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Number of inequivalent (ordered) solutions to n^2 = sum of 5 squares of integers >= 0.
4

%I #10 Feb 17 2015 09:54:24

%S 1,1,2,2,3,4,5,6,7,8,13,12,13,17,25,22,27,31,35,38,46,49,61,61,61,73,

%T 92,83,112,106,118,127,147,138,185,175,178,198,239,212,254,262,298,

%U 294,341,304,404,376,385,432,483,441,539,517,560,551,680,587,745,693,698

%N Number of inequivalent (ordered) solutions to n^2 = sum of 5 squares of integers >= 0.

%H Alois P. Heinz, <a href="/A065459/b065459.txt">Table of n, a(n) for n = 0..750</a>

%e a(5)=4 because 25 produces {0,0,0,0,5}, {0,0,0,3,4}, {0,1,2,2,4}, {2,2,2,2,3}.

%t Length/@Table[SumOfSquaresRepresentations[5, (k)^2], {k, 72}]

%Y Cf. A063014, A016727.

%Y Column k=5 of A255212.

%K nonn

%O 0,3

%A _Wouter Meeussen_, Nov 18 2001

%E a(0)=1 prepended by _Alois P. Heinz_, Feb 17 2015