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%I #5 Sep 20 2017 10:57:37
%S 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,
%T 29,27,26,31,31,33,78,33,76,37,43,39,92,45,95,37,32,39,86,35,89,49,49,
%U 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
%N 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).
%C 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.
%C 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.
%F a(n) = min { k in IN | A171102(k) + A171102(n) in A171102 }.
%e The smallest pandigital number A171102(1) = A050278(1) = 1023456789, added to itself, yields again a pandigital number, 2046913578. Therefore, a(1) = 1.
%e Similarly, A171102(1) = 1023456789 added to the second pandigital number A171102(2) = 1023456798, yields the pandigital number 2046913587. Therefore also a(2) = 1.
%e 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.
%o (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.
%Y Cf. A292569 (the actual pandigital number to be added), A171102, A050278.
%K nonn,base
%O 1,3
%A _M. F. Hasler_, Sep 19 2017
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