A302644 - OEIS

%I #23 Jul 12 2018 00:47:56

%S 1,2,6,20,70,252,896,2976,8955,24310,60038,136500,289016,575680,

%T 1087920,1964384,3408789,5712426,9282070,14674100,22635690,34153988,

%U 50514256,73368000,104812175,147480606,204648822,280353556,379528220,508155720,673440032,883998016

%N a(n) = (n+2)*(n+1)*(n^6-12*n^5+70*n^4-210*n^3+409*n^2-378*n+360)/720.

%C The limit as q->1^- of the unimodal polynomial [q^(4k+3)(1-q^n)( q-q^n)-q^(3k)q(1+q)( 1-q^n)( q-q^2+q^5-q^n)-q^(2k)(q^(nk)(q^(2n)(q^(9)- q^(8)-q^(7)+q^(6)+ q^(5)-q^(3)+q)-q^n(q^(10)-q^(8)+q^(6)+q^(5))+q^(10))-q^(2n)+q^n(q^(5)+q^(4)-q^(2)+1)-q^(9)+q^(7)-q^(5)-q^(4)+q^(3)+q^(2)-q)+q^k q^(nk)q^(3) ( 1+q ) ( 1-q^n )  ( q^5-q^n+q^(n+3)-q^(n+4))-q^(nk)q^6(1-q^n)( q-q^n)]/[(1-q)^2(1-q^2)^2(1-q^3)(1-q^(n-1))(1-q^n)q^(2k+1)] after making the simplification k=n. The unimodal polynomial is from O'Hara's proof of unimodality of q-binomials after making the restriction to partitions of size <=3. See G_3(n,k) from arXiv:1711.11252.

%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="/A302644/b302644.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_09">Index entries for linear recurrences with constant coefficients</a>, signature (9,-36,84,-126,126,-84,36,-9,1).

%F a(n) = (n+2)*(n+1)*(n^6-12*n^5+70*n^4-210*n^3+409*n^2-378*n+360)/720.

%F From _Colin Barker_, Apr 11 2018: (Start)

%F G.f.: (1 - 7*x + 24*x^2 - 46*x^3 + 64*x^4 - 36*x^5 + 56*x^6) / (1 - x)^9.

%F a(n) = 9*a(n-1) - 36*a(n-2) + 84*a(n-3) - 126*a(n-4) + 126*a(n-5) - 84*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9) for n>8.

%F (End)

%e For n=4, G_3(4,4)=q^16+q^15+2*q^14+3*q^13+5*q^12+5*q^11+7*q^10+7*q^9+8*q^8+7*q^7+7*q^6+5*q^5+5*q^4+3*q^3+2*q^2+q+1 (using the formula in the comments). Then substituting q=1 yields 70.

%o (PARI) Vec((1 - 7*x + 24*x^2 - 46*x^3 + 64*x^4 - 36*x^5 + 56*x^6) / (1 - x)^9 + O(x^40)) \\ _Colin Barker_, Apr 11 2018

%Y Cf. A000984, A002522, A302612, A302645, A302646.

%K nonn,easy

%O 0,2

%A _Bryan T. Ek_, Apr 10 2018

%E More terms from _Colin Barker_, Apr 11 2018