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%I #32 Apr 13 2020 01:41:35
%S 0,0,0,0,0,0,0,0,0,0,2,0,1,2,42,4,3,0,1,0,0,1,17,0,131,26,0,16,11,0,1,
%T 2,37,1,1,0,1,2,21,0,3,0,7,8,0,6,83,0,1,0,89,8,26,0,97,142783940,3,1,
%U 1,0,4,8,0,14,37,49994,380,20,17,0,65,0,62,1,3,-1,29,46,235,0,0,18,29,0,1,53
%N a(n) is the minimal value of k >= 0, such that the concatenation of the decimal digits of n,n+1,...,n+k is divisible by the digit sum of the concatenation, or -1 if no such k is known.
%C As with A332580 a heuristic argument based on the divergent sum of reciprocals which approximates the probability that the digit sum of the concatenation of n+1,n+2,...,n+k will divide the concatenation suggests that k should always exist. However in the first one thousand terms there are currently fourteen terms which are unknown and have a k value of at least 10^9. These are n = 76, 250, 273, 546, 585, 663, 695, 744, 749, 760, 790, 866, 867, 983. The largest known k value in this range is k = 600747353 for n = 693, which has a corresponding digit sum of 23123615211.
%C See the companion sequence A333830 for the corresponding digit sum for each value of n.
%C The author acknowledges Joseph Myers whose algorithm to find terms in A332580 was modified and used to find the large k values in this sequence.
%H Scott R. Shannon, <a href="/A333687/b333687.txt">Table of n, a(n) for n = 1..1000</a>
%e a(1) = 0 as 1 is divisible by its digit sum 1 so no concatenation of additional numbers is required. This is also true for n = 2 to 10.
%e a(11) = 2 as 11 requires the concatenation of two more numbers, 12 and 13, to form 111213, which is divisible by its digit sum 9.
%e a(12) = 0 as 12 is divisible by its digit sum 3.
%e a(16) = 4 as 16 requires the concatenation of four more numbers, 17,18,19 and 20, to form 1617181920, which is divisible by its digit sum 36.
%Y Cf. A333830, A007953, A332580, A332558, A332563, A332542, A332830, A332867, A005349.
%K sign,base
%O 1,11
%A _Scott R. Shannon_, Apr 02 2020
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