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A061912 a(n) is the smallest m for which sqrt(sum of digits of m^2) = n.

%I A061912 #40 Jun 22 2024 08:16:36
%S A061912 0,1,2,3,13,67,264,1667,16667,94863,1643167,29983327,706399164,
%T A061912 31144643167,1296109172867,62441868958167,6927459779738887,
%U A061912 447213595487659543,77453069648658793167,14104963594032775808167,3146266035952345970972687
%N A061912 a(n) is the smallest m for which sqrt(sum of digits of m^2) = n.
%C A061912 a(15) <= 62441868958167. - _Donovan Johnson_, Jul 10 2012
%C A061912 a(21) <= 29999999949999914454883190583. a(22) <= 948566760423324122079007168333. - _Zhining Yang_, Jun 21 2024
%H A061912 Mathematical Reflections, <a href="https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2014-05/mr_4_2014_solutions.pdf">Solution to Problem J307</a>, Issue 5, 2014, p. 1.
%e A061912 Sum of digits of 13^2 = sum of digits of 169 = 16 is the first occurrence of 4^2, so a(4) = 13.
%p A061912 f := []: a := 1: for i from 1 to 10 do for j from 1 do if sqrt(convert(convert(j^2,base,10),`+`)) = i then f := [op(f),j]; a := j; break fi; od; od; f;
%t A061912 t={}; m=0; Do[While[Sqrt[Total[IntegerDigits[m^2]]] != n, m++]; AppendTo[t, m], {n,0,9}]; t (* _Jayanta Basu_, May 06 2013 *)
%o A061912 (PARI) a(n) = my(k=0); while(sumdigits(k^2) != n^2, k++); k; \\ _Michel Marcus_, Jan 07 2017
%Y A061912 Cf. A007953, A004159, A061909, A061910, A061911, A067072.
%Y A061912 Cf. A067075, A286650.
%K A061912 nonn,base,more
%O A061912 0,3
%A A061912 _Asher Auel_, May 17 2001
%E A061912 a(11) from _John W. Layman_, Jan 10 2002
%E A061912 a(12) from _Ryan Propper_, Jul 07 2005
%E A061912 a(13) from _Zak Seidov_, Jan 27 2011
%E A061912 a(14) from _Donovan Johnson_, Jul 10 2012
%E A061912 a(15)-a(20) from _Zhining Yang_, Jun 21 2024