A289321 a(n) = Sum_{k >= 0}(-1)^k*binomial(n,5*k+1).

0, 1, 2, 3, 4, 5, 5, 0, -20, -75, -200, -450, -900, -1625, -2625, -3625, -3625, 0, 13125, 47500, 124375, 278125, 556250, 1006250, 1628125, 2250000, 2250000, 0, -8140625, -29453125, -77109375, -172421875, -344843750, -623828125, -1009375000, -1394921875

Offset: 0

Author

Vladimir Shevelev, Jul 02 2017

Keywords

• sign

Comments

a(n) = 0 for n == 7 (mod 10). - Robert Israel, Jul 12 2017

{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 Shevelev link respectively. - Vladimir Shevelev, Jul 24 2017

Links

• Robert Israel, Table of n, a(n) for n = 0..3579
• Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5).

Crossrefs

Cf. A139398, A133476, A139714, A139748, A139761, A289306.

Programs

Maple

f:= gfun:-rectoproc({5*a(n)-10*a(n+1)+10*a(n+2)-5*a(n+3)+a(n+4), a(0)=0,
a(1)=1, a(2)=2, a(3) = 3, a(4)=4}, a(n), remember):
map(f, [$0..40]); # Robert Israel, Jul 11 2017

Mathematica

Table[Sum[(-1)^k*Binomial[n, 5 k + 1], {k, 0, n}], {n, 0, 35}] (* or *)
CoefficientList[Series[((-1 + x)^3 x)/((-1 + x)^5 - x^5), {x, 0, 35}], x] (* Michael De Vlieger, Jul 04 2017 *)
LinearRecurrence[{5,-10,10,-5},{0,1,2,3,4},40] (* Harvey P. Dale, Dec 25 2022 *)

PARI

a(n) = sum(k=0, (n-1)\5, (-1)^k*binomial(n, 5*k+1)); \\ Michel Marcus, Jul 03 2017

Formula

G.f.: ((-1+x)^3 x)/((-1+x)^5-x^5). - Peter J. C. Moses, Jul 02 2017

For n>=1, a(n) = (2/5)*(phi+2)^(n/2)*(cos(Pi*(n-2)/10) + (phi-1)^n* cos (3*Pi*(n-2)/10)), where phi is the golden ratio.

a(n+m) = a(n)*K_1(m) + K_1(n)*a(m) - K_5(n)*K_3(m) - K_4(n)*K_4(m) - K_3(n)*K_5(m), where K_1 is A289306, K_3 is A289387, K_4 is A289388, K_5 is A289389. - Vladimir Shevelev, Jul 24 2017