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A300389
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The number of paths of length 13*n from the origin to the line y = 2*x/11 with unit East and North steps that stay below the line or touch it.
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9
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1, 6, 593, 87143, 15149546, 2891511017, 585739005066, 123655688922720, 26908765569970320, 5993187329634638043, 1359541058523676017369, 313029501692713279534165, 72965556751635426636633639, 17184586991024424745328563477, 4083065013894860643162116395527
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OFFSET
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0,2
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COMMENTS
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Equivalent to nonnegative walks from (0,0) to (13*n,0) with step set [1,2], [1,-11].
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LINKS
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FORMULA
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G.f. satisfies: f = f^78*t^6 + 5*f^67*t^5 - f^66*t^5 + 6*f^65*t^5 + 10*f^56*t^4 - 4*f^55*t^4 + 20*f^54*t^4 - 5*f^53*t^4 + 15*f^52*t^4 + 10*f^45*t^3 - 6*f^44*t^3 + 24*f^43*t^3 - 12*f^42*t^3 + 30*f^41*t^3 - 10*f^40*t^3 + 20*f^39*t^3 + 5*f^34*t^2 - 4*f^33*t^2 + 12*f^32*t^2 - 9*f^31*t^2 + 18*f^30*t^2 - 12*f^29*t^2 + 20*f^28*t^2 - 10*f^27*t^2 + 15*f^26*t^2 + f^23*t - f^22*t + 2*f^21*t - 2*f^20*t + 3*f^19*t - 3*f^18*t + 4*f^17*t - 4*f^16*t + 5*f^15*t - 5*f^14*t + 6*f^13*t + 1.
O.g.f.: A(x) = exp( Sum_{n >= 1} (1/13)*binomial(13*n, 2*n)*x^n/n ) - Bizley.
Recurrence: a(0) = 1 and a(n) = (1/n) * Sum_{k = 0..n-1} (1/13)*binomial(13*n-13*k, 2*n-2*k)*a(k) for n >= 1. (End)
The sequence defined by b(n) := [x^n] A(x)^n begins [1, 6, 1222, 282993, 69239846, 17468997381, 4494716943847, 1172353182893367, ...] and conjecturally satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 5 except for p = 11 and p = 13 (checked up to p = 101). - Peter Bala, Sep 14 2021
a(n) ~ c * 13^(13*n) / (n^(3/2) * 2^(2*n) * 11^(11*n)), where c = 0.0250562444901910770802983936320823301923793538303930752981380507191770309... - Vaclav Kotesovec, Sep 16 2021
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EXAMPLE
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For n=1, the possible walks are EEEEEEEEEEENN, EEEEEEEEEENEN, EEEEEEEEENEEN, EEEEEEEENEEEN, EEEEEEEENEEEEN, EEEEEEENEEEEN.
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MATHEMATICA
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m = 15;
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CROSSREFS
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KEYWORD
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nonn,walk,easy
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AUTHOR
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STATUS
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approved
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