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.

A287896 a(n) = A002487(n)*A001511(n).

Original entry on oeis.org

1, 2, 2, 3, 3, 4, 3, 4, 4, 6, 5, 6, 5, 6, 4, 5, 5, 8, 7, 9, 8, 10, 7, 8, 7, 10, 8, 9, 7, 8, 5, 6, 6, 10, 9, 12, 11, 14, 10, 12, 11, 16, 13, 15, 12, 14, 9, 10, 9, 14, 12, 15, 13, 16, 11, 12, 10, 14, 11, 12, 9, 10, 6, 7, 7, 12, 11, 15, 14, 18, 13, 16, 15, 22, 18, 21, 17, 20, 13, 15, 14, 22, 19, 24, 21, 26, 18, 20, 17
Offset: 1

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Author

I. V. Serov, Jun 02 2017

Keywords

Comments

Proposed name: N-fusc.
Each number n>0 appears in this sequence exactly n times.
From Yosu Yurramendi, Apr 08 2019: (Start)
The terms (n>0) may be written as a left-justified array with rows of length 2^m:
1,
2, 2,
3, 3, 4, 3,
4, 4, 6, 5, 6, 5, 6, 4,
5, 5, 8, 7, 9, 8, 10, 7, 8, 7, 10, 8, 9, 7, 8, 5,
6, 6, 10, 9, 12, 11, 14, 10, 12, 11, 16, 13, 15, 12, 14, 9, 10, 9, ...
...
as well as right-justified fashion:
1,
2, 2,
3, 3, 4, 3,
4, 4, 6, 5, 6, 5, 6, 4,
5, 5, 8, 7, 9, 8, 10, 7, 8, 7, 10, 8, 9, 7, 8, 5,
... 14, 9, 10, 9, 14, 12, 15, 13, 16, 11, 12, 10, 14, 11, 12, 9, 10, 6,
From these two dispositions interesting properties can be induced (see FORMULA section)
(End)

Crossrefs

Programs

  • Mathematica
    Table[Block[{a = 1, b = 0, m = n}, While[m > 0, If[OddQ@ m, b = a + b, a = a + b]; m = Floor[m/2]]; b] IntegerExponent[2 n, 2], {n, 89}] (* Michael De Vlieger, Jun 14 2017, after Jean-François Alcover at A002487 *)
  • Python
    from functools import reduce
    def A287896(n): return (n&-n).bit_length()*sum(reduce(lambda x,y:(x[0],x[0]+x[1]) if int(y) else (x[0]+x[1],x[1]),bin(n)[-1:2:-1],(1,0))) # Chai Wah Wu, Jul 14 2022

Formula

a(1) = 1; for n>1: a(n) = (A002487(n-1) + A002487(n) + A002487(n+1))/2.
a(n) = A007306(n) - A288002(n).
From Yosu Yurramendi, Apr 08 2019: (Start)
For m >= 0, 0 <= k < 2^m, a(2^(m+1)+k) - a(2^m+k) = a(k). a(0) = 1 is needed.
For m >= 0, 0 <= k < 2^m, a(2^(m+1)-1-k) - a(2^(m)-1-k) = a(k).
(End)