cp's OEIS Frontend

A292684 a(n) is the number of positive integers k not divisible by 10 such that f(kN) = f(N) for N = A292683(n) and f(x) = x / (x without its first digit: A217657(x)).

%I A292684 #7 Oct 18 2017 13:32:58
%S A292684 9,4,1,9,9,4,3,3,3,3,1,1,9,9,9,7,4,9,9,9,3,3,3,3,3,3,1,1,1,1,9,9,9,9,
%T A292684 9,1,1,1,1,1,1,9,9,9,9,3,9,9,9,9,9,7,3,3,3,3,3,3,3,3,3,1,9,9,9,9,9,9,9
%N A292684 a(n) is the number of positive integers k not divisible by 10 such that f(kN) = f(N) for N = A292683(n) and f(x) = x / (x without its first digit: A217657(x)).
%C A292684 Sequence A292683 lists the numbers n which are divisible by A217657(n), i.e., by n with its first digit removed.
%C A292684 We exclude k with trailing 0's (just like in A292683) because if k*N has the property, then 10*k*N trivially also has the property.
%C A292684 Is there any number for which there are more than 9 possible k-values?
%C A292684 All of the k-values are listed in the table A292685.
%e A292684 For A292683(1) = 11, we have k = 1, ..., 9 satisfying 11*k / A217657(11*k) = 11.
%e A292684 For A292683(2) = 12, we have k = 1, 2, 3, 4 satisfying 12*k / A217657(12*k) = 6.
%e A292684 For A292683(3) = 15, we have only k = 1 satisfying 15*k / A217657(15*k) = 3.
%e A292684 For A292683(4) = 21, we have k = 1, 2, 3, 4, 5, 15, 25, 35 and 45 satisfying 21*k / A217657(21*k) = 2.
%o A292684 (PARI) (A217657(n)=n%10^logint(n,10)); A292684(n,N=A292683(n),r=N/A217657(N),a=[1])={for(k=2,oo,k%10||next;k>10*a[#a]&&break;A217657(k*N)*r==k*N&&a=concat(a,k));#a} \\ Instead of the 1st arg. n, one can directly give N (= A292683(n) by default) as 2nd arg. One could store only the last 'a' (and increase a counter) instead of storing all 'a's.
%Y A292684 Cf. A292683, A292685, A217657, A000030.
%K A292684 nonn,base
%O A292684 1,1
%A A292684 _M. F. Hasler_, Oct 17 2017