A193530 Expansion of (1 - 2*x - 2*x^2 + 3*x^3 + x^5)/((1-x)*(1-2*x-x^2)*(1-2*x^2-x^4)).

1, 1, 2, 3, 7, 13, 31, 66, 159, 363, 876, 2065, 4985, 11915, 28765, 69156, 166957, 402373, 971414, 2343519, 5657755, 13654969, 32966011, 79577190, 192116331, 463786191, 1119678912, 2703086893, 6525829037, 15754607063, 38034986041, 91824246216, 221683340569, 535190123593, 1292063254826

Offset: 0

Author

F. Chapoton and N. J. A. Sloane, Jul 29 2011

Keywords

• nonn
• easy

Comments

This sequence was initially confused with A003120, but they are different sequences. The g.f. used here as the definition was found by Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Gy. Tasi and F. Mizukami, Quantum algebraic-combinatoric study of the conformational properties of n-alkanes, J. Math. Chemistry, 25, 1999, 55-64 (see p. 63).
• Index entries for linear recurrences with constant coefficients, signature (3,1,-7,3,-1,1,1).

Crossrefs

Cf. A000129, A001333, A002203, A135153.

Programs

Magma

m:=40; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-2*x-2*x^2 +3*x^3+x^5)/((1-x)*(1-2*x-x^2)*(1-2*x^2-x^4)) )); // Vincenzo Librandi, Aug 28 2016

Maple

f:=n->if n mod 2 = 0 then (1/4)*(A001333(n-2)+A001333((n-2)/2)+A001333((n-4)/2)+1) else (1/4)*(A001333(n-2)+A001333((n-1)/2)+A001333((n-3)/2)+1); fi; # produces the sequence with a different offset

Mathematica

LinearRecurrence[{3,1,-7,3,-1,1,1}, {1,1,2,3,7,13,31}, 40] (* Vincenzo Librandi, Aug 28 2016 *)
Table[(2 +LucasL[n, 2] +2*(1+(-1)^n)*Fibonacci[(n+2)/2, 2] + 2*(1-(-1)^n)*Fibonacci[(n+1)/2, 2])/8, {n, 0, 40}] (* G. C. Greubel, May 21 2021 *)

Sage

@CachedFunction
def Pell(n): return n if (n<2) else 2*Pell(n-1) + Pell(n-2)
def A193530(n): return (1 + Pell(n+1) - Pell(n) + (1 + (-1)^n)*Pell((n+2)/2) + (1-(-1)^n)*Pell((n+1)/2) )/4
[A193530(n) for n in (0..40)] # G. C. Greubel, May 21 2021

Formula

a(n) = 1 + A005409(floor((n+3)/2)) + A107769(n).

From G. C. Greubel, May 21 2021: (Start)

a(n) = (1 + A001333(n) + A135153(n+2))/4.

a(n) = (2 + Q(n) + 2*(1+(-1)^n)*Pell((n+2)/2) + 2*(1-(-1)^n)*Pell((n+1)/2))/8.

a(2*n) = (2 + Q(2*n) + 4*Pell(n+1))/8.

a(2*n+1) = (2 + Q(2*n+1) + 4*Pell(n+1))/8, where Pell(n) = A000129(n), and Q(n) = A002203. (End)