

A292570


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).


1



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

1,3


COMMENTS

The first 9*9! pandigital numbers (having each digit 09 exactly once) are listed in A050278, which is extended to the infinite sequence A171102 of pandigital numbers having each digit 09 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 10digit solution a(n) < 10^10.


LINKS



FORMULA



EXAMPLE

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.


PROG

(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.


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



