A385868 Number of ways to tile a hexagonal strip made up of n equilateral triangles, using triangles, diamonds, and trapezoids.
1, 1, 2, 4, 7, 13, 39, 66, 110, 200, 604, 1032, 1741, 3149, 9476, 16202, 27337, 49461, 148841, 254466, 429308, 776774, 2337580, 3996430, 6742361, 12199339, 36711974, 62764458, 105889743, 191592331, 576566591, 985724436, 1663012914, 3008983882, 9055057632, 15480937786
Offset: 0
Examples
Here is one of the a(13) = 3149 possible tilings for this strip of 13 triangular cells: ____________ / /\ \ \ /__ /__\ \ __\ \ /\ /\ \____/__\/__\.
Links
- Index entries for linear recurrences with constant coefficients, signature (0,0,0,15,0,0,0,9,0,0,0,32,0,0,0,9,0,0,0,1,0,0,0,-1).
Programs
-
Mathematica
a[0] = 1; a[1] = 1; a[2] = 2; a[3] = 4; a[4] = 7; a[5] = 13; a[6] = 39; a[7] = 66; a[n_] := a[n] = Switch[Mod[n, 4], 0, a[n-1] + a[n-3] + 2 a[n-4] + 3 a[n-5] + 2 a[n-6] + a[n-7], 1, a[n-1] + a[n-2] + a[n-4] + a[n-5] + a[n-6], 2, a[n-1] + a[n-2] + 3 a[n-3] + a[n-4] + 3 a[n-5] + 2 a[n-6] + a[n-7], 3, a[n-1] + a[n-2] + a[n-3] + a[n-4] + a[n-5] + a[n-6]]; Table[a[n], {n, 0, 40}]
Formula
a(n) = 15*a(n-4) + 9*a(n-8) + 32*a(n-12) + 9*a(n-16) + 1*a(n-20) - 1*a(n-24).
a(4*n+2) = t(2*n+1)^2 + t(2*n)^2 + 2*t(2*n)*t(2*n-1) + a(4*n) + 2*a(4*n-1) + Sum_{i=1..n-1} a(4*i)*(t(2*(n-i))^2 + 2*t(2*(n-i))*t(2*(n-i)-1)) + Sum_{i=1..n-1} a(4*i-1)*2*t(2*(n-i))^2, for t(n) = A000073(n+2).
G.f.: (x^18-x^17+x^16-x^14+4*x^12-6*x^11-x^10+4*x^9+4*x^8-6*x^7-9*x^6+2*x^5+8*x^4-4*x^3 -2*x^2-x-1) / (-x^24+x^20+9*x^16+32*x^12+9*x^8+15*x^4-1). - Alois P. Heinz, Jul 21 2025
Comments