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A369955 a(n) is the least integer m such that the sum of the digits of m^2 is 9*(k+n) where k is the number of digits of m.

%I A369955 #42 Feb 24 2024 21:21:21
%S A369955 3,63,3114,8937,94863,5477133,82395381,706399164,9380293167,
%T A369955 99497231067,4472135831667,62441868958167,836594274358167,
%U A369955 9983486364492063,435866837461509417,707106074079263583,77453069648658793167,754718284918279954614,8882505274864168010583
%N A369955 a(n) is the least integer m such that the sum of the digits of m^2 is 9*(k+n) where k is the number of digits of m.
%C A369955 3|a(n).
%H A369955 Shouen Wang, <a href="http://bbs.mathchina.com/bbs/forum.php?mod=viewthread&amp;tid=2060076">Chinese BBS: How many of these A's are there?</a>
%e A369955 a(2)=3114 because 3114 is the least 4-digit integer whose square has digit sum 9*(4+2) = 9*6 = 54: 3114^2 = 9696996 and 9+6+9+6+9+9+6 = 54.
%t A369955 n=0;For[k=0,k<10^8/3,k++,If[Total[IntegerDigits[9k^2]]==9*(n+Ceiling@Log10@(3k)),Print[{n,3k}];n++]]
%o A369955 (Python)
%o A369955 def sd(n):
%o A369955     return sum(int(d) for d in str(n*n))
%o A369955 n=0
%o A369955 for k in range(0,10**8,3):
%o A369955     if sd(k)==9*(len(str(k))+n):
%o A369955         print([n,k])
%o A369955         n+=1
%o A369955 (PARI) a(n) = my(m=1); while (sumdigits(m^2) != 9*(#Str(m)+n), m++); m; \\ _Michel Marcus_, Feb 10 2024
%Y A369955 Cf. A004159, A067179, A369953.
%K A369955 nonn,base
%O A369955 0,1
%A A369955 _Zhining Yang_, Feb 06 2024
%E A369955 a(9)-a(18) from _Zhao Hui Du_, Feb 19 2024