A002104 -id:A002104 - cp's OEIS frontend

A345454 E.g.f.: log(1 - log(1 - x) * exp(x)).

0, 1, 2, 1, -5, 3, 141, 348, -1938, 3013, 274327, 1583338, -4613476, 41135339, 3201505997, 33153080054, 49123558416, 2360520208825, 133442956587099, 2109709010976874, 14751973018988252, 338170133891984663, 15120630911878380457, 324654726628159335686

Offset: 0

Ilya Gutkovskiy, Jul 18 2021

sign

Index entries for sequences related to logarithmic numbers

Cf. A002104, A009321, A009324, A191365, A346427.

nmax = 23; CoefficientList[Series[Log[1 - Log[1 - x] Exp[x]], {x, 0, nmax}], x] Range[0, nmax]!
A002104[n_] := A002104[n] = n! Sum[1/((n - k) k!), {k, 0, n - 1}]; a[0] = 0; a[n_] := a[n] = A002104[n] - (1/n) Sum[Binomial[n, k] A002104[n - k] k a[k], {k, 1, n - 1}]; Table[a[n], {n, 0, 23}]

my(x='x+O('x^25)); concat(0, Vec(serlaplace(log(1 - log(1 - x) * exp(x))))) \\ Michel Marcus, Jul 19 2021

a(0) = 0; a(n) = A002104(n) - (1/n) * Sum_{k=1..n-1} binomial(n,k) * A002104(n-k) * k * a(k).

A346405 a(n) = (n!)^2 * Sum_{k=0..n-1} 1 / ((n-k)^2 * k!).

Original entry on oeis.org

0, 1, 5, 31, 268, 3476, 70656, 2202432, 98622336, 5954736384, 463100042880, 44924476970880, 5308404719823360, 749930460864929280, 124754522068412651520, 24129984694192721971200, 5368254991077002482483200, 1360938718277588430567014400, 389980903967231535140578099200

Offset: 0

Ilya Gutkovskiy, Jul 15 2021

nonn

Cf. A001819, A002104, A002720, A061573, A346409.

Table[(n!)^2 Sum[1/((n - k)^2 k!), {k, 0, n - 1}], {n, 0, 18}]
nmax = 18; CoefficientList[Series[PolyLog[2, x] Exp[x], {x, 0, nmax}], x] Range[0, nmax]!^2

Sum_{n>=0} a(n) * x^n / (n!)^2 = polylog(2,x) * exp(x).

A346410 a(n) = (n!)^2 * Sum_{k=0..n-1} 1 / ((n-k) * k!)^2.

Original entry on oeis.org

0, 1, 5, 22, 152, 2001, 45097, 1527506, 71864928, 4466430513, 353828600029, 34770661312190, 4148422395161464, 590479899466175681, 98824492409739430401, 19209838771051338898234, 4291488438323868507946880, 1091819942877526843993466529, 313819508664449992611846900981

Offset: 0

Ilya Gutkovskiy, Jul 15 2021

nonn

Cf. A002104, A006040, A066998, A193563, A336291, A346411.

Table[(n!)^2 Sum[1/((n - k) k!)^2, {k, 0, n - 1}], {n, 0, 18}]
nmax = 18; CoefficientList[Series[PolyLog[2, x] BesselI[0, 2 Sqrt[x]], {x, 0, nmax}], x] Range[0, nmax]!^2

Sum_{n>=0} a(n) * x^n / (n!)^2 = polylog(2,x) * BesselI(0,2*sqrt(x)).

A348482 Triangle read by rows: T(n,k) = (Sum_{i=k..n} i!)/(k!) for 0 <= k <= n.

Original entry on oeis.org

1, 2, 1, 4, 3, 1, 10, 9, 4, 1, 34, 33, 16, 5, 1, 154, 153, 76, 25, 6, 1, 874, 873, 436, 145, 36, 7, 1, 5914, 5913, 2956, 985, 246, 49, 8, 1, 46234, 46233, 23116, 7705, 1926, 385, 64, 9, 1, 409114, 409113, 204556, 68185, 17046, 3409, 568, 81, 10, 1

Offset: 0

Werner Schulte, Oct 20 2021

The matrix inverse M = T^(-1) has terms M(n,n) = 1 for n >= 0, M(n,n-1) = -(n+1) for n > 0, and M(n,n-2) = n for n > 1, otherwise 0.

			The triangle T(n,k) for 0 <= k <= n starts:
n\k :       0       1       2      3      4     5    6   7   8  9
=================================================================
  0 :       1
  1 :       2       1
  2 :       4       3       1
  3 :      10       9       4      1
  4 :      34      33      16      5      1
  5 :     154     153      76     25      6     1
  6 :     874     873     436    145     36     7    1
  7 :    5914    5913    2956    985    246    49    8   1
  8 :   46234   46233   23116   7705   1926   385   64   9   1
  9 :  409114  409113  204556  68185  17046  3409  568  81  10  1
  etc.

Sela Fried, On a sum involving factorials, 2024.

Cf. A109398, A094587, A002104 (row sums), A173184 (alt. row sums), A000012 (main diagonal), A000027(1st subdiagonal), A000290 (2nd subdiagonal), A081437 (3rd subdiagonal), A192398 (4th subdiagonal), A003422 (column 0), A007489 (column 1), A345889 (column 2), A143122.

T[n_, k_] := Sum[i!, {i, k, n}]/k!; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, Oct 20 2021 *)

T(n,n) = 1 and T(2*n,n) = A109398(n) for n >= 0; T(n,n-1) = n+1 for n > 0; T(n,n-2) = n^2 for n > 1.
T(n,k) - T(n-1,k) = (n!) / (k!) = A094587(n,k) for 0 <= k < n.
T(n,k) = (k+2) * (T(n,k+1) - T(n,k+2)) for 0 <= k < n-1.
T(n,k) = (T(n,k-1) - 1) / k for 0 < k <= n.
T(n,k) * T(n-1,k-1) - T(n-1,k) * T(n,k-1) = (n!) / (k!) for 0 < k < n.
T(n,1) = T(n,0)-1 = Sum_{k=0..n-1} T(n,k)/(k+2) for n > 0 (conjectured).
Sum_{k=0..n} binomial(k+r,k) * (1-k) * T(n+r,k+r) = binomial(n+r+1,n) for n >= 0 and r >= 0.
Sum_{k=0..n} (-1)^k * (k+1) * T(n,k) = (1 + (-1)^n) / 2 for n >= 0.
Sum_{k=0..n} (-1)^k * (k!) * T(n,k) = Sum_{k=0..n} (k!) * (1+(-1)^k) / 2 for n >= 0.
The row polynomials p(n,x) = Sum_{k=0..n} T(n,k) * x^k for n >= 0 satisfy the following equations:
  (a) p(n,x) - p'(n,x) = (x^(n+1)-1) / (x-1) for n >= 0, where p' is the first derivative of p;
  (b) p(n,x) - (n+1) * p(n-1,x) + n * p(n-2,x) = x^n for n > 1.
  (c) p(n,x) = (x+1) * p(n-1,x) + 1 + Sum_{i=1..n-1} (d/dx)^i p(n-1,x) for n > 0 (conjectured).
Row sums p(n,1) equal A002104(n+1) for n >= 0.
Alternating row sums p(n,-1) equal A173184(n) for n >= 0 (conjectured).
The three conjectures stated above are true. See links. - Sela Fried, Jul 11 2024.
From Peter Luschny, Jul 11 2024: (Start)
T(n, k) = (t(k) - t(n + 1)) / k!, where t(n) = (-1)^(n + 1) * Gamma(n + 1) * Subfactorial(-(n + 1)).
T(n, k) = A143122(n, k) / k!. (End)

A352149 a(n) = Sum_{k=0..n-1} binomial(n,k)^2 * (n-k-1)!.

Original entry on oeis.org

0, 1, 5, 20, 90, 499, 3395, 27474, 256984, 2720169, 32080501, 416574212, 5900292266, 90461885331, 1491788697451, 26318520300986, 494449968500832, 9852544385880961, 207497251731808341, 4604297325494524516, 107348917822006139114, 2623224641748615607715, 67035139167875735937219

Offset: 0

Ilya Gutkovskiy, Mar 06 2022

nonn

Cf. A002104, A002720, A352150.

Table[Sum[Binomial[n, k]^2 (n - k - 1)!, {k, 0, n - 1}], {n, 0, 22}]
nmax = 22; Assuming[x > 0, CoefficientList[Series[BesselI[0, 2 Sqrt[x]] (ExpIntegralEi[x] - Log[x] - EulerGamma), {x, 0, nmax}], x]] Range[0, nmax]!^2

a(n) = sum(k=0, n-1, binomial(n,k)^2 * (n-k-1)!); \\ Michel Marcus, Mar 06 2022

Sum_{n>=0} a(n) * x^n / (n!)^2 = BesselI(0,2*sqrt(x)) * (Ei(x) - log(x) - gamma).
From Vaclav Kotesovec, Mar 06 2022: (Start)
Recurrence: n*(n^3 - 12*n^2 + 46*n - 50)*a(n) = (2*n^5 - 24*n^4 + 95*n^3 - 123*n^2 + 37*n - 2)*a(n-1) - (n^6 - 12*n^5 + 51*n^4 - 95*n^3 + 121*n^2 - 139*n + 74)*a(n-2) + (n-2)*(2*n^5 - 27*n^4 + 140*n^3 - 344*n^2 + 409*n - 173)*a(n-3) - (n-3)^2*(n-2)*(n^3 - 9*n^2 + 25*n - 15)*a(n-4).
a(n) ~ exp(2*sqrt(n) - n - 1/2) * n^(n - 3/4) / sqrt(2). (End)

A354419 Expansion of e.g.f. log(1+4x) exp(x)/4.

Original entry on oeis.org

0, 1, -2, 23, -276, 4509, -91190, 2205587, -62104168, 1995807993, -72089029802, 2891304481999, -127498010037244, 6131189086886421, -319320539953144158, 17905976286288568267, -1075611833288214177232, 68909527979479961534705

Offset: 0

Seiichi Manyama, May 27 2022

sign

Cf. A002104, A353546, A353547, A353548, A353549.

my(N=20, x='x+O('x^N)); concat(0, Vec(serlaplace(log(1+4*x)*exp(x)/4)))

a(n) = n!*sum(k=0, n-1, (-4)^(n-1-k)/((n-k)*k!));

a_vector(n) = my(v=vector(n+1, i, if(i==2, 1, 0))); for(i=2, n, v[i+1]=(-4*i+5)*v[i]+4*(i-1)*v[i-1]+1); v;

a(n) = n! * Sum_{k=0..n-1} (-4)^(n-1-k) / ((n-k) * k!).
a(0) = 0, a(1) = 1, a(n) = (-4 * n + 5) * a(n-1) + 4 * (n-1) * a(n-2) + 1.
a(n) ~ -(-1)^n * (n-1)! * 4^(n-1) / exp(1/4). - Vaclav Kotesovec, Jun 08 2022

A356926 E.g.f. satisfies A(x)^A(x) = 1/(1 - x)^exp(x).

Original entry on oeis.org

1, 1, 2, 3, 10, 35, 121, 1092, 5216, 39321, 558643, 2433508, 48144944, 688652549, 2176310995, 145742587616, 1334993574032, 5551320939809, 799648465754835, 1049695714507276, 90069170433616208, 6281942689646504501, -53282051261767839293, 2356158301117802408472

Offset: 0

Seiichi Manyama, Sep 04 2022

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Eric Weisstein's World of Mathematics, Lambert W-Function.

Cf. A191365, A356925.
Cf. A356912, A356913.
Cf. A002104, A177885.
Cf. A141209.

nmax = 23; A[_] = 1;
Do[A[x_] = ((1 - x)^(-Exp[x]))^(1/A[x]) + O[x]^(nmax+1) // Normal, {nmax}];
CoefficientList[A[x], x]*Range[0, nmax]! (* Jean-François Alcover, Mar 04 2024 *)

my(N=30, x='x+O('x^N)); Vec(serlaplace(sum(k=0, N, (-k+1)^(k-1)*(-exp(x)*log(1-x))^k/k!)))

my(N=30, x='x+O('x^N)); Vec(serlaplace(exp(lambertw(-exp(x)*log(1-x)))))

my(N=30, x='x+O('x^N)); Vec(serlaplace(-exp(x)*log(1-x)/lambertw(-exp(x)*log(1-x))))

E.g.f.: A(x) = Sum_{k>=0} (-k+1)^(k-1) * (-exp(x) * log(1-x))^k / k!.
E.g.f.: A(x) = exp( LambertW(-exp(x) * log(1-x)) ).
E.g.f.: A(x) = -exp(x) * log(1-x)/LambertW(-exp(x) * log(1-x)).

A381016 Expansion of e.g.f. -log(1-x) * sin(x).

Original entry on oeis.org

0, 0, 2, 3, 4, 20, 110, 651, 4520, 36000, 322618, 3213595, 35226860, 421419492, 5463436134, 76301056755, 1142009233872, 18236159031584, 309463272791538, 5561354285804115, 105510576441518164, 2107380222724155540, 44200537412519181278, 971311172969442165883

Offset: 0

Seiichi Manyama, Feb 12 2025

nonn

Cf. A002104, A009410, A009416, A177699.

my(N=30, x='x+O('x^N)); concat([0, 0], Vec(serlaplace(-log(1-x)*sin(x))))

a(n) = -sum(k=1, n\2, (-1)^k*(n-2*k)!*binomial(n, 2*k-1));

a(n) = -Sum_{k=1..floor(n/2)} (-1)^k * (n-2*k)! * binomial(n,2*k-1).

A320838 a(0) = 0, a(n) is the number of x such that a(x) = a(n-1) and there exists no y such that x < y < n and a(y) > a(n-1).

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 3, 2, 1, 1, 2, 2, 3, 3, 4, 1, 1, 2, 1, 1, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 3, 2, 1, 1, 2, 2, 3, 3, 4, 2, 1, 1, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 3, 2, 1, 1, 2, 2, 3, 3, 4, 3, 1, 1, 2, 1, 1, 2, 2, 3, 2, 1, 1, 2, 2, 3, 3, 4, 4, 5

Offset: 0

Thomas Anton, Oct 21 2018

nonn

After the n-th occurrence (with n <= m) of m after a number larger than n comes the segment of the sequence from the first appearance of n to the first appearance of m.

The first appearance of n in this sequence is given by A002104(n).

			Start with a(0) = 0.
No larger number has occurred yet, and the number of 0's since the start of the sequence is 1, so a(1) = 1.
No larger number has occurred yet, and the number of 1's since the start of the sequence is 1, so a(2) = 1.
Still no larger number has occurred, and the number of 1's since the start of the sequence is 2, so a(3) = 2.
No larger number has occurred yet, and the number of 2's since the start of the sequence is 1, so a(4) = 1.
The number of 1's that have occurred since the last appearance of a larger number is 1, so a(5) = 1.
Etc.

Cf. A002104.

a[0] = 0; a[n_] := a[n] = Count[Range[0, n-1], x_ /; a[x] == a[n - 1] && ! AnyTrue[ Range[x+1, n-1], a[#] > a[n-1] &]]; a /@ Range[0, 89] (* Giovanni Resta, Oct 22 2018 *)

lista(nn) = {va = vector(nn); va[1] = 0; for (n=2, nn, nb = 0; forstep (k=n-1, 1, -1, if (va[k] == va[n-1], nb++); if (va[k] > va[n-1], break);); va[n] = nb;); va;} \\ Michel Marcus, Oct 22 2018

Let s(n) be the first time n appears in the sequence, then s(n) = Sum_{k=0...n-1} (s(n-1)-s(k)+1).

cp's OEIS Frontend

A345454 E.g.f.: log(1 - log(1 - x) * exp(x)).

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A346405 a(n) = (n!)^2 * Sum_{k=0..n-1} 1 / ((n-k)^2 * k!).

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A346410 a(n) = (n!)^2 * Sum_{k=0..n-1} 1 / ((n-k) * k!)^2.

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A348482 Triangle read by rows: T(n,k) = (Sum_{i=k..n} i!)/(k!) for 0 <= k <= n.

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A352149 a(n) = Sum_{k=0..n-1} binomial(n,k)^2 * (n-k-1)!.

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A354419 Expansion of e.g.f. log(1+4*x) * exp(x)/4.

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A356926 E.g.f. satisfies A(x)^A(x) = 1/(1 - x)^exp(x).

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A381016 Expansion of e.g.f. -log(1-x) * sin(x).

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A320838 a(0) = 0, a(n) is the number of x such that a(x) = a(n-1) and there exists no y such that x < y < n and a(y) > a(n-1).

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A354419 Expansion of e.g.f. log(1+4x) exp(x)/4.