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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.

%I A266578 #17 Nov 15 2022 12:20:34
%S A266578 0,8714,51,0,251,21,0,86,351,0,9209,86073,0,926,93,0,9635,6012,0,8714,
%T A266578 6,0,9017,651,0,401,81,0,3701,51,0,926,40611,0,41,936,0,3251,6882,0,
%U A266578 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
%N 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.
%C A266578 See A266586 for the variant where repeated digits are not counted.
%C A266578 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, ...).
%C A266578 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.
%H A266578 David W. Wilson, <a href="/A266578/b266578.txt">Table of n, a(n) for n = 1..10000</a>
%F A266578 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.
%e A266578 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.
%e A266578 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.
%e A266578 Since 3*51 = 153 has the same digits than concat(3,51), and 51 is the smallest such number, a(3) = 153.
%e A266578 For n = 4 there is again no N with the desired property, thus a(4) = 0.
%e A266578 Since 5*251 = 1255 has the same digits (with repetition) than "5" and "251" together, a(5) = 1255.
%t A266578 Table[If[Mod[n, 3] == 1, 0, k = 1; While[Sort@ IntegerDigits[n k] != Sort@ Join[IntegerDigits@ n, IntegerDigits@ k], k++]; k], {n, 66}]
%o A266578 (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))
%o A266578 (Python)
%o A266578 from itertools import count
%o A266578 from collections import Counter as Ctr
%o A266578 def a(n):
%o A266578     r = n%3
%o A266578     if r == 1: return 0
%o A266578     s = str(n)
%o A266578     return next(N for N in count(r, 3) if Ctr(str(n*N)) == Ctr(s+str(N)))
%o A266578 print([a(n) for n in range(1, 67)]) # _Michael S. Branicky_, Nov 15 2022
%K A266578 nonn,base
%O A266578 1,2
%A A266578 _David W. Wilson_ and _M. F. Hasler_, Jan 01 2016