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%I #63 Apr 19 2023 09:55:44
%S 0,0,0,1,4,10,20,34,48,48,0,-164,-560,-1352,-2704,-4616,-6528,-6528,0,
%T 22288,76096,183712,367424,627232,887040,887040,0,-3028544,-10340096,
%U -24963200,-49926400,-85229696,-120532992,-120532992,0,411525376,1405035520,3392055808
%N Expansion of x^3 / (1 - 4x + 6x^2 - 4x^3 + 2x^4).
%C {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 04 2017
%D A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.
%H G. C. Greubel, <a href="/A099589/b099589.txt">Table of n, a(n) for n = 0..1000</a>
%H 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 Vladimir Shevelev, <a href="http://list.seqfan.eu/oldermail/seqfan/2017-July/017798.html">Coefficient of x^k in ((x+1)^n modulo x^N+1)</a>, seqfan, Thu Jul 20 2017.
%H G. Tollisen and T. Lengyel, <a href="http://www.emis.de/journals/INTEGERS/papers/e4/e4.Abstract.html">A congruential identity and the 2-adic order of lacunary sums of binomial coefficients</a>, Integers 4 (2004), #A4.
%H Maran van Heesch, <a href="https://research.tue.nl/en/studentTheses/the-multiplicative-complexity-of-symmetric-functions-over-a-field">The multiplicative complexity of symmetric functions over a field with characteristic p</a>, Thesis, 2014.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-2).
%F G.f.: x^3/((1-x)^4 + x^4), the binomial transform of x^3/(1+x^4). - _Paul Barry_, Apr 01 2005
%F Coefficient of x^3 in (1+x)^n mod (1 + x^4).
%F a(n) = (1/(2*sqrt(2)))*((2-sqrt(2))^(n/2)*(cos(3*Pi*n/8) + sin(3*Pi*n/8)) + (2+sqrt(2))^(n/2)*(sin(Pi*n/8) - cos(Pi*n/8))). - _Paul Barry_, Apr 01 2005
%F From _Colin Barker_, Nov 08 2015: (Start)
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - 2*a(n-4) for n > 4.
%F G.f.: x^3 / (2*x^4 - 4*x^3 + 6*x^2 - 4*x + 1). (End)
%F From _Vladimir Shevelev_, Jul 04 2017: (Start)
%F a(n) = Sum_{t >= 0} (-1)^t*binomial(n,4*t+3).
%F a(n) = round((2+sqrt(2))^(n/2)*cos(Pi*(n-6)/8)/2), where round(x) is the integer nearest to x.
%F a(n+m) = a(n)*K_1(m) + K_3(n)*K_2(m) + K_2(n)*K_3(m) + K_1(n)*a(m), where
%F K_1 is A099586, K_2 is A099587, K_3 is A099588. (End)
%t Round@Table[(1/(2*sqrt(2)))*(2-sqrt(2))^(n/2)*(Cos(3*Pi*n/8) + Sin(3*Pi*n/8)) + (2+sqrt(2))^(n/2)*(Sin(Pi*n/8) - Cos(Pi*n/8))), {n, 1, 200}] (* _G. C. Greubel_, Nov 07 2015 *)
%t RecurrenceTable[{a[n]== 4*a[n-1] - 6*a[n-2] + 4*a[n-3] - 2*a[n-4], a[1]=0, a[2]=0, a[3]=1, a[4]=4}, a, {n, 1, 250}] (* _G. C. Greubel_, Nov 10 2015 *)
%t Table[Sum[(-1)^k*Binomial[n, 4 k + 3], {k, 0, n}], {n, 0, 37}] (* _Michael De Vlieger_, Jun 30 2017 *)
%t a[n_] := n*(n-1)*(n-2)/6 HypergeometricPFQ[{(3-n)/4, (4-n)/4, (5-n)/4, (6-n)/4}, {5/4, 3/2, 7/4}, -1]; Array[a, 40, 0] (* _Jean-François Alcover_, Jul 20 2017, from _Vladimir Shevelev_'s first formula *)
%o (PARI) a(n) = polcoeff(((1+x)^n)%(x^4+1),3)
%o (PARI) concat([0, 0], Vec(x^3/((1-x)^4+x^4) + O(x^50))) \\ _Altug Alkan_, Nov 08 2015
%o (PARI) a(n) = sum(t=0, (n-3)\4, (-1)^t*binomial(n,4*t+3)); \\ _Michel Marcus_, Jun 30 2017
%Y Cf. A099586, A099587, A099588.
%K sign,easy
%O 0,5
%A _Ralf Stephan_, Oct 24 2004
%E a(0)=0 added by _N. J. A. Sloane_, Jul 04 2017
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