|
%I #22 Jan 20 2022 12:43:09
%S 1,3025,25502500,250500250000,2500500025000000,25000500002500000000,
%T 250000500000250000000000,2500000500000025000000000000,
%U 25000000500000002500000000000000,250000000500000000250000000000000000,2500000000500000000025000000000000000000
%N a(n) = Sum_{i=0..10^n} i^3.
%C These terms k = x.y satisfy Diophantine equation x.y = (x+y)^2, when x and y have the same number of digits, "." means concatenation, and y may not begin with 0. So, this is a subsequence of A350870 and A238237. - _Bernard Schott_, Jan 20 2022
%F a(n) = (10^n+1)^2 * 10^(2*n) / 4.
%F From _Bernard Schott_, Jan 20 2022: (Start)
%F a(n) = A037156(n)^2.
%F a(n) = A350869(n) + 10^(3*n). (End)
%e a(1) = Sum_{i=0..10} i^3 = (Sum_{i=0..10} i)^2 = 3025.
%o (PARI) sumcu(n) = for(x=0,n,y=10^x;z=y^2*(y+1)^2/4;(print1(z","))) - _Cino Hilliard_, Jun 18 2007
%Y Cf. A037156, A238237, A350869, A350870.
%K easy,nonn
%O 0,2
%A _Marvin Ray Burns_
%E Edited by _N. J. A. Sloane_, Jul 02 2008 at the suggestion of _R. J. Mathar_
| 