A075001 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.

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

Offset: 1

Author

Amarnath Murthy, Aug 31 2002

Keywords

• nonn
• base

Comments

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

a(n)=1 iff n is in A029455. - 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

Examples

			a(11) = 9 as 910111213141516171819 the concatenation of 11 numbers from 9 to 19 is divisible by 11 (11*82737383012865106529).
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000 (first 2500 terms from Robert G. Wilson v)

Crossrefs

Cf. A058183, A074991, A074992, A074993, A074994, A074995, A074996, A036377, A073086, A074999, A075000, A077306, A075000.

Cf. A029455 (indices of 1's).

Programs

Mathematica

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 *)

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

Extensions

More terms from Rick L. Shepherd, Sep 03 2002