A238846 Expansion of (1-2*x)/(1-3*x+x^2)^2.
1, 4, 13, 40, 120, 354, 1031, 2972, 8495, 24110, 68016, 190884, 533293, 1484020, 4115185, 11375764, 31358376, 86223942, 236540915, 647556620, 1769374931, 4826148314, 13142564448, 35736448200, 97037995225, 263156279524, 712795854421, 1928547574912, 5212430732760
Offset: 0
Examples
a(0) = 1*1 = 1; a(1) = 1*3 + 1*1 = 4; a(2) = 1*8 + 1*3 + 2*1 = 13; a(3) = 1*21 + 1*8 + 2*3 + 5*1 = 40; a(4) = 1*55 + 1*21 + 2*8 + 5*3 + 13*1 = 120; etc. (from first recurrence formula). a(0) = 3*0 - 0 + 1 = 1; a(1) = 3*1 - 0 + 1 = 4; a(2) = 3*4 - 1 + 2 = 13; a(3) = 3*13 - 4 + 5 = 40; a(4) = 3*40 - 13 + 13 = 120; etc (from second recurrence formula). G.f. = 1 + 4*x + 13*x^2 + 40*x^3 + 120*x^4 + 354*x^5 + 1031*x^6 + ... - _Michael Somos_, Nov 23 2021
Links
- Michael De Vlieger, Table of n, a(n) for n = 0..2385
- Sergi Elizalde, Rigoberto Flórez, and José Luis Ramírez, Enumerating symmetric peaks in non-decreasing Dyck paths, Ars Mathematica Contemporanea (2021).
- David Eppstein, Non-crossing Hamiltonian Paths and Cycles in Output-Polynomial Time, arXiv:2303.00147 [cs.CG], 2023, p. 20.
- Rigoberto Flórez, Leandro Junes, Luisa M. Montoya, and José L. Ramírez, Counting Subwords in Non-Decreasing Dyck Paths, J. Int. Seq. (2025) Vol. 28, Art. No. 25.1.6. See pp. 6, 15, 17, 19.
- Valentin Ovsienko, Shadow sequences of integers, from Fibonacci to Markov and back, arXiv:2111.02553 [math.CO], 2021.
- Index entries for linear recurrences with constant coefficients, signature (6,-11,6,-1).
Programs
-
Mathematica
LinearRecurrence[{6, -11, 6, -1}, {1, 4, 13, 40}, 30] (* Bruno Berselli, Mar 06 2014 *) a[ n_] := If[n < 0, SeriesCoefficient[ x^3*(2 - x)/(1 - 3*x + x^2)^2, {x, 0, -n}], SeriesCoefficient[ (1 - 2*x)/(1 - 3*x + x^2)^2, {x, 0, n}]]; (* Michael Somos, Nov 23 2021 *)
-
PARI
{a(n) = if(n<0, polcoeff( x^3*(2-x)/(1-3*x+x^2)^2 + x*O(x^-n), -n), polcoeff( (1-2*x)/(1-3*x+x^2)^2 + x*O(x^n), n))}; /* Michael Somos, Nov 23 2021 */
Formula
a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4) for n>3, a(0)=1, a(1)=4, a(2)=13, a(3)=40.
a(n) = 3*a(n-1) - a(n-2) + A001519(n) for n>1, a(0)=1, a(1)=4.
a(n) = A238731(n+1, 1).
a(n) = -A124037(n+1, 1).
a(n) = (-1)^n*A126126(n+1, 1).
a(n) = ( (3 + sqrt(5))^(1+n)*(8 - (1 - sqrt(5))*(13 + 5*n)) + (3 - sqrt(5))^(1+n)*(8 - (1 + sqrt(5))*(13 + 5*n)) ) / (25*2^(2+n)). - Bruno Berselli, Mar 06 2014
From Philippe Deléham, Mar 06 2014: (Start)
0 = 2 + a(n)*(a(n+1) - a(n+3)) + a(n+1)*(-6*a(n+1) + 12*a(n+2)) + a(n+2)*(-6*a(n+2) + a(n+3)) for all n in Z. - Michael Somos, Nov 23 2021
E.g.f.: exp(3*x/2)*(5*(5 + 4*x)*cosh(sqrt(5)*x/2) + sqrt(5)*(17 + 10*x)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Mar 04 2025
Comments