OFFSET
1,2
COMMENTS
These are also the numbers that cannot be written as i*j + i + j (i,j >= 1). - Rainer Rosenthal, Jun 24 2001; Henry Bottomley, Jul 06 2002
The values of k for which Sum_{j=0..n} ((-1)^j*binomial(k, j)*binomial(k-1-j, n-j)/(j+1) produces an integer for all n such that n < k. Setting k=10 yields [0, 1, 4, 11, 19, 23, 19, 11, 4, 1, 0] for n = [ -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], so 10 is in the sequence. Setting k=3 yields [0, 1, 1/2, 1/2] for n = [ -1, 0, 1, 2], so 3 is not in the sequence. - Dug Eichelberger (dug(AT)mit.edu), May 14 2001
n such that x^n + x^(n-1) + x^(n-2) + ... + x + 1 is irreducible. - Robert G. Wilson v, Jun 22 2002
Records for Euler totient function phi.
Together with 0, n such that (n+1) divides (n!+1). - Benoit Cloitre, Aug 20 2002; corrected by Charles R Greathouse IV, Apr 20 2010
n such that phi(n^2) = phi(n^2 + n). - Jon Perry, Feb 19 2004
Numbers having only the trivial perfect partition consisting of a(n) 1's. - Lekraj Beedassy, Jul 23 2006
Numbers n such that the sequence {binomial coefficient C(k,n), k >= n } contains exactly one prime. - Artur Jasinski, Dec 02 2007
From Reinhard Zumkeller, Jul 10 2009: (Start)
The first N terms can be generated by the following sieving process:
start with {1, 2, 3, 4, ..., N - 1, N};
for i := 1 until SQRT(N) do
(if (i is not striked out) then
(for j := 2 * i + 1 step i + 1 until N do
(strike j from the list)));
remaining numbers = {a(n): a(n) <= N}. (End)
a(n) = partial sums of A075526(n-1) = Sum_{1..n} A075526(n-1) = Sum_{1..n} (A008578(n+1) - A008578(n)) = Sum_{1..n} (A158611(n+2) - A158611(n+1)) for n >= 1. - Jaroslav Krizek, Aug 04 2009
A171400(a(n)) = 1 for n <> 2: subsequence of A171401, except for a(2) = 2. - Reinhard Zumkeller, Dec 08 2009
Numerator of (1 - 1/prime(n)). - Juri-Stepan Gerasimov, Jun 05 2010
Numbers n such that A002322(n+1) = n. This statement is stronger than repeating the property of the entries in A002322, because it also says in reciprocity that this sequence here contains no numbers beyond the Carmichael numbers with that property. - Michel Lagneau, Dec 12 2010
prime(a(n)) + prime(k) < prime(a(k) + k) for at least one k <= a(n): A212210(a(n),k) < 0. - Reinhard Zumkeller, May 05 2012
Except for the first term, numbers n such that the sum of first n natural numbers does not divide the product of first n natural numbers; that is, n*(n + 1)/2 does not divide n!. - Jayanta Basu, Apr 24 2013
BigOmega(a(n)) equals BigOmega(a(n)*(a(n) + 1)/2), where BigOmega = A001222. Rationale: BigOmega of the product on the right hand side factorizes as BigOmega(a/2) + Bigomega(a+1) = BigOmega(a/2) + 1 because a/2 and a + 1 are coprime, because BigOmega is additive, and because a + 1 is prime. Furthermore Bigomega(a/2) = Bigomega(a) - 1 because essentially all 'a' are even. - Irina Gerasimova, Jun 06 2013
Record values of A060681. - Omar E. Pol, Oct 26 2013
Deficiency of n-th prime. - Omar E. Pol, Jan 30 2014
Conjecture: All the sums Sum_{k=s..t} 1/a(k) with 1 <= s <= t are pairwise distinct. In general, for any integers d >= -1 and m > 0, if Sum_{k=i..j} 1/(prime(k)+d)^m = Sum_{k=s..t} 1/(prime(k)+d)^m with 0 < i <= j and 0 < s <= t then we must have (i,j) = (s,t), unless d = m = 1 and {(i,j),(s,t)} = {(4,4),(8,10)} or {(4,7),(5,10)}. (Note that 1/(prime(8)+1)+1/(prime(9)+1)+1/(prime(10)+1) = 1/(prime(4)+1) and Sum_{k=5..10} 1/(prime(k)+1) = 1/(prime(4)+1) + Sum_{k=5..7} 1/(prime(k)+1).) - Zhi-Wei Sun, Sep 09 2015
Numbers n such that (prime(i)^n + n) is divisible by (n+1), for all i >= 1, except when prime(i) = n+1. - Richard R. Forberg, Aug 11 2016
a(n) is the period of Fubini numbers (A000670) over the n-th prime. - Federico Provvedi, Nov 28 2020
REFERENCES
Archimedeans Problems Drive, Eureka, 40 (1979), 28.
Harvey Dubner, Generalized Fermat primes, J. Recreational Math., 18 (1985): 279-280.
M. Gardner, The Colossal Book of Mathematics, pp. 31, W. W. Norton & Co., NY, 2001.
M. Gardner, Mathematical Circus, pp. 251-2, Alfred A. Knopf, NY, 1979.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
T. D. Noe, Table of n, a(n) for n = 1..10000
Thomas F. Bloom, Unit fractions with shifted prime denominators, arXiv:2305.02689 [math.NT], 2023.
R. P. Boas & N. J. A. Sloane, Correspondence, 1974
Harvey Dubner, Generalized Fermat primes, J. Recreational Math. 18.4 (1985-1986), 279. (Annotated scanned copy)
Armel Mercier, Problem E 3065, American Mathematical Monthly, 1984, p. 649.
Armel Mercier, S. K. Rangarajan, J. C. Binz and Dan Marcus, Problem E 3065, American Mathematical Monthly, No. 4, 1987, pp. 378.
Poo-Sung Park, Additive uniqueness of PRIMES-1 for multiplicative functions, arXiv:1708.03037 [math.NT], 2017.
J. R. Rickard and J. J. Hitchcock, Problem Drive 4, Archimedeans Problems Drive, Eureka, 40 (1979), 28-29, 40. (Annotated scanned copy)
FORMULA
a(n) = (p-1)! mod p where p is the n-th prime, by Wilson's theorem. - Jonathan Sondow, Jul 13 2010
a(n) = A000040(n) - 1. - Omar E. Pol, Oct 26 2013
MAPLE
for n from 2 to 271 do if (n! mod n^2 = n*(n-1) and (n<>4) then print(n-1) fi od; # Gary Detlefs, Sep 10 2010
# alternative
A006093 := proc(n)
ithprime(n)-1 ;
end proc:
seq(A006093(n), n=1..100) ; # R. J. Mathar, Feb 06 2019
MATHEMATICA
Table[Prime[n] - 1, {n, 1, 30}] (* Vladimir Joseph Stephan Orlovsky, Apr 27 2008 *)
a[ n_] := If[ n < 1, 0, -1 + Prime @ n] (* Michael Somos, Jul 17 2011 *)
Prime[Range[60]] - 1 (* Alonso del Arte, Oct 26 2013 *)
PROG
(PARI) isA006093(n) = isprime(n+1) \\ Michael B. Porter, Apr 09 2010
(PARI) A006093(n) = prime(n)-1 \\ Michael B. Porter, Apr 09 2010
(PARI) \\ Sieve as described in Rainer Rosenthal's comment.
m=270; s=vector(m); for(i=1, m, for(j=i, m, k=i*j+i+j; if(k<=m, s[k]=1))); for(k=1, m, if(s[k]==0, print1(k, ", "))); \\ Hugo Pfoertner, May 14 2019
(Haskell)
a006093 = (subtract 1) . a000040 -- Reinhard Zumkeller, Mar 06 2012
(Magma) [NthPrime(n)-1: n in [1..100]]; // Vincenzo Librandi, Nov 17 2015
(GAP) Filtered([1..280], IsPrime)-1; # Muniru A Asiru, Nov 25 2018
(Python)
from sympy import prime
for n in range(1, 100): print(prime(n)-1, end=', ') # Stefano Spezia, Nov 30 2018
CROSSREFS
a(n) = K(n, 1) and A034693(K(n, 1)) = 1 for all n. The subscript n refers to this sequence and K(n, 1) is the index in A034693. - Labos Elemer
Essentially the same as A039915.
KEYWORD
nonn,easy,nice
AUTHOR
EXTENSIONS
Obfuscating comments removed by Joerg Arndt, Mar 11 2010
Edited by Charles R Greathouse IV, Apr 20 2010
STATUS
approved