A254411 Limit of f(f(f(...f(0))...) modulo n as the number of iterations of f(x)=2^x+1 grows.

0, 1, 0, 1, 3, 3, 2, 1, 0, 3, 9, 9, 6, 9, 3, 1, 3, 9, 0, 13, 9, 9, 7, 9, 18, 19, 0, 9, 20, 3, 9, 1, 9, 3, 23, 9, 32, 19, 6, 33, 34, 9, 40, 9, 18, 7, 35, 33, 23, 43, 3, 45, 42, 27, 53, 9, 0, 49, 32, 33, 54, 9, 9, 1, 58, 9, 44, 37, 30, 23, 30, 9, 2, 69, 18, 57, 9, 45, 65, 33, 0, 75, 25, 9, 3, 83, 78, 9, 68, 63, 58, 53, 9, 35, 38, 33, 71, 23, 9, 93

Offset

1,5

Comments

Also, limit of f(f(f(...f(m))...) modulo n for any integer m>=0.

Links

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Formula

a(n) = limit of A254429(m) mod n as m grows.

a(n) = A254429(A227944(n)+k) mod n for any k>=1. In particular, a(n) = A254429(n) mod n.

Prog

(PARI) { A254411(m) = my(g); if(m==1, return(0)); g=2^valuation(m, 2); m\=g; lift( chinese(Mod(0, g), Mod(2, m)^A254411(eulerphi(m)) ) + 1) }

Crossrefs

Cf. A245970, A254410

Sequence in context: A106449 A256522 A097278 * A056223 A261143 A100013

Adjacent sequences: A254408 A254409 A254410 * A254412 A254413 A254414

Keyword

nonn,easy,nice,look

Author

Max Alekseyev, Jan 30 2015

Status

approved