%I #39 Jul 12 2018 00:48:04
%S 0,1,2,6,20,70,252,924,3432,12705,45430,152438,472836,1352078,3578680,
%T 8827080,20439984,44745513,93185994,185640070,355452020,656846190,
%U 1175604980,2044130980,3462303000,5725877625,9264588606,14692562262,22874204836,35009334470
%N Values of unimodal polynomial analogous to A302612 and A302644 arising from a partition size <= 4 restriction.
%C Consider the unimodal polynomial from O'Hara's proof of unimodality of q-binomials after making the restriction to partitions of size <=4. See G_4(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="/A302645/b302645.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_12">Index entries for linear recurrences with constant coefficients</a>, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).
%F a(n) = n*(n+2)*(n+1)*(n^8-32*n^7+462*n^6-3836*n^5+20013*n^4-66836*n^3+140804*n^2-171216*n+100800)/120960.
%F From _Colin Barker_, Apr 19 2018: (Start)
%F G.f.: x*(1 - 10*x + 48*x^2 - 140*x^3 + 281*x^4 - 390*x^5 + 430*x^6 - 220*x^7 + 330*x^8) / (1 - x)^12.
%F a(n) = 12*a(n-1) - 66*a(n-2) + 220*a(n-3) - 495*a(n-4) + 792*a(n-5) - 924*a(n-6) + 792*a(n-7) - 495*a(n-8) + 220*a(n-9) - 66*a(n-10) + 12*a(n-11) - a(n-12) for n>11.
%F (End)
%e For n=6, G_4(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 - 10*x + 48*x^2 - 140*x^3 + 281*x^4 - 390*x^5 + 430*x^6 - 220*x^7 + 330*x^8) / (1 - x)^12 + O(x^40))) \\ _Colin Barker_, Apr 19 2018
%Y Cf. A000984, A002522, A302612, A302644, A302646.
%K nonn,easy
%O 0,3
%A _Bryan T. Ek_, Apr 10 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