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A258440
Number of squares of all sizes in 3*n*(n+1)/2-ominoes in form of three-quarters of Aztec diamonds.
3
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
1,1
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.
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
EXAMPLE
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.
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 *)
PROG
(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
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Luce ETIENNE, May 30 2015
EXTENSIONS
Typo in data fixed by Colin Barker, Jun 01 2015
Name edited by Michel Marcus, Dec 22 2020
STATUS
approved