A257784 Numbers n such that the sum of the digits squared times the sum of the digits of n to some power equals n.

0, 1, 512, 2511, 4913, 5832, 17576, 19683, 24624, 32144, 37000, 111616, 382360, 415000, 420224, 2219400, 14041600, 16328000, 19300032, 30681423, 39203125, 62025728, 78535423, 186836625, 214292000, 432265248, 1120141312, 3479669440, 18529084125, 25342447725

Offset: 1

Author

Pieter Post, May 08 2015

Keywords

• base
• nonn

Comments

When the power is 1 the numbers are the cubes of their digit sum (A061209).

There are no 2-digit and 18-digit terms. - Chai Wah Wu, Jan 11 2016

Examples

			For power 2: 24624 = (2+4+6+2+4)^2*(2^2+4^2+6^2+2^2+4^2).
For power 3: 111616 = (1+1+1+6+1+6)^2*(1^3+1^3+1^3+6^3+1^3+6^3).
		

Links

• Giovanni Resta and Chai Wah Wu, Table of n, a(n) for n = 1..80 n = 1..43 from Giovanni Resta

Crossrefs

Cf. A061209, A115518, A130680.

Programs

Python

# WARNING: this prints numbers in the sequence, but not in increasing order.
def moda(n,a):
    kk = 0
    while n > 0:
        kk= kk+(n%10)**a
        n =int(n//10)
    return kk
def sod(n):
    kk = 0
    while n > 0:
        kk= kk+(n%10)
        n =int(n//10)
    return kk
for a in range (1, 10):
    for c in range (1, 10**8):
        if c==sod(c)**2*moda(c,a):
            print(c, end=",")

Extensions

a(16)-a(30) from Giovanni Resta, May 09 2015