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Square root of sum defined in A007475(n) and A001032(n).
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%I #17 Feb 24 2018 12:07:23

%S 1,5,77,92,70,195,143,3854,357,245,413,4088,2257,2222,652,679,278949,

%T 3366,1281,67963,1612,8555,1518,63412,1159158,2619,2725,13862,60973,

%U 3069,10790,3128,4620,5083,42918,3406

%N Square root of sum defined in A007475(n) and A001032(n).

%C 6a(n)^2 is divisible by A001032(n). Proof: Let s = A007475(n), n = A001032(n), then a(n)^2 = sum(k=s, s+n-1, k^2) = n/6*(2n^2+(6s-3)n+6s^2-6s+1).

%H Christopher E. Thompson, <a href="/A076215/b076215.txt">Table of n, a(n) for n = 1..10438</a>

%e A001032(3)=11, A007475(3)=18, so 18^2+19^2+...+28^2 (11 terms) = 77^2.

%Y Cf. A001032, A007475.

%K nonn

%O 1,2

%A _Ralf Stephan_, Nov 03 2002

%E Offset corrected to 1 (to match A001032) by _M. F. Hasler_, Feb 02 2016