(n+1)) - a(2*n). - Aleksander Bosek, Mar 04 2019
a(n+12) = a(n) + 2*a(n+4) + a(n+11);
a(n+16) = a(n) + 4*a(n+9) + a(n+13);
a(n+18) = a(n) + 2*a(n+6) + 5*a(n+12);
a(n+21) = a(n) + 2*a(n+12) + 6*a(n+14);
a(n+27) = a(n) + 3*a(n+9) + 4*a(n+22). (End)
a(n) = Sum_{j=0..floor((n-g)/(2*g))} 2*n/(n-2*(g-2)*j-(g-2)) * Hypergeometric2F1([-(n-2g*j-g)/2, -(2j+1)], [1], 1), g = 3 and n an odd integer. - Richard Turk, Oct 14 2019
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EXAMPLE
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We note that if a + b + c = 0 then:
1) a^3 + b^3 + c^3 = 3*a*b*c,
2) a^4 + b^4 + c^4 = 2*((a^2 + b^2 + c^2)/2)^2,
3) (a^5 + b^5 + c^5)/5 = (a^3 + b^3 + c^3)/3 * (a^2 +
b^2 + c^2)/2,
4) (a^7 + b^7 + c^7)/7 = (a^5 + b^5 + c^5)/5 * (a^2 + b^2 + c^2)/2 = 2*(a^3 + b^3 + c^3)/3 * (a^4 + b^4 + c^4)/4,
5) (a^7 + b^7 + c^7)/7 * (a^3 + b^3 + c^3)/3 = ((a^5 + b^5 + c^5)/5)^2.
Hence, by the Binet formula for a(n) we obtain the relations: a(3) = 3, a(4) = 2*(a(2)/2)^2 = 2, a(5)/5 = a(3)/3 * a(2)/2, i.e., a(5) = 5, and similarly that a(7) = 7. (End)
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MAPLE
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A001608 :=proc(n) option remember; if n=0 then 3 elif n=1 then 0 elif n=2 then 2 else procname(n-2)+procname(n-3); fi; end proc;
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MATHEMATICA
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LinearRecurrence[{0, 1, 1}, {3, 0, 2}, 50] (* Harvey P. Dale, Jun 26 2011 *)
per = Solve[x^3 - x - 1 == 0, x]; f[n_] := Floor @ Re[N[ per[[1, -1, -1]]^n + per[[2, -1, -1]]^n + per[[3, -1, -1]]^n]]; Array[f, 46, 0] (* Robert G. Wilson v, Jun 29 2010 *)
CoefficientList[Series[(3 - x^2)/(1 - x^2 - x^3), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 03 2015 *)
Table[RootSum[-1 - # + #^3 &, #^n &], {n, 0, 20}] (* Eric W. Weisstein, Mar 30 2017 *)
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PROG
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(PARI) a(n)=if(n<0, 0, polsym(x^3-x-1, n)[n+1])
(Haskell)
a001608 n = a000931_list !! n
a001608_list = 3 : 0 : 2 : zipWith (+) a001608_list (tail a001608_list)
(Python)
A001608_list, a, b, c = [3, 0, 2], 3, 0, 2
for _ in range(100):
a, b, c = b, c, a+b
(GAP) a:=[3, 0, 2];; for n in [4..20] do a[n]:=a[n-2]+a[n-3]; od; a; # Muniru A Asiru, Jul 12 2018
(Magma) I:=[3, 0, 2]; [n le 3 select I[n] else Self(n-2) +Self(n-3): n in [1..50]]; // G. C. Greubel, Mar 18 2019
(Sage) ((3-x^2)/(1-x^2-x^3)).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Mar 18 2019
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CROSSREFS
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Cf. A013998 (Unrestricted Perrin pseudoprimes).
Cf. A018187 (Restricted Perrin pseudoprimes).
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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Additional comments from Mike Baker, Oct 11 2005
Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021
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STATUS
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approved
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A095263
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a(n+3) = 3*a(n+2) - 2*a(n+1) + a(n).
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+10
28
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1, 3, 7, 16, 37, 86, 200, 465, 1081, 2513, 5842, 13581, 31572, 73396, 170625, 396655, 922111, 2143648, 4983377, 11584946, 26931732, 62608681, 145547525, 338356945, 786584466, 1828587033, 4250949112, 9882257736, 22973462017, 53406819691
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OFFSET
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1,2
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COMMENTS
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a(n+1) = number of n-tuples over {0,1,2} without consecutive digits. For the general case see A096261.
Diagonal sums of Riordan array (1/(1-x)^3, x/(1-x^3)), A127893. - Paul Barry, Jan 07 2008
The signed variant (-1)^(n+1)*a(n+1) is the bottom right entry of the n-th power of the matrix [[0,1,0],[0,0,1],[-1,-2,-3]]. - Roger L. Bagula, Jul 01 2007
a(n) is the number of generalized compositions of n+1 when there are i^2/2-i/2 different types of i, (i=1,2,...). - Milan Janjic, Sep 24 2010
Dedrickson (Section 4.1) gives a bijection between colored compositions of n, where each part k has one of binomial(k,2) colors, and 0,1,2 strings of length n-2 without sequential digits (i.e., avoiding 01 and 12). Cf. A052529. - Peter Bala, Sep 17 2013
Except for the initial 0, this is the p-INVERT of (1,1,1,1,1,...) for p(S) = 1 - S^2 - S^3; see A291000. - Clark Kimberling, Aug 24 2017
For n>1, a(n-1) is the number of ways to split [n] into an unspecified number of intervals and then choose 2 blocks (i.e., subintervals) from each interval. For example, for n=6, a(5)=37 since the number of ways to split [6] into intervals and then select 2 blocks from each interval is C(6,2) + C(4,2)*C(2,2) + C(3,2)*C(3,2) + C(2,2)*C(4,2) + C(2,2)*C(2,2)*C(2,2). - Enrique Navarrete, May 20 2022
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LINKS
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FORMULA
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Let M = the 3 X 3 matrix [0 1 0 / 0 0 1 / 1 -2 3]; then M^n *[1 0 0] = [a(n-2) a(n-1) a(n)].
a(n)/a(n-1) tends to 2.3247179572..., an eigenvalue of M and a root of the characteristic polynomial. [Is that constant equal to 1 + A060006? - Michel Marcus, Oct 11 2014] [Yes, the limit is the root of the equation -1 + 2*x - 3*x^2 + x^3 = 0, after substitution x = y + 1 we have the equation for y: -1 - y + y^3 = 0, y = A060006. - Vaclav Kotesovec, Jan 27 2015]
Related to the Padovan sequence A000931 as follows : a(n)=A000931(3n+4). Also the binomial transform of A000931(n+4).
a(n) = Sum_{k=0..floor((n+1)/2)} binomial(n+k, n-2*k+1).
a(n) = Sum_{k=0..floor((n+1)/2)} binomial(n+k, 3*k-1). (End)
G.f.: x/(1 -3*x +2*x^2 -x^3).
a(n) = Sum_{k=0..floor(n/2)} binomial(n+k+2,3*k+2).
a(n) = Sum_{k=0..n} binomial(n,k) * Sum_{j=0..floor((k+4)/2)} binomial(j,k-2j+4). (End)
If p[i]=i(i-1)/2 and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=2, a(n-1)=det A. - Milan Janjic, May 02 2010
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EXAMPLE
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a(9) = 1081 = 3*465 - 2*200 + 86.
M^9 * [1 0 0] = [a(7) a(8) a(9)] = [200 465 1081].
G.f. = x + 3*x^2 + 7*x^3 + 16*x^4 + 37*x^5 + 86*x^6 + 200*x^7 + ...
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MAPLE
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A:= gfun:-rectoproc({a(n+3)=3*a(n+2)-2*a(n+1)+a(n), a(1)=1, a(2)=3, a(3)=7}, a(n), remember):
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MATHEMATICA
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a[1]=1; a[2]=3; a[3]=7; a[n_]:= a[n]= 3a[n-1] -2a[n-2] +a[n-3]; Table[a[n], {n, 22}] (* Or *)
a[n_]:= (MatrixPower[{{0, 1, 2, 3}, {1, 2, 3, 0}, {2, 3, 0, 1}, {3, 0, 1, 2}}, n].{{1}, {0}, {0}, {0}})[[2, 1]]; Table[ a[n], {n, 22}] (* Robert G. Wilson v, Jun 16 2004 *)
RecurrenceTable[{a[1]==1, a[2]==3, a[3]==7, a[n+3]==3a[n+2]-2a[n+1]+a[n]}, a, {n, 30}] (* Harvey P. Dale, Sep 17 2022 *)
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PROG
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(Magma) I:=[1, 3, 7]; [n le 3 select I[n] else 3*Self(n-1) -2*Self(n-2) +Self(n-3): n in [1..30]]; // G. C. Greubel, Apr 12 2021
(Sage) [sum( binomial(n+k+1, 3*k+2) for k in (0..(n-1)//2)) for n in (1..30)] # G. C. Greubel, Apr 12 2021
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A075778
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Decimal expansion of the real root of x^3 + x^2 - 1.
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+10
24
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7, 5, 4, 8, 7, 7, 6, 6, 6, 2, 4, 6, 6, 9, 2, 7, 6, 0, 0, 4, 9, 5, 0, 8, 8, 9, 6, 3, 5, 8, 5, 2, 8, 6, 9, 1, 8, 9, 4, 6, 0, 6, 6, 1, 7, 7, 7, 2, 7, 9, 3, 1, 4, 3, 9, 8, 9, 2, 8, 3, 9, 7, 0, 6, 4, 6, 0, 8, 0, 6, 5, 5, 1, 2, 8, 0, 8, 1, 0, 9, 0, 7, 3, 8, 2, 2, 7, 0, 9, 2, 8, 4, 2, 2, 5, 0, 3, 0, 3, 6, 4, 8, 3, 7, 7
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OFFSET
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0,1
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COMMENTS
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Also decimal expansion of the root of x^(1/sqrt(x+1)) = (1/sqrt(x+1))^x. The root of (1/x)^(1/sqrt(x+1)) = (1/sqrt(x+1))^(1/x) is the golden ratio. - Michel Lagneau, Apr 17 2012
The following decomposition holds true: X^3 + X^2 - 1 = (X - r)*(X + i * e^(-i*a) * r^(-1/2))*(X - i * e^(i*a) * r^(-1/2)), where a = arcsin(1/(2*r^(3/2))), see A218197 for the decimal expansion of a and the paper of Witula et al. for details. - Roman Witula, Oct 22 2012
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REFERENCES
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Roman Witula, E. Hetmaniok, D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, submitted to Proceedings of the 15th International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012.
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LINKS
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FORMULA
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Let 0 < a < 1 be any real number. Then a is the lesser and 1 is the greater and a^2/1 = 1/(a+1) and a^3 + a^2 - 1 = 0. Solving this using PARI we have 0.7548776662466927600495088964... . The general cubic can also be solved in radicals.
Equals -(1/3) + (1/3)*(25/2 - (3*sqrt(69))/2)^(1/3) + (1/3)*((1/2)*(25 + 3*sqrt(69)))^(1/3).
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EXAMPLE
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0.7548776662466927600495088963585286918946066...
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MAPLE
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1/3-root[3](25/2-3*sqrt(69)/2)/3 -root[3](25/2+3*sqrt(69)/2)/3;
-% ;
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MATHEMATICA
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RealDigits[N[Solve[x^3 + x^2 - 1 == 0, x] [[1]] [[1, 2]], 111]] [[1]]
RealDigits[x /. FindRoot[x^3 + x^2 == 1, {x, 1}, WorkingPrecision -> 120]][[1]] (* Harvey P. Dale, Nov 23 2012 *)
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PROG
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(PARI) solve(x=0, 1, x^3+x^2-1)
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A092526
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Decimal expansion of (2/3)*cos( (1/3)*arccos(29/2) ) + 1/3, the real root of x^3 - x^2 - 1.
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+10
24
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1, 4, 6, 5, 5, 7, 1, 2, 3, 1, 8, 7, 6, 7, 6, 8, 0, 2, 6, 6, 5, 6, 7, 3, 1, 2, 2, 5, 2, 1, 9, 9, 3, 9, 1, 0, 8, 0, 2, 5, 5, 7, 7, 5, 6, 8, 4, 7, 2, 2, 8, 5, 7, 0, 1, 6, 4, 3, 1, 8, 3, 1, 1, 1, 2, 4, 9, 2, 6, 2, 9, 9, 6, 6, 8, 5, 0, 1, 7, 8, 4, 0, 4, 7, 8, 1, 2, 5, 8, 0, 1, 1, 9, 4, 9, 0, 9, 2, 7, 0, 0, 6, 4, 3, 8
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OFFSET
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1,2
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COMMENTS
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This is the limit x of the ratio N(n+1)/N(n) for n -> infinity of the Narayana sequence N(n) = A000930(n). The real root of x^3 - x^2 - 1. See the formula section. - Wolfdieter Lang, Apr 24 2015
This is the fourth smallest Pisot number. - Iain Fox, Oct 13 2017
Sometimes called the supergolden ratio or Narayana's cows constant, and denoted by the symbol psi. - Ed Pegg Jr, Feb 01 2019
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REFERENCES
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S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.2.3.
Paul J. Nahin, The Logician and the Engineer, How George Boole and Claude Shannon Created the Information Age, Princeton University Press, Princeton and Oxford, 2013, Chap. 7: Some Combinational Logic Examples, Section 7.1: Channel Capacity, Shannon's Theorem, and Error-Detection Theory, page 120.
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LINKS
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FORMULA
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The only real irrational root of x^4-x^2-x-1 (-1 is also a root). [Nahim]
Equals (2/3)*cos( (1/3)*arccos(29/2) ) + 1/3.
Equals (1/6)*(116+12*sqrt(93))^(1/3) + 2/(3*(116+12*sqrt(93))^(1/3)) + 1/3. - Vaclav Kotesovec, Dec 18 2014
Equals (1 + 1/r + r)/3 where r = ((29 + sqrt(837))/2)^(1/3). - Peter Luschny, Apr 04 2020
Equals (1/3)*(1 + ((1/2)*(29 + (3*sqrt(93))))^(1/3) + ((1/2)*(29 - 3*sqrt(93)))^(1/3)). See A075778. - Wolfdieter Lang, Aug 17 2022
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EXAMPLE
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1.46557123187676802665673122521993910802557756847228570164318311124926...
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MATHEMATICA
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RealDigits[(2 Cos[ ArcCos[ 29/2]/3] + 1)/3, 10, 111][[1]] (* Robert G. Wilson v, Apr 12 2004 *)
RealDigits[ Solve[ x^3 - x^2 - 1 == 0, x][[1, 1, 2]], 10, 111][[1]] (* Robert G. Wilson v, Oct 10 2013 *)
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PROG
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(PARI) allocatemem(932245000); default(realprecision, 20080); x=solve(x=1, 2, x^3 - x^2 - 1); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b092526.txt", n, " ", d)); \\ Harry J. Smith, Jun 21 2009
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A086106
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Decimal expansion of positive root of x^4 - x^3 - 1 = 0.
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+10
17
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1, 3, 8, 0, 2, 7, 7, 5, 6, 9, 0, 9, 7, 6, 1, 4, 1, 1, 5, 6, 7, 3, 3, 0, 1, 6, 9, 1, 8, 2, 2, 7, 3, 1, 8, 7, 7, 8, 1, 6, 6, 2, 6, 7, 0, 1, 5, 5, 8, 7, 6, 3, 0, 2, 5, 4, 1, 1, 7, 7, 1, 3, 3, 1, 2, 1, 1, 2, 4, 9, 5, 7, 4, 1, 1, 8, 6, 4, 1, 5, 2, 6, 1, 8, 7, 8, 6, 4, 5, 6, 8, 2, 4, 9, 0, 3, 5, 5, 0, 9, 3, 7
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OFFSET
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1,2
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COMMENTS
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Also the growth constant of the Fibonacci 3-numbers A003269 [Stakhov et al.]. - R. J. Mathar, Nov 05 2008
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LINKS
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FORMULA
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Equals (1 + (A^2 + sqrt(A^4 - 16*u*A^2 + 2*A))/A)/4 with A = sqrt(8*u + 3/2), u = (-(Bp/2)^(1/3) + (Bm/2)^(1/3)*(1 - sqrt(3)*i)/2 - 3/8)/6, with Bp = 27 + 3*sqrt(3*283), Bm = 27 - 3*sqrt(3*283), and i = sqrt(-1). (Standard computation of a quartic.) The other (negative) real root -A230151 is obtained by using in the first formula the negative square root. The other two complex roots are obtained by replacing A by -A in these two formulas. - Wolfdieter Lang, Aug 19 2022
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EXAMPLE
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1.380277569...
The four solutions are the present one, -A230151, and the two complex ones 0.2194474721... - 0.9144736629...*i and its complex conjugate. - Wolfdieter Lang, Aug 19 2022
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MATHEMATICA
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RealDigits[Root[ -1 - #1^3 + #1^4 &, 2], 10, 110][[1]]
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PROG
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(PARI) default(realprecision, 20080); x=solve(x=1, 2, x^4 - x^3 - 1); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b086106.txt", n, " ", d)); \\ Iain Fox, Oct 23 2017
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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