A185076 a(n) is the least number k such that (sum of digits of k^2) + (number of digits of k^2) = n, or 0 if no such k exists.

0, 1, 0, 10, 2, 100, 11, 1000, 4, 3, 6, 8, 19, 35, 7, 16, 34, 106, 13, 41, 24, 17, 37, 107, 323, 43, 124, 317, 67, 113, 63, 114, 134, 343, 83, 133, 367, 1024, 167, 374, 264, 314, 386, 1043, 313, 583, 1303, 3283, 707, 1183, 3316, 836, 1333, 3286, 10133

Offset: 1

Author

Carmine Suriano, Feb 23 2011

Keywords

• nonn
• base

Comments

a(n) < sqrt(10^(n-1)). 0 < a(2m) <= 10^(m-1) with the upper bound reached for 1<=m<=4. - Chai Wah Wu, Mar 15 2023

Examples

			a(7)=11 since 7 = sumdigits(121) + numberdigits(121) = 4 + 3.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..185

Crossrefs

Cf. A004159, A185679.

Programs

Mathematica

Table[k=1; While[d=IntegerDigits[k^2]; n>Length[d] && n != Total[d] + Length[d], k++]; If[Length[d] >= n, k=0]; k, {n, 50}]

Python

from itertools import count
def A185076(n):
    for k in count(1):
        if n == (t:=len(s:=str(k**2)))+sum(map(int,s)):
            return k
        if t >= n:
            return 0 # Chai Wah Wu, Mar 15 2023

Formula

n = A004159(a(n)) + A185679(a(n)).