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%I #19 May 07 2017 05:30:47
%S 6,231,20400,2003001,200045352,20000567352,1959085094400,
%T 200000030000001,20118337236261000,1999999999505541852,
%U 200000000030000000001,19994255180823548693100,1959183673472326530612252,200000000000105810631542400,20118343160415860069040000000
%N Number of representations of 10^n as sum of 6 triangular numbers.
%C a(n) is nearly 2*10^(2*n) because a(n) is almost (4*10^n+3)^2 / 8.
%H Seiichi Manyama, <a href="/A286314/b286314.txt">Table of n, a(n) for n = 0..17</a>
%F a(n) = A008440(10^n).
%F a(n) = 1/8 * (Sum_{d|4*10^n+3, d == 3 mod 4} d^2 - Sum_{d|4*10^n+3, d == 1 mod 4} d^2).
%e a(0) = 1/8 * (Sum_{d|7, d == 3 mod 4} d^2 - Sum_{d|7, d == 1 mod 4} d^2) = 1/8 * (7^2 - 1^2) = 6.
%e a(1) = 1/8 * (Sum_{d|43, d == 3 mod 4} d^2 - Sum_{d|43, d == 1 mod 4} d^2) = 1/8 * (43^2 - 1^2) = 231.
%e a(2) = 1/8 * (Sum_{d|403, d == 3 mod 4} d^2 - Sum_{d|403, d == 1 mod 4} d^2) = 1/8 * (403^2 + 31^2 - 13^2 - 1^2) = 20400.
%Y Cf. A008440, A286315.
%K nonn
%O 0,1
%A _Seiichi Manyama_, May 06 2017
%E More terms from _Seiichi Manyama_, May 07 2017
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