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A373407
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Smallest positive integer k such that no more than n numbers (formed by multiplying k by a digit) are anagrams of k, or -1 if no such number exists.
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0
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OFFSET
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1,2
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COMMENTS
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For n = 2..6 all terms are divisible by 9.
For n >= 4, a(n) must be divisible by 9, or a(n) = -1, because all anagrams d*k of k for d = 2, 3, 5, 6, 8 and 9 are divisible by 9. Thus there are only 3 values of d, i.e., 1, 4 and 7, for which k*d must not be divisible by 9.
If a(n) exists for n > 1 then 9|a(n). Holds for n = 2 and n = 3 by inspection. Proof for n >= 4: if k*d is an anagram of k where 2 <= d <= 9 then k*d - k = k*(d-1) is a multiple of 9. For this to be true, k must be a multiple of 9 as d is not of the form 1 (mod 3) for all d. - David A. Corneth, Jun 04 2024
The following were constructed from multiples of cyclic numbers (cf. A180340, Wikipedia):
a(6) = 142857 = (10^6 - 1) / 7;
a(7) <= 1304347826086956521739 = 3*(10^22 - 1) / 23;
a(8) <= 1176470588235294 = 2*(10^16 - 1) / 17;
a(9) <= 105263157894736842 = 2*(10^18 - 1) / 19. (End)
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LINKS
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EXAMPLE
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a(2) = 1035, because 1035 * 1 = 1035 and 1035 * 3 = 3105 are anagrams of 1035, and no other number 1035 * i with digit i is an anagram of 1035, and no lesser number verifies this property.
Table n, k, set of multipliers.
1 1 [1]
2 1035 [1, 3]
3 123876 [1, 3, 7]
4 1402857 [1, 2, 3, 5]
5 1037520684 [1, 2, 4, 5, 8]
6 142857 [1, 2, 3, 4, 5, 6]
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PROG
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(PARI) isok(k, n) = my(d=vecsort(digits(k))); sum(i=1, 9, vecsort(digits(k*i)) == d) == n; \\ Michel Marcus, Jun 04 2024
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CROSSREFS
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KEYWORD
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more,nonn,base,fini
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AUTHOR
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STATUS
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approved
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