A239904 a(n) = n - wt(n) + (number of times 11 appears in binary expansion of n).

0, 0, 1, 2, 3, 3, 5, 6, 7, 7, 8, 9, 11, 11, 13, 14, 15, 15, 16, 17, 18, 18, 20, 21, 23, 23, 24, 25, 27, 27, 29, 30, 31, 31, 32, 33, 34, 34, 36, 37, 38, 38, 39, 40, 42, 42, 44, 45, 47, 47, 48, 49, 50, 50, 52, 53, 55, 55, 56, 57, 59, 59, 61, 62, 63, 63, 64, 65, 66, 66, 68, 69, 70, 70, 71, 72, 74, 74, 76

Offset: 0

Author

N. J. A. Sloane, Apr 06 2014

Keywords

• nonn
• base

Comments

This is G_{2, 1/4}(n) in Prodinger's notation.

Links

• Gheorghe Coserea, Table of n, a(n) for n = 0..10000
• Helmut Prodinger, Generalizing the sum of digits function, SIAM J. Algebraic Discrete Methods 3 (1982), no. 1, 35--42. MR0644955 (83f:10009).

Crossrefs

Cf. A000120, A014081, A239906, A239907.

Programs

Maple

A000120 := proc(n) add(i, i=convert(n, base, 2)) end:
# A014081:
cn := proc(v, k) local n, s, nn, i, j, som, kk;
som := 0;
kk := convert(cat(seq(1, j = 1 .. k)),string);
n := convert(v, binary);
s := convert(n, string);
nn := length(s);
for i to nn - k + 1 do
if substring(s, i .. i + k - 1) = kk then som := som + 1 fi od;
som; end;
[seq(n-A000120(n)+cn(n,2), n=0..100)];

Mathematica

cn[n_, k_] := Count[Partition[IntegerDigits[n, 2], k, 1], Table[1, {k}]]; Table[n - DigitCount[n, 2, 1] + cn[n, 2], {n, 0, 78}] (* Michael De Vlieger, Sep 18 2015 *)

PARI

a(n) = n - hammingweight(n) + hammingweight(bitand(n, n>>1));
vector(79, i, a(i-1))  \\ Gheorghe Coserea, Sep 24 2015

Python

def A239904(n): return n-n.bit_count()+(n&(n>>1)).bit_count() # Chai Wah Wu, Feb 12 2023

Formula

a(n) = n - A000120(n) + A014081(n).