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A333687
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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.
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2
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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, 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, 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
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OFFSET
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1,11
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COMMENTS
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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.
See the companion sequence A333830 for the corresponding digit sum for each value of n.
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.
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LINKS
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EXAMPLE
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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.
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.
a(12) = 0 as 12 is divisible by its digit sum 3.
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.
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CROSSREFS
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KEYWORD
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sign,base
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AUTHOR
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STATUS
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approved
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