A320528 Number of chiral pairs of color patterns (set partitions) in a row of length n using exactly 5 colors (subsets).
0, 0, 0, 0, 0, 6, 64, 508, 3428, 21132, 123050, 688850, 3752350, 20032446, 105372624, 548066568, 2826316248, 14478890712, 73794322750, 374602205590, 1895629599050, 9568906539786, 48208435317284, 242500368793628, 1218342441784468, 6115097961883092, 30669103347259650, 153720181809997530, 770100204404335350, 3856500105221902326
Offset: 1
Examples
For a(6)=6, the chiral pairs are AABCDE-ABCDEE, ABACDE-ABCDED, ABCADE-ABCDEC, ABCDAE-ABCDEB, ABBCDE-ABCDDE, and ABCBDE-ABCDCE.
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (13,-48,-36,551,-683,-1542,3546,80,-4280,2400).
Programs
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Magma
I:=[0,0,0,0,0,6,64,508,3428,21132]; [n le 10 select I[n] else 13*Self(n-1)-48*Self(n-2)-36*Self(n-3)+551*Self(n-4)-683*Self(n-5) -1542*Self(n-6)+3546*Self(n-7)+80*Self(n-8)-4280*Self(n-9) +2400*Self(n-10): n in [1..30]]; // G. C. Greubel, Oct 20 2018
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Mathematica
k=5; Table[(StirlingS2[n,k] - If[EvenQ[n], 3StirlingS2[n/2+2,5] - 11StirlingS2[n/2+1,5] + 6StirlingS2[n/2,5], StirlingS2[(n+5)/2,5] - 3StirlingS2[(n+3)/2,5]])/2, {n,30}] Ach[n_, k_] := Ach[n, k] = If[n<2, Boole[n==k && n>=0], k Ach[n-2,k] + Ach[n-2,k-1] + Ach[n-2,k-2]] (* A304972 *) k = 5; Table[(StirlingS2[n, k] - Ach[n, k])/2, {n, 1, 30}] LinearRecurrence[{13, -48, -36, 551, -683, -1542, 3546, 80, -4280, 2400}, {0, 0, 0, 0, 0, 6, 64, 508, 3428, 21132}, 30]
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PARI
m=30; v=concat([0,0,0,0,0,6,64,508,3428,21132], vector(m-10)); for(n=11, m, v[n] = 13*v[n-1]-48*v[n-2]-36*v[n-3]+551*v[n-4]-683*v[n-5] -1542*v[n-6] +3546*v[n-7] +80*v[n-8] -4280*v[n-9] +2400*v[n-10]); v \\ G. C. Greubel, Oct 20 2018
Formula
a(n) = (S2(n,k) - A(n,k))/2, where k=5 is the number of colors (sets), S2 is the Stirling subset number A008277 and A(n,k) = [n>1] * (k*A(n-2,k) + A(n-2,k-1) + A(n-2,k-2)) + [n<2 & n==k & n>=0].
G.f.: (x^5 / Product_{k=1..5} (1 - k*x) - x^5 (1 + x) (1 - 3 x^2) (1 + 2 x - 2 x^2) / Product_{k=1..5} (1 - k*x^2)) / 2.
a(n) = 13*a(n-1) - 48*a(n-2) - 36*a(n-3) + 551*a(n-4) - 683*a(n-5) - 1542*a(n-6) + 3546*a(n-7) + 80*a(n-8) - 4280*a(n-9) + 2400*a(n-10) for n>10. - Colin Barker, May 12 2020
Comments