A028839 -id:A028839 - cp's OEIS frontend

A257786 Numbers n such that the square root of the sum of the digits times the sum of the digits of n in some power equal n.

0, 1, 27, 376, 13131, 234595324075, 54377519037479592374299, 8326623359858152426050700, 1513868951125582592290131113769528

Offset: 1

Pieter Post, May 08 2015

It appears that this sequence is finite.

			376 = sqrt(3+7+6)*(3^2+7^2+6^2).
13131 = sqrt(1+3+1+3+1)*(1^7+3^7+1^7+3^7+1^7).

Cf. A028839, A061209, A115518, A130680.

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:
        k= kk+(n%10)
        n =int(n//10)
    return kk
for a in range (1, 10):
    for c in range (1, 10**8):
        if c**2==sod(c)*moda(c,a)**2:
            print (a,c, sod(c),moda(c,a))

a(6) from Giovanni Resta, May 09 2015

a(7)-a(9) from Chai Wah Wu, Nov 29 2015

A331203 Numbers k such that k/(digsum(k)) is an integer cube.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 72, 243, 320, 486, 512, 640, 704, 832, 960, 1000, 1088, 1125, 2000, 2401, 3000, 3430, 4000, 4116, 4802, 5000, 5145, 5831, 6000, 6174, 6517, 6860, 7000, 7546, 8000, 8575, 8918, 9000, 9216, 9947, 19683, 35152, 35937, 41743, 43940, 46137

Offset: 1

K. D. Bajpai, Jan 12 2020

If m belongs to the sequence, then 1000*m also belongs to the sequence. - Rémy Sigrist, Jan 12 2020

			a(11) = 243: 243/(2 + 4 + 3) = 27 = 3^3.
a(12) = 320: 320/(3 + 2 + 0) = 64 = 4^3.

Cf. A001102, A005349, A007953, A028839, A059094.

[n : n in[1 .. 1000] | IsIntegral((n/(&+Intseq(n)))^(1/3))];

Select[Range[100000], IntegerQ[CubeRoot[#/Total[IntegerDigits[#]]]] &]

is(n) = my (k=n/sumdigits(n)); type(k)==type(42) && ispower(k,3) \\ Rémy Sigrist, Jan 12 2020

cp's OEIS Frontend

A257786 Numbers n such that the square root of the sum of the digits times the sum of the digits of n in some power equal n.

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