A276502 Least k > 0 such that A045876(n) divides A045876(n*10^k).

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 6

Offset: 1

Author

Altug Alkan, Sep 10 2016

Keywords

• nonn
• base

Comments

Corresponding values of A045876(n*10^a(n))/A045876(n) are 11, 11, 11, 11, 11, 11, 11, 11, 11, 101, 303, 303, 303, 303, 303, 303, 303, 303, 303, 101, 303, 303, 303, 303, 303, 303, 303, 303, 303, 101, ...

From Charlie Neder, Jul 16 2018: (Start)

From the formula for A045876(n) we make the following modifications:

- A (the mean of the digits) becomes S/D (sum of digits / # of digits)

- N (# of arrangements of digits) becomes R*Z (# of arrangements of nonzero digits * # of ways to insert the proper number of zeros)

Appending zeros to n does not change S or R, so if (S*R*Z*I/D)(n) divides (S*R*Z*I/D)(n*10^k), then (Z*I/D)(n) divides (Z*I/D)(n*10^k). However, Z, I, and D are completely determined by the number of digits of n and the number of those digits which are zero, so a(n) = a(A136400(n)). (End)

Examples

			a(10) = 2 because A045876(10) = 1+10 = 11 does not divide A045876(100) = 1+10+100 = 111 and 11 divides A045876(1000) = 1+10+100+1000 = 1111.
		

Crossrefs

Cf. A045876.

Programs

Mathematica

A045876[n_] := Total[FromDigits /@ Permutations[IntegerDigits[n]]]; a[n_] := For[k = 1, True, k++, If[Divisible[A045876[n*10^k], A045876[n]], Return[k] ] ]; Array[a, 101] (* Jean-François Alcover, Jul 26 2017 *)

PARI

A047726(n) = n=eval(Vec(Str(n))); (#n)!/prod(i=0, 9, sum(j=1, #n, n[j]==i)!);
A055642(n) = #Str(n);
A007953(n) = sumdigits(n);
A045876(n) = ((10^A055642(n)-1)/9)*(A047726(n)*A007953(n)/A055642(n));
a(n) = {my(k = 1); while (A045876(n*(10^k)) % A045876(n), k++); k; }