A343447 Smallest m such that alternating integer 101...101 = A094028(m) is a multiple of A045572(n), (i.e., integers coprime with 10).

0, 2, 2, 8, 10, 2, 7, 8, 2, 10, 26, 13, 14, 32, 2, 2, 4, 20, 22, 20, 23, 12, 8, 28, 29, 8, 32, 32, 34, 3, 32, 12, 80, 40, 41, 21, 2, 14, 47, 98, 1, 16, 52, 53, 2, 55, 8, 23, 120, 14, 20, 20, 64, 8, 3, 22, 68, 32, 20, 73, 74, 71, 38, 38, 32, 80, 82, 38, 8, 42

Offset: 1

Author

Bernard Schott, Apr 15 2021

Keywords

• nonn
• base

Comments

Every number coprime with 10 has a smallest multiple that is repunit (A099679).

Every positive number has a smallest multiple consisting of a succession of 1's followed by a succession of 0's (A052983).

Every number coprime with 10 has a smallest multiple that is alternating of the form 1010...0101 (this sequence).

Examples

			A045572(3) = 7, the smallest alternating multiple of 7 in A094028 is A094028(2) = 10101 because 1443*7 = 10101, as 1 and 101 are not divisible by 7, so a(3) = 2.
		

Crossrefs

Cf. A030141, A045572, A052983, A094028, A099679.

Programs

Mathematica

a[n_] := Module[{k = (5*n + (Mod[3*n + 2, 4] - 4))/2, m = 0}, While[! Divisible[1 + 100*(100^m - 1)/99, k], m++]; m]; Array[a, 100] (* Amiram Eldar, Apr 15 2021 *)

PARI

a045572(n)=10*(n>>2)+[-1,1,3,7][n%4+1] \\ after Charles R Greathouse IV in A045572
a094028(n) = 1+100*(100^n-1)/99
a(n) = for(m=0, oo, if(a094028(m)%a045572(n)==0, return(m))) \\ Felix Fröhlich, Apr 15 2021

Extensions

More terms from Felix Fröhlich, Apr 15 2021