A248323 - OEIS

A248323

Numbers n which appear at least twice in A037278(n), concatenation of their divisors written in base 10.

11, 12, 42, 84, 124, 135, 248, 325, 366, 510, 550, 555, 624, 650, 714, 1010, 1111, 1525, 1734, 2510, 3913, 4020, 5100, 5500, 5610, 5625, 8040, 11111, 13515, 16575, 21175, 24104, 25500, 28160, 34170, 35250, 35610, 36800, 37444, 44919, 50100, 51020, 51102, 51250, 52000

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OFFSET

1,1

COMMENTS

Overlapping is allowed, so a(1) = 11 is in the sequence, with concatenated divisors A037278(11) = "111".

All repunits (10^k-1)/9 = A000042(k) = A002275(k) with even k = number of digits (as to be divisible by 11) but not multiples of 3, i.e., k in A047235, have divisors {1, 11, ..., 1010...101, 1111...111} and therefore are in this sequence.

Numbers n = floor(10^(8+3k)/7), k>=0, also belong to this sequence; for k>=2m, the number n appears (at least) m+2 times in A037278(n). [Found by extending results from Hans Havermann.]

The smallest terms that appear more than twice in the concatenation are 1111, 400200, 800400, 28571428, all 3 times, and 42857142, 4 times. - Hans Havermann, Oct 05 2014

LINKS

Chai Wah Wu, Table of n, a(n) for n = 1..340

E. Angelini, Divisors showing a string, SeqFan list, Oct 03 2014

EXAMPLE

The divisors of 42 are 1, 2, 3, 6, 7, 14, 21, 42, and "42" appears twice in their concatenation A037278(42) = "12367142142".

PROG

(PARI) is(n)={d=concat(apply(digits, divisors(n))); n=digits(n); for(j=0, #d-#n-1, for(i=1, #n, d[i+j]==n[i]||next(2)); return(1))}

(Python)

from sympy import divisors

import re

A248323_list = [n for n in range(1, 10**7) if len(list(re.finditer('(?='+str(n)+')', ''.join([str(d) for d in divisors(n)])))) > 1]

# Chai Wah Wu, Nov 01 2014