A135027 Numbers k such that the sum of the digits of k^2 is 10. Multiples of 10 are omitted.

8, 19, 35, 46, 55, 71, 145, 152, 179, 251, 332, 361, 449, 451, 548, 649, 4499, 20249, 20251, 24499, 100549, 114499, 316261

Offset: 1

Author

Zak Seidov, Feb 10 2008

Keywords

• base
• more
• nonn

Comments

A subsequence of A056020. - R. J. Mathar, Feb 10 2008

Next term > 10000000. - R. J. Mathar, Oct 20 2009

If it exists, a(24) > 10^10. - Hugo Pfoertner, May 17 2021

If it exists, a(24) > 10^29. - Michael S. Branicky, May 30 2021

Examples

			Corresponding squares are 64, 361, 1225, 2116, 3025, 5041, 21025, 23104, 32041, 63001, 110224, 130321, 201601, 203401, 300304, 421201, 20241001, 410022001, 410103001, 600201001, 10110101401, 13110021001, 100021020121.
8^2 = 64 and 6+4 = 10. 316261^2 = 100021020121 and 1+0+0+0+2+1+0+2+0+1+2+1 = 10. - _Zak Seidov_, Aug 26 2009
		

Links

• Michael S. Branicky, Python program

Crossrefs

Cf. A007953, A056020.

Programs

Mathematica

s={};Do[If[Mod[n,10]>0&&10==Total[IntegerDigits[n^2]],AppendTo[s,n]], {n,10^8}];s (* Zak Seidov, Aug 26 2009 *)

PARI

is(n) = sumdigits(n^2)==10 && n%10 > 0 \\ Felix Fröhlich, May 17 2021

Python

def A007953(n):
    a=0
    sh=n
    while sh > 0:
        a += sh % 10
        sh //= 10
    return a
def isA135027(n):
    if n % 10 == 0:
        return False
    else:
        return A007953(n**2) == 10
for n in range(70000):
    if isA135027(n):
        print(n)
# R. J. Mathar, Oct 20 2009

Python

# See linked program to go to large numbers
def ok(n): return n%10 != 0 and sum(map(int, str(n*n))) == 10
print(list(filter(ok, range(316262)))) # Michael S. Branicky, May 30 2021