A099587 a(n) = coefficient of x in (1+x)^n mod (1+x^4).

0, 1, 2, 3, 4, 4, 0, -14, -48, -116, -232, -396, -560, -560, 0, 1912, 6528, 15760, 31520, 53808, 76096, 76096, 0, -259808, -887040, -2141504, -4283008, -7311552, -10340096, -10340096, 0, 35303296, 120532992, 290992384, 581984768

Offset: 0

Author

Ralf Stephan, Oct 24 2004

Keywords

• sign
• easy

Comments

{A099586, A099587, A099588, A099589} is the difference analog of the trigonometric functions {k_1(x), k_2(x), k_3(x), k_4(x)} of order 4.

For the definition, see [Erdelyi] and the Shevelev link. - Vladimir Shevelev, Jul 03 2017

References

• A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
• Vladimir Shevelev, Coefficient of x^k in ((x+1)^n modulo x^N+1), seqfan, Thu Jul 20 2017.
• G. Tollisen and T. Lengyel, A Congruential Identity and the 2-adic Order of Lacunary Sums of Binomial Coefficients, Integers 4 (2004), #A4.
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-2).

Crossrefs

Cf. A099586, A099588, A099589.

Cf. A057681, A099586, A139398, A133476, A139714, A139748, A139761.

Programs

Mathematica

RecurrenceTable[{a[1]=1, a[2]=2, a[3]=3, a[4]=4, a[n] = 4*a[n-1] - 6*a[n-2] + 4*a[n-3] - 2*a[n-4]}, a, {n, 1, 100}] (* G. C. Greubel, Nov 09 2015 *)
a[n_] := n*HypergeometricPFQ[{(1-n)/4, (2-n)/4, (3-n)/4, (4-n)/4}, {1/2, 3/4, 5/4}, -1]; Array[a, 40, 0] (* Jean-François Alcover, Jul 20 2017, from Vladimir Shevelev's first formula *)
LinearRecurrence[{4,-6,4,-2},{0,1,2,3},50] (* Harvey P. Dale, Mar 27 2022 *)

PARI

a(n) = polcoeff(((1+x)^n)%(x^4+1),1)

Formula

G.f.: x*(x-1)^2 / (2*x^4-4*x^3+6*x^2-4*x+1). - Colin Barker, Jul 15 2013

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - 2*a(n-4). - G. C. Greubel, Nov 09 2015

From Vladimir Shevelev, Jun 29 2017: (Start)

a(n) = Sum_{k >= 0}(-1)^k*binomial(n,4*k+1).

a(n) = round((2+sqrt(2))^(n/2)*cos(Pi*(n-2)/8)/2), where round(x) is the integer nearest to x.

a(n+m) = a(n)*K_1(m) + K_1(n)*a(m) - K_4(n)*K_3(m) - K_3(n)*K_4(m), where K_1 is A099586, K_3=A099588, and K_4=A099589.

(End)

a(n) = A099589(n+2)-2*A099589(n+1)+A099589(n). - R. J. Mathar, Jun 28 2025

Extensions

a(0)=0 added by N. J. A. Sloane, Jun 30 2017