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A254411
Limit of f(f(f(...f(0))...) modulo n as the number of iterations of f(x)=2^x+1 grows.
4
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
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
KEYWORD
nonn,easy,nice,look
AUTHOR
Max Alekseyev, Jan 30 2015
STATUS
approved