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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.
3
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
1,2
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.
LINKS
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.
EXAMPLE
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.
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}]
PROG
(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
CROSSREFS
Sequence in context: A031877 A222815 A233679 * A031852 A115614 A165630
KEYWORD
nonn,base
AUTHOR
David W. Wilson and M. F. Hasler, Jan 01 2016
STATUS
approved