A246353 If n = Sum 2^e_i, e_i distinct, then a(n) = Position of (product prime_{e_i+1}) among squarefree numbers (A005117).

1, 2, 3, 5, 4, 7, 11, 19, 6, 10, 14, 28, 23, 44, 65, 129, 8, 15, 21, 41, 34, 69, 101, 203, 48, 94, 144, 283, 233, 470, 703, 1405, 9, 17, 26, 49, 40, 80, 120, 236, 57, 111, 168, 334, 279, 554, 833, 1661, 89, 176, 261, 521, 438, 873, 1304, 2610, 609, 1217, 1827, 3650, 3046, 6091, 9131

Offset: 0

Author

Antti Karttunen, Aug 23 2014

Keywords

• nonn

Comments

This is an inverse function to A048672. Note the indexing: here the domain starts from 0, but the range starts from 1, while in A048672 it is the opposite.

Sequence is obtained when the range of A019565 is compacted so that it becomes surjective on N, thus the logarithmic scatter plots look very similar. (Same applies to A064273). Compare also to the plot of A005940.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..478
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Cf. A000040, A005940, A013928, A019565, A048672, A064273.

Programs

PARI

allocatemem(234567890);
default(primelimit, 2^22)
uplim_for_13928 = 13123111;
v013928 = vector(uplim_for_13928); A013928(n) = v013928[n];
v013928[1]=0; n=1; while((n < uplim_for_13928), if(issquarefree(n), v013928[n+1] = v013928[n]+1, v013928[n+1] = v013928[n]); n++);
A019565(n) = {factorback(Mat(vector(if(n, #n=vecextract(binary(n), "-1..1")), j, [prime(j), n[j]])~))}; \\ This function from M. F. Hasler
A246353(n) = 1+A013928(A019565(n));
for(n=0, 478, write("b246353.txt", n, " ", A246353(n)));

Python

from math import prod, isqrt
from sympy import prime, mobius
def A246353(n):
    m = prod(prime(i) for i,j in enumerate(bin(n)[-1:1:-1],1) if j=='1')
    return int(sum(mobius(k)*(m//k**2) for k in range(1, isqrt(m)+1))) # Chai Wah Wu, Feb 22 2025

Scheme

(definec (A246353 n) (let loop ((n n) (i 1) (p 1)) (cond ((zero? n) (A013928 (+ 1 p))) ((odd? n) (loop (/ (- n 1) 2) (+ 1 i) (* p (A000040 i)))) (else (loop (/ n 2) (+ 1 i) p)))))

Formula

a(n) = A013928(1+A019565(n)) = 1 + A013928(A019565(n)).

a(n) = A064273(n) + 1.

For all n >= 0, A048672(a(n)) = n.

For all n >= 1, a(A048672(n)) = n.