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%I #11 Nov 24 2019 21:43:32
%S 1,5,2,4,1,3,3,4,15,4,2,3,1,2,5,2,4,8,8,25,6,2,2,6,1,5,3,7,2,3,5,8,13,
%T 2,35,8,2,3,5,4,1,4,9,4,9,1,3,4,2,9,12,18,6,45,10,1,2,4,3,5,1,14,2,3,
%U 5,11,1,2,3,5,7,8,12,16,23,34,55,12,1,1,3,4
%N Square array T(n, k) read by antidiagonals upwards, n > 0 and k > 0: T(n, k) is the least m > 0 such that m * n contains k as a substring in its decimal representation.
%C This sequence is well defined: for any n > 0 and k > 0:
%C - ceiling(k * 10^A055642(n)/n) * n starts with k,
%C - hence T(n, k) <= ceil(k * 10^A055642(n)/n) <= 10 * k,
%C - and every column is bounded,
%C - the conjectured maximum values for the first 9 columns are: 5, 12, 17, 32, 25, 24, 35, 32, 72.
%F T(1, k) = k.
%F T(n, n) = 1.
%F T(n, 1) = A317173(n).
%e Array T(n, k) begins:
%e n\k| 1 2 3 4 5 6 7 8 9 10 11 12
%e ---+------------------------------------------------------------
%e 1| 1 2 3 4 5 6 7 8 9 10 11 12
%e 2| 5 1 15 2 25 3 35 4 45 5 55 6
%e 3| 4 4 1 8 5 2 9 6 3 34 37 4
%e 4| 3 3 8 1 13 4 18 2 23 25 28 3
%e 5| 2 4 6 8 1 12 14 16 18 2 22 24
%e 6| 2 2 5 4 9 1 12 3 15 17 19 2
%e 7| 2 3 5 2 5 8 1 4 7 15 16 16
%e 8| 2 3 4 3 7 2 9 1 12 13 14 14
%e 9| 2 3 4 5 5 4 3 2 1 12 13 14
%e 10| 1 2 3 4 5 6 7 8 9 1 11 12
%o (PARI) T(n, k, base=10) = { my (w=base^#digits(k, base)); for (m=1, oo, my (mn=m*n); while (mn >= k, if (mn % w == k, return (m), mn \= base))) }
%Y Cf. A055642, A317173.
%K nonn,base,tabl
%O 1,2
%A _Rémy Sigrist_, Jul 23 2018
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