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A069862
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Smallest k such that n divides the concatenation of numbers from (n+1) to (n+k), where (n+1) is on the most significant side.
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10
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1, 2, 2, 2, 5, 2, 9, 4, 8, 10, 10, 8, 22, 16, 5, 4, 2, 8, 3, 20, 20, 10, 17, 12, 25, 22, 26, 16, 25, 20, 110, 20, 11, 2, 20, 8, 998, 52, 38, 20, 60, 20, 4, 32, 35, 42, 50, 20, 96, 50, 2, 96, 93, 26, 10, 20, 3, 50, 44, 20, 46, 40, 45, 40, 50, 32, 86, 32, 17, 20, 75, 72, 26, 926, 50
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,2
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COMMENTS
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Minimum number of consecutive subsequent integers after n that must be concatenated together in ascending order such that n divides the concatenated term.
Concatenation always begins at n+1. Note that multiples of 11 seems to require more terms than any other number. 385 requires 9860. 451 requires 100270 terms be concatenated together into a 495,000 digit number. - Chuck Seggelin (barkeep(AT)plastereddragon.com), Oct 29 2003; corrected by Chai Wah Wu, Oct 19 2014
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LINKS
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EXAMPLE
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a(7) = 9 as 7 divides 8910111213141516 the concatenation of numbers from 8(= 7+1) to 16 (= 7+9).
a(5) = 5 because 5 will divide the number formed by concatenating the 5 integers after 5 in ascending order (i.e. 678910). a(385) = 9860 because 385 will divide the concatenation of 386,387,388,...,10245.
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MAPLE
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c[1] := 1:for n from 2 to 172 do k := 1:g := (n+k) mod n:while(true) do k := k+1:b := convert(n+k, base, 10):g := (g*10^nops(b)+n+k) mod n: if((g mod n)=0) then c[n] := k:break:fi:od:od:seq(c[l], l=1..172);
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MATHEMATICA
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f[n_] := Block[{k = n + 1}, d = k; While[ !IntegerQ[d/n], k++; d = d*10^Floor[Log[10, k] + 1] + k]; k - n]; Table[ f[n], {n, 1, 75}] (* Robert G. Wilson v, Nov 04 2003 *)
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PROG
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(Python)
....nk, kr, r = n+1, 1, 1 if n > 1 else 0
....while r:
........nk += 1
........kr = (kr + 1) % n
........r = (r*(10**len(str(nk)) % n)+kr) % n
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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