A162899 - OEIS

%I #10 Sep 26 2020 12:55:13

%S 1,10,14,30,39,64,80,116,141,190,226,290,339,420,484,584,665,786,886,

%T 1030,1151,1320,1464,1660,1829,2054,2250,2506,2731,3020,3276,3600,

%U 3889,4250,4574,4974,5335,5776,6176,6660,7101,7630,8114,8690,9219,9844,10420

%N Partial sums of [A052938(n)^2].

%C Another plausible solution, besides A115391 and A116955, to A115603: Each additional term of the partial sums here is the square of a number that alternately differs +2, -1, +2, -1, ..., from the previous number that is squared: a(3) = 30 = 1^2 + 3^2 + 2^2 + 4^2, where 1, 3, 2, 4 display this pattern.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2,1,-4,1,2,-1).

%F a(n) = sum(k=0..n, A052938(n)^2).

%F a(n) = (60-36*(-1)^n+(109-9*(-1)^n)*n+24*n^2+2*n^3)/24. G.f.: (4*x^4-4*x^3-7*x^2+8*x+1) / ((x-1)^4*(x+1)^2). - _Colin Barker_, Jul 18 2013

%t LinearRecurrence[{2,1,-4,1,2,-1},{1,10,14,30,39,64},50] (* _Harvey P. Dale_, Sep 26 2020 *)

%o (PARI) a(n) = sum(k=0, n, (if(k%2==0, k+2, k+5)/2)^2)

%Y Cf. A115603, A115391, A116955.

%K nonn,easy

%O 0,2

%A _Rick L. Shepherd_, Jul 16 2009