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%I #18 Oct 27 2023 22:00:47
%S 1,2,3,3,3,1,2,3,4,4,3,3,2,3,4,4,3,4,3,4,3,3,3,4,4,3,5,4,5,5,4,5,3,5,
%T 4,1,2,3,4,4,3,2,3,4,5,3,4,5,4,4,4,4,4,3,5,5,5,4,5,3,5,4,5,5,2,3,4,4,
%U 3,4,5,4,5,4,4,5,4,4,4,3,5,5,6,4,5,5,5,5,5,5,3,4,4,4,5,3,4,3,5,4,5,4,5,4,3
%N Square root of the sum of the digits of k^2 when this sum is a square.
%F a(n) = sqrt(A004159(A061910(n))) = sqrt(A007953((A061910(n))^2)). - _Zak Seidov_, Jul 04 2012
%e 6^2 = 36 and 3+6 = 9 is a square, thus 3 is in the sequence. 13^2 = 169 and 1+6+9 = 16 is a square, thus 4 is in the sequence.
%p readlib(issqr): f := []: for n from 1 to 200 do if issqr(convert(convert(n^2,base,10),`+`)) then f := [op(f),sqrt(convert(convert(n^2,base,10),`+`))] fi; od; f;
%t Select[Table[Sqrt[Total[IntegerDigits[n^2]]],{n,350}],IntegerQ] (* _Jayanta Basu_, May 06 2013 *)
%Y Cf. A007953, A004159, A061909, A061910, A061912.
%K nonn,base
%O 1,2
%A _Asher Auel_, May 17 2001
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