A260918 Number of squares of all sizes in polyominoes obtained by union of two pyramidal figures (A092498) with intersection equals A002623.

0, 1, 5, 15, 33, 60, 100, 154, 224, 313, 423, 555, 713, 898, 1112, 1358, 1638, 1953, 2307, 2701, 3137, 3618, 4146, 4722, 5350, 6031, 6767, 7561, 8415, 9330, 10310, 11356, 12470, 13655, 14913, 16245, 17655, 19144, 20714, 22368, 24108, 25935, 27853, 29863

Offset: 0

Author

Luce ETIENNE, Aug 04 2015

Keywords

• nonn
• easy

Comments

The resulting polyforms are n*(3*n-1)/2-polyominoes.

Also they are 6*n-gons with n>1.

Schäfli's notation for figure corresponding to a(1): 4.

Examples

			a(1)=1, a(2)=5, a(3)=12+3=15, a(4)=22+9+2=33, a(5)=35+18+7=60, a(6)=51+30+15+4=100.
		

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Luce ETIENNE, Illustration of initial terms
• Index entries for linear recurrences with constant coefficients, signature (2,0,-1,-1,0,2,-1).

Crossrefs

Cf. A002623, A092498, A173196, A000212, A258440.

Programs

Magma

[(52*n^3+90*n^2+20*n-3*(32*Floor((n+1)/3)+3*(1-(-1)^n)))/144: n in [0..50]]; // Vincenzo Librandi, Aug 12 2015

Mathematica

Table[(52 n^3 + 90 n^2 + 20 n - 3 (32 Floor[(n + 1) / 3] + 3 (1 - (-1)^n))) / 144, {n, 0, 45}] (* Vincenzo Librandi, Aug 12 2015 *)

PARI

concat(0, Vec(x*(4*x^3+5*x^2+3*x+1)/((x-1)^4*(x+1)*(x^2+x+1)) + O(x^100))) \\ Colin Barker, Aug 08 2015

Formula

a(n) = A258440(n) - A000212(n+1).

a(n) = (1/8)*((Sum_{i=0..floor(2*n/3)} (4*n+1-6*i-(-1)^i)*(4*n-1-6*i+(-1)^i)) - (Sum_{j=0..(2*n-1+(-1)^n)/4} (2*n+1-(-1)^n-4*j)*(2*n+1+(-1)^n-4*j))).

a(n) = (52*n^3+90*n^2+20*n-3*(32*floor((n+1)/3)+3*(1-(-1)^n)))/144.

G.f.: x*(4*x^3+5*x^2+3*x+1) / ((x-1)^4*(x+1)*(x^2+x+1)). - Colin Barker, Aug 08 2015

E.g.f.: (3*exp(x)*x*(65 + x*(123 + 26*x)) + 32*sqrt(3)*exp(-x/2)*sin(sqrt(3)*x/2) - 27*sinh(x))/216. - Stefano Spezia, Nov 15 2024

Extensions

Two repeated terms deleted by Colin Barker, Aug 08 2015