A341944 Next larger integer with same number of runs in binary expansion as n.

3, 4, 7, 6, 9, 8, 15, 12, 11, 18, 13, 14, 17, 16, 31, 24, 19, 20, 23, 22, 37, 26, 25, 28, 27, 34, 29, 30, 33, 32, 63, 48, 35, 36, 39, 38, 41, 40, 47, 44, 43, 74, 45, 46, 53, 50, 49, 56, 51, 52, 55, 54, 69, 58, 57, 60, 59, 66, 61, 62, 65, 64, 127, 96, 67, 68

Offset: 1

Author

Rémy Sigrist, Feb 24 2021

Keywords

• nonn
• base

Comments

Number of runs in binary expansion is given by A005811.

This is a permutation of A107907.

Examples

			The first terms in decimal and in binary, alongside A005811(n), are:
  n   a(n)  bin(n)  bin(a(n))  A005811(n)
  --  ----  ------  ---------  ----------
   1     3       1         11           1
   2     4      10        100           2
   3     7      11        111           1
   4     6     100        110           2
   5     9     101       1001           3
   6     8     110       1000           2
   7    15     111       1111           1
   8    12    1000       1100           2
   9    11    1001       1011           3
  10    18    1010      10010           4
  11    13    1011       1101           3
  12    14    1100       1110           2
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..8191
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A000975, A005811, A057168, A107907.

Programs

PARI

a(n) = my (r=hammingweight(bitxor(n, n>>1))); for (k=n+1, oo, if (r==hammingweight(bitxor(k, k>>1)), return (k)))

Python

def runs(n): return bin(n^(n>>1)).count('1')
def a(n):
  nruns, m = runs(n), n + 1
  while runs(m) != nruns: m += 1
  return m
print([a(n) for n in range(1, 67)]) # Michael S. Branicky, Feb 24 2021

Formula

A005811(a(n)) = A005811(n).

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