OFFSET
0,1
COMMENTS
Number of tilings of the 0-mod-4 pillow of order n is a perfect square times a(n). [Propp, 1999, p. 271]
REFERENCES
J. Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see Problem 12).
LINKS
A.H.M. Smeets, Table of n, a(n) for n = 0..2169
J. Propp, Updated article
J. Propp, Enumeration of matchings: problems and progress, in L. J. Billera et al. (eds.), New Perspectives in Algebraic Combinatorics
Index entries for linear recurrences with constant coefficients, signature (2,2,2,-1).
FORMULA
G.f.: (5 + 3*x + x^2 -x^3) / (1 - 2*x - 2*x^2 - 2*x^3 + x^4).
a(-n) = a(-5 + n). a(-1) = a(-2) = 1. a(n) = 2*a(n-1) + 2*a(n-2) + 2*a(n-3) - a(n-4). - Michael Somos, Dec 15 2011
A112835(2*n + 2) = a(n).
Lim_{n -> inf} a(n)/a(n-1) = A318605. - A.H.M. Smeets, Sep 12 2018
EXAMPLE
G.f. = 5 + 13*x + 37*x^2 + 109*x^3 + 313*x^4 + 905*x^5 + 2617*x^6 + 7561*x^7 + ...
MAPLE
seq(coeff(series((5+3*x+x^2-x^3)/(1-2*x-2*x^2-2*x^3+x^4), x, n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Sep 12 2018
MATHEMATICA
CoefficientList[Series[(5 + 3*x + x^2 -x^3)/(1 - 2*x - 2*x^2 - 2*x^3 + x^4), {x, 0, 50}], x] (* Stefano Spezia, Sep 12 2018 *)
LinearRecurrence[{2, 2, 2, -1}, {5, 13, 37, 109}, 30] (* Harvey P. Dale, Sep 03 2021 *)
PROG
(PARI) {a(n) = my(m = n+2); if( m < 0, m = -1 - m); polcoeff( (1 - x + x^2 - x^3) / (1 - 2*x - 2*x^2 - 2*x^3 + x^4) + x * O(x^m), m)}; /* Michael Somos, Dec 15 2011 */
(PARI) x='x+O('x^33); Vec((5+3*x+x^2-x^3)/(1-2*x-2*x^2-2*x^3+x^4)) \\ Altug Alkan, Sep 12 2018
(GAP) a:=[5, 13, 37, 109];; for n in [5..30] do a[n]:=2*a[n-1]+2*a[n-2]+2*a[n-3]-a[n-4]; od; a; # Muniru A Asiru, Sep 12 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, May 28 2002
STATUS
approved