A001819 -id:A001819 - cp's OEIS frontend

A092043 a(n) = numerator(n!/n^2).

1, 1, 2, 3, 24, 20, 720, 630, 4480, 36288, 3628800, 3326400, 479001600, 444787200, 5811886080, 81729648000, 20922789888000, 19760412672000, 6402373705728000, 6082255020441600, 115852476579840000, 2322315553259520000

Offset: 1

Ralf Stephan, Mar 28 2004

Numerator of expansion of dilog(x) = Li_2(x) = -Integral_{t=0..x} (log(1-t)/t)*dt. See the Weisstein link.

E.g.f. of {a(n)/A014973(n)}_{n>=1} is Li_2(x) (with 0 for n=0).

Vincenzo Librandi, Table of n, a(n) for n = 1..200
A. N. Kirillov, Dilogarithm identities, arXiv:hep-th/9408113, 1994.
Eric Weisstein's World of Mathematics, Dilogarithm

Denominator is in A014973.

Cf. A000142, A001819, A181569.

[Numerator(Factorial(n)/n^2): n in [1..30]]; // Vincenzo Librandi, Apr 15 2014

Table[Numerator[n!/n^2], {n, 1, 40}] (* Vincenzo Librandi, Apr 15 2014 *)
Table[(n-1)!/n,{n,30}]//Numerator (* Harvey P. Dale, Apr 03 2018 *)

PARI
```
a(n)=numerator(n!/n^2)
```

a(n)=numerator(polcoeff(serlaplace(dilog(x)),n))

From Wolfdieter Lang, Apr 28 2017: (Start)
a(n) = numerator(n!/n^2) = numerator((n-1)!/n), n >= 1. See the name.
E.g.f. {a(n)/A014973(n)}_{n>=1} with 0 for n=0 is  Li_2(x). See the comment.
(-1)^n*a(n+1)/A014973(n+1) = (-1)^n*n!/(n+1) = Sum_{k=0..n} Stirling1(n, k)*Bernoulli(k), with Stirling1 = A048994 and Bernoulli(k) = A027641(k)/A027642(k), n >= 0. From inverting the formula for B(k) in terms of Stirling2 = A048993.(End)
From Wolfdieter Lang, Oct 26 2022: (Start)
a(n) = (n-1)!/gcd(n,(n-1)!) = A000142(n-1)/A181569(n-1), n >= 1.
The expansion of (1+x)*exp(x) has coefficients A014973(n+1)/a(n+1), for n >= 0. (End)

Comment rewritten by Wolfdieter Lang, Apr 28 2017

A160562 Triangle of scaled central factorial numbers, T(n,k) = A008958(n,n-k).

Original entry on oeis.org

1, 1, 1, 1, 10, 1, 1, 91, 35, 1, 1, 820, 966, 84, 1, 1, 7381, 24970, 5082, 165, 1, 1, 66430, 631631, 273988, 18447, 286, 1, 1, 597871, 15857205, 14057043, 1768195, 53053, 455, 1, 1, 5380840, 397027996, 704652312, 157280838, 8187608, 129948, 680, 1

Offset: 0

Jonathan Vos Post, May 19 2009

This is table 4 on page 12 of Gelineau and Zeng, read downwards by columns.
Reversing rows gives A008958.
Apparently the table can also be obtained by deleting each second row and column of A136630.

			Triangle starts:
  1;
  1,     1;
  1,    10,      1;
  1,    91,     35,      1;
  1,   820,    966,     84,     1;
  1,  7381,  24970,   5082,   165,   1;
  1, 66430, 631631, 273988, 18447, 286, 1;
  ...

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows n = 0..150, flattened)
Qi Fang, Ya-Nan Feng, and Shi-Mei Ma, Alternating runs of permutations and the central factorial numbers, arXiv:2202.13978 [math.CO], 2022.
Yoann Gelineau and Jiang Zeng, Combinatorial Interpretations of the Jacobi-Stirling Numbers, arXiv:0905.2899 [math.CO], May 18 2009.

Cf. A002452 (column k=1), A002453 (column k=2), A000447 (right column k=n-1), A185375 (right column k=n-2).

Cf. A001819, A008275, A008277, A008958, A039599, A136630.

A160562 := proc(n,k) npr := 2*n+1 ; kpr := 2*k+1 ; sinh(t*sinh(x)) ; npr!*coeftayl(%,x=0,npr) ; coeftayl(%,t=0,kpr) ; end: seq(seq(A160562(n,k),k=0..n),n=0..15) ; # R. J. Mathar, Sep 09 2009

T[n_, k_] := Sum[(-1)^(k - m)*(2m + 1)^(2n + 1)*Binomial[2k, k + m]/(k + m + 1), {m, 0, k}]/(4^k*(2k)!);
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 22 2017 *)

T(n,k) = (1/((2*k)!*4^k)) * Sum_{m=0..k} (-1)^(k-m)*A039599(k,m)*(2*m+1)^(2*n). - Werner Schulte, Nov 01 2015

T(n,k) = ((-1)^(n-k)*(2*n+1)!/(2*k+1)!) * [x^(2*n+1)]sin(x)^(2*k+1) = ((2*n+1)!/(2*k+1)!) * [x^(2*n+1)]sinh(x)^(2*k+1). Note that sin(x)^(2*k+1) = (Sum_{i=0..k} (-1)^i*binomial(2*k+1,k-i)*sin((2*i+1)*x))/(2^(2*k)). - Jianing Song, Oct 29 2023

More terms from R. J. Mathar, Sep 09 2009

A291505 a(n) = (n!)^7 * Sum_{i=1..n} 1/i^7.

Original entry on oeis.org

0, 1, 129, 282251, 4624680320, 361307736471424, 101143400834944548864, 83296040059942781485105152, 174684539610200377980575079727104, 835510910973061065615656036610946891776, 8355109938323553617123838798161699143680000000

Offset: 0

Seiichi Manyama, Aug 25 2017

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..92

Cf. A000254 (k=1), A001819 (k=2), A066989 (k=3), A099827 (k=5), A291456 (k=6), this sequence (k=7), A291506 (k=8), A291507 (k=9), A291508 (k=10).

Column k=7 of A291556.

Table[(n!)^7 * Sum[1/i^7, {i, 1, n}], {n, 0, 15}] (* Vaclav Kotesovec, Aug 27 2017 *)

a(n) = n!^7*sum(i=1, n, 1/i^7); \\ Michel Marcus, Aug 26 2017

a(0) = 0, a(1) = 1, a(n+1) = (n^7+(n+1)^7)*a(n) - n^14*a(n-1) for n > 0.
a(n) ~ zeta(7) * (2*Pi)^(7/2) * n^(7*n+7/2) / exp(7*n). - Vaclav Kotesovec, Aug 27 2017
Sum_{n>=0} a(n) * x^n / (n!)^7 = polylog(7,x) / (1 - x). - Ilya Gutkovskiy, Jul 15 2020

A291506 a(n) = (n!)^8 * Sum_{i=1..n} 1/i^8.

Original entry on oeis.org

0, 1, 257, 1686433, 110523752704, 43173450975314176, 72514862031522895036416, 418033821374598847702425993216, 7013444132843374500928464765799366656, 301905779820559925981495987360836056017534976

Offset: 0

Seiichi Manyama, Aug 25 2017

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..83

Cf. A000254 (k=1), A001819 (k=2), A066989 (k=3), A099827 (k=5), A291456 (k=6), A291505 (k=7), this sequence (k=8), A291507 (k=9), A291508 (k=10).

Column k=8 of A291556.

Table[(n!)^8 * Sum[1/i^8, {i, 1, n}], {n, 0, 15}] (* Vaclav Kotesovec, Aug 27 2017 *)

a(n) = n!^8*sum(i=1, n, 1/i^8); \\ Michel Marcus, Aug 26 2017

a(0) = 0, a(1) = 1, a(n+1) = (n^8+(n+1)^8)*a(n) - n^16*a(n-1) for n > 0.
a(n) ~ 8 * Pi^12 * n^(8*n+4) / (4725 * exp(8*n)). - Vaclav Kotesovec, Aug 27 2017
Sum_{n>=0} a(n) * x^n / (n!)^8 = polylog(8,x) / (1 - x). - Ilya Gutkovskiy, Jul 15 2020

A291507 a(n) = (n!)^9 * Sum_{i=1..n} 1/i^9.

Original entry on oeis.org

0, 1, 513, 10097891, 2647111616000, 5170142516807540224, 52103129720841632885243904, 2102549272223560776918400601161728, 282199388424234851655058321255905292713984, 109329825340451764123791003609208862665771818418176

Offset: 0

Seiichi Manyama, Aug 25 2017

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..75

Cf. A000254 (k=1), A001819 (k=2), A066989 (k=3), A099827 (k=5), A291456 (k=6), A291505 (k=7), A291506 (k=8), this sequence (k=9), A291508 (k=10).

Column k=9 of A291556.

Table[(n!)^9 * Sum[1/i^9, {i, 1, n}], {n, 0, 12}] (* Vaclav Kotesovec, Aug 27 2017 *)

a(n) = n!^9*sum(i=1, n, 1/i^9); \\ Michel Marcus, Aug 26 2017

a(0) = 0, a(1) = 1, a(n+1) = (n^9+(n+1)^9)*a(n) - n^18*a(n-1) for n > 0.
a(n) ~ zeta(9) * (2*Pi)^(9/2) * n^(9*n+9/2) / exp(9*n). - Vaclav Kotesovec, Aug 27 2017
Sum_{n>=0} a(n) * x^n / (n!)^9 = polylog(9,x) / (1 - x). - Ilya Gutkovskiy, Jul 15 2020

A291508 a(n) = (n!)^10 * Sum_{i=1..n} 1/i^10.

Original entry on oeis.org

0, 1, 1025, 60526249, 63466432537600, 619789443653380965376, 37476298202061058687475122176, 10586126703664512292193022557971021824, 11366767006463449393869821987386636472445566976, 39633465899293694663690352980684333029782095493517541376

Offset: 0

Seiichi Manyama, Aug 25 2017

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..69

Cf. A000254 (k=1), A001819 (k=2), A066989 (k=3), A099827 (k=5), A291456 (k=6), A291505 (k=7), A291506 (k=8), A291507 (k=9), this sequence (k=10).

Column k=10 of A291556.

Table[(n!)^10 * Sum[1/i^10, {i, 1, n}], {n, 0, 12}] (* Vaclav Kotesovec, Aug 27 2017 *)

a(n) = n!^10*sum(i=1, n, 1/i^10); \\ Michel Marcus, Aug 26 2017

a(0) = 0, a(1) = 1, a(n+1) = (n^10+(n+1)^10)*a(n) - n^20*a(n-1) for n > 0.
a(n) ~ 32 * Pi^15 * n^(10*n+5) / (93555 * exp(10*n)). - Vaclav Kotesovec, Aug 27 2017
Sum_{n>=0} a(n) * x^n / (n!)^10 = polylog(10,x) / (1 - x). - Ilya Gutkovskiy, Jul 15 2020

A142996 a(0) = 0, a(1) = 1, a(n+1) = (2n^2+2n+7)a(n) - n^4a(n-1), n >= 1.

Original entry on oeis.org

0, 1, 11, 193, 5092, 189916, 9541872, 622179216, 51129292032, 5172077028096, 631719119232000, 91679469784704000, 15596136686979072000, 3074102117690701824000, 695050625746441101312000

Offset: 0

Peter Bala, Jul 18 2008

This is the case m = 2 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n^2+2*n+m^2+m+1 )*a(n) - n^4*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series sum {k = 1..inf} 1/k^2 for the constant zeta(2). See A142995 for remarks on the general case.

Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Cf. A001819, A003215, A142995, A142997, A142998.

p := n -> 3*n^2+3*n+1: a := n -> n!^2*p(n)*sum (1/(k^2*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..20);

RecurrenceTable[{a[0]==0,a[1]==1,a[n+1]==(2n^2+2n+7)a[n]-n^4 a[n-1]},a,{n,20}] (* Harvey P. Dale, Jun 27 2017 *)

a(n) = n!^2*p(n)*sum {k = 1..n} 1/(k^2*p(k-1)*p(k)), where p(n) = 3*n^2+3*n+1 = A003215(n) is the polynomial that gives the crystal ball sequence for the A_2 lattice. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n^2+2*n+7)*a(n) - n^4*a(n-1). The sequence b(n):= n!^2*p(n) satisfies the same recurrence with b(0) = 1, b(1) = 7. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(7-1^4/(11-2^4/(19-3^4/(31-...-(n-1)^4/(2*n^2-2*n+7))))), for n >=2. Thus the behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = 1/(7-1^4/(11-2^4/(19-3^4/(31-...-n^4/((2n^2+2n+7)-...))))) = sum {k = 1..inf} 1/(k^2*(9*k^4-3*k^2+1)) = zeta(2)-2*(1-1/4). The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 31] (replace x by 2x+1 in the corollary and apply Entry 14).

A142997 a(0) = 0, a(1) = 1, a(n+1) = (2n^2+2n+13)a(n) - n^4a(n-1).

Original entry on oeis.org

0, 1, 17, 409, 13756, 624364, 36981072, 2777988240, 258456976128, 29199105421056, 3939691125888000, 625956978121344000, 115709065165486080000, 24625602280458786816000

Offset: 0

Peter Bala, Jul 18 2008, Oct 16 2008

This is the case m = 3 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n^2+2*n+m^2+m+1)*a(n) - n^4*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series sum {k = 1..inf} 1/k^2 for the constant zeta(2). See A142995 for remarks on the general case.

Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Cf. A001819, A005902, A142995, A142996, A142998.

p := n -> (10*n^3+15*n^2+11*n+3)/3: a := n -> n!^2*p(n)*sum (1/(k^2*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..20);

a(n) = n!^2*p(n)*sum {k = 1..n} 1/(k^2*p(k-1)*p(k)), where p(n) = (10*n^3+15*n^2+11*n+3)/3 = A005902(n) is the polynomial that gives the crystal ball sequence for the A_3 lattice. Recurrence: a(0) = 1, a(1) = 1, a(n+1) = (2*n^2+2*n+13)*a(n) - n^4*a(n-1). The sequence b(n):= n!^2*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 13. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(13-1^4/(17-2^4/(25-3^4/(37-...-(n-1)^4/(2*n^2-2*n+13))))), for n >= 2. Thus the behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = 1/(13-1^4/(17-2^4/(25-3^4/(37-...-n^4/((2*n^2+2*n+13)-...))))) = sum {k = 1..inf} 1/(k^2*p(k-1)*p(k)) = 2*(1-1/4+1/9) - zeta(2). The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 31] (replace x by 2x+1 in the corollary and apply Entry 14).

A142998 a(0) = 0, a(1) = 1, a(n+1) = (2n^2+2n+21)a(n) - n^4a(n-1).

Original entry on oeis.org

0, 1, 25, 809, 34380, 1890076, 131608656, 11369370384, 1196133878016, 150793148779776, 22461588531072000, 3905311348190592000, 784153616550893568000, 180142618195367442432000

Offset: 0

Peter Bala, Jul 18 2008

This is the case m = 4 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n^2+2*n+m^2+m+1)*a(n) - n^4*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series sum {k = 1..inf} 1/k^2 for the constant zeta(2). See A142995 for remarks on the general case.

Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Cf. A008384, A001819, A142995, A142996, A142997.

p := n -> (35*n^4+70*n^3+85*n^2+50*n+12)/12: a := n -> n!^2*p(n)*sum (1/(k^2*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..20);

a(n) = n!^2*p(n)*sum {k = 1..n} 1/(k^2*p(k-1)*p(k)), where p(n) = (35*n^4+70*n^3+85*n^2+50*n+12)/12 = A008384(n) is the polynomial that gives the crystal ball sequence for the A_4 lattice. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n^2+2*n+21)*a(n) - n^4*a(n-1). The sequence b(n):= n!^2*p(n) satisfies the same recurrence with the initial conditions b(0) = 0, b(1) = 21. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(21-1^4/(25-2^4/(33-3^4/(45-...-(n-1)^4/(2*n^2-2*n+21))))), for n >=2. Thus the behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = 1/(21-1^4/(25-2^4/(33-3^4/(45-...-n^4/((2*n^2+2*n+21)-...))))) = sum {k = 1..inf} 1/(k^2*p(k-1)*p(k)) = zeta(2) - 2*(1-1/4+1/9-1/16). The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 31] (replace x by 2x+1 in the corollary and apply Entry 14).

A160563 Table of the number of (n,k)-Riordan complexes, read by rows.

Original entry on oeis.org

1, 1, 1, 9, 10, 1, 225, 259, 35, 1, 11025, 12916, 1974, 84, 1, 893025, 1057221, 172810, 8778, 165, 1, 108056025, 128816766, 21967231, 1234948, 28743, 286, 1, 18261468225, 21878089479, 3841278805, 230673443, 6092515, 77077, 455, 1, 4108830350625, 4940831601000

Offset: 0

Jonathan Vos Post, May 19 2009

From Table 4, right-hand side, of Gelineau and Zeng.

Essentially a row-reversal of A008956. - R. J. Mathar, May 20 2009

			Triangle starts:
  [0]         1;
  [1]         1,          1;
  [2]         9,         10,        1;
  [3]       225,        259,       35,        1;
  [4]     11025,      12916,     1974,       84,     1;
  [5]    893025,    1057221,   172810,     8778,   165,    1;
  [6] 108056025,  128816766, 21967231,  1234948, 28743,  286, 1;
.
For row 3: F(x) := 1/cos(x). Then 225*F(x) + 259*(d/dx)^2(F(x)) + 35*(d/dx)^4(F(x)) + (d/dx)^6(F(x)) = 720*(1/cos(x))^7, where F^(r) denotes the r-th derivative of F(x).

Yoann Gelineau and Jiang Zeng, Combinatorial Interpretations of the Jacobi-Stirling Numbers, arXiv:0905.2899 [math.CO], May 2009.
W. Zhang, Some identities involving the Euler and the central factorial numbers, The Fibonacci Quarterly, Vol. 36, Number 2, May 1998.

Cf. A001819, A008275, A008277, A160562, A091885, A182867.

t := proc(n,k) option remember ; expand(x*mul(x+n/2-i,i=1..n-1)) ; coeftayl(%,x=0,k) ; end:
v := proc(n,k) option remember ; 4^(n-k)*t(2*n+1,2*k+1) ; end:
A160563 := proc(n,k) abs(v(n,k)) ; end: for n from 0 to 10 do for k from 0 to n do printf("%d,",A160563(n,k)) ; od: od: # R. J. Mathar, May 20 2009
# Using a bivariate generating function (albeit generating signed terms):
gf := (t + sqrt(1 + t^2))^x: ser := series(gf, t, 20):
ct := n -> coeff(ser, t, n): T := (n, k) -> n!*coeff(ct(n), x, k):
OddPart := (T, len) -> local n, k;
seq(print(seq(T(n, k), k = 1..n, 2)), n = 1..2*len, 2):
OddPart(T, 6);  # Peter Luschny, Mar 03 2024

t[, 0] = 1; t[n, n_] := t[n, n] = ((2*n - 1)!!)^2; t[n_, k_] := t[n, k] = (2*n - 1)^2*t[n - 1, k - 1] + t[n - 1, k];
T[n_, k_] := t[n, n - k];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 28 2017, after R. J. Mathar's comment *)

a(n,k) = |v(n,k)| where v(n,k) = v(n-1,k-1) - (2n-1)^2*v(n-1,k); eq (4.2).
Let F(x) = 1/cos(x). Then (2*n)!*(1/cos(x))^(2*n+1) = Sum_{k=0..n} T(n,k)*F^(2*k)(x), where F^(r) denotes the r-th derivative of F(x) (Zhang 1998). An example is given below. - Peter Bala, Feb 06 2012
Given a (0, 0)-based triangle U we call the triangle [U(n, k), k=1..n step 2, n=1..len step 2] the 'odd subtriangle' of U. This triangle is the odd subtriangle of U(n, k) = n! * [x^(n-k)] [t^n] (t + sqrt(1 + t^2))^x, albeit with signed terms. See A182867 for the even subtriangle. - Peter Luschny, Mar 03 2024

Extended by R. J. Mathar, May 20 2009

cp's OEIS Frontend

A092043 a(n) = numerator(n!/n^2).

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A160562 Triangle of scaled central factorial numbers, T(n,k) = A008958(n,n-k).

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A291505 a(n) = (n!)^7 * Sum_{i=1..n} 1/i^7.

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A291506 a(n) = (n!)^8 * Sum_{i=1..n} 1/i^8.

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A291507 a(n) = (n!)^9 * Sum_{i=1..n} 1/i^9.

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A291508 a(n) = (n!)^10 * Sum_{i=1..n} 1/i^10.

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A142996 a(0) = 0, a(1) = 1, a(n+1) = (2*n^2+2*n+7)*a(n) - n^4*a(n-1), n >= 1.

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A142997 a(0) = 0, a(1) = 1, a(n+1) = (2*n^2+2*n+13)*a(n) - n^4*a(n-1).

A142996 a(0) = 0, a(1) = 1, a(n+1) = (2n^2+2n+7)a(n) - n^4a(n-1), n >= 1.

A142997 a(0) = 0, a(1) = 1, a(n+1) = (2n^2+2n+13)a(n) - n^4a(n-1).

A142998 a(0) = 0, a(1) = 1, a(n+1) = (2n^2+2n+21)a(n) - n^4a(n-1).