login
floor(s*t/(s+t)), where s(n) are the squares, t(n) the triangular numbers.
0

%I #12 Oct 07 2015 11:55:27

%S 0,1,3,6,9,13,17,23,28,35,42,50,59,68,78,88,100,111,124,137,151,166,

%T 181,197,213,231,248,267,286,306,327,348,370,392,416,439,464,489,515,

%U 542,569,597,625,655,684,715,746,778,811,844,878,912,948,983,1020,1057

%N floor(s*t/(s+t)), where s(n) are the squares, t(n) the triangular numbers.

%H <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,0,0,0,0,0,0,1,-2,1).

%F a(n) = floor(s*t/(s+t)) where s = A000290(n) = n^2, t = A000217(n) = n*(n+1)/2. a(n) = floor((n^3+n^2) / (3*n+1)).

%F G.f.: (-x^10 + 2*x^9 - x^8 + 2*x^7 + x^5 + x^3 + x^2 + x)/((1-x)^2*(1-x^9)). - _Ralf Stephan_, Jan 15 2014

%t With[{nn=60},Floor[Times@@#/Total[#]]&/@Thread[{Range[nn]^2,Accumulate[ Range[ nn]]}]] (* or *) LinearRecurrence[{2,-1,0,0,0,0,0,0,1,-2,1},{0,1,3,6,9,13,17,23,28,35,42},60] (* _Harvey P. Dale_, Oct 07 2015 *)

%o (Python)

%o for n in range(1, 99):

%o s = n*n

%o t = n*(n+1)/2

%o print str(s*t//(s+t))+',',

%Y Cf. A000217, A000290.

%K nonn,easy

%O 1,3

%A _Alex Ratushnyak_, Jan 05 2014