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A075001
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Smallest k such that the concatenation of n consecutive numbers starting with k (from k to n+k-1) is a multiple of n; or 0 if no such number exists.
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5
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1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 9, 1, 4, 7, 1, 5, 23, 1, 14, 1, 9, 9, 13, 5, 1, 21, 1, 13, 12, 1, 36, 21, 9, 3, 41, 1, 34, 33, 9, 21, 12, 9, 33, 9, 1, 13, 28, 5, 48, 1, 23, 21, 3, 1, 11, 13, 14, 41, 28, 1, 114, 115, 9, 41, 21, 9, 23, 69, 1, 61, 73, 5, 14, 43, 1, 145, 13, 9, 127, 41, 9, 95
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OFFSET
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1,4
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COMMENTS
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Conjecture: For every n there exists a k.
First occurrence of k where a(n)=k: 1, 103, 4, 13, 8, 105, 14, 87, 11, 699, 55, 29, 23, 19, 114, 261, 102, 97, 178, 219, 26, 121, 17, 151, 92, ..., . - Robert G. Wilson v
Increasing a(n)'s: 1, 3, 5, 9, 23, 36, 41, 48, 114, 115, 145, 166, 175, 221, 251, ..., at n = 1, 4, 8, 11, 17, 31, 35, 49, 61, 62, 76, 85, 122, 133, 170, 179, 217, 229, ..., . - Robert G. Wilson v
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LINKS
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EXAMPLE
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a(11) = 9 as 910111213141516171819 the concatenation of 11 numbers from 9 to 19 is divisible by 11 (11*82737383012865106529).
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MATHEMATICA
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f[n_] := Block[{c = 1, id = Range@n}, While[k = FromDigits@Flatten@IntegerDigits@id/n; ! IntegerQ@k, id++; c++ ]; c]; Array[f, 82] (* Robert G. Wilson v, Oct 20 2007 *)
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PROG
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(PARI) /* The following program assumes the conjecture is true. */ /* It has found nonzero a(n) for n up to 500. */ {for(n=1, 500, k=0; until(s%n==0, k++; s=0; for(m=k, k+n-1, s=s*(10^length(Str(m)))+m)); print1(k, ", "))}
(PARI) a(n) = {my(ld = 1, hd = n, qd, m = Mod(1, n), pow10, qdn = #digits(n), t=log(10*n+.5)\log(10)); qd = n*t+t-10^t\9; pow10 = Mod(10, n)^(qd-1); for(i = 2, n, m = m * Mod(10, n)^#digits(i) + i; ); while(1, if(lift(m) == 0, return(ld)); m -= ld * pow10; hd++; m = m * Mod(10, n)^#digits(hd) + hd; ld++; pow10*=10^(#digits(hd) - #digits(ld)); ) } \\ David A. Corneth, Aug 23 2020
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CROSSREFS
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Cf. A058183, A074991, A074992, A074993, A074994, A074995, A074996, A036377, A073086, A074999, A075000, A077306, A075000.
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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