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.
3, 63, 3114, 8937, 94863, 5477133, 82395381, 706399164, 9380293167, 99497231067, 4472135831667, 62441868958167, 836594274358167, 9983486364492063, 435866837461509417, 707106074079263583, 77453069648658793167, 754718284918279954614, 8882505274864168010583
Offset: 0
Examples
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.
Links
- Shouen Wang, Chinese BBS: How many of these A's are there?
Programs
-
Mathematica
n=0;For[k=0,k<10^8/3,k++,If[Total[IntegerDigits[9k^2]]==9*(n+Ceiling@Log10@(3k)),Print[{n,3k}];n++]]
-
PARI
a(n) = my(m=1); while (sumdigits(m^2) != 9*(#Str(m)+n), m++); m; \\ Michel Marcus, Feb 10 2024
-
Python
def sd(n): return sum(int(d) for d in str(n*n)) n=0 for k in range(0,10**8,3): if sd(k)==9*(len(str(k))+n): print([n,k]) n+=1
Extensions
a(9)-a(18) from Zhao Hui Du, Feb 19 2024
Comments