(Translated by https://www.hiragana.jp/)
A038504 - OEIS
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A038504
Sum of every 4th entry of row n in Pascal's triangle, starting at "n choose 1".
23
0, 1, 2, 3, 4, 6, 12, 28, 64, 136, 272, 528, 1024, 2016, 4032, 8128, 16384, 32896, 65792, 131328, 262144, 523776, 1047552, 2096128, 4194304, 8390656, 16781312, 33558528, 67108864, 134209536, 268419072, 536854528, 1073741824
OFFSET
0,3
COMMENTS
Number of strings over Z_2 of length n with trace 1 and subtrace 0.
Same as number of strings over GF(2) of length n with trace 1 and subtrace 0.
From Gary W. Adamson, Mar 13 2009: (Start)
M^n * [1,0,0,0] = [A038503(n), A000749(n), A038505(n), a(n)]; where
M = a 4x4 matrix [1,1,0,0; 0,1,1,0; 0,0,1,1; 1,0,0,1]. Sum of terms = 2^n
Example: M^6 * [1,0,0,0] = [16, 20, 16, 12], Sum = 2^6 = 64. (End)
{A038503, A038504, A038505, A000749} is the difference analog of the hyperbolic functions {h_1(x), h_2(x), h_3(x), h_4(x)} of order 4. For the definitions of {h_i(x)} and the difference analog {H_i (n)} see [Erdelyi] and the Shevelev link respectively. - Vladimir Shevelev, Jul 31 2017
REFERENCES
A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.
FORMULA
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3), n > 3. - Paul Curtz, Mar 01 2008
G.f.: x*(1-x)^2/((1-2*x)*(1-2*x+2*x^2)).
From Paul Barry, Jul 25 2004: (Start)
Binomial transform of x/(1-x^4).
G.f.: x*(1-x)^2/((1-x)^4 - x^4) = x/(1-2*x) - x^3/((1-x)^4 - x^4).
a(n) = Sum_{k=0..floor(n/4)} binomial(n, 4*k+1).
a(n) = Sum_{k=0..n} binomial(n, k)*(sin(Pi*k/2)/2 + (1 - (-1)^k)/4).
a(n) = 2^(n-2) + 2^((n-2)/2)*sin(Pi*n/4) - 0^n/4. (End)
a(n; t, s) = a(n-1; t, s) + a(n-1; t+1, s+t+1) where t is the trace and s is the subtrace.
(1, 2, 3, 4, 6, ...) is the binomial transform of (1, 1, 0, 0, 1, 1, ...). - Gary W. Adamson, May 15 2007
From Vladimir Shevelev, Jul 31 2017: (Start)
For n >= 1, {H_i(n)} are linearly dependent sequences: a(n) = H_2(n) = H_1(n) + H_3(n) - H_4(n);
a(n+m) = a(n)*H_1(m) + H_1(n)*a(m) + H_4(n)*H_3(m) + H_3(n)*H_4(m), where H_1 = A038503, H_3 = A038505, H_4 = A000749.
For proofs, see Shevelev's link, Theorems 2, 3. (End)
a(n) = (1/4)*(2^((n+1)/2)*ChebyshevU(n-1, 1/sqrt(2)) + 2^n - [n=0]). - G. C. Greubel, Apr 20 2023
EXAMPLE
a(2;1,0) = 3 since the two binary strings of trace 1, subtrace 0 and length 2 are { 10, 01 }.
MATHEMATICA
CoefficientList[Series[x(1-x)^2/((1-2x)(1-2x+2x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 22 2012 *)
LinearRecurrence[{4, -6, 4}, {0, 1, 2, 3}, 40] (* Harvey P. Dale, Aug 23 2017 *)
PROG
(Magma) [0] cat [n le 3 select n else 4*Self(n-1) -6*Self(n-2) + 4*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jun 22 2012
(SageMath)
@CachedFunction
def a(n): # a = A038504
if (n<4): return n
else: return 4*a(n-1) - 6*a(n-2) + 4*a(n-3)
[a(n) for n in range(51)] # G. C. Greubel, Apr 20 2023
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
STATUS
approved