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A292570
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Least k > 0 such that A171102(n) + A171102(k) is again a term of A171102, the pandigital numbers (having each digit from '0' to '9' at least once).
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1
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1, 1, 3, 5, 3, 2, 7, 7, 9, 20, 9, 23, 13, 19, 15, 6, 21, 4, 13, 8, 15, 17, 11, 14, 25, 25, 27, 29, 27, 26, 31, 31, 33, 78, 33, 76, 37, 43, 39, 92, 45, 95, 37, 32, 39, 86, 35, 89, 49, 49, 51, 98, 51, 101, 55, 55, 57, 18, 57, 16, 61, 104, 63, 24, 107, 22, 61, 115, 63, 10, 117, 12, 73, 97, 75, 30, 99, 28, 79, 103, 81, 116, 105, 119, 85, 44, 87, 102, 47, 100, 109, 38, 111, 113, 41, 110, 73, 50, 75, 77, 53, 74, 79, 56, 81, 62, 59, 65
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OFFSET
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1,3
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COMMENTS
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The first 9*9! pandigital numbers (having each digit 0-9 exactly once) are listed in A050278, which is extended to the infinite sequence A171102 of pandigital numbers having each digit 0-9 at least once.
For all n, a(n) is well defined, because to any pandigital number N = A171102(n) we can add the number M(N) = 123456789*10^k with k = # digits of N, which is pandigital (in the above extended sense) as well as is the sum N + M(N). In practice, there are much smaller solutions. We conjecture that there is always a 10-digit solution a(n) < 10^10.
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LINKS
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FORMULA
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EXAMPLE
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The smallest pandigital number A171102(1) = A050278(1) = 1023456789, added to itself, yields again a pandigital number, 2046913578. Therefore, a(1) = 1.
Similarly, A171102(1) = 1023456789 added to the second pandigital number A171102(2) = 1023456798, yields the pandigital number 2046913587. Therefore also a(2) = 1.
Considering the third pandigital number A171102(3) = 1023456879, we have to add itself in order to get a pandigital number, 2046913758. (Adding A171102(1) or A171102(2) yields 2046913668 and 2046913677, respectively, which are not pandigital.) Therefore a(3) = 3.
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PROG
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(PARI) a(n)={n=A171102(n); for(k=1, 9e9, #Set(digits(n+A171102(k))>9&&return(k))} \\ For illustrational purpose ; not optimized for efficiency.
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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