A235145 a(n) = Number of steps to reach a fixed point or 2-cycle, when iterating A235027 starting from value n.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 2

Offset: 0

Author

Antti Karttunen, Jan 03 2014

Keywords

• nonn
• base

Comments

Equally, a(n) = minimum number of steps needed to repeat k = A235027(k)(starting from k = n) until A001222(A235027(k)) = A001222(k).

Or in other words, how many times are needed to repeatedly factorize the number, to reverse the bits of each odd prime factor (with A056539) and factorize and bit-reverse the reversed factors again, until the number of prime divisors no more grows, meaning that we have found either a fixed point or entered a cycle of two.

Examples

			19, '10011' in binary, when reversed, yields '11001' = 25, when factored, yields 5 * 5, ('101' * '101' in binary), which divisors stay same when reversed, thus it took one iteration step to reach a point where the number of prime divisors no more grows. Thus a(19)=1.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..10001

Crossrefs

A235146 gives the positions of records. Cf. A001222, A056539, A074832, A235027.

Programs

PARI

revbits(n) = fromdigits(Vecrev(binary(n)), 2);
a235027(n) = {f = factor(n); for (k=1, #f~, if (f[k,1] != 2, f[k,1] = revbits(f[k,1]););); factorback(f);}
find(v, newn) = {for (k=1, #v, if (v[#v -k + 1] == newn, return (k));); return (0);}
a(n) = {ok = 0; v = [n]; while (! ok, newn = a235027(n); ind = find(v, newn); if (ind, ok = 1, v = concat(v, newn); n = newn);); #v - ind;} \\ Michel Marcus, Aug 06 2017

Formula

If A235027(A235027(n)) = n, a(n)=0, otherwise 1+a(A235027(n)).

Equally, if A001222(A235027(n)) = A001222(n), a(n)=0, otherwise 1+a(A235027(n)).

a(2n) = a(n), and in general, for composite values a(u * v) = max(a(u),a(v)).

For composite n, a(n) = Max_{p|n} a(p). [The above reduces to this: select the maximal value from all values a(p) computed for primes p dividing n]

For prime p, a(p) = 0 if A056539(p) is also prime (p is 2 or in A074832), otherwise a(p) = 1+a(A056539(p)).