cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A014311 Numbers with exactly 3 ones in binary expansion.

Original entry on oeis.org

7, 11, 13, 14, 19, 21, 22, 25, 26, 28, 35, 37, 38, 41, 42, 44, 49, 50, 52, 56, 67, 69, 70, 73, 74, 76, 81, 82, 84, 88, 97, 98, 100, 104, 112, 131, 133, 134, 137, 138, 140, 145, 146, 148, 152, 161, 162, 164, 168, 176, 193, 194, 196, 200, 208, 224, 259, 261, 262, 265, 266, 268, 273, 274, 276, 280, 289, 290, 292, 296, 304
Offset: 1

Views

Author

Al Black (gblack(AT)nol.net)

Keywords

Comments

Equivalently, sums of three distinct powers of 2.
Appears to give all n such that 64 is the highest power of 2 dividing A005148(n). - Benoit Cloitre, Jun 22 2002
From Gus Wiseman, Oct 05 2020: (Start)
These are numbers k such that the k-th composition in standard order has length 3. 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 sequence together with the corresponding standard compositions begins:
7: (1,1,1) 44: (2,1,3) 97: (1,5,1)
11: (2,1,1) 49: (1,4,1) 98: (1,4,2)
13: (1,2,1) 50: (1,3,2) 100: (1,3,3)
14: (1,1,2) 52: (1,2,3) 104: (1,2,4)
19: (3,1,1) 56: (1,1,4) 112: (1,1,5)
21: (2,2,1) 67: (5,1,1) 131: (6,1,1)
22: (2,1,2) 69: (4,2,1) 133: (5,2,1)
25: (1,3,1) 70: (4,1,2) 134: (5,1,2)
26: (1,2,2) 73: (3,3,1) 137: (4,3,1)
28: (1,1,3) 74: (3,2,2) 138: (4,2,2)
35: (4,1,1) 76: (3,1,3) 140: (4,1,3)
37: (3,2,1) 81: (2,4,1) 145: (3,4,1)
38: (3,1,2) 82: (2,3,2) 146: (3,3,2)
41: (2,3,1) 84: (2,2,3) 148: (3,2,3)
42: (2,2,2) 88: (2,1,4) 152: (3,1,4)
(End)

Crossrefs

Cf. A038465 (base 3), A038471 (base 4), A038475 (base 5).
Cf. A081091 (primes), A212190 (squares), A212192 (triangular numbers), A173589 (Fibbinary).
Cf. A057168.
Cf. A000079, A018900, A014311, A014312, A014313, A023688, A023689, A023690, A023691 (Hammingweight = 1, 2, ..., 9).
A000217(n-2) counts compositions into three parts.
A001399(n-3) = A069905(n) = A211540(n+2) counts the unordered case.
A001399(n-6) = A069905(n-3) = A211540(n-1) counts the unordered strict case.
A001399(n-6)*6 = A069905(n-3)*6 = A211540(n-1)*6 counts the strict case.
A014612 is an unordered version, with strict case A007304.
A337453 is the strict case.
A337461 counts the coprime case.
A033992 lists numbers divisible by exactly three different primes.
A323024 lists numbers with exactly three different prime multiplicities.

Programs

  • C
    unsigned hakmem175(unsigned x) {
        unsigned s, o, r;
        s = x & -x;  r = x + s;
        o = r ^ x;  o = (o >> 2) / s;
        return r | o;
    }
    unsigned A014311(int n) {
        if (n == 1) return 7;
        return hakmem175(A014311(n - 1));
    }  // Peter Luschny, Jan 01 2014
    
  • Haskell
    a014311 n = a014311_list !! (n-1)
    a014311_list = [2^x + 2^y + 2^z |
                    x <- [2..], y <- [1..x-1], z <- [0..y-1]]
    -- Reinhard Zumkeller, May 03 2012
    
  • Mathematica
    Select[Range[200], (Count[IntegerDigits[#, 2], 1] == 3)&]
    nn = 8; Flatten[Table[2^i + 2^j + 2^k, {i, 2, nn}, {j, 1, i - 1}, {k, 0, j - 1}]] (* T. D. Noe, Nov 05 2013 *)
  • PARI
    for(n=0,10^3,if(hammingweight(n)==3,print1(n,", "))); \\ Joerg Arndt, Mar 04 2014
    
  • PARI
    print1(t=7);for(i=2,50,print1(","t=A057168(t))) \\ M. F. Hasler, Aug 27 2014
    
  • Python
    A014311_list = [2**a+2**b+2**c for a in range(2,6) for b in range(1,a) for c in range(b)] # Chai Wah Wu, Jan 24 2021
    
  • Python
    from itertools import islice
    def A014311_gen(): # generator of terms
        yield (n:=7)
        while True: yield (n:=n^((a:=-n&n+1)|(a>>1)) if n&1 else ((n&~(b:=n+(a:=n&-n)))>>a.bit_length())^b)
    A014311_list = list(islice(A014311_gen(),20)) # Chai Wah Wu, Mar 10 2025
    
  • Python
    from math import isqrt, comb
    from sympy import integer_nthroot
    def A014311(n): return (1<<(r:=n-1-comb((m:=integer_nthroot(6*n,3)[0])+(t:=(n>comb(m+2,3)))+1,3))-comb((k:=isqrt(b:=r+1<<1))+(b>k*(k+1)),2))+(1<<(a:=isqrt(s:=n-comb(m-(t^1)+2,3)<<1))+((s<<2)>(a<<2)*(a+1)+1))+(1<Chai Wah Wu, Mar 10 2025

Formula

A000120(a(n)) = 3. - Reinhard Zumkeller, May 03 2012
Start with A084468. If n is in sequence, then 2n is too. - Ralf Stephan, Aug 16 2013
a(n+1) = A057168(a(n)). - M. F. Hasler, Aug 27 2014
a(n) = 2^A056558(n-1) + 2^A194848(n-1) + 2^A194847(n-1). - Ridouane Oudra, Sep 06 2020
Sum_{n>=1} 1/a(n) = A367110 = 1.428591545852638123996854844400537952781688750906133068397189529775365950039... (calculated using Baillie's irwinSums.m, see Links). - Amiram Eldar, Feb 14 2022

Extensions

Extension and program by Olivier Gérard