A306549 - OEIS

%I #30 Jun 01 2024 15:05:14

%S 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,

%T 20,4,5,1,720,120,144,24,180,30,36,6,240,40,48,8,60,10,12,2,360,60,72,

%U 12,90,15,18,3,120,20,24,4,30,5,6,1,5040,720,840,120,1008

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

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

%H Rémy Sigrist, <a href="/A306549/b306549.txt">Table of n, a(n) for n = 0..16384</a>

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

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

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

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

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

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

%e The first terms, alongside the positions of zeros and the binary representation of n, are:

%e   n   a(n)  Pos.zeros  bin(n)

%e   --  ----  ---------  ------

%e    0     1  {1}             0

%e    1     1  {}              1

%e    2     2  {2}            10

%e    3     1  {}             11

%e    4     6  {2,3}         100

%e    5     2  {2}           101

%e    6     3  {3}           110

%e    7     1  {}            111

%e    8    24  {2,3,4}      1000

%e    9     6  {2,3}        1001

%e   10     8  {2,4}        1010

%e   11     2  {2}          1011

%e   12    12  {3,4}        1100

%e   13     3  {3}          1101

%e   14     4  {4}          1110

%e   15     1  {}           1111

%t A306549[n_] := Times @@ Flatten[Position[IntegerDigits[n, 2], 0]];

%t Array[A306549, 100, 0] (* _Paolo Xausa_, Jun 01 2024 *)

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

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

%o (Python)

%o from math import prod

%o def a(n): return prod(i for i, bi in enumerate(bin(n)[2:], 1) if bi == "0")

%o print([a(n) for n in range(70)]) # _Michael S. Branicky_, Jun 01 2024

%Y Cf. A070939, A124773, A306286.

%K nonn,base,easy

%O 0,3

%A _Rémy Sigrist_, May 04 2019