cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A175402 a(n) is the number of iterations of {r -> (((D_1^D_2)^D_3)^...)^D_k, where D_k is the k-th decimal digit of r} needed to reach a one-digit number, starting at r = n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 2, 3, 4, 1, 1, 1, 3, 2, 3, 3, 3, 3, 2, 1, 1, 2, 3, 3, 2, 2, 2, 3, 3, 1, 1, 3, 2, 3, 3, 2, 3, 2, 2, 1, 1, 4, 4, 2, 3, 3, 3, 2, 2, 1, 1, 4, 4, 2, 2, 2, 3, 2, 2, 1, 1, 3, 4, 2, 3, 3, 2, 2, 2, 1, 1, 2, 3, 3, 2, 3, 3, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Jaroslav Krizek, May 01 2010

Keywords

Comments

Conjecture: max(a(n)) = 4.
Assuming that A020665(2) = 86, A020665(3) = 68, A020665(5) = 58, and A020665(7) = 35, this conjecture is true, since in that case the largest power of a decimal digit that has no 0 is 7^35, and of those powers p none have a(p) > 3. The only n for which a(n) = 4 are those where one iteration goes to 6^2, 7^2, 6^3, 7^3, or 2^9.

Examples

			For n = 29: a(29) = 4 because for the number 29 there are 4 steps of defined iteration: {2^9 = 512}, {(5^1)^2 = 25}, {2^5 = 32}, {3^2 = 9}.
		

Crossrefs

Programs

  • PARI
    iter(n)=my(v=eval(Vec(Str(n))));v[1]^prod(i=2,#v,v[i])
    a(n)=my(k=0);while(n>9,k++;n=iter(n));k

Extensions

Corrected, extended, comment, and program from Charles R Greathouse IV, Aug 03 2010