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 A333574 Number of  Hamiltonian paths in the n X 2 grid graph which start at any of the n vertices on left side of the graph and terminate at any of the n vertices on the right side. 2
 1, 2, 4, 6, 10, 14, 20, 26, 34, 42, 52, 62, 74, 86, 100, 114, 130, 146, 164, 182, 202, 222, 244, 266, 290, 314, 340, 366, 394, 422, 452, 482, 514, 546, 580, 614, 650, 686, 724, 762, 802, 842, 884, 926, 970, 1014, 1060, 1106, 1154, 1202, 1252, 1302, 1354, 1406, 1460 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 LINKS Colin Barker, Table of n, a(n) for n = 1..1000 Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1). FORMULA G.f.: x*(1+2*x*(1-x^2+x^3)/((1+x)*(1-x)^3)). From Colin Barker, Mar 27 2020: (Start) a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n>5. a(n) = (9 + (-1)^(1+n) - 4*n + 2*n^2) / 4 for n>1. (End) EXAMPLE a(1) = 1;    +--+ a(2) = 2;    +  +   *--*    |  |   |  |    *--*   +  + a(3) = 4;    +  +   +--*   *--+   *--*    |  |      |   |      |  |    *  *   *--*   *--*   *  *    |  |   |         |   |  |    *--*   *--+   +--*   +  + PROG (PARI) N=66; x='x+O('x^N); Vec(x*(1+2*x*(1-x^2+x^3)/((1+x)*(1-x)^3))) (Python) # Using graphillion from graphillion import GraphSet import graphillion.tutorial as tl def A(start, goal, n, k):     universe = tl.grid(n - 1, k - 1)     GraphSet.set_universe(universe)     paths = GraphSet.paths(start, goal, is_hamilton=True)     return paths.len() def A333571(n, k):     if n == 1: return 1     s = 0     for i in range(1, n + 1):         for j in range(k * n - n + 1, k * n + 1):             s += A(i, j, k, n)     return s def A333574(n):     return A333571(n, 2) print([A333574(n) for n in range(1, 25)]) CROSSREFS Column k=2 of A333571. Cf. A333510. Sequence in context: A094589 A071425 A115065 * A008804 A001307 A322010 Adjacent sequences:  A333571 A333572 A333573 * A333575 A333576 A333577 KEYWORD nonn,easy AUTHOR Seiichi Manyama, Mar 27 2020 STATUS approved

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Last modified September 28 17:43 EDT 2020. Contains 337393 sequences. (Running on oeis4.)