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a(n) = Sum_{i=1..n} C(i+4,5)^2.
22

%I #17 Sep 08 2022 08:45:11

%S 1,37,478,3614,19490,82994,296438,923702,2580071,6588075,15606084,

%T 34685508,72976852,146387476,281597860,521971876,936053677,1629533233,

%U 2761788434,4568378450,7391175350,11718183750,18235516650,27894475050,41997225075,62305185111

%N a(n) = Sum_{i=1..n} C(i+4,5)^2.

%H T. D. Noe, <a href="/A086025/b086025.txt">Table of n, a(n) for n = 1..1000</a>

%H John Engbers and Christopher Stocker, <a href="http://epublications.marquette.edu/mscs_fac/456/">Two Combinatorial Proofs of Identities Involving Sums of Powers of Binomial Coefficients</a>, Integers 16 (2016), #A58.

%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 From _R. J. Mathar_, Jun 16 2010: (Start)

%F G.f.: x*(1+x)*(x^4+24*x^3+76*x^2+24*x+1)/(x-1)^12.

%F a(n) = n*(2*n+5)*(n+5)*(n+4)*(n+3)*(n+2)*(n+1)*(63*n^4 +630*n^3 +1855*n^2 +1400*n +12) / 19958400. (End)

%t Table[n*(2*n+5)*(n+5)*(n+4)*(n+3)*(n+2)*(n+1)*(63*n^4 +630*n^3 +1855*n^2 +1400*n +12)/19958400, {n,1,30}] (* _G. C. Greubel_, Nov 22 2017 *)

%o (PARI) for(n=1,30, print1(sum(i=1,n, binomial(i+4,5)^2), ", ")) \\ _G. C. Greubel_, Nov 22 2017

%o (Magma) [n*(2*n+5)*(n+5)*(n+4)*(n+3)*(n+2)*(n+1)*(63*n^4 +630*n^3 +1855*n^2 +1400*n +12)/19958400: n in [1..30]]; // _G. C. Greubel_, Nov 22 2017

%Y Cf. A087127, A024166, A085438, A085439, A085440, A085441, A085442, A086020, A086021, A086022, A086023, A086024, A086026, A086027, A086028, A086029, A086030.

%K easy,nonn

%O 1,2

%A _André F. Labossière_, Jul 11 2003

%E More terms from _R. J. Mathar_, Jun 16 2010