A306286 -id:A306286 - cp's OEIS frontend

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

Rémy Sigrist, May 04 2019

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

			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

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

Cf. A070939, A124773, A306286.

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

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

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

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

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.

A307218 Numbers x with k digits in base 2 (MSD(x)_2 = d_1, LSD(x)_2 = d_k) that are equal to the product of the positions of 1's (see examples and formula).

Original entry on oeis.org

1, 1350, 47520, 1995840, 59376240

Offset: 1

Paolo P. Lava, Mar 29 2019

If 0's instead of 1's are considered, then the sequence starts with 2, 12, 504, ...
If the positions are counted with MSD(x)_2 = d_k and LSD(x)_2 = d_1 then the sequence starts with 1, 2, 6, 12, 576000, ... (see A161324).
Next value greater than 10^12. - Giovanni Resta, Mar 29 2019

			1350 in base 2 is 10101000110. The 1's are in positions 1, 3, 5, 9, 10 and 1*3*5*9*10 = 1350.
47520 in base 2 is 1011100110100000. The 1's are in positions 1, 3, 4, 5, 8, 9, 11 and 1*3*4*5*8*9*11 = 47520.
1995840 in base 2 is 111100111010001000000. The 1's are in positions 1, 2, 3, 4, 7, 8, 9, 11, 15 and 1*2*3*4*7*8*9*11*15 = 1995840.

Cf. A000030, A007088, A010879, A161324, A306286.

P:=proc(q) local a,b,k,n; for n from 1 to q do
a:=convert(n,base,2); b:=1; for k from 1 to nops(a) do
if a[k]=1 then b:=b*(nops(a)-k+1); fi; od; if b=n then print(n);
fi; od; end: P(10^9);

b(n)=fromdigits(binary(n), 10); \\ A007088
is(n)={k=1;v=digits(b(n));for(j=2,#v,if(v[j]==1,k=k*j));k==n;} \\ Jinyuan Wang, Mar 29 2019

Solutions of the equation x = Product_{j=1..k} ((1/2)*(1+j+(-1)^d_j*(1-j))), where d_j are the digits of x in base 2.

cp's OEIS Frontend

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Python

Formula

A307218 Numbers x with k digits in base 2 (MSD(x)_2 = d_1, LSD(x)_2 = d_k) that are equal to the product of the positions of 1's (see examples and formula).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

PARI

Formula