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A156887 a(n) = Sum_{k=0..n} C(n,k)*C(4*n+k,k).

%I A156887 #24 Jan 09 2023 13:34:43
%S A156887 1,6,64,768,9708,126386,1676956,22548168,306167324,4188703512,
%T A156887 57649462164,797294161824,11071026740964,154250752864812,
%U A156887 2155368246401224,30192512693210888,423859798484668188,5961793387214958792,83998039356129372448,1185277027372535468544
%N A156887 a(n) = Sum_{k=0..n} C(n,k)*C(4*n+k,k).
%C A156887 a(n)=[x^n] (1+6x+14x^2+16x^3+9x^4+2x^5)^n. The coefficients (1,6,14,16,9,2) are the 6th row of A029635.
%H A156887 Seiichi Manyama, <a href="/A156887/b156887.txt">Table of n, a(n) for n = 0..862</a>
%H A156887 P. Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL16/Barry2/barry231.html">A Note on a Family of Generalized Pascal Matrices Defined by Riordan Arrays</a>, Journal of Integer Sequences, 16 (2013), #13.5.4.
%F A156887 Conjecture: 576*n*(32901928701*n-65877527665)*(4*n-3)*(2*n-1)*(4*n-1)*a(n) +(-8795436181229177*n^5 +35251410418024655*n^4 -47934714902592853*n^3 +29414167990853161*n^2 -9060238526902314*n +1466702211905280)*a(n-1) +8*(10299715469615*n^5 -136961193094719*n^4 +872530072905392*n^3 -2699499511785411*n^2 +3902106377543903*n -2123717948975100)*a(n-2) -64*(2*n-5)*(4*n-9)*(n-2)*(27741827*n-2925269736)*(4*n-11)*a(n-3)=0. - _R. J. Mathar_, Feb 25 2015
%F A156887 From _Peter Bala_, Feb 11 2018: (Start)
%F A156887 a(n) = Sum_{k = 0..n} (-1)^(n-k)*C(n,k)*C(4*n+k,n)*2^k.
%F A156887 a(n) = Sum_{k = 0..n} C(n,k)*C(4*n,k)*2^(n-k).
%F A156887 8*(4*n)*(4*n-1)*(4*n-2)**(4*n-3)*(4633*n^3-19662*n^2+27593*n-12804)*a(n) = (137604733*n^7-859190528*n^6+2179882848*n^5-2890753162*n^4+2144669963*n^3-880916550*n^2+182941416*n-14515200)*a(n-1) - (4*n-4)*(4*n-5)*(4*n-6)*(4*n-7)*(4633*n^3-5763*n^2+2168*n-240)*a(n-2). A proof of Mathar's conjectured third-order recurrence above follows easily using this second-order recurrence.  (End)
%F A156887 a(n) ~ sqrt(5 + 33/sqrt(41)) * ((29701 + 4633*sqrt(41)))^n / (sqrt(Pi*n) * 2^(12*n + 2)). - _Vaclav Kotesovec_, Jan 09 2023
%p A156887 A156887 := proc(n)
%p A156887     add(binomial(n,k)*binomial(4*n+k,k),k=0..n) ;
%p A156887 end proc: # _R. J. Mathar_, Feb 25 2015
%t A156887 Table[Sum[Binomial[n,k]Binomial[4n+k,k],{k,0,n}],{n,0,30}] (* _Harvey P. Dale_, Jul 24 2018 *)
%o A156887 (PARI) {a(n) = sum(k=0, n, binomial(n, k)*binomial(4*n+k, k))} \\ _Seiichi Manyama_, Feb 02 2019
%Y A156887 Cf. A001850, A114496, A156886, A359646.
%K A156887 easy,nonn
%O A156887 0,2
%A A156887 _Paul Barry_, Feb 17 2009