A322823 a(n) = 0 if n is 1 or a Fermi-Dirac prime (A050376), otherwise a(n) = 1 + a(A300840(n)).

0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 2, 0, 1, 2, 0, 0, 1, 0, 3, 2, 1, 0, 2, 0, 1, 2, 3, 0, 2, 0, 1, 2, 1, 4, 3, 0, 1, 2, 3, 0, 2, 0, 3, 4, 1, 0, 2, 0, 1, 2, 3, 0, 2, 4, 3, 2, 1, 0, 3, 0, 1, 5, 3, 4, 2, 0, 3, 2, 4, 0, 3, 0, 1, 2, 3, 5, 2, 0, 4, 0, 1, 0, 3, 4, 1, 2, 3, 0, 4, 5, 3, 2, 1, 4, 2, 0, 1, 6, 3, 0, 2, 0, 3, 4

Offset: 1

Author

Antti Karttunen, Dec 29 2018

Keywords

• nonn

Comments

For n > 1, a(n) gives the number of edges needed to traverse from n to reach the leftmost branch (where the terms of A050376 are located) in the binary tree illustrated in A052330.

Links

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

Crossrefs

Cf. A050376, A052330, A052331, A302777, A300840, A322822.

Programs

PARI

up_to = 10000;
ispow2(n) = (n && !bitand(n,n-1));
A302777(n) = ispow2(isprimepower(n));
A050376list(up_to) = { my(v=vector(up_to), i=0); for(n=1,oo,if(A302777(n), i++; v[i] = n); if(i == up_to,return(v))); };
v050376 = A050376list(up_to);
A050376(n) = v050376[n];
A052330(n) = { my(p=1,i=1); while(n>0, if(n%2, p *= A050376(i)); i++; n >>= 1); (p); };
A052331(n) = { my(s=0,e); while(n > 1, fordiv(n, d, if(((n/d)>1)&&ispow2(isprimepower(n/d)), e = vecsearch(v050376, n/d); if(!e, print("v050376 too short!"); return(1/0)); s += 2^(e-1); n = d; break))); (s); };
A300840(n) = A052330(A052331(n)>>1);
A322823(n) = if((1==n)||(1==A302777(n)),0,1+A322823(A300840(n)));

Formula

a(1) = 0; for n > 1, if A302777(n) == 1, a(n) = 0, otherwise a(n) = 1 + a(A300840(n)).