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%I #17 Apr 16 2020 18:51:54
%S 1,13,48,118,235,411,658,988,1413,1945,2596,3378,4303,5383,6630,8056,
%T 9673,11493,13528,15790,18291,21043,24058,27348,30925,34801,38988,
%U 43498,48343,53535,59086,65008,71313,78013,85120,92646,100603,109003,117858,127180
%N a(n) = (4*n^3 + 11*n^2 + 9*n + 2)/2.
%C Binomial transform yields 1,12,23,12,0,0,0,0,0,0,.. - _R. J. Mathar_, Apr 21 2008
%D J. H. Conway and R. K. Guy, The Book of Numbers, p. 83.
%H G. C. Greubel, <a href="/A135712/b135712.txt">Table of n, a(n) for n = 0..1000</a>
%H M. E. Larsen, <a href="https://www.jstor.org/stable/2686923">The eternal triangle - a history of a counting problem</a>, College Math. J., 20 (1989), 370-392.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F G.f.: (1 + 9*x + 2*x^2) / (1-x)^4. - _R. J. Mathar_, Apr 21 2008
%F From _G. C. Greubel_, Oct 29 2016: (Start)
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
%F E.g.f.: (1/2)*(2 + 24*x + 23*x^2 + 4*x^3)*exp(x). (End)
%F a(n) = ((2*n+1)*(2*n+3)*(4*n+3) - 1)/8 = (n+1)*(4*n^2 + 7*n + 2)/2, for n >= 0. See the Conway and Guy reference. - _Wolfdieter Lang_, Apr 16 2020
%t Table[(4*n^3 + 11*n^2 + 9*n + 2)/2,{n,0,25}] (* or *) LinearRecurrence[{4,-6,4,-1}, {1,13,48,118}, 25] (* _G. C. Greubel_, Oct 29 2016 *)
%Y Bisection of A002717 (odd part).
%Y Partial sums of A033570. - _Bruno Berselli_, Nov 28 2013
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_, Mar 05 2008
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