A256757 Number of iterations of A007733 required to reach 1.

0, 1, 2, 1, 2, 2, 3, 1, 3, 2, 3, 2, 3, 3, 2, 1, 2, 3, 4, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 2, 3, 1, 3, 2, 3, 3, 4, 4, 3, 2, 3, 3, 4, 3, 3, 4, 5, 2, 4, 3, 2, 3, 4, 4, 3, 3, 4, 4, 5, 2, 3, 3, 3, 1, 3, 3, 4, 2, 4, 3, 4, 3, 4, 4, 3, 4, 3, 3, 4, 2, 5, 3, 4, 3, 2, 4, 4, 3, 4, 3, 3, 4, 3, 5, 4, 2, 3, 4, 3, 3

Offset: 1

Author

Ivan Neretin, Apr 09 2015

Keywords

• nonn
• easy

Comments

In other words, the minimal height (not counting k) of the power tower 2^(2^(...^(2^k)...)) required to make it eventually constant modulo n (=A245970(n)) for sufficiently large k.

a(n) <= A227944(n) + 1. - Max Alekseyev, Oct 11 2016

Links

• Ivan Neretin, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A007733, A256607 (second iteration), A256758 (positions of records), A003434, A227944 (similarly built upon the totient function).

Programs

Haskell

a256757 n = fst $ until ((== 1) . snd)
            (\(i, x) -> (i + 1, fromIntegral $ a007733 x)) (0, n)
-- Reinhard Zumkeller, Apr 13 2015

Mathematica

A007733 = Function[n, MultiplicativeOrder[2, n/(2^IntegerExponent[n, 2])]];
a = Function[n, k = 0; m = n; While[m > 1, m = A007733[m]; k++]; k];
Table[a[n], {n, 100}] (* Ivan Neretin, Apr 13 2015 *)

PARI

a(n) = {if (n==1, return(0)); nb = 1; while((n = znorder(Mod(2, n/2^valuation(n, 2)))) != 1, nb++); nb;} \\ Michel Marcus, Apr 11 2015

Formula

For n>1, a(n) = a(A007733(n)) + 1.