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Numbers that are the sum of 3 squares > 1.
1

%I #11 Dec 17 2021 07:36:26

%S 12,17,22,24,27,29,33,34,36,38,41,43,44,45,48,49,50,54,56,57,59,61,62,

%T 65,66,67,68,69,70,72,74,75,76,77,78,81,82,83,84,86,88,89,90,93,94,96,

%U 97,98,99,101,102,104,105,106,107,108,109,110,113,114,115,116,117,118,120,121,122,123,125,126,129

%N Numbers that are the sum of 3 squares > 1.

%H Michael S. Branicky, <a href="/A302359/b302359.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Su#ssq">Index entries for sequences related to sums of squares</a>

%e 33 is in the sequence because 33 = 2^2 + 2^2 + 5^2.

%t max = 130; f[x_] := Sum[x^(k^2), {k, 2, 20}]^3; Exponent[#, x] & /@ List @@ Normal[Series[f[x], {x, 0, max}]]

%t With[{nn=15},Select[Union[Total/@Tuples[Range[2,nn]^2,3]],#<=nn^2+8&]] (* _Harvey P. Dale_, Jul 05 2021 *)

%o (Python)

%o from itertools import count, takewhile, combinations_with_replacement as mc

%o def aupto(N):

%o sqrs = list(takewhile(lambda x: x<=N, (i**2 for i in count(2))))

%o sum3 = set(sum(c) for c in mc(sqrs, 3) if sum(c) <= N)

%o return sorted(sum3)

%o print(aupto(129)) # _Michael S. Branicky_, Dec 17 2021

%Y Cf. A000378, A000408, A085989, A214514, A281155.

%K nonn

%O 1,1

%A _Ilya Gutkovskiy_, Apr 06 2018