A292377 a(1) = 0, and for n > 1, a(n) = a(A252463(n)) + [n == 3 (mod 4)].

0, 0, 1, 0, 1, 1, 2, 0, 0, 1, 3, 1, 3, 2, 2, 0, 3, 0, 4, 1, 1, 3, 5, 1, 0, 3, 1, 2, 5, 2, 6, 0, 2, 3, 3, 0, 6, 4, 4, 1, 6, 1, 7, 3, 1, 5, 8, 1, 0, 0, 4, 3, 8, 1, 2, 2, 3, 5, 9, 2, 9, 6, 2, 0, 2, 2, 10, 3, 4, 3, 11, 0, 11, 6, 1, 4, 3, 4, 12, 1, 0, 6, 13, 1, 4, 7, 6, 3, 13, 1, 3, 5, 5, 8, 5, 1, 13, 0, 3, 0, 13, 4, 14, 3, 2

Offset: 1

Author

Antti Karttunen, Sep 17 2017

Keywords

• nonn

Comments

For numbers > 1, iterate the map x -> A252463(x) which divides even numbers by 2 and shifts every prime in the prime factorization of odd n one index step towards smaller primes. a(n) counts the numbers of the form 4k+3 encountered until 1 has been reached. The count includes also n itself if it is of the form 4k+3 (A004767).

In other words, locate the node which contains n in binary tree A005940 and traverse from that node towards the root, counting all numbers of the form 4k+3 that occur on the path.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16384
• Index entries for sequences computed from indices in prime factorization

Crossrefs

Cf. A000120, A005940, A061395, A064989, A252463, A292375, A292376, A292378, A292383.

Programs

Mathematica

a[1] = 0; a[n_] := a[n] = a[Which[n == 1, 1, EvenQ@ n, n/2, True, Times @@ Power[Which[# == 1, 1, # == 2, 1, True, NextPrime[#, -1]] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger@ n]] + Boole[Mod[n, 4] == 3]; Array[a, 105]

Formula

a(1) = 0, and for n > 1, a(n) = a(A252463(n)) + floor((n mod 4)/3).

Equivalently, a(2n) = a(n), and for odd numbers n > 1, a(n) = a(A064989(n)) + [n == 3 (mod 4)].

a(n) = A000120(A292383(n)).

Other identities. For n >= 1:

a(n) >= A292376(n).

a(A000040(n)) = A267098(n).

1 + a(n) - A292375(n) = A292378(n).

For n >= 2, a(n) + A292375(n) = A061395(n).