A335884 The length of a longest path from n to a power of 2, when applying the nondeterministic maps k -> k - k/p and k -> k + k/p, where p can be any of the odd prime factors of k, and the maps can be applied in any order.

0, 0, 1, 0, 2, 1, 2, 0, 2, 2, 3, 1, 3, 2, 3, 0, 3, 2, 3, 2, 3, 3, 4, 1, 4, 3, 3, 2, 4, 3, 4, 0, 4, 3, 4, 2, 4, 3, 4, 2, 4, 3, 4, 3, 4, 4, 5, 1, 4, 4, 4, 3, 4, 3, 5, 2, 4, 4, 5, 3, 5, 4, 4, 0, 5, 4, 5, 3, 5, 4, 5, 2, 5, 4, 5, 3, 5, 4, 5, 2, 4, 4, 5, 3, 5, 4, 5, 3, 5, 4, 5, 4, 5, 5, 5, 1, 5, 4, 5, 4, 5, 4, 5, 3, 5

Offset: 1

Author

Antti Karttunen, Jun 29 2020

Keywords

• nonn

Comments

The length of a longest path from n to a power of 2, when using the transitions x -> A171462(x) and x -> A335876(x).

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537

Crossrefs

Cf. A052126, A171462, A335875, A335876, A335881, A335885, A335904, A335908.

Cf. A335883 (position of the first occurrence of each n).

Programs

PARI

A335884(n) = { my(f=factor(n)); sum(k=1,#f~,if(2==f[k,1],0,f[k,2]*(1+max(A335884(f[k,1]-1),A335884(f[k,1]+1))))); };

PARI

\\ Or empirically as:
A171462(n) = if(1==n,0,(n-(n/vecmax(factor(n)[, 1]))));
A335876(n) = if(1==n,2,(n+(n/vecmax(factor(n)[, 1]))));
A209229(n) = (n && !bitand(n,n-1));
A335884(n) = if(A209229(n),0,my(xs=Set([n]),newxs,a,b,u); for(k=1,oo, newxs=Set([]); if(!#xs, return(k-1)); for(i=1,#xs,u = xs[i]; a = A171462(u); if(!A209229(a), newxs = setunion([a],newxs)); b = A335876(u); if(!A209229(b), newxs = setunion([b],newxs))); xs = newxs));

Formula

Fully additive with a(2) = 0, and a(p) = 1+max(a(p-1), a(p+1)), for odd primes p.

For all n >= 1, A335904(n) >= a(n) >= A335881(n) >= A335875(n) >= A335885(n).

For all n >= 0, a(A335883(n)) = n.