A306361 -id:A306361 - cp's OEIS frontend

A306360 Numbers k such that A101337(k)/k is an integer.

1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 459, 1634, 8208, 9474, 13598, 48495, 54748, 92727, 93084, 119564, 174961, 306979, 548834, 1741725, 3194922, 4210818, 9800817, 9926315, 12720569, 24678050, 24678051, 88593477, 144688641, 146511208

Offset: 1

Ctibor O. Zizka, Feb 10 2019

A005188 is a subsequence of this sequence.

Sequence is finite. In particular, a(n) < 10^60. If k >= 10^60, then A101337(k) < k. - Chai Wah Wu, Feb 26 2019

			For k = 1, (1^1)/1 = 1;
for k = 459, (4^3 + 5^3 + 9^3) / 459 = 2.

Giovanni Resta, Table of n, a(n) for n = 1..109 (full sequence; first 56 terms from Chai Wah Wu)
Barry Fagin, Idempotent Factorizations of Square-free Integers, Preprints (2019).

Cf. A005188, A101337, A306354, A306361.

Select[Range[10^6], IntegerQ[Total[IntegerDigits[#]^IntegerLength[#]]/#] &] (* Michael De Vlieger, Aug 01 2019 *)

isok(n) = frac(A101337(n)/n) == 0; \\ Michel Marcus, Feb 11 2019

select( is(n)=!(A101337(n)%n), [0..999]) \\ M. F. Hasler, Nov 17 2019

A306360_list, k = [], 1
while k < 10**9:
    s = str(k)
    l, c = len(s), 0
    for i in range(l):
        c = (c + int(s[i])**l) % k
    if c == 0:
        A306360_list.append(k)
    k += 1 # Chai Wah Wu, Feb 26 2019

a(22)-a(37) from Daniel Suteu, Feb 10 2019

A306354 a(n) = gcd(n, A101337(n)).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 2, 1, 4, 1, 2, 1, 4, 1, 3, 1, 1, 3, 1, 1, 9, 1, 1, 3, 8, 1, 2, 1, 4, 1, 2, 1, 16, 1, 25, 1, 1, 1, 1, 5, 1, 1, 1, 1, 12, 1, 2, 9, 4, 1, 6, 1, 4, 3, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 16, 1, 2, 1, 4, 1, 2, 1, 8, 1, 9

Offset: 1

Ctibor O. Zizka, Feb 09 2019

A101337(n) / n = r, r an integer, gives A306360. A101337(n) / n = 1 gives A005188. n / A101337(n) = s, s an integer, gives A306361. The motivation for this sequence was the question as to which numbers n have the property A101337(n) / n = r and the property n / A101337(n) = s?

			For n = 24, a(24) = gcd(24, 2*2 + 4*4) = gcd(24,20) = 4, thus a(24) = 4;
for n = 153, a(153) = gcd(153, 1*1*1 + 5*5*5 + 3*3*3) = gcd(153,153) = 153, thus a(153) = 153.

Cf. A005188, A066750, A101337.

Array[GCD[#1, Total[#2^Length[#2]]] & @@ {#, IntegerDigits@ #} &, 90] (* Michael De Vlieger, Feb 09 2019 *)

a(n) = my(d=digits(n)); gcd(n, sum(i=1, #d, d[i]^#d)); \\ Michel Marcus, Feb 12 2019

from math import gcd
def A306354(n): return gcd(n,sum(int(d)**len(str(n)) for d in str(n))) # Chai Wah Wu, Jan 26 2022

A334601 Positive integers m such that sum of cubes of the digits of m, t=A055012(m), is a multiple of m (m/A055012(m) is an integer >= 1).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 24, 27, 37, 48, 153, 370, 371, 407, 459

Offset: 1

Zak Seidov, May 07 2020 and May 12 2020

Corresponding values of t: 1, 8, 27, 64, 125, 216, 343, 512, 729, 72, 351, 370, 576, 153, 370, 371, 407, 918 (first 9 terms are all cubes).
Corresponding values of t/m: 1, 4, 9, 16, 25, 36, 49, 64, 81, 3, 13, 10, 12, 1, 1, 1, 1, 2 (first 9 terms are all squares).
The subsequence of numbers m such that sum of cubes of its digits is equal to m is A046197 \ {0}. - Bernard Schott, May 11 2020

			m = 459, t = 4^3 + 5^3 + 9^3 = 918, t/m = 2.

Cf. A005188, A046197, A046459, A055012, A055642, A101337, A306354, A306360, A306361, A329291.

Select[Range[500], Divisible[Plus @@ (IntegerDigits[#]^3), #] &] (* Amiram Eldar, May 11 2020 *)

isok(m) = my(d=digits(m)); sum(k=1, #d, d[k]^3) % m == 0; \\ Michel Marcus, May 14 2020

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A306360 Numbers k such that A101337(k)/k is an integer.

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A306354 a(n) = gcd(n, A101337(n)).

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A334601 Positive integers m such that sum of cubes of the digits of m, t=A055012(m), is a multiple of m (m/A055012(m) is an integer >= 1).

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