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.

Showing 1-2 of 2 results.

A269565 Array read by antidiagonals: T(n,m) is the number of (directed) Hamiltonian paths in K_n X K_m.

Original entry on oeis.org

1, 2, 2, 6, 8, 6, 24, 60, 60, 24, 120, 816, 1512, 816, 120, 720, 17520, 83520, 83520, 17520, 720, 5040, 550080, 8869680, 22394880, 8869680, 550080, 5040, 40320, 23839200, 1621680480, 13346910720, 13346910720, 1621680480, 23839200, 40320
Offset: 1

Views

Author

Andrew Howroyd, Feb 29 2016

Keywords

Comments

Equivalently, the number of directed Hamiltonian paths on the n X m rook graph.
Conjecture: T(n,m) mod n!*m! = 0. - Mikhail Kurkov, Feb 08 2019
The above conjecture is true since a path defines an ordering on the rows and columns by the order in which they are first visited by the path. Every permutation of rows and columns therefore gives a different path. - Andrew Howroyd, Feb 08 2021

Examples

			Array begins:
===========================================================
n\m|    1      2        3            4               5
---+-------------------------------------------------------
1  |    1,     2,       6,          24,            120, ...
2  |    2,     8,      60,         816,          17520, ...
3  |    6,    60,    1512,       83520,        8869680, ...
4  |   24,   816,   83520,    22394880,    13346910720, ...
5  |  120, 17520, 8869680, 13346910720, 50657369241600, ...
...
		

Crossrefs

Main diagonal is A096970.
Columns 2..3 are A096121, A329319.

Formula

From Andrew Howroyd, Oct 20 2019: (Start)
T(n,m) = T(m,n).
T(n,1) = n!. (End)

A329320 a(n) = Sum_{k=0..floor(log_2(n))} 1 - A035263(1 + floor(n/2^k)).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 3, 2, 3, 1, 2, 2, 2, 2, 3, 2, 3, 1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 2, 3, 3, 3, 1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 2, 3, 3, 3, 1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 4, 3
Offset: 0

Views

Author

Mikhail Kurkov, Nov 10 2019 [verification needed]

Keywords

Comments

Sequence which arise from attempts to simplify computing of A329319.
For all positive integers k, the subsequence a(2^k) to a(3*2^(k-1)-1) is identical to the subsequence a(3*2^(k-1)) to a(2^(k+1)-1). Also subsequences a(2^k) to a(3*2^(k-1)-1) and a(0) to a(2^(k-1)-1) always differ by 1.

Crossrefs

Programs

  • PARI
    a(n) = if (n==0, 0, a(floor(n/2)) + valuation(n+1, 2) %  2); \\ Michel Marcus, Nov 13 2019
    
  • PARI
    a(n)=my(s,t); while(n, n>>=valuation(n,2); t=valuation(n+1,2); s+=(t+1)\2; n>>=t); s \\ Charles R Greathouse IV, Oct 14 2021

Formula

a(n) = a(floor(n/2)) + 1 - A035263(n+1) for n>0 with a(0)=0.
a(2^m+k) = a(k mod 2^(m-1)) + 1 for 0<=k<2^m, m>0 with a(0)=0, a(1)=1.
Showing 1-2 of 2 results.