A362905 Array read by antidiagonals: T(n,k) is the number of n element multisets of length k vectors over GF(2) that sum to zero.

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 4, 2, 1, 1, 1, 8, 5, 3, 1, 1, 1, 16, 15, 11, 3, 1, 1, 1, 32, 51, 50, 14, 4, 1, 1, 1, 64, 187, 276, 99, 24, 4, 1, 1, 1, 128, 715, 1768, 969, 232, 30, 5, 1, 1, 1, 256, 2795, 12496, 11781, 3504, 429, 45, 5, 1, 1, 1, 512, 11051, 93600, 162877, 73440, 10659, 835, 55, 6, 1

Offset: 0

Author

Andrew Howroyd, May 27 2023

Keywords

• nonn
• tabl

Comments

Equivalently, T(n,k) is the number multisets with n elements drawn from {0..2^k-1} such that the bitwise-xor of all the elements gives zero.

T(n,k) is the number of equivalence classes of n X k binary matrices with an even number of 1's in each column under permutation of rows.

T(n,k) is the number of equivalence classes of n X k binary matrices under permutation of rows and complementation of columns.

Examples

			Array begins:
=========================================
n/k| 0 1  2   3     4      5        6 ...
---+-------------------------------------
0  | 1 1  1   1     1      1        1 ...
1  | 1 1  1   1     1      1        1 ...
2  | 1 2  4   8    16     32       64 ...
3  | 1 2  5  15    51    187      715 ...
4  | 1 3 11  50   276   1768    12496 ...
5  | 1 3 14  99   969  11781   162877 ...
6  | 1 4 24 232  3504  73440  1878976 ...
7  | 1 4 30 429 10659 394383 18730855 ...
  ...
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..1325 (first 51 antidiagonals).

Crossrefs

Columns k=0..4 are A000012, A004526(n+2), A053307, A362906, A363350.

Rows n=2..3 are A000079, A007581.

Main diagonal is A363351.

Cf. A054724, A340030, A340312, A363349.

Programs

Mathematica

A362905[n_,k_]:=(Binomial[2^k+n-1,n]+If[EvenQ[n],(2^k-1)Binomial[2^(k-1)+n/2-1,n/2],0])/2^k;Table[A362905[n-k,k],{n,0,15},{k,n,0,-1}] (* Paolo Xausa, Nov 19 2023 *)

PARI

T(n,k)={(binomial(2^k+n-1, n) + if(n%2==0, (2^k-1)*binomial(2^(k-1)+n/2-1,n/2)))/2^k}

Formula

T(n,k) = binomial(2^k+n-1, n)/2^k for odd n;

T(n,k) = (binomial(2^k+n-1, n) + (2^k-1)*binomial(2^(k-1)+n/2-1, n/2))/2^k for even n.

G.f. of column k: (1/(1-x)^(2^k) + (2^k-1)/(1-x^2)^(2^(k-1)))/2^k.