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.

A299595 T(n,k)=Number of nXk 0..1 arrays with every element unequal to 0, 1 or 4 horizontally, vertically or antidiagonally adjacent elements, with upper left element zero.

Original entry on oeis.org

1, 2, 2, 3, 1, 3, 5, 3, 3, 5, 8, 5, 5, 5, 8, 13, 8, 11, 11, 8, 13, 21, 15, 19, 23, 19, 15, 21, 34, 26, 35, 53, 53, 35, 26, 34, 55, 45, 65, 121, 113, 121, 65, 45, 55, 89, 80, 120, 250, 256, 256, 250, 120, 80, 89, 144, 140, 220, 533, 541, 576, 541, 533, 220, 140, 144, 233, 245
Offset: 1

Views

Author

R. H. Hardin, Feb 13 2018

Keywords

Comments

Table starts
..1..2...3....5....8...13....21....34....55.....89....144....233.....377
..2..1...3....5....8...15....26....45....80....140....245....431.....756
..3..3...5...11...19...35....65...120...220....404....744...1369....2517
..5..5..11...23...53..121...250...533..1162...2490...5327..11465...24641
..8..8..19...53..113..256...541..1148..2488...5349..11453..24617...52916
.13.15..35..121..256..576..1225..2601..5625..12100..25921..55696..119716
.21.26..65..250..541.1225..2583..5488.11899..25570..54750.117697..252985
.34.45.120..533.1148.2601..5488.11656.25269..54309.116285.249968..537305
.55.80.220.1162.2488.5625.11899.25269.54738.117679.252010.541648.1164265

Examples

			Some solutions for n=5 k=4
..0..0..1..0. .0..1..0..0. .0..1..0..0. .0..0..0..0. .0..0..0..0
..0..0..0..0. .0..0..0..0. .0..0..0..0. .0..0..0..1. .1..0..0..0
..0..0..0..0. .0..0..0..0. .0..0..0..1. .0..0..0..0. .0..0..0..0
..1..0..0..1. .1..0..0..0. .1..0..0..0. .1..0..0..0. .0..0..0..1
..0..0..0..0. .0..0..0..0. .0..0..0..0. .0..0..1..0. .0..0..0..0
		

Crossrefs

Column 1 is A000045(n+1).
Column 2 is A193147.

Formula

Empirical for diagonal: [linear recurrence of order 15] for n>18
Empirical for column k:
k=1: a(n) = a(n-1) +a(n-2)
k=2: a(n) = a(n-1) +2*a(n-3) +a(n-5) for n>6
k=3: a(n) = a(n-1) +2*a(n-3) +a(n-4) +a(n-5) for n>8
k=4..99: a(n) = a(n-1) +a(n-2) +3*a(n-3) +a(n-4) -a(n-5) -a(n-6) for n>9