A292376 a(1) = 0, a(2n) = 0, and for odd numbers n > 1, a(n) = a(A064989(n)) + [n == 3 (mod 4)].

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

Offset: 1

Author

Antti Karttunen, Sep 17 2017

Keywords

• nonn

Comments

For odd numbers > 1, iterate the map x -> A064989(x), which shifts every prime in the prime factorization of n one index step towards smaller primes. a(n) counts the numbers of the form 4k+3 encountered until the first number which is even has been reached. This count includes also n itself if it is of the form 4k+3 (A004767).

In other words, locate the position where n is in square array A246278 and moving up by that column, count all numbers of the form 4k+3 before an even number at the top of the column is reached.

Links

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

Crossrefs

Cf. A055396, A064989, A246271, A246278, A292374, A292377.

Cf. also A038802 (odd bisection of a(n) + A292374(n)).

Programs

Mathematica

a[1] = 0; a[n_] := a[n] = If[EvenQ@ n, 0, a[Times @@ Power[Which[# == 1, 1, # == 2, 1, True, NextPrime[#, -1]] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger@ n] + Boole[Mod[n, 4] == 3]]; Array[a, 105] (* Michael De Vlieger, Sep 17 2017 *)

Formula

a(1) = 0, a(2n) = 0, and for odd numbers n > 1, a(n) = a(A064989(n)) + floor((n mod 4)/3).

Other identities and observations. For n >= 1.

a(n) <= A292377(n).

For n >= 2, a(n) + A292374(n) + 1 = A055396(n).