A145561 Alternating row sums of triangle A049029 (S2(5)).
1, 4, 31, 359, 5546, 107249, 2492701, 67693534, 2103854581, 73651161959, 2868077514776, 122980857764819, 5758029769553101, 292305762924889804, 15992593021331060611, 938143525674896325299, 58739433900424758545186, 3910020681156059085488189
Offset: 1
Programs
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Mathematica
Table[DifferenceRoot[Function[{y, k}, {-32 k (1 + k) (1 + 2 k) (1 + 4 k) (3 + 4 k) y[k] + (1679 + 5920 k + 8080 k^2 + 5120 k^3 + 1280 k^4) y[1 + k] + (-2550 - 4580 k - 2880 k^2 - 640 k^3) y[2 + k] + (675 + 640 k + 160 k^2) y[3 + k] + (-50 - 20 k) y[4 + k] + y[5 + k] == 0, y[0] == -1, y[1] == 1, y[2] == 4, y[3] == 31, y[4] == 359}]][n], {n, 1, 20}] (* Benedict W. J. Irwin, Jul 12 2017 *)
Formula
a(n) = Sum_{m=1..n} (-1)^(m+1)*A049029(n,m), n>=1.
E.g.f.: (from Jabotinsky structure): 1-exp(1-1/(1-4*x)^(1/4)).
a(n) = y(n), where y(0) = -1, y(1) = 1, y(2) = 4, y(3) = 31, y(4) = 359, and -32*k*(1 + k)*(1 + 2 k)*(1 + 4 k)*(3 + 4 k)*y(k) + (1679 + 5920 k + 8080 k^2 + 5120 k^3 + 1280 k^4)*y(k+1) + (-2550 - 4580 k - 2880 k^2 - 640 k^3)*y(k+2) + (675 + 640 k + 160 k^2)*y(k+3) + (-50 - 20 k)*y(k+4) + y(k+5) = 0. - Benedict W. J. Irwin, Jul 12 2017
From Seiichi Manyama, Jan 20 2025: (Start)
a(n) = -Sum_{k=0..n} 4^(n-k) * |Stirling1(n,k)| * A000587(k).
a(n) = -e * (-4)^n * n! * Sum_{k>=0} (-1)^k * binomial(-k/4,n)/k!. (End)
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