A334190
a(n) = exp(1/2) * Sum_{k>=0} (2*k + 1)^n / ((-2)^k * k!).
Original entry on oeis.org
1, 0, -2, -4, 4, 64, 248, 48, -6512, -51200, -171296, 830400, 17870400, 144684032, 441316224, -5976726784, -119879356160, -1123892297728, -3962230563328, 70410917051392, 1686366492509184, 19578100126072832, 101728414306826240, -1258662784047370240, -42727186269262737408
Offset: 0
-
nmax = 24; CoefficientList[Series[1/(1 - x) Sum[(-x/(1 - x))^k/Product[(1 - 2 j x/(1 - x)), {j, 1, k}], {k, 0, nmax}], {x, 0, nmax}], x]
nmax = 24; CoefficientList[Series[Exp[x + (1 - Exp[2 x])/2], {x, 0, nmax}], x] Range[0, nmax]!
Table[Sum[Binomial[n, k] * 2^k * BellB[k, -1/2], {k, 0, n}], {n, 0, 24}] (* Vaclav Kotesovec, Apr 18 2020 *)
A334191
a(n) = exp(1/3) * Sum_{k>=0} (3*k + 1)^n / ((-3)^k * k!).
Original entry on oeis.org
1, 0, -3, -9, 0, 189, 1377, 4374, -26001, -560601, -4999482, -18631053, 235966365, 5966310960, 71037580689, 407585191059, -3965310883512, -157871090202975, -2631946996862451, -24922384546473810, 45577755305571339, 7795795206234609027, 192159735553383097014
Offset: 0
-
nmax = 22; CoefficientList[Series[1/(1 - x) Sum[(-x/(1 - x))^k/Product[(1 - 3 j x/(1 - x)), {j, 1, k}], {k, 0, nmax}], {x, 0, nmax}], x]
nmax = 22; CoefficientList[Series[Exp[x + (1 - Exp[3 x])/3], {x, 0, nmax}], x] Range[0, nmax]!
Table[Sum[Binomial[n, k] * 3^k * BellB[k, -1/3], {k, 0, n}], {n, 0, 22}] (* Vaclav Kotesovec, Apr 18 2020 *)
A334193
a(0) = 1; thereafter a(n) = exp(1/n) * Sum_{k>=0} (n*k + 1)^n / ((-n)^k * k!).
Original entry on oeis.org
1, 0, -2, -9, -16, 625, 21384, 571438, 13471744, 188661555, -9794500000, -1476328587789, -134710712340480, -10664210861777200, -744650964057237888, -37832162051689453125, 831929248561267474432, 725944099523076464203157, 167435684777981700601449984
Offset: 0
-
Table[SeriesCoefficient[1/(1 - x) Sum[(-x/(1 - x))^k/Product[(1 - n j x/(1 - x)), {j, 1, k}], {k, 0, n}], {x, 0, n}], {n, 0, 18}]
Join[{1}, Table[n! SeriesCoefficient[Exp[x + (1 - Exp[n x])/n], {x, 0, n}], {n, 1, 18}]]
Join[{1}, Table[Sum[Binomial[n, k]*n^k*BellB[k, -1/n], {k, 0, n}], {n, 1, 18}]] (* Vaclav Kotesovec, Apr 18 2020 *)
Showing 1-3 of 3 results.