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A093917 a(n) = n^3+n for odd n, (n^3+n)*3/2 for even n: Row sums of A093915.

%I A093917 #13 Jan 21 2019 19:01:55
%S A093917 2,15,30,102,130,333,350,780,738,1515,1342,2610,2210,4137,3390,6168,
%T A093917 4930,8775,6878,12030,9282,16005,12190,20772,15650,26403,19710,32970,
%U A093917 24418,40545,29822,49200,35970,59007,42910,70038,50690,82365,59358
%N A093917 a(n) = n^3+n for odd n, (n^3+n)*3/2 for even n: Row sums of A093915.
%C A093917 Initially defined as sum of the n-th row of the triangle A093915, constructed by trial and error. Namely, this row should contain n consecutive integers [x,x+1,...,x+n-1], listed in A093915, and have its sum a(n) = n*x+n(n-1)/2 equal to the least possible strict (>1) multiple of the sum of the indices of these elements in A093915, which equals A006003(n) = (n^3+n)/2. For odd n, a(n) = 2 A006003(n) is obtained for x = A093916(n). For even n, the sum a(n) cannot equal 2 A006003(n), but it does equal 3 A006003(n) for x = A093916(n). Hence this simple explicit definition of a(n). - _M. F. Hasler_, Apr 04 2009
%H A093917 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0,4,0,-6,0,4,0,-1).
%F A093917 a(n) = n*A093916(n)+n(n-1)/2. - _M. F. Hasler_, Apr 04 2009
%F A093917 a(2n-1) = 2*(2n-1)*(2n^2 -2n +1), a(2n) = 3*n*(4n^2 +1).
%F A093917 G.f.: x*(2+15*x+22*x^2+42*x^3+22*x^4+15*x^5+2*x^6) / ( (x-1)^4*(1+x)^4 ). - _R. J. Mathar_, Mar 21 2016
%Y A093917 Cf. A093915, A093916, A093918.
%K A093917 nonn,easy
%O A093917 1,1
%A A093917 _Amarnath Murthy_, Apr 25 2004
%E A093917 More terms from _Jorge Coveiro_, Jul 25 2006
%E A093917 Edited by _M. F. Hasler_, Apr 04 2009