A336950
E.g.f.: 1 / (1 - x * exp(2*x)).
Original entry on oeis.org
1, 1, 6, 42, 392, 4600, 64752, 1063216, 19952256, 421227648, 9880951040, 254960721664, 7176891675648, 218857588139008, 7187394935347200, 252897556424140800, 9491754142468702208, 378509920569294684160, 15982018774576565649408, 712306819507400060502016
Offset: 0
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nmax = 19; CoefficientList[Series[1/(1 - x Exp[2 x]), {x, 0, nmax}], x] Range[0, nmax]!
Join[{1}, Table[n! Sum[(2 (n - k))^k/k!, {k, 0, n}], {n, 1, 19}]]
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k] k 2^(k - 1) a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 19}]
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seq(n)={ Vec(serlaplace(1 / (1 - x*exp(2*x + O(x^n))))) } \\ Andrew Howroyd, Aug 08 2020
A336951
E.g.f.: 1 / (1 - x * exp(3*x)).
Original entry on oeis.org
1, 1, 8, 69, 780, 11145, 191178, 3823785, 87406056, 2247785073, 64228084110, 2018771719569, 69221032558956, 2571290056399545, 102860527370221026, 4408690840306136505, 201557641172689004112, 9790792086366911655009, 503570143277542340304534
Offset: 0
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nmax = 18; CoefficientList[Series[1/(1 - x Exp[3 x]), {x, 0, nmax}], x] Range[0, nmax]!
Join[{1}, Table[n! Sum[(3 (n - k))^k/k!, {k, 0, n}], {n, 1, 18}]]
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k] k 3^(k - 1) a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 18}]
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seq(n)={ Vec(serlaplace(1 / (1 - x*exp(3*x + O(x^n))))) } \\ Andrew Howroyd, Aug 08 2020
A336952
E.g.f.: 1 / (1 - x * exp(4*x)).
Original entry on oeis.org
1, 1, 10, 102, 1336, 22200, 443664, 10334128, 275060608, 8236914048, 274069953280, 10031110907136, 400520747437056, 17324601073921024, 807023462798608384, 40278407730378332160, 2144307919689898491904, 121291661335680615284736, 7264376142168665821741056
Offset: 0
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nmax = 18; CoefficientList[Series[1/(1 - x Exp[4 x]), {x, 0, nmax}], x] Range[0, nmax]!
Join[{1}, Table[n! Sum[(4 (n - k))^k/k!, {k, 0, n}], {n, 1, 18}]]
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k] k 4^(k - 1) a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 18}]
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seq(n)={ Vec(serlaplace(1 / (1 - x*exp(4*x + O(x^n))))) } \\ Andrew Howroyd, Aug 08 2020
A351791
Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = n! * Sum_{j=0..n} (-k * (n-j))^j/j!.
Original entry on oeis.org
1, 1, 1, 1, 1, 2, 1, 1, 0, 6, 1, 1, -2, -3, 24, 1, 1, -4, -6, -4, 120, 1, 1, -6, -3, 40, 25, 720, 1, 1, -8, 6, 132, 120, 114, 5040, 1, 1, -10, 21, 248, -375, -1872, -287, 40320, 1, 1, -12, 42, 364, -2120, -8298, -3920, -4152, 362880, 1, 1, -14, 69, 456, -5655, -12144, 86121, 155776, -1647, 3628800
Offset: 0
Square array begins:
1, 1, 1, 1, 1, 1, ...
1, 1, 1, 1, 1, 1, ...
2, 0, -2, -4, -6, -8, ...
6, -3, -6, -3, 6, 21, ...
24, -4, 40, 132, 248, 364, ...
120, 25, 120, -375, -2120, -5655, ...
Main diagonal gives (-1)^n *
A302398(n).
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T[n_, k_] := n!*(1 + Sum[(-k*(n - j))^j/j!, {j, 1, n}]); Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Feb 19 2022 *)
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T(n, k) = n!*sum(j=0, n, (-k*(n-j))^j/j!);
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T(n, k) = if(n==0, 1, n*sum(j=0, n-1, (-k)^(n-1-j)*binomial(n-1, j)*T(j, k)));
Showing 1-4 of 4 results.