A111927 - OEIS

A111927

Expansion of x^3 / ((x-1)*(2*x-1)*(x^2-x+1)).

0, 0, 0, 1, 4, 10, 21, 42, 84, 169, 340, 682, 1365, 2730, 5460, 10921, 21844, 43690, 87381, 174762, 349524, 699049, 1398100, 2796202, 5592405, 11184810, 22369620, 44739241, 89478484, 178956970, 357913941, 715827882, 1431655764, 2863311529, 5726623060

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OFFSET

0,5

COMMENTS

Binomial transform of sequence (0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0). Note: the binomial transform of the sequence (0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0) is A111926; the binomial transform of the sequence (0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0) is A024495 (disregarding first two terms, which are both zero).

The sequence relates the calculation of the logarithm of the Twin Prime Constants of order 3 to the sequence of prime zeta functions, see definition 7 in arXiv:0903.2514. - R. J. Mathar, Mar 28 2009

LINKS

Colin Barker, Table of n, a(n) for n = 0..1000

Antoine-Augustin Cournot, Solution d'un problème d'analyse combinatoire, Bulletin des Sciences Mathématiques, Physiques et Chimiques, item 34, volume 11, 1829, pages 93-97. Also at Google Books. Page 97 case p=3 formula y^(0) = a(n). (But misprint "- (2/3)*cos" should be "+ (2/3)*cos".)

R. J. Mathar, Hardy-Littlewood constants embedded into infinite products over all positive integers, arXiv:0903.2514 [math.NT], 2009-2011.

Christian Ramus, Solution générale d'un problème d'analyse combinatoire, Journal für die Reine und Angewandte Mathematik (Crelle's journal), volume 11, 1834, pages 353-355. Page 353 case p=3 formula y^(0) = a(n). (But misprint "+ (1/3)*cos" should be "+ (2/3)*cos", per the general case equation A page 354.)

Kevin Ryde, Iterations of the Terdragon Curve, section Lines, quantity Lines_k(2) = a(k+1).

Index entries for linear recurrences with constant coefficients, signature (4,-6,5,-2).

FORMULA

a(n+2) - a(n+1) + a(n) = A000225(n).

a(n) - a(n-1) = A024495(n-1).

From Colin Barker, Feb 10 2017: (Start)

a(n) = 2^n/3 + 2*cos((Pi*n)/3)/3 - 1. [Cournot]

a(n) = 4*a(n-1) - 6*a(n-2) + 5*a(n-3) - 2*a(n-4) for n > 3.

(End)

a(n) = (2^n+A087204(n))/3 - 1. - R. J. Mathar, Aug 07 2017

MAPLE

seq(sum(binomial(n, k*3), k=1..n), n=0..33); # Zerinvary Lajos, Oct 23 2007