A260719 a(n) = A091222(A260735(n)): number of irreducible factors (in ring GF(2)[X]) of the binary encoded polynomial obtained after the n-th iteration of A234742, when starting with the initial value 455.

5, 5, 3, 4, 3, 5, 9, 2, 7, 2, 6, 6, 2, 2, 2, 3, 8, 8, 6, 5, 5, 7, 6, 4, 5, 6, 2, 7, 6, 4, 5, 4, 5, 4, 5, 9, 4, 10, 3, 4, 7, 4, 4, 3, 4, 3, 5, 8, 6, 4, 7, 5, 3, 7, 3, 3, 3, 3, 3, 7, 3, 5, 6, 6, 9, 4, 9, 3, 5, 6, 3, 4, 5, 7, 7, 4, 5, 2, 10, 11, 6, 6, 7, 13, 4, 6, 5, 10, 6, 4, 7, 4, 10, 8, 3, 7, 7, 4, 5, 5, 2, 4, 8, 3, 4, 3, 7, 4, 6, 3, 15, 3, 4, 7, 6, 6, 6, 5, 5, 8, 4

Offset: 0

Author

Antti Karttunen, Aug 04 2015

Keywords

• nonn

Comments

Records occur in positions 0, 6, 37, 79, 83, 110, 329, 554, 1019, 1318, 2027, and they are 5, 9, 10, 11, 13, 15, 16, 17, 20, 21, 23.

First 2's occur at positions 7, 9, 12, 13, 14, 26, 77, 100, 127, 158, 161, 173, 183, 193, 201, 208, 442, 447, 528, 544, 642, 706, 1033, 1089, 1222, 1831.

Note that if this sequence ever obtains value 1, then the rest of terms are also 1's, as then A260735 has attained as its constant value one of the terms of A091214 (which is a subsequence of A235035, the fixed points of A234742).

Examples

			See example in A260735. This sequence gives the number of those irreducible factors (in ring GF(2)[X], not necessarily all primes in Z) that are multiplied together (in ordinary way) to get the next term of A260735. For example, a(0) = 5 (for 3 * 3 * 7 * 7 * 7), a(1) = 5 (for 3 * 7 * 7 * 13 * 13).
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..2049

Crossrefs

Cf. A091214, A091222, A234742, A235035, A260735, A260720.

Programs

PARI

allocatemem((2^30));
{my(n=455, fm); for(i=0,2049, fm=factor(Pol(binary(n))*Mod(1, 2)); write("b260719.txt", i, " ", sum(k=1, #fm~, fm[k, 2])); n = factorback(subst(lift(fm),x,2))); };

Scheme

(define (A260719 n) (A091222 (A260735 n)))

Formula

a(n) = A091222(A260735(n)).