A038374 -id:A038374 - cp's OEIS frontend

A223909 Numbers for which the maximal run of 1's in their binary representation contains odd number of 1's.

1, 2, 4, 5, 7, 8, 9, 10, 14, 16, 17, 18, 20, 21, 23, 28, 29, 31, 32, 33, 34, 36, 37, 39, 40, 41, 42, 46, 55, 56, 57, 58, 59, 62, 64, 65, 66, 68, 69, 71, 72, 73, 74, 78, 80, 81, 82, 84, 85, 87, 92, 93, 95, 103, 110, 112, 113, 114, 115, 116, 117, 118, 119, 124

Offset: 1

Vladimir Shevelev, Mar 29 2013

We call these numbers "maxodious".

If a(n) is in the sequence, then 2^k*a(n) is in the sequence. If a(n)==0 (mod 4) is in the sequence, then a(n)+1 is in the sequence. If a(n)==0 (mod 8) is in the sequence, then a(n)+1, a(n)+2 are in the sequence.

Peter J. C. Moses, Table of n, a(n) for n = 1..10000

Cf. A000069, A001969, A038374.

Select[Range[500],OddQ[Max[Map[Count[#,1]&,Split[IntegerDigits[#,2]]]]]&] (* Peter J. C. Moses, Mar 29 2013 *)

A038374(n)=n>>=valuation(n,2);if(n<2,return(n)); my(e=valuation(n+1,2)); max(e, A038374(n>>e))
is(n)=A038374(n)%2 \\ Charles R Greathouse IV, Jan 12 2014

Numbers n such that A038374(n) is odd. - Charles R Greathouse IV, Jan 12 2014

A300655 a(n) is the length of the longest contiguous block of ones in the binary expansion of 1/n.

Original entry on oeis.org

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

Offset: 1

Rémy Sigrist, Mar 10 2018

This sequence has similarities with A038374: here we consider the binary expansion of 1/n, there the binary expansion of n.

			The first terms, alongside the binary representation of 1/n, are:
  n   a(n)  bin(1/n) with repeating digits in parentheses
  --  ----  ---------------------------------------------
   1     1  1.(0)
   2     1  0.1(0)
   3     1  0.(01)
   4     1  0.01(0)
   5     2  0.(0011)
   6     1  0.0(01)
   7     1  0.(001)
   8     1  0.001(0)
   9     3  0.(000111)
  10     2  0.0(0011)
  11     3  0.(0001011101)
  12     1  0.00(01)
  13     3  0.(000100111011)
  14     1  0.0(001)
  15     1  0.(0001)
  16     1  0.0001(0)
  17     4  0.(00001111)
  18     3  0.0(000111)
  19     4  0.(000011010111100101)
  20     2  0.00(0011)

Cf. A038374, A300630.

a(n) = my (w=1, s=Set(), f=1/max(n,2)); while (!setsearch(s,f), while (floor(f*2^(w+1))==2^(w+1)-1, w++); s=setunion(s,Set(f)); f=frac(f*2)); return (w)

a(2*n) = a(n).
a(2^k + 1) = k for any k > 0.
a(n) = 1 iff n belongs to A300630.

A092607 Length of longest contiguous block of ones in binary representation of n!.

Original entry on oeis.org

1, 1, 1, 2, 2, 4, 2, 3, 3, 2, 5, 2, 6, 3, 3, 3, 3, 6, 5, 3, 5, 5, 5, 6, 4, 6, 6, 6, 4, 5, 6, 3, 3, 7, 6, 5, 5, 12, 8, 10, 8, 5, 5, 7, 6, 5, 8, 9, 8, 6, 4, 6, 7, 7, 7, 6, 8, 5, 8, 8, 7, 5, 9, 8, 8, 7, 7, 11, 7, 7, 8, 14, 10, 9, 7, 7, 7, 6, 7, 8, 10, 8, 10, 9, 7, 7, 5, 7, 7, 7, 9, 8, 7, 8, 10, 6, 8, 7

Offset: 0

Reinhard Zumkeller, Apr 11 2004

			n = 10: 10! = 3628800 = '1101110101111100000000' = '.........11111........': a(10) = 5.

Cf. A000142, A007088, A036603, A038374.

a[n_] := Max[Length /@ Split[IntegerDigits[n!, 2]][[1 ;; -1 ;; 2]]]; Array[a, 100, 0] (* Amiram Eldar, Jul 29 2025 *)

a(n) = A038374(A000142(n)).

A245536 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=k-r-1, or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=(k-r-1)*a(j).

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane, Jul 25 2014

nonn

Defined by the recurrence given in A245196, taking G(n)=n (n>=0) and m=1.

Changing G from [0,1,2,3,4,...] to [1,2,3,4,5,6,...] produces A038374.

Cf. A245196, A038374.

G:=[seq(n,n=0..30)];
m:=1;
f:=proc(n) option remember; global m,G; local k,r,j,np;
   k:=1+floor(log[2](n)); np:=2^k-n;
   if np=1 then r:=0; j:=0; else r:=1+floor(log[2](np-1)); j:=2^r-np; fi;
   if j=0 then G[k-r-1+1]; else m*G[k-r-1+1]*f(j); fi;
end;
[seq(f(n),n=1..120)];

A307503 Least prime containing at least n consecutive 1's in its binary representation.

Original entry on oeis.org

2, 2, 3, 7, 31, 31, 127, 127, 1021, 3583, 4093, 6143, 8191, 8191, 81919, 131071, 131071, 131071, 524287, 524287, 4194301, 14680063, 16777213, 67108859, 536870909, 536870909, 536870909, 536870909, 2147483647, 2147483647, 2147483647, 2147483647, 21474836479

Offset: 0

John Mason, Apr 11 2019

For n > 0, a(n) = A000040(m) for the lowest m such that A090000(m) >= n.

a(n) = A087522(n) for n = 0 through 7, and in all other cases when a(n) is a base 2 repunit (Mersenne) prime.

			a(0) = 2, the smallest prime containing >= 0 1's.
a(1) = 2, the smallest prime containing >= 1 consecutive 1's.
a(2) = 3, the smallest prime containing >= 2 consecutive 1's.

Chai Wah Wu, Table of n, a(n) for n = 0..3314

Cf. A038374, A090000, A087522, A201914.

Cf. A090593 (with exactly n consecutive ones).

nbo(n)=if (n==0, return (0)); n>>=valuation(n, 2); if(n<2, return(n)); my(e=valuation(n+1, 2)); max(e, nbo(n>>e)); \\ A038374
a(n) = my(p=2); while(nbo(p) < n, p=nextprime(p+1)); p; \\ Michel Marcus, Apr 14 2019

a(n) <= A201914(n). - Rémy Sigrist, Apr 11 2019

a(n) = min_{k>=n} A090593(k). - Chai Wah Wu, Apr 26 2019

a(28)-a(32) from Chai Wah Wu, Apr 26 2019

A383270 Length of the longest sequence of contiguous 1s in the binary expansion of n after flipping at most one 0-bit to 1.

Original entry on oeis.org

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

Offset: 0

Darío Clavijo, Apr 21 2025

There is only one allowed bit flip from 0 to 1 for each term, unless n is of the form 2^k-1 in which case a(n) = k.

At most one bit flip from 0 to 1 is allowed. If the original binary representation already contains the longest run of 1s, no flip is required.

			a(3) = 2 because 3 = 11_2 and there is no need to flip any bit.
a(1775) = 8 because 1775 = 11011101111_2 and if we flip the 7th bit we get 1101111111_2 which has the longest sequence of contiguous 1s of length 8.

Cf. A000079, A000120, A007088, A070939, A038374.

a(n) = if (n==0, return(1)); my(b=binary(n), vz = select(x->(x==0), b, 1), m=0); if (#vz <= 1, return (#b)); vz = concat(0, concat(Vec(vz), #b+1)); for (i=1, #vz-2, m = max(m, vz[i+2]-vz[i]-1)); m; \\ Michel Marcus, May 02 2025

def a(n):
    if n == 0: return 1
    if n.bit_length() == n.bit_count(): return n.bit_length()
    c = p = m = 0
    while n:
        if n & 1: c += 1
        else:
            p = c * ((n & 2) > 0)
            c = 0
        if (pc := p + c) > m: m = pc
        n >>= 1
    return m + 1
print([a(n) for n in range(1,88)])

a(n) <= 1 + floor(log_2(n)).
a(2^k-1) = k.
a(2^k) = 2 for k > 0.
a(2^k+1) = 2.
A038374(n) <= a(n) <= A000120(n)+1. - Michael S. Branicky, Apr 21 2025

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A223909 Numbers for which the maximal run of 1's in their binary representation contains odd number of 1's.

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A300655 a(n) is the length of the longest contiguous block of ones in the binary expansion of 1/n.

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A092607 Length of longest contiguous block of ones in binary representation of n!.

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A245536 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=k-r-1, or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=(k-r-1)*a(j).

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A307503 Least prime containing at least n consecutive 1's in its binary representation.

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A383270 Length of the longest sequence of contiguous 1s in the binary expansion of n after flipping at most one 0-bit to 1.

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