A266578 Least number N such that the product n*N has the same digits as the concatenation (n,N) (counting repetitions), or 0 if no such number exists.

0, 8714, 51, 0, 251, 21, 0, 86, 351, 0, 9209, 86073, 0, 926, 93, 0, 9635, 6012, 0, 8714, 6, 0, 9017, 651, 0, 401, 81, 0, 3701, 51, 0, 926, 40611, 0, 41, 936, 0, 3251, 6882, 0, 35, 678, 0, 9203, 3141, 0, 371, 2913, 0, 251, 3, 0, 635, 846, 0, 2171, 834, 0, 845, 21, 0, 1814, 585, 0, 281, 9843

Offset: 1

Author

David W. Wilson and M. F. Hasler, Jan 01 2016

Keywords

• nonn
• base

Comments

See A266586 for the variant where repeated digits are not counted.

One has a(3k-1) = 3*b(k)-1 with b = (2905, 84, 29, 3070, 309, 3212, 2905, ...), and a(3k) = 3*c(k) with c = (17, 7, 117, 28691, 31, 2004, ...).

If n = 1 (mod 3), then the sum of digits of n*N = N (mod 3) is always different from the sum of digits of concat(n,N) which is 1+N (mod 3), therefore a(3k+1) = 0 for all k.

Examples

			For n = 1 there cannot be a number N such that n*N (= N) has the same digits as concat(n,N) (= "1N"), therefore a(1)=0.
For n = 2 and N = 8714 we have n*N = 17428 which has the same digits (1,2,4,7,8) as concat(n,N) = 28714. This N is the smallest such number, therefore a(2) = 8714.
Since 3*51 = 153 has the same digits than concat(3,51), and 51 is the smallest such number, a(3) = 153.
For n = 4 there is again no N with the desired property, thus a(4) = 0.
Since 5*251 = 1255 has the same digits (with repetition) than "5" and "251" together, a(5) = 1255.
		

Links

• David W. Wilson, Table of n, a(n) for n = 1..10000

Programs

Mathematica

Table[If[Mod[n, 3] == 1, 0, k = 1; While[Sort@ IntegerDigits[n k] != Sort@ Join[IntegerDigits@ n, IntegerDigits@ k], k++]; k], {n, 66}]

PARI

a(n,L=if(n%3!=1,9e9),d=digits(n))=for(k=2,L,vecsort(digits(k*n))==vecsort(concat(d,digits(k)))&&return(k))

Python

from itertools import count
from collections import Counter as Ctr
def a(n):
    r = n%3
    if r == 1: return 0
    s = str(n)
    return next(N for N in count(r, 3) if Ctr(str(n*N)) == Ctr(s+str(N)))
print([a(n) for n in range(1, 67)]) # Michael S. Branicky, Nov 15 2022

Formula

a(3k+1) = 0 for all k >= 0; a(3k+2) = 2 (mod 3) for all k >= 0; a(3k) = 0 (mod 3) for all k >= 1.