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 1-3 of 3 results.

A330805 Number of squares and rectangles in the interior of the square with vertices (n,0), (0,n), (-n,0) and (0,-n) in a square (x,y)-grid.

Original entry on oeis.org

0, 9, 51, 166, 410, 855, 1589, 2716, 4356, 6645, 9735, 13794, 19006, 25571, 33705, 43640, 55624, 69921, 86811, 106590, 129570, 156079, 186461, 221076, 260300, 304525, 354159, 409626, 471366, 539835, 615505, 698864, 790416, 890681, 1000195, 1119510, 1249194, 1389831
Offset: 0

Views

Author

Luce ETIENNE, Jan 01 2020

Keywords

Comments

Collection: 2*n*(n+1)-ominoes.
Number of squares (all sizes): (8*n^3 + 24*n^2 + 22*n - 3*(-1)^n + 3)/12.
Number of rectangles (all sizes): (8*n^4 + 24*n^3 + 22*n^2 + 3*(-1)^n - 3)/12.

Examples

			a(1) = 4*1+5 = 9; a(2) = 4*5+31 = 51; a(3) = 4*15 + 106 = 166; a(4) = 4*36 + 270 = 410.
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{5,-10,10,-5,1},{0,9,51,166,410},40] (* Harvey P. Dale, Jun 27 2020 *)

Formula

G.f.: x*(x + 3)^2/(1 - x)^5.
E.g.f.: (1/6)*exp(x)*x*(54 + 99*x + 40*x^2 + 4*x^3). - Stefano Spezia, Jan 01 2020
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = n*(n + 1)*(4*n^2 + 12*n + 11)/6.
a(n) = 4*A000332(n+3) + A212523(n+1).
a(n) = 9*A000332(n+3) + 6*A000332(n+2) + A000332(n+1). - Mircea Dan Rus, Aug 26 2020
a(n) = 3*A004320(n) + A004320(n-1). - Mircea Dan Rus, Aug 26 2020

A258440 Number of squares of all sizes in 3*n*(n+1)/2-ominoes in form of three-quarters of Aztec diamonds.

Original entry on oeis.org

3, 11, 25, 49, 84, 132, 196, 278, 379, 503, 651, 825, 1028, 1262, 1528, 1830, 2169, 2547, 2967, 3431, 3940, 4498, 5106, 5766, 6481, 7253, 8083, 8975, 9930, 10950, 12038, 13196, 14425, 15729, 17109, 18567, 20106, 21728, 23434, 25228, 27111, 29085, 31153, 33317, 35578, 37940, 40404, 42972, 45647, 48431
Offset: 1

Views

Author

Luce ETIENNE, May 30 2015

Keywords

Comments

These polyominoes are 6*n-gons, and thus their number of vertices is n*(3*n+7).
Schäfli's notation for figure corresponding to a(1): 4.4.4.

Examples

			a(1)=3, a(2)=9+2=11, a(3)=18+7=25, a(4)=30+15+4=49, a(5)=45+26+11+2=84.
		

Crossrefs

Programs

  • Magma
    [(52*n^3+186*n^2+212*n-3*(32*Floor(n/3)+3*(1-(-1)^n)))/144: n in [1..50]]; // Vincenzo Librandi, Jun 02 2015
  • Maple
    A258440:=n->(52*n^3+186*n^2+212*n-3*(32*floor(n/3)+3*(1-(-1)^n)))/144: seq(A258440(n), n=1..100); # Wesley Ivan Hurt, Jul 10 2015
  • Mathematica
    Table[(52 n^3 + 186 n^2 + 212 n - 3 (32 Floor[n/3] + 3 (1 - (-1)^n)))/144, {n, 45}] (* Vincenzo Librandi, Jun 02 2015 *)
  • PARI
    Vec(x*(2*x^3+3*x^2+5*x+3)/((x-1)^4*(x+1)*(x^2+x+1)) + O(x^100)) \\ Colin Barker, Jun 01 2015
    

Formula

a(n) = 2*A241526(n) - A173196(n+1).
a(n) = (1/8)*(Sum_{i=0..(n-1-floor(n/3)}(4*n+1-6*i-(-1)^i)*(4*n+3-6*i+(-1)^i)- Sum_{j=0..(2*n-1+(-1)^n)}(2*n+1+(-1)^n-4*j)*(2*n+1-(-1)^n-4*j)).
a(n) = (52*n^3+186*n^2+212*n-3*(32*floor(n/3)+3*(1-(-1)^n)))/144.
a(n) = 2*a(n-1)-a(n-3)-a(n-4)+2*a(n-6)-a(n-7) for n>7. - Colin Barker, Jun 01 2015
G.f.: x*(2*x^3+3*x^2+5*x+3) / ((x-1)^4*(x+1)*(x^2+x+1)). - Colin Barker, Jun 01 2015

Extensions

Typo in data fixed by Colin Barker, Jun 01 2015
Name edited by Michel Marcus, Dec 22 2020

A111500 Number of squares in an n X n grid of squares with diagonals.

Original entry on oeis.org

1, 10, 31, 72, 137, 234, 367, 544, 769, 1050, 1391, 1800, 2281, 2842, 3487, 4224, 5057, 5994, 7039, 8200, 9481, 10890, 12431, 14112, 15937, 17914, 20047, 22344, 24809, 27450, 30271, 33280, 36481, 39882, 43487, 47304, 51337, 55594, 60079, 64800, 69761, 74970
Offset: 0

Views

Author

Floor van Lamoen, Nov 16 2005

Keywords

Comments

This sequence is the sum of the number of squares with horizontal/vertical sides (whose length is a positive integer), which is equal to Sum_{j=1..n} j^2 = (n*(n + 1)*(2*n + 1))/6, and the number of squares with diagonal sides (whose length is a multiple of sqrt(2)/2), which is Sum_{j=1..n} (A111746(n - 1)) = floor((n*(4*n^2 - 1))/6). - Marco Ripà, Jan 14 2024

Crossrefs

Programs

  • Maple
    seq(n^3+n^2/2-1/4+1/4*(-1)^n,n=1..65);
  • PARI
    Vec((x^3+3*x^2+7*x+1) / ((x-1)^4*(x+1)) + O(x^100)) \\ Colin Barker, May 28 2015

Formula

a(n) = n^3 + n^2/2 - 1/4 + (1/4)*(-1)^n.
a(n) = 3*a(n-1) - 2*a(n-2) - 2*a(n-3) + 3*a(n-4) - a(n-5) for n > 4. - Colin Barker, May 28 2015
G.f.: (x^3 + 3*x^2 + 7*x + 1) / ((x-1)^4*(x+1)). - Colin Barker, May 28 2015
From Marco Ripà, Jan 14 2024: (Start)
a(n) = Sum_{j=1..n} (A000290(j) + A111746(j-1)).
a(n) = floor(n^3 + n^2/2). (End)
Showing 1-3 of 3 results.