A292375 a(1) = 1, and for n > 1, a(n) = a(A252463(n)) + [n == 1 (mod 4)].

1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 2, 1, 1, 4, 2, 4, 2, 3, 2, 4, 1, 3, 3, 1, 2, 5, 1, 5, 1, 3, 4, 1, 2, 6, 4, 2, 2, 7, 3, 7, 2, 2, 4, 7, 1, 4, 3, 3, 3, 8, 1, 3, 2, 5, 5, 8, 1, 9, 5, 2, 1, 4, 3, 9, 4, 5, 1, 9, 2, 10, 6, 2, 4, 2, 2, 10, 2, 2, 7, 10, 3, 3, 7, 4, 2, 11, 2, 3, 4, 6, 7, 3, 1, 12, 4, 2, 3, 13, 3, 13, 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+1 encountered until 1 has been reached, which is also included in the count. The count includes also n itself if it is of the form 4k+1 (A016813), thus a(1) = 1.

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+1 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, A001248, A005940, A061395, A064989, A252463, A292374, A292377, A292378, A292381, A292583.

Programs

Mathematica

a[1] = 1; 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] == 1]; Array[a, 105] (* Michael De Vlieger, Sep 17 2017 *)

PARI

A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
A292375(n) = if(1==n,n,if(!(n%2),A292375(n/2),(if(1==(n%4),1,0)+A292375(A064989(n)))));

Scheme

;; With memoization-macro definec.
(definec (A292375 n) (if (= 1 n) 1 (+ (if (= 1 (modulo n 4)) 1 0) (A292375 (A252463 n)))))

Formula

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

a(n) = A000120(A292381(n)).

Other identities and observations. For n >= 1:

a(n) >= A292374(n).

a(A000040(n))-1 = A267097(n).

1 + A292377(n) - a(n) = A292378(n).

For n >= 2, a(n) + A292377(n) = A061395(n).

From Antti Karttunen, Apr 22 2022: (Start)

For n >= 2, a(n^2) = A061395(n). [Because A292377(n^2) = 0]

For n >= 1, a(A001248(n)) = n. [See comments in A292583]

(End)