cp's OEIS Frontend

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

%I A343447 #16 Apr 24 2021 03:25:03
%S A343447 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,
%T A343447 32,34,3,32,12,80,40,41,21,2,14,47,98,1,16,52,53,2,55,8,23,120,14,20,
%U A343447 20,64,8,3,22,68,32,20,73,74,71,38,38,32,80,82,38,8,42
%N A343447 Smallest m such that alternating integer 101...101 = A094028(m) is a multiple of A045572(n), (i.e., integers coprime with 10).
%C A343447 Every number coprime with 10 has a smallest multiple that is repunit (A099679).
%C A343447 Every positive number has a smallest multiple consisting of a succession of 1's followed by a succession of 0's (A052983).
%C A343447 Every number coprime with 10 has a smallest multiple that is alternating of the form 1010...0101 (this sequence).
%e A343447 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.
%t A343447 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 *)
%o A343447 (PARI) a045572(n)=10*(n>>2)+[-1,1,3,7][n%4+1] \\ after _Charles R Greathouse IV_ in A045572
%o A343447 a094028(n) = 1+100*(100^n-1)/99
%o A343447 a(n) = for(m=0, oo, if(a094028(m)%a045572(n)==0, return(m))) \\ _Felix Fröhlich_, Apr 15 2021
%Y A343447 Cf. A030141, A045572, A052983, A094028, A099679.
%K A343447 nonn,base
%O A343447 1,2
%A A343447 _Bernard Schott_, Apr 15 2021
%E A343447 More terms from _Felix Fröhlich_, Apr 15 2021