This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A289387 #73 Dec 23 2024 14:53:45
%S A289387 0,0,1,3,6,10,15,20,20,0,-75,-275,-725,-1625,-3250,-5875,-9500,-13125,
%T A289387 -13125,0,47500,171875,450000,1006250,2012500,3640625,5890625,8140625,
%U A289387 8140625,0,-29453125,-106562500,-278984375,-623828125,-1247656250,-2257031250
%N A289387 a(n) = Sum_{k>=0} (-1)^k*binomial(n, 5*k+2).
%C A289387 {A289306, A289321, A289387, A289388, A289389} is the difference analog of the trigonometric functions {k_1(x), k_2(x), k_3(x), k_4(x), k_5(x)} of order 5. For the definitions of {k_i(x)} and the difference analog {K_i (n)} see [Erdelyi] and the first Shevelev link respectively.
%C A289387 From _Robert Israel_, Jul 11 2017: (Start)
%C A289387 a(n)=0 for n == 9 (mod 10).
%C A289387 A112765(a(10*k)) = (5/2)*k - 3/4 - (-1)^k/4.
%C A289387 A112765(a(10*k+2)) = (5/2)*k - 1/4 + (-1)^k/4.
%C A289387 A112765(a(10*k+3)) = A112765(a(10*k+4)) = (5/2)*k + 1/4 - (-1)^k/4.
%C A289387 A112765(a(10*k+5)) = A112765(a(10*k+6)) = (5/2)*k + 3/4 + (-1)^k/4.
%C A289387 A112765(a(10*k+7)) = A112765(a(10*k+8)) = (5/2)*k + 5/4 - (-1)^k/4. (End)
%C A289387 Note that from author's formula (see below) we have that, except for zeros in the sequence mentioned by _Robert Israel_, there are only a(0) = a(1) = 0. Indeed, otherwise for some value of n we should have the equality (phi-1)^n = -cos(Pi*(n-4)/10)/cos(3*Pi*(n-4)/10). However, the absolute value of the right hand side takes the six distinct values only: 1, phi, phi^2, phi^(-1), phi^(-2), 1/3 (the last value we have when n == 9 (mod 10), since lim_{x->Pi/2}cos(x)/cos(3*x)= -1/3). Thus for n>=3, we have (phi-1)^n = phi^(-n) < |cos(Pi*(n-4)/10)/cos(3*Pi*(n-4)/10)|. - _Vladimir Shevelev_, Jul 15 2017
%D A289387 A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.
%H A289387 Vladimir Shevelev, <a href="https://arxiv.org/abs/1706.01454">Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n</a>, arXiv:1706.01454 [math.CO], 2017.
%H A289387 Vladimir Shevelev, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2017-July/017799.html">A matrix property of sums of binomial coefficients</a>, Seqfan, Fri Jul 21 2017.
%H A289387 Wikipedia, <a href="http://en.wikipedia.org/wiki/Circulant_matrix">Circulant matrix</a>
%F A289387 G.f.: -((-1+x)^2*x^2)/((-1+x)^5 - x^5). - _Peter J. C. Moses_, Jul 05 2017
%F A289387 For n>=1, a(n) = (2/5)*(phi+2)^(n/2)*(cos(Pi*(n-4)/10) + (phi-1)^n*cos(3* Pi*(n-4)/10)), where phi is the golden ratio.
%F A289387 a(n+m) = a(n)*K_1(m) + K_2(n)*K_2(m) + K_1(n)*a(m) - K_5(n)*K_4(m) - K_4(n)*K_5(m), where K_1 is A289306, K_2 is A289321, K_4 is A289388, K_5 is A289389.
%F A289387 For every n>=1, the determinant of circulant matrix of order 5 (see [Wikipedia]) with the entries (-1)^(i-1)* K_i(n), i=1..5, is 0. Here K_1, K_2, K_4 and K_5 are the same as in the previous formula, while K_3(n) = a(n). For a proof and a generalization see the second Shevelev link that also contains two unsolved problems. - _Vladimir Shevelev_, Jul 26 2017
%p A289387 f:= gfun:-rectoproc({5*a(n)-10*a(n+1)+10*a(n+2)-5*a(n+3)+a(n+4), a(0)=0,
%p A289387 a(1)=0, a(2)=1, a(3) = 3,a(4)=6},a(n),remember):
%p A289387 map(f, [$0..40]); # _Robert Israel_, Jul 11 2017
%t A289387 Table[Sum[(-1)^k*Binomial[n, 5 k + 2], {k, 0, n}], {n, 0, 35}] (* or *)
%t A289387 CoefficientList[Series[-((-1 + x)^2 x^2)/((-1 + x)^5 - x^5), {x, 0, 35}], x] (* _Michael De Vlieger_, Jul 10 2017 *)
%o A289387 (PARI) a(n) = sum(k=0, (n-2)\5, (-1)^k*binomial(n, 5*k+2)); \\ _Michel Marcus_, Jul 05 2017
%Y A289387 Cf. A139398, A133476, A139714, A139748, A139761, A289306, A289321, A289388, A289389.
%K A289387 sign,easy
%O A289387 0,4
%A A289387 _Vladimir Shevelev_, Jul 05 2017
%E A289387 More terms from _Peter J. C. Moses_, Jul 05 2017