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%I #27 May 24 2017 06:45:55
%S 0,7,38,111,244,455,762,1183,1736,2439,3310,4367,5628,7111,8834,10815,
%T 13072,15623,18486,21679,25220,29127,33418,38111,43224,48775,54782,
%U 61263,68236,75719,83730,92287,101408,111111,121414,132335,143892,156103,168986
%N a(n) = n*(3*n^2 + 3*n + 1).
%C The series Sum a(n)/A007559(n+1)^3 has the sum 1/9 (cf. A249352), analogous to Sum_{n=1..oo} A000217(n)/A001147(n+1)^2 = 1/8 (cf. A249348 and A249349).
%C Also, nonnegative numbers m such that 9*m + 1 is a cube. - _Bruno Berselli_, May 23 2017
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F G.f.: x*(7+10*x+x^2) / (x-1)^4 . - _R. J. Mathar_, Oct 28 2014
%F a(n) = n*A003215(n). - _R. J. Mathar_, Oct 28 2014
%t Table[n (3 n^2 + 3 n + 1), {n, 0, 38}] (* or *)
%t CoefficientList[Series[x (7 + 10 x + x^2)/(x - 1)^4, {x, 0, 38}], x] (* _Michael De Vlieger_, May 23 2017 *)
%o (PARI) A249354(n)=3*n^3+3*n^2+n
%Y Cf. A132355: numbers m such that 9*m + 1 is a square.
%K nonn,easy
%O 0,2
%A _M. F. Hasler_, Oct 26 2014
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