A235146 a(n) = Least integer k such that it takes n iterations of "factor and reverse bits of odd prime divisors" (A235027) before a fixed point or cycle of 2 is reached; records in A235145.

0, 19, 139, 719, 4793, 23773, 260863, 2375231, 21793843

Offset: 0

Author

Antti Karttunen, Jan 03 2014

Keywords

• nonn
• base
• more
• hard

Comments

Note, as for all composite values A235145(u * v) = max(A235145(u), A235145(v)) which can be further reduced as A235145(n) = Max_{p|n} A235145(p), and because for any odd prime p, lpf(A056539(p)) >= 3 (where lpf = A020639, the least prime dividing n) while 1/2 < A056539(n)/n < 2, it follows that this sequence gives also the positions of the records in A235145, as its new values must appear in order.

Also, because of that multiplicativity criterion, all terms (after zero) must be primes, and specifically, the terms are a subset of A235030 (i.e., of A204219).

Conjecture: additional property is that the primes here belong to that subset of p in A204219 for which A056539(p) > p. The list of such primes begins as: 19, 79, 103, 137, 139, 149, 157, 179, 191, 239, 271, 281, 293, 311, 317, 347, 367, 379, 439, 523, 541, 547, 557, 563, 569, 587, 607, 613, 647, 659, 719, 733, 743, 751, 787, ...

Crossrefs

A subset of A235030 and A204219.

Cf. A235145, A235027, A056539.

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);}
a235145(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;}
a(n) = {k = 0; while (a235145(k) != n, k = nextprime(k+1)); k;}
lista(nn) = {kprec = 0; for (n=0, nn, k = kprec; while (a235145(k) != n, k = nextprime(k+1)); print1(k, ", "); kprec = k;);} \\ Michel Marcus, Aug 06 2017

Extensions

a(5)-a(8) from Michel Marcus, Aug 06 2017