A023416 -id:A023416 - cp's OEIS frontend

A372433 Binary weight (number of ones in binary expansion) of the n-th squarefree number.

1, 1, 2, 2, 2, 3, 2, 3, 3, 3, 4, 2, 3, 3, 3, 4, 3, 4, 4, 5, 2, 2, 3, 3, 3, 4, 3, 3, 4, 4, 5, 4, 4, 5, 4, 4, 5, 5, 5, 2, 2, 3, 3, 3, 4, 3, 3, 4, 4, 5, 3, 4, 4, 4, 5, 4, 5, 5, 5, 6, 3, 4, 4, 5, 4, 4, 5, 5, 5, 6, 4, 4, 5, 5, 6, 5, 6, 7, 2, 2, 3, 3, 3, 3, 3, 4, 4

Offset: 1

Gus Wiseman, May 04 2024

Harvey P. Dale, Table of n, a(n) for n = 1..1000
MathOverflow, Are there primes of every Hamming weight?
Wikipedia, Hamming weight.

Restriction of A000120 to A005117.
For prime instead of squarefree we have A014499, zeros A035103.
Counting zeros instead of ones gives A372472, cf. A023416, A372473.
For binary length instead of weight we have A372475.
A003714 lists numbers with no successive binary indices.
A030190 gives binary expansion, reversed A030308.
A048793 lists positions of ones in reversed binary expansion, sum A029931.
A145037 counts ones minus zeros in binary expansion, cf. A031443, A031444, A031448, A097110.
A371571 lists positions of zeros in binary expansion, sum A359359.
A371572 lists positions of ones in binary expansion, sum A230877.
A372515 lists positions of zeros in reversed binary expansion, sum A359400.
A372516 counts ones minus zeros in binary expansion of primes, cf. A177718, A177796, A372538, A372539.
Cf. A039004, A049093, A049094, A059015, A069010, A070939, A073642, A211997, A368494, A372474.

DigitCount[Select[Range[100],SquareFreeQ],2,1]
Total[IntegerDigits[#,2]]&/@Select[Range[200],SquareFreeQ] (* Harvey P. Dale, Feb 14 2025 *)

from math import isqrt
from sympy import mobius
def A372433(n):
    def f(x): return n+x-sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1))
    m, k = n, f(n)
    while m != k:
        m, k = k, f(k)
    return int(m).bit_count() # Chai Wah Wu, Aug 02 2024

a(n) = A000120(A005117(n)).

a(n) + A372472(n) = A372475(n) = A070939(A005117(n)).

A073138 Largest number having in its binary representation the same number of 0's and 1's as n.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 6, 7, 8, 12, 12, 14, 12, 14, 14, 15, 16, 24, 24, 28, 24, 28, 28, 30, 24, 28, 28, 30, 28, 30, 30, 31, 32, 48, 48, 56, 48, 56, 56, 60, 48, 56, 56, 60, 56, 60, 60, 62, 48, 56, 56, 60, 56, 60, 60, 62, 56, 60, 60, 62, 60, 62, 62, 63, 64, 96, 96, 112, 96, 112, 112

Offset: 0

Reinhard Zumkeller, Jul 16 2002

From Trevor Green (green(AT)snoopy.usask.ca), Nov 26 2003: (Start)
a(n)/n has an accumulation point at x exactly when x is in the interval [1, 2]. Proof: Clearly n <= a(n) < 2n. Let b(n) = a(n)/n, then b(n) must always lie in [1,2) and all the accumulation points of the sequence must lie in [1,2]. We shall show that every such number is an accumulation point.
First, consider any d-bit integer n. Suppose that z of these bits are 0. Let n' be the (d+z)-bit integer whose first d bits are the same as those of n and whose remaining bits are all 1. Then a(n') will have to be the (d+z)-bit integer whose first d bits are all 1 and whose last z bits are all 0.
Thus n' = (n+1)*2^z-1; a(n') = (2^d-1)2^z; and b(n') = (2^d-1)/(n+1) + epsilon, where 0 < epsilon < 2^(1-d). So to get an accumulation point x, we just choose n(d) to be the d-bit integer such that (2^d-1)/(n(d)+1) < x <= (2^d-1)/n(d), or equivalently, n(d) = floor((2^d-1)/x). If x lies in [1,2), then n(d) will always be a d-bit number for sufficiently large d.
Then n'(d) yields an increasing subsequence of the integers for which b(n'(d)) converges to x. For x = 2, choose n(d) = 2^(d-1), which is always a d-bit number; then b(n'(d)) = (2^d-1)/(2^(d-1)+1) + epsilon = 2 + epsilon', where epsilon' also heads for 0 as d blows up. This proves the claim.
(End)

			a(20)=24, as 20='10100' and 24 is the greatest number having two 1's and three 0's: 17='10001', 18='10010', 20='10100' and 24='11000'.

Seiichi Manyama, Table of n, a(n) for n = 0..8191 (terms 0..1023 from T. D. Noe)
Index entries for sequences related to binary expansion of n

Cf. A007088, A073137, A000523, A073139, A073140, A073141, A319650.
Cf. A030109.
Cf. A038573.
Decimal equivalent of A221714. - N. J. A. Sloane, Jan 26 2013

a073138 n = a038573 n * a080100 n  -- Reinhard Zumkeller, Jan 16 2012

a:= n-> Bits[Join](sort(Bits[Split](n))):
seq(a(n), n=0..100);  # Alois P. Heinz, Jun 26 2021

f[n_] := Module[{idn=IntegerDigits[n, 2], o, l}, l=Length[idn]; o=Count[idn, 1]; FromDigits[Join[Table[1, {o}], Table[0, {l-o}]], 2]]; Table[f[i], {i, 0, 70}]
ln[n_] := Module[{idn=IntegerDigits[n, 2], len, zer}, len=Length[idn]; zer=Count[idn, 0]; FromDigits[Join[Table[1, {len-zer}], Table[0, {zer}]], 2]]; Table[ln[i], {i, 0, 70}]
a[z_] := 2^(Floor[Log[2, z]] + 1) * (1 - 2^(-Sum[k, {k, IntegerDigits[n, 2]}])) Column[Table[a[p], {p, 500}], Right] (* Trevor G. Hyde (thyde12(AT)amherst.edu), Jul 14 2008 *)
Table[FromDigits[ReverseSort[IntegerDigits[n,2]],2],{n,0,70}] (* Harvey P. Dale, Mar 13 2023 *)

a(n) = fromdigits(vecsort(binary(n),,4), 2); \\ Michel Marcus, Sep 26 2018

def a(n): return int("".join(sorted(bin(n)[2:], reverse=True)), 2)
print([a(n) for n in range(71)]) # Michael S. Branicky, Jun 27 2021

def A073138(n): return (m:=1<>n.bit_count()) # Chai Wah Wu, Aug 18 2025

a(n+1) = a(floor(n/2))*2 + (n mod 2)*(2^floor(log_2(n)) - a(floor(n/2))); a(0)=0.
A023416(a(n)) = A023416(n), A000120(a(n)) = A000120(n).
a(0)=0, a(1)=1, a(2n) = 2a(n), a(2n+1) = a(n) + 2^floor(log_2(n)). - Ralf Stephan, Oct 05 2003
a(n) = 2^(floor(log_2(n)) + 1) * (1 - 2^(-d(n))) where d(n) = digit sum of base-2 expansion of n. - Trevor G. Hyde (thyde12(AT)amherst.edu), Jul 14 2008
a(n) = A038573(n) * A080100(n). - Reinhard Zumkeller, Jan 16 2012
n <= a(n) < 2n. - Charles R Greathouse IV, Aug 07 2024

A080100 a(n) = 2^(number of 0's in binary representation of n).

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 2, 1, 8, 4, 4, 2, 4, 2, 2, 1, 16, 8, 8, 4, 8, 4, 4, 2, 8, 4, 4, 2, 4, 2, 2, 1, 32, 16, 16, 8, 16, 8, 8, 4, 16, 8, 8, 4, 8, 4, 4, 2, 16, 8, 8, 4, 8, 4, 4, 2, 8, 4, 4, 2, 4, 2, 2, 1, 64, 32, 32, 16, 32, 16, 16, 8, 32, 16, 16, 8, 16, 8, 8, 4, 32, 16, 16, 8, 16, 8, 8, 4, 16, 8, 8, 4, 8, 4

Offset: 0

Reinhard Zumkeller, Jan 28 2003

Number of numbers k, 0<=k<=n, such that (k AND n) = 0 (bitwise logical AND): a(n) = #{k : T(n,k)=n, 0<=k<=n}, where T is defined as in A080099.

Same parity as the Catalan numbers (A000108). - Paul D. Hanna, Nov 14 2012

Reinhard Zumkeller, Table of n, a(n) for n = 0..8191
George Beck and Karl Dilcher, A Matrix Related to Stern Polynomials and the Prouhet-Thue-Morse Sequence, arXiv:2106.10400 [math.CO], 2021.
Ralf Stephan, Divide-and-conquer generating functions. I. Elementary sequences, arXiv:math/0307027 [math.CO], 2003.

Cf. A001316.
Cf. A002487.
This is Guy Steele's sequence GS(5, 3) (see A135416).
Cf. A048896.
Cf. A023416, A080791.

import Data.List (transpose)
a080100 n = a080100_list !! n
a080100_list =  1 : zs where
   zs =  1 : (concat $ transpose [map (* 2) zs, zs])
-- Reinhard Zumkeller, Aug 27 2014, Mar 07 2011

a:= n-> 2^add(1-i, i=Bits[Split](n)):
seq(a(n), n=0..93);  # Alois P. Heinz, Aug 18 2025

f[n_] := 2^DigitCount[n, 2, 0]; f[0] = 1; Array[f, 94, 0] (* Robert G. Wilson v *)

PARI
```
a(n)=if(n<1,n==0,(2-n%2)*a(n\2))
```

a(n)=local(A,m); if(n<0,0,m=1; A=1+O(x); while(m<=n,m*=2; A=subst(A,x,x^2)*(2+x)-1); polcoeff(A,n))

def A080100(n): return 1<Chai Wah Wu, Aug 18 2025

G.f. satisfies: F(x^2) = (1+F(x))/(x+2). - Ralf Stephan, Jun 28 2003
a(2n) = 2a(n), n>0. a(2n+1) = a(n). - Ralf Stephan, Apr 29 2003
a(n) = 2^A080791(n). a(n)=2^A023416(n), n>0.
a(n) = sum(k=0, n, C(n+k, k) mod 2). - Benoit Cloitre, Mar 06 2004
a(n) = sum(k=0, n, C(2n-k, n) mod 2). - Paul Barry, Dec 13 2004
G.f. satisfies: A(x) = Sum_{n>=0} [A(x)^n (mod 2)]*x^n, where A(x)^n (mod 2) reduces all coefficients modulo 2 to {0,1}. - Paul D. Hanna, Nov 14 2012

Keyword base added by Rémy Sigrist, Jan 18 2018

A359359 Sum of positions of zeros in the binary expansion of n, where positions are read starting with 1 from the left (big-endian).

Original entry on oeis.org

1, 0, 2, 0, 5, 2, 3, 0, 9, 5, 6, 2, 7, 3, 4, 0, 14, 9, 10, 5, 11, 6, 7, 2, 12, 7, 8, 3, 9, 4, 5, 0, 20, 14, 15, 9, 16, 10, 11, 5, 17, 11, 12, 6, 13, 7, 8, 2, 18, 12, 13, 7, 14, 8, 9, 3, 15, 9, 10, 4, 11, 5, 6, 0, 27, 20, 21, 14, 22, 15, 16, 9, 23, 16, 17, 10

Offset: 0

Gus Wiseman, Jan 03 2023

			The binary expansion of 100 is (1,1,0,0,1,0,0), with zeros at positions {3,4,6,7}, so a(100) = 20.

The number of zeros is A023416, partial sums A059015.
For positions of 1's we have A230877, reversed A029931.
The reversed version is A359400.
A003714 lists numbers with no successive binary indices.
A030190 gives binary expansion.
A039004 lists the positions of zeros in A345927.
Cf. A000120, A048793, A065359, A069010, A070939, A073642, A083652, A328594, A328595, A359402, A359495.

Table[Total[Join@@Position[IntegerDigits[n,2],0]],{n,0,100}]

a(n>0) = binomial(A029837(n)+1,2) - A230877(n).

A102683 Number of digits 9 in decimal representation of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 0, 0, 0, 0, 0

Offset: 0

N. J. A. Sloane, Feb 03 2005

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Cf. A007095, A011539, A027868, A054899, A055640, A055641, A102669-A102685, A117804, A122840, A122841, A160093, A160094, A196563, A196564, A235049.
Cf. A102684 (partial sums).
Cf. A000120, A000788, A023416, A059015 (for base 2).

a102683 =  length . filter (== '9') . show
-- Reinhard Zumkeller, Dec 29 2011

p:=proc(n) local b,ct,j: b:=convert(n,base,10): ct:=0: for j from 1 to nops(b) do if b[j]>=9 then ct:=ct+1 else ct:=ct fi od: ct: end: seq(p(n),n=0..116); # Emeric Deutsch, Feb 23 2005

a[n_] := DigitCount[n, 10, 9]; Array[a, 100, 0] (* Amiram Eldar, Jul 24 2023 *)

a(A007095(n)) = 0; a(A011539(n)) > 0. - Reinhard Zumkeller, Dec 29 2011
From Hieronymus Fischer, Jun 10 2012: (Start)
a(n) = Sum_{j=1..m+1} (floor(n/10^j + 1/10) - floor(n/10^j)), where m=floor(log_10(n)).
G.f.: g(x) = (1/(1-x))*Sum_{j>=0} (x^(9*10^j) - x^(10*10^j))/(1-x^10^(j+1)). (End)
a(A235049(n)) = 0. - Reinhard Zumkeller, Apr 16 2014

More terms from Emeric Deutsch, Feb 23 2005

A333632 Rotational period of the k-th composition in standard order; a(0) = 0.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 3, 2, 3, 3, 1, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 1, 1, 2, 2, 3, 1, 3, 3, 4, 2, 3, 1, 4, 3, 2, 4, 5, 2, 3, 3, 4, 3, 4, 2, 5, 3, 4, 4, 5, 4, 5, 5, 1, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4

Offset: 0

Gus Wiseman, Apr 12 2020

nonn

A composition of n is a finite sequence of positive integers summing to n. The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The a(299) = 5 rotations:
  (1,1,3,2,2)
  (1,3,2,2,1)
  (3,2,2,1,1)
  (2,2,1,1,3)
  (2,1,1,3,2)
The a(9933) = 4 rotations:
  (1,2,1,3,1,2,1,3)
  (1,3,1,2,1,3,1,2)
  (2,1,3,1,2,1,3,1)
  (3,1,2,1,3,1,2,1)

Aperiodic compositions are counted by A000740.
Aperiodic binary words are counted by A027375.
The orderless period of prime indices is A052409.
Numbers whose binary expansion is periodic are A121016.
Periodic compositions are counted by A178472.
The version for binary expansion is A302291.
Numbers whose prime signature is aperiodic are A329139.
Compositions by number of distinct rotations are A333941.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Necklaces are A065609.
- Sum is A070939.
- Equal runs are counted by A124767.
- Rotational symmetries are counted by A138904.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Lyndon compositions are A275692.
- Co-Lyndon compositions are A326774.
- Aperiodic compositions are A328594.
- Rotational period is A333632 (this sequence).
- Co-necklaces are A333764.
- Reversed necklaces are A333943.
Cf. A000031, A001037, A008965, A019536, A211100, A328595, A328596, A329312, A329313, A329326.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[Length[Union[Array[RotateRight[stc[n],#]&,DigitCount[n,2,1]]]],{n,0,100}]

a(n) = A000120(n)/A138904(n) = A302291(n) - A023416(n)/A138904(n).

A359400 Sum of positions of zeros in the reversed binary expansion of n, where positions in a sequence are read starting with 1 from the left.

Original entry on oeis.org

1, 0, 1, 0, 3, 2, 1, 0, 6, 5, 4, 3, 3, 2, 1, 0, 10, 9, 8, 7, 7, 6, 5, 4, 6, 5, 4, 3, 3, 2, 1, 0, 15, 14, 13, 12, 12, 11, 10, 9, 11, 10, 9, 8, 8, 7, 6, 5, 10, 9, 8, 7, 7, 6, 5, 4, 6, 5, 4, 3, 3, 2, 1, 0, 21, 20, 19, 18, 18, 17, 16, 15, 17, 16, 15, 14, 14, 13

Offset: 0

Gus Wiseman, Jan 05 2023

			The reversed binary expansion of 100 is (0,0,1,0,0,1,1), with zeros at positions {1,2,4,5}, so a(100) = 12.

The number of zeros is A023416, partial sums A059015.
Row sums of A368494.
For positions of 1's we have A029931, non-reversed A230877.
The non-reversed version is A359359.
A003714 lists numbers with no successive binary indices.
A030190 gives binary expansion, reverse A030308.
A039004 lists the positions of zeros in A345927.
Cf. A000120, A048793, A069010, A070939, A073642, A328594, A328595, A344618, A359402, A359495.

long A359400(long n) {
  long result = 0, counter = 1;
  do {
    if (n % 2 == 0)
      result += counter;
    counter++;
    n /= 2;
  } while (n > 0);
  return result; } // Frank Hollstein, Jan 06 2023

Table[Total[Join@@Position[Reverse[IntegerDigits[n,2]],0]],{n,0,100}]

def a(n): return sum(i for i, bi in enumerate(bin(n)[:1:-1], 1) if bi=='0')
print([a(n) for n in range(78)]) # Michael S. Branicky, Jan 09 2023

a(n) = binomial(A029837(n)+1, 2) - A029931(n), for n>0.

A048967 Number of even entries in row n of Pascal's triangle (A007318).

Original entry on oeis.org

0, 0, 1, 0, 3, 2, 3, 0, 7, 6, 7, 4, 9, 6, 7, 0, 15, 14, 15, 12, 17, 14, 15, 8, 21, 18, 19, 12, 21, 14, 15, 0, 31, 30, 31, 28, 33, 30, 31, 24, 37, 34, 35, 28, 37, 30, 31, 16, 45, 42, 43, 36, 45, 38, 39, 24, 49, 42, 43, 28, 45, 30, 31, 0, 63, 62, 63, 60, 65, 62, 63, 56, 69, 66, 67

Offset: 0

Brian Galebach

In rows 2^k - 1 all entries are odd.
a(n) = 0 (all the entries in the row are odd) iff n = 2^m - 1 for some m >= 0 and then n belongs to sequence A000225. - Avi Peretz (njk(AT)netvision.net.il), Apr 21 2001
Also number of zeros in n-th row of Sierpiński's triangle (cf. A047999): a(n) = A023416(A001317(n)). - Reinhard Zumkeller, Nov 24 2012

			Row 4 is 1 4 6 4 1 with 3 even entries so a(4)=3.

Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 0..1000 from T. D. Noe)

Cf. A007318, A001316, A000120, A000225, A038573.

Cf. A249304.

import Data.List (transpose)
a048967 n = a048967_list !! n
a048967_list = 0 : xs where
   xs = 0 : concat (transpose [zipWith (+) [1..] xs, map (* 2) xs])
-- Reinhard Zumkeller, Nov 14 2014, Nov 24 2012

Table[n + 1 - Sum[ Mod[ Binomial[n, k], 2], {k, 0, n} ], {n, 0, 100} ]
a[n_] := n + 1 - 2^DigitCount[n, 2, 1]; Array[a, 100, 0] (* Amiram Eldar, Jul 27 2023 *)

a(n)=if(n<1,0,if(n%2==0,a(n/2)+n/2,2*a((n-1)/2)))

def A048967(n): return n+1-(1<Chai Wah Wu, May 03 2023

a(n) = n+1 - A001316(n) = n+1 - 2^A000120(n) = n+1 - Sum_{k=0..n} (C(n, k) mod 2) = Sum_{k=0..n} ((1 - C(n, k)) mod 2).
a(2n) = a(n) + n, a(2n+1) = 2a(n). - Ralf Stephan, Oct 07 2003
a(n) = row sums in A219463 = A000120(A219843(n)). - Reinhard Zumkeller, Nov 30 2012
A249304(n+1) = a(n+1) + a(n). - Reinhard Zumkeller, Nov 14 2014
G.f.: 1/(1 - x)^2 - Product_{k>=0} (1 + 2*x^(2^k)). - Ilya Gutkovskiy, Jul 19 2019

A087117 Number of zeros in the longest string of consecutive zeros in the binary representation of n.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 1, 0, 3, 2, 1, 1, 2, 1, 1, 0, 4, 3, 2, 2, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 0, 5, 4, 3, 3, 2, 2, 2, 2, 3, 2, 1, 1, 2, 1, 1, 1, 4, 3, 2, 2, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 0, 6, 5, 4, 4, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 4, 3, 2, 2, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 5, 4, 3, 3, 2, 2

Offset: 0

Reinhard Zumkeller, Aug 14 2003

The following four statements are equivalent: a(n) = 0; n = 2^k - 1 for some k > 0; A087116(n) = 0; A023416(n) = 0.

The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. Then a(k) is the maximum part of this composition, minus one. The maximum part is A333766(k). - Gus Wiseman, Apr 09 2020

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for sequences related to binary expansion of n

Cf. A023416, A007088, A007814.
Cf. A025480, A038374, A090046, A090047, A090048, A090049, A090050.
Positions of zeros are A000225.
Positions of terms <= 1 are A003754.
Positions of terms > 0 are A062289.
Positions of first appearances are A131577.
The version for prime indices is A252735.
The proper maximum is A333766.
The version for minimum is A333767.
Maximum prime index is A061395.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Sum is A070939.
- Runs are counted by A124767.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Runs-resistance is A333628.
- Weakly decreasing compositions are A114994.
- Weakly increasing compositions are A225620.
- Strictly decreasing compositions are A333255.
- Strictly increasing compositions are A333256.
Cf. A029931, A048793, A061395, A087117, A228351, A328594, A333217, A333218, A333219, A333767, A333768.

import Data.List (unfoldr, group)
a087117 0       = 1
a087117 n
  | null $ zs n = 0
  | otherwise   = maximum $ map length $ zs n where
  zs = filter ((== 0) . head) . group .
       unfoldr (\x -> if x == 0 then Nothing else Just $ swap $ divMod x 2)
-- Reinhard Zumkeller, May 01 2012

A087117 := proc(n)
    local d,l,zlen ;
    if n = 0 then
        return 1 ;
    end if;
    d := convert(n,base,2) ;
    for l from nops(d)-1 to 0 by -1 do
        zlen := [seq(0,i=1..l)] ;
        if verify(zlen,d,'sublist') then
            return l ;
        end if;
    end do:
    return 0 ;
end proc; # R. J. Mathar, Nov 05 2012

nz[n_]:=Max[Length/@Select[Split[IntegerDigits[n,2]],MemberQ[#,0]&]]; Array[nz,110,0]/.-\[Infinity]->0 (* Harvey P. Dale, Sep 05 2017 *)

h(n)=if(n<2, return(0)); my(k=valuation(n,2)); if(k, max(h(n>>k), k), n++; n>>=valuation(n,2); h(n-1))
a(n)=if(n,h(n),1) \\ Charles R Greathouse IV, Apr 06 2022

a(n) = max(A007814(n), a(A025480(n-1))) for n >= 2. - Robert Israel, Feb 19 2017
a(2n+1) = a(n) (n>=1); indeed, the binary form of 2n+1 consists of the binary form of n with an additional 1 at the end - Emeric Deutsch, Aug 18 2017
For n > 0, a(n) = A333766(n) - 1. - Gus Wiseman, Apr 09 2020

A243499 Product of parts of integer partitions as enumerated in the table A125106.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 4, 1, 4, 3, 6, 2, 9, 4, 8, 1, 5, 4, 8, 3, 12, 6, 12, 2, 16, 9, 18, 4, 27, 8, 16, 1, 6, 5, 10, 4, 15, 8, 16, 3, 20, 12, 24, 6, 36, 12, 24, 2, 25, 16, 32, 9, 48, 18, 36, 4, 64, 27, 54, 8, 81, 16, 32, 1, 7, 6, 12, 5, 18, 10, 20, 4, 24, 15, 30, 8, 45, 16, 32, 3

Offset: 0

Antti Karttunen, Jun 28 2014

This sequence and A341392 have the same set of values on intervals from 2^m to 2^(m+1) - 1 for m >= 0. - Mikhail Kurkov, Jun 18 2021 [verification needed]

Antti Karttunen, Table of n, a(n) for n = 0..8191

Cf. A125106, A161511 (gives the corresponding sums), A227184, A003963, A243504, A006068, A005940, A163511, A000110, A007814, A023416, A053645, A329369 (similar recurrence), A341392.

(define (A243499 n) (let loop ((n n) (i 1) (p 1)) (cond ((zero? n) p) ((even? n) (loop (/ n 2) (+ i 1) p)) (else (loop (/ (- n 1) 2) i (* p i))))))

Can also be obtained by mapping with an appropriate permutation from the products of parts of each partition computed for other enumerations similar to A125106:
a(n) = A227184(A006068(n)).
a(n) = A003963(A005940(n+1)).
a(n) = A243504(A163511(n)).
From Mikhail Kurkov, Jul 11 2021: (Start)
a(n) = (1 + A023416(n))*a(A053645(n)) for n > 0 with a(0) = 1.
a(2n+1) = a(n) for n >= 0.
a(2n) = A341392(2*A059894(n)) = a(n - 2^f(n)) + a(2n - 2^f(n)) = (2 + f(n))*a(n - 2^f(n)) for n > 0 with a(0)=1 where f(n) = A007814(n).
Sum_{k=0..2^n - 1} a(k) = A000110(n+1) for n >= 0.
a((4^n - 1)/3) = n! for n >= 0.
a(2^m*(2^n - 1)) = (m+1)^n for n >= 0, m >= 0. (End) [verification needed]

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