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%I #7 Sep 20 2017 10:57:26
%S 1023456789,1023456789,1023456879,1023456978,1023456879,1023456798,
%T 1023457689,1023457689,1023457869,1023459687,1023457869,1023459867,
%U 1023458679,1023459678,1023458769,1023456987,1023459768,1023456897,1023458679,1023457698,1023458769
%N Least pandigital number which sums up with the n-th pandigital number A171102(n) to another pandigital number.
%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) (equal to the concatenation of 123456789 and 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) = A171102(A292570(n)).
%F a(n) = min { N in A171102 | N + 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) = A171102(1) = 1023456789.
%e Similarly, A171102(1) = 1023456789, added to the second pandigital number A171102(2) = 1023456798, yields the pandigital number 2046913587. Therefore also a(2) = A171102(1) = 1023456789.
%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) = A171102(3) = 1023456879.
%o (PARI) a(n)={n=A171102(n);for(k=1,9e9,#Set(digits(n+A171102(k)))>9&&return(A171102(k)))} \\ For illustrational purpose only; not optimized for efficiency.
%Y Cf. A292570 (index of a(n) within A171102), A171102, A050278.
%K nonn,base
%O 1,1
%A _M. F. Hasler_, Sep 19 2017
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