A306549 a(n) is the product of the positions of the zeros in the binary expansion of n (the most significant bit having position 1).

1, 1, 2, 1, 6, 2, 3, 1, 24, 6, 8, 2, 12, 3, 4, 1, 120, 24, 30, 6, 40, 8, 10, 2, 60, 12, 15, 3, 20, 4, 5, 1, 720, 120, 144, 24, 180, 30, 36, 6, 240, 40, 48, 8, 60, 10, 12, 2, 360, 60, 72, 12, 90, 15, 18, 3, 120, 20, 24, 4, 30, 5, 6, 1, 5040, 720, 840, 120, 1008

Offset: 0

Author

Rémy Sigrist, May 04 2019

Keywords

• nonn
• base
• easy

Comments

Apparently, the variant where the least significant bit has position 1 corresponds to A124773.

Examples

			The first terms, alongside the positions of zeros and the binary representation of n, are:
  n   a(n)  Pos.zeros  bin(n)
  --  ----  ---------  ------
   0     1  {1}             0
   1     1  {}              1
   2     2  {2}            10
   3     1  {}             11
   4     6  {2,3}         100
   5     2  {2}           101
   6     3  {3}           110
   7     1  {}            111
   8    24  {2,3,4}      1000
   9     6  {2,3}        1001
  10     8  {2,4}        1010
  11     2  {2}          1011
  12    12  {3,4}        1100
  13     3  {3}          1101
  14     4  {4}          1110
  15     1  {}           1111
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..16384

Crossrefs

Cf. A070939, A124773, A306286.

Programs

Mathematica

A306549[n_] := Times @@ Flatten[Position[IntegerDigits[n, 2], 0]];
Array[A306549, 100, 0] (* Paolo Xausa, Jun 01 2024 *)

PARI

a(n) = my (b=binary(n)); prod(k=1, #b, if (b[k]==0, k, 1))

PARI

a(n) = vecprod(Vec(select(x->(x==0), binary(n), 1)));

Python

from math import prod
def a(n): return prod(i for i, bi in enumerate(bin(n)[2:], 1) if bi == "0")
print([a(n) for n in range(70)]) # Michael S. Branicky, Jun 01 2024

Formula

a(n) = A070939(n)! / A306286(n).

a(2*n) = a(n) * (1+A070939(n)).

a(2*n+1) = a(n).

a(2^k) = (k+1)! for any k >= 0.

a(2^k-1) = 1 for any k >= 0.

a(2^k-2) = k for any k >= 1.