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

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

Author

Ctibor O. Zizka, Feb 09 2019

Keywords

• nonn
• base

Comments

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?

Examples

			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.
		

Crossrefs

Cf. A005188, A066750, A101337.

Programs

Mathematica

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

PARI

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

Python

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