A276502 - OEIS

%I #25 Aug 30 2018 15:51:51

%S 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,

%T 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,

%U 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

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

%C 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, ...

%C From _Charlie Neder_, Jul 16 2018: (Start)

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

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

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

%C 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)

%e 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.

%t 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 *)

%o (PARI) A047726(n) = n=eval(Vec(Str(n))); (#n)!/prod(i=0, 9, sum(j=1, #n, n[j]==i)!);

%o A055642(n) = #Str(n);

%o A007953(n) = sumdigits(n);

%o A045876(n) = ((10^A055642(n)-1)/9)*(A047726(n)*A007953(n)/A055642(n));

%o a(n) = {my(k = 1); while (A045876(n*(10^k)) % A045876(n), k++); k; }

%Y Cf. A045876.

%K nonn,base

%O 1,10

%A _Altug Alkan_, Sep 10 2016