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%I #30 Jan 06 2023 03:22:13
%S 1,4,12,44,144,528,1808,6676,23536,87568,315136,1180680,4314560,
%T 16263896,60138816,227899484,850600944,3238194560,12177384544,
%U 46542879384,176110444736,675431779856,2568878867200,9882068082112,37747540858240,145593279888736,558190182662144
%N Number of walks of length n on square lattice, starting at origin, staying in first and third quadrants.
%C Is there a formula analogous to the (conjectured) formula for A060900?
%C Could be broken into the number of walks that are constrained to a quadrant and the number that cross the origin. (I.e., 2*A005566(n) + 2*A005566(n-2)*A005568(1) + 2*A005566(n-4)*A005568(2) + ... + All terms that cross the origin twice + three times + ... + Cross floor(n/2) times.) - _Benjamin Phillabaum_, Mar 13 2011
%H Alois P. Heinz, <a href="/A060897/b060897.txt">Table of n, a(n) for n = 0..750</a> (first 251 terms from Sean A. Irvine)
%F G.f.: 1 + 2*(B(x)-1)/(2 - C(x^2)) where B(x) is the g.f. of A005566 and C(x) is the g.f. of A005568. - _Andrew Howroyd_, Jan 05 2023
%o (PARI) \\ here B is A005566 and C is aerated A005568 as g.f.'s.
%o B(n)={sum(n=0, n, x^n*binomial(n, n\2)*binomial(n+1, (n+1)\2), O(x*x^n))}
%o C(n)={sum(n=0, (n+1)\2, x^(2*n)*binomial(2*n,n)*binomial(2*n+2,n+1)/((n+1)*(n+2)), O(x*x^n))}
%o seq(n) = {Vec( 1 + 2*(B(n)-1)/(2-C(n)) )} \\ _Andrew Howroyd_, Jan 05 2023
%Y Cf. A005566, A005568, A001700, A060898, A060899, A060900.
%K nonn
%O 0,2
%A _David W. Wilson_, May 05 2001
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