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%I #34 Jul 12 2018 00:48:13
%S 0,1,2,6,20,70,252,924,3432,12870,48620,183755,683046,2443168,8263360,
%T 26184420,77558760,215182923,561542454,1385168400,3245959640,
%U 7260395142,15567955260,32124894880,64016082000,123566718600,231661933776,422854091037,753068219386
%N Values of unimodal polynomial analogous to A302612 and A302644 arising from a partition size <= 5 restriction.
%C Consider the unimodal polynomial from O'Hara's proof of unimodality of q-binomials after making the restriction to partitions of size <=5. See G_5(n,k) from arXiv:1711.11252. If we make the simplification k=n and take the limit as q->1^-, we obtain the listed polynomial.
%C As the size restriction s increases, G_s->G_infinity=G: the q-binomials. Then substituting k=n and q=1 yields the central binomial coefficients: A000984.
%H Colin Barker, <a href="/A302646/b302646.txt">Table of n, a(n) for n = 0..1000</a>
%H Bryan Ek, <a href="https://arxiv.org/abs/1711.11252">q-Binomials and related symmetric unimodal polynomials</a>, arXiv:1711.11252 [math.CO], 2017-2018.
%H Bryan Ek, <a href="https://arxiv.org/abs/1804.05933">Unimodal Polynomials and Lattice Walk Enumeration with Experimental Mathematics</a>, arXiv:1804.05933 [math.CO], 2018.
%H <a href="/index/Rec#order_15">Index entries for linear recurrences with constant coefficients</a>, signature (15,-105,455,-1365,3003,-5005,6435,-6435,5005,-3003,1365,-455,105,-15,1).
%F a(n) = n*(n+3)*(n+2)*(n+1)*(n^10-50*n^9+1140*n^8-15420*n^7+136533*n^6-824370*n^5+3436190*n^4-9762880*n^3+18198936*n^2-20242080*n+10886400)/43545600.
%F From _Colin Barker_, Apr 19 2018: (Start)
%F G.f.: x*(1 - 13*x + 81*x^2 - 315*x^3 + 855*x^4 - 1701*x^5 + 2583*x^6 - 2961*x^7 + 2835*x^8 - 1365*x^9 + 2002*x^10) / (1 - x)^15.
%F a(n) = 15*a(n-1) - 105*a(n-2) + 455*a(n-3) - 1365*a(n-4) + 3003*a(n-5) - 5005*a(n-6) + 6435*a(n-7) - 6435*a(n-8) + 5005*a(n-9) - 3003*a(n-10) + 1365*a(n-11) - 455*a(n-12) + 105*a(n-13) - 15*a(n-14) + a(n-15) for n>14.
%F (End)
%e For n=6, G_5(6,6)=q^36+q^35+2*q^34+3*q^33+5*q^32+7*q^31+11*q^30+13*q^29+18*q^28+22*q^27+28*q^26+32*q^25+39*q^24+42*q^23+48*q^22+51*q^21+55*q^20+55*q^19+58*q^18+55*q^17+55*q^16+51*q^15+48*q^14+42*q^13+39*q^12+32*q^11+28*q^10+22*q^9+18*q^8+13*q^7+11*q^6+7*q^5+5*q^4+3*q^3+2*q^2+q+1 (using the formula in the referenced paper). Then substituting q=1 yields 924.
%o (PARI) concat(0, Vec(x*(1 - 13*x + 81*x^2 - 315*x^3 + 855*x^4 - 1701*x^5 + 2583*x^6 - 2961*x^7 + 2835*x^8 - 1365*x^9 + 2002*x^10) / (1 - x)^15 + O(x^40))) \\ _Colin Barker_, Apr 19 2018
%Y Cf. A000984, A002522, A302612, A302644, A302645.
%K nonn,easy
%O 0,3
%A _Bryan T. Ek_, Apr 11 2018
%E More terms from _Colin Barker_, Apr 11 2018
%E 0 prepended to the sequence and formulas adjusted accordingly by _Colin Barker_, Apr 19 2018
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