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a(n) = binomial(n+5, n)*binomial(n+8, 5).
1

%I #23 Sep 01 2022 07:13:34

%S 56,756,5292,25872,99792,324324,924924,2378376,5621616,12388376,

%T 25729704,50791104,95938752,174350232,306211752,521694096,864913896,

%U 1399125420,2213431220,3431347920,5221616400,7811703900,11504509380,16698853080,23914406880,33821804016

%N a(n) = binomial(n+5, n)*binomial(n+8, 5).

%H T. D. Noe, <a href="/A105940/b105940.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

%F G.f.: -28*(x+2)*(2*x+1) / (x-1)^11. - _Colin Barker_, Jan 28 2013

%F From _Amiram Eldar_, Sep 01 2022: (Start)

%F Sum_{n>=0} 1/a(n) = 580367/1764 - 100*Pi^2/3.

%F Sum_{n>=0} (-1)^n/a(n) = 74537/588 - 1280*log(2)/7. (End)

%e a(0) = C(0+5,0)*C(0+8,5) = C(5,0)*C(8,5) = 1*56 = 56

%e a(6) = C(6+5,6)*C(6+8,5) = C(11,6)*C(14,5) = 462*2002 = 924924.

%p with(combinat); for i from 0 to 25 do print(i,numbcomb(i+5,i)*numbcomb(i+8,5)); end; # _Jim Nastos_, Oct 26 2005

%t a[n_] := Binomial[n + 5, 5] * Binomial[n + 8, 5]; Array[a, 25, 0] (* _Amiram Eldar_, Sep 01 2022 *)

%Y Cf. A062145.

%K easy,nonn

%O 0,1

%A _Zerinvary Lajos_, Apr 27 2005

%E More terms from _Jim Nastos_, Oct 26 2005