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a(n) = C(n+7,n) * C(n+10,7).
1

%I #22 Sep 06 2022 03:01:09

%S 120,2640,28512,205920,1132560,5096520,19631040,66745536,204787440,

%T 576438720,1507608960,3700494720,8593371072,19004570640,40244973120,

%U 81980500800,161264274600,307350735120,569168028000,1026681084000,1807851474000,3113521983000

%N a(n) = C(n+7,n) * C(n+10,7).

%H T. D. Noe, <a href="/A105943/b105943.txt">Table of n, a(n) for n = 0..1000</a>

%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 G.f.: (24*(-5*x^4-35*x^3-63*x^2-35*x-5))/(x-1)^15. - _Harvey P. Dale_, Nov 14 2011

%F From _Amiram Eldar_, Sep 06 2022: (Start)

%F Sum_{n>=0} 1/a(n) = 114905939/6480 - 5390*Pi^2/3.

%F Sum_{n>=0} (-1)^n/a(n) = 14336*log(2)/9 - 3577279/3240. (End)

%e If n=0 then C(0+7,0)*C(0+10,7) = C(7,0)*C(10,7) = 1*120 = 120.

%e If n=6 then C(6+7,6)*C(6+10,7) = C(13,6)*C(16,7) = 1716*11440 = 19631040.

%p A105943:=n->binomial(n+7,n)*binomial(n+10,7): seq(A105943(n), n=0..40); # _Wesley Ivan Hurt_, Apr 18 2017

%t Table[Binomial[n+7,n]Binomial[n+10,7],{n,0,30}] (* _Harvey P. Dale_, Nov 14 2011 *)

%o (Python)

%o A105943_list, m = [], [3432, -3432, 1320, 0]+[120]*11

%o for _ in range(10**2):

%o A105943_list.append(m[-1])

%o for i in range(14):

%o m[i+1] += m[i] # _Chai Wah Wu_, Jan 24 2016

%Y Cf. A062145.

%K easy,nonn

%O 0,1

%A _Zerinvary Lajos_, Apr 27 2005

%E More terms from _Harvey P. Dale_, Nov 14 2011