A066989 -id:A066989 - cp's OEIS frontend

A143003 a(0) = 0, a(1) = 1, a(n+1) = (2n+1)(n^2+n+5)a(n) - n^6a(n-1).

0, 1, 21, 1091, 114520, 21298264, 6410456640, 2923097201856, 1920450126458880, 1747596822651334656, 2133806329230225408000, 3405545462439659704320000, 6950705677729940374290432000, 17807686090745585163974737920000

Offset: 0

Peter Bala, Jul 19 2008

This is the case m = 1 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+2*m^2+2*m+1)*a(n) - n^6*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series Sum_{k>=1} 1/k^3 for Apery's constant zeta(3). For other cases see A066989 (m=0), A143004 (m=2), A143005 (m=3) and A143006 (m=4).
The solution to the general recurrence may be expressed as a sum: a(n) = n!^3*p_m(n)*Sum_{k = 1..n} 1/(k^3*p_m(k-1)*p_m(k)), where p_m(x) = Sum_{k = 0..n} C(2*k,k)^2*C(n+k,2*k)*C(x+k,2*k) is a polynomial in x of degree 2*m.
The first few are p_0(x) = 1, p_1(x) = 2*x^2 + 2*x + 1, p_2(x) = (3*x^4 + 6*x^3 + 9*x^2 + 6*x + 2)/2 and p_3(x) = (10*x^6 + 30*x^5 + 85*x^4 + 120*x^3 + 121*x^2 + 66*x + 18)/18. For fixed n, the sequence [p_n(k)]k>=0 is the crystal ball sequence for the product lattice A_n x A_n. See A143007 for the table of values [p_n(k)] n,k >= 0. Observe that [p_n(n)] n >= 0 is the sequence of Apery numbers A005259.
The reciprocity law p_m(n) = p_n(m) holds for nonnegative integers m and n. In particular we have p_m(1) = 2*m^2 + 2*m + 1 and p_m(2) = (3*m^4 + 6*m^3 + 9*m^2 + 6*m + 2)/2.
The polynomial p_m(x) is the unique polynomial solution of the difference equation (x+1)^3*f(x+1) + x^3*f(x-1) = (2*x+1)*(x^2+x+2*m^2+2*m+1)*f(x), normalized so that f(0) = 1. The reciprocity law now yields the Apery-like recursion m^3*p_m(x) + (m-1)^3*p_(m-2)(x) = (2*m-1)*(m^2-m+1+2*x^2+2*x)*p_(m-1)(x).
The polynomial functions p_m(x) have their zeros on the vertical line Re x = -1/2 in the complex plane; that is, the polynomials p_m(x-1), m = 1,2,3,..., satisfy a Riemann hypothesis (adapt the proof of the lemma on p. 4 of [BUMP et al.]).
The general recurrence in the first paragraph above has a second solution b(n) = n!^3*p_m(n) with initial conditions b(0) = 1, b(1) = 2*m^2+2*m+1. Hence the behavior of a(n) for large n is given by lim_{n -> infinity} a(n)/b(n) = Sum_{k>=1} 1/(k^3*p_m(k-1)*p_m(k)) = 1/((2*m^2+2*m+1) - 1^6/(3*(2*m^2+2*m+3) - 2^6/(5*(2*m^2+2*m+7) - 3^6/(7*(2*m^2+2*m+13) - ...)))) = Sum_{k>=1} 1/(m+k)^3. The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii)].
For the corresponding results for the constant zeta(2) see A142995. For corresponding results for the constant log(2) see A142979 and A142992.

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

Seiichi Manyama, Table of n, a(n) for n = 0..181
D. Bump, K. Choi, P. Kurlberg and J. Vaaler, A local Riemann hypothesis, I, Math. Zeit. 233, (2000), 1-19.

Cf. A066989, A008288, A108625, A142979, A142992, A142995, A143004, A143005, A143006, A143007.

The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)

p := n -> 2*n^2+2*n+1: a := n -> n!^3*p(n)*sum (1/(k^3*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..14)

RecurrenceTable[{a[0]==0,a[1]==1,a[n+1]==(2n+1)(n^2+n+5)a[n]- n^6 a[n-1]}, a[n],{n,15}] (* Harvey P. Dale, Jun 20 2011 *)

a(n) = n!^3*p(n)*Sum_{k = 1..n} 1/(k^3*p(k-1)*p(k)), where p(n) = 2*n^2 + 2*n + 1 = A001844(n).
Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+5)*a(n) - n^6*a(n-1).
The sequence b(n):= n!^3*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 5. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(5 - 1^6/(21 - 2^6/(55 - 3^6/(119 - ... - (n-1)^6/((2*n-1)*(n^2-n+5)))))), for n >= 2. The behavior of a(n) for large n is given by lim_{n -> infinity} a(n)/b(n) = Sum_{k>=1} 1/(k^3*(4*k^4 + 1)) = 1/(5 - 1^6/(21 - 2^6/(55 - 3^6/(119 - ... - n^6/((2*n+1)*(n^2+n+5) - ...))))) = zeta(3) - 1, where the final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii) at x = 1].

A001819 Central factorial numbers: second right-hand column of triangle A008955.

Original entry on oeis.org

0, 1, 5, 49, 820, 21076, 773136, 38402064, 2483133696, 202759531776, 20407635072000, 2482492033152000, 359072203696128000, 60912644957448192000, 11977654199703478272000, 2702572249389834608640000

Offset: 0

N. J. A. Sloane

Coefficient of x^2 in Product_{k=0..n}(x + k^2). - Ralf Stephan, Aug 22 2004
p divides a(p-1) for prime p > 3. p divides a((p-1)/2) for prime p > 3. For prime p, p^2 divides a(n) for n > 2*p+1. - Alexander Adamchuk, Jul 11 2006; last comment corrected by Michel Marcus, May 20 2020
The ratio a(n)/A001044(n) is the partial sum of the reciprocals of squares. E.g., a(4)/A001044(4) = 820/576 = 1/1 + 1/4 + 1/9 + 1/16. - Pierre CAMI, Oct 30 2006
a(n) is the (n-1)-st elementary symmetric function of the squares of the first n numbers. - Anton Zakharov, Nov 06 2016
Primes p such that p^2 | a(p-1) are the Wolstenholme primes A088164. - Amiram Eldar and Thomas Ordowski, Aug 08 2019

J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Seiichi Manyama, Table of n, a(n) for n = 0..253 (terms 0..50 from T. D. Noe)
Takao Komatsu, Convolution identities of poly-Cauchy numbers with level 2, arXiv:2003.12926 [math.NT], 2020.
Mircea Merca, A Special Case of the Generalized Girard-Waring Formula, J. Integer Sequences, Vol. 15 (2012), Article 12.5.7.
Index entries for sequences related to factorial numbers

Cf. A000254, A001044, A002455, A007406, A007407, A066989, A024167, A142995.
Second right-hand column of triangle A008955.
Equals row sums of A162990(n)/(n+1)^2 for n >= 1.

Table[Sum[1/i^2,{i,1,n}]/Product[1/i^2,{i,1,n}],{n,1,40}] (* Alexander Adamchuk, Jul 11 2006 *)
Table[n!^2*HarmonicNumber[n, 2], {n, 0, 15}] (* Jean-François Alcover, May 09 2012, after Joe Keane *)

a(n)=n!^2*sum(k=1,n,1/k^2) \\ Charles R Greathouse IV, Nov 06 2016

a_n = (n!)^2 * Sum_{k=1..n} 1/k^2. - Joe Keane (jgk(AT)jgk.org)
a(n) ~ (1/3)*Pi^3*n*e^(-2*n)*n^(2*n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
Sum_{n>=0} a(n)*x^n/n!^2 = polylog(2, x)/(1-x). - Vladeta Jovovic, Jan 23 2003
a(n) = Sum_{i=1..n} 1/i^2 / Product_{i=1..n} 1/i^2. - Alexander Adamchuk, Jul 11 2006
a(0) = 0, a(n) = a(n-1)*n^2 + A001044(n-1). E.g., a(1) = 0*1 + 1 = 1 since A001044(0) = 1; a(2) = 1*2^2 + 1 = 5 since A001044(1) = 1; a(3) = 5*3^2 + 4 = 49 since A001044(2) = 4; and so on. - Pierre CAMI, Oct 30 2006
Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n^2 + 2*n + 1)*a(n) - n^4*a(n-1). The sequence b(n) = n!^2 satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 1. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(1 - 1^4/(5 - 2^4/(13 - 3^4/(25 - ... -(n-1)^4/((2*n^2 - 2*n + 1)))))), leading to the infinite continued fraction expansion zeta(2) = 1/(1-1^4/(5 - 2^4/(13 - 3^4/(25 - ... - n^4/((2*n^2 + 2*n + 1) - ...))))). Compare with A142995. Compare also with A024167 and A066989. - Peter Bala, Jul 18 2008
a(n)/(n!)^2 -> zeta(2) = A013661 as n -> infinity, rewriting the Keane formula. - Najam Haq (njmalhq(AT)yahoo.com), Jan 13 2010
a(n) = s(n+1,2)^2 - 2*s(n+1,1)*s(n+1,3), where s(n,k) are Stirling numbers of the first kind, A048994. - Mircea Merca, Apr 03 2012

Minor edits by Vaclav Kotesovec, Jan 28 2015

A291556 Square array A(n,k), n>=0, k>=0, read by antidiagonals: A(n,k) = (n!)^k * Sum_{i=1..n} 1/i^k.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 3, 3, 0, 1, 5, 11, 4, 0, 1, 9, 49, 50, 5, 0, 1, 17, 251, 820, 274, 6, 0, 1, 33, 1393, 16280, 21076, 1764, 7, 0, 1, 65, 8051, 357904, 2048824, 773136, 13068, 8, 0, 1, 129, 47449, 8252000, 224021776, 444273984, 38402064, 109584, 9

Offset: 0

Seiichi Manyama, Aug 26 2017

			Square array begins:
   0,  0,   0,     0,      0, ...
   1,  1,   1,     1,      1, ...
   2,  3,   5,     9,     17, ...
   3, 11,  49,   251,   1393, ...
   4, 50, 820, 16280, 357904, ...

Seiichi Manyama, Antidiagonals n = 0..59, flattened

Columns k=0-10 give: A001477, A000254, A001819, A066989, A203229, A099827, A291456, A291505, A291506, A291507, A291508.
Rows n=0-3 give: A000004, A000012, A000051, A074528.
Main diagonal gives A060943.

A:= (n, k)-> n!^k * add(1/i^k, i=1..n):
seq(seq(A(n, d-n), n=0..d), d=0..10);  # Alois P. Heinz, Aug 26 2017

A[0, ] = 0; A[1, ] = 1; A[n_, k_] := A[n, k] = ((n-1)^k + n^k) A[n-1, k] - (n-1)^(2k) A[n-2, k];
Table[A[n-k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, May 11 2019 *)

A(0, k) = 0, A(1, k) = 1, A(n+1, k) = (n^k+(n+1)^k)*A(n, k) - n^(2*k)*A(n-1, k).

A291456 a(n) = (n!)^6 * Sum_{i=1..n} 1/i^6.

Original entry on oeis.org

0, 1, 65, 47449, 194397760, 3037656102976, 141727869124448256, 16674281388691716870144, 4371079210518164503303028736, 2322975003299339366419974718488576, 2322977286679362958150790503464960000000

Offset: 0

Seiichi Manyama, Aug 24 2017

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..104

Column k=6 of A291556.
Cf. A000254 (k=1), A001819 (k=2), A066989 (k=3), A203229 (k=4), A099827 (k=5).
Cf. A008516, A103345, A103346.

Table[(n!)^6 * Sum[1/i^6, {i, 1, n}], {n, 0, 15}] (* Vaclav Kotesovec, Aug 27 2017 *)

a(0) = 0, a(1) = 1, a(n+1) = (n^6 + (n+1)^6)*a(n) - n^12*a(n-1) for n > 0.
a(n) ~ 8 * Pi^9 * n^(6*n+3) / (945 * exp(6*n)). - Vaclav Kotesovec, Aug 27 2017
a(n) = (n!)^6 * A103345(n)/A103346(n). - Petros Hadjicostas, May 10 2020
Sum_{n>=0} a(n) * x^n / (n!)^6 = polylog(6,x) / (1 - x). - Ilya Gutkovskiy, Jul 15 2020

A143004 a(0) = 0, a(1) = 1, a(n+1) = (2n+1)(n^2+n+13)a(n) - n^6a(n-1).

Original entry on oeis.org

0, 1, 45, 4211, 704120, 191875384, 79755181632, 48072816950976, 40372248180436992, 45735898093934800896, 68049684624570789888000, 130036437291331549384704000, 313117351023401464093212672000

Offset: 0

Peter Bala, Jul 19 2008

This is the case m = 2 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+2*m^2+2*m+1 )*a(n) - n^6*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^3 for Apery's constant zeta(3). For remarks on the general theory see A143003 (m=1). For other cases see A066989 (m=0), A143005 (m=3) and A143006 (m=4).

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

Seiichi Manyama, Table of n, a(n) for n = 0..180

Cf. A066989, A143003, A143005, A143006, A143007.

p := n -> (3*n^4+6*n^3+9*n^2+6*n+2)/2: a := n -> n!^3*p(n)*sum (1/(k^3*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..14)

RecurrenceTable[{a[0]==0,a[1]==1,a[n+1]==(2n+1)(n^2+n+13)a[n]-n^6 a[n-1]}, a,{n,20}] (* Harvey P. Dale, Jan 23 2012 *)

a(n) = n!^3*p(n)*sum {k = 1..n} 1/(k^3*p(k-1)*p(k)), where p(n) = (3*n^4+6*n^3+9*n^2+6*n+2)/2. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+13)*a(n) - n^6*a(n-1). The sequence b(n):= n!^3*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- 1^6/(45- 2^6/(95- 3^6/(175-...- (n-1)^6/((2*n-1)*(n^2-n+13)))))), for n >=2. The behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = sum {k = 1..inf} 1/(k^3*p(k-1)*p(k)) = 1/(13- 1^6/(45- 2^6/(95- 3^6/(175-...- n^6/((2*n+1)*(n^2+n+13)-...))))) = zeta(3) - (1 + 1/2^3), where the final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii) at x = 2].

A143005 a(0) = 0, a(1) = 1, a(n+1) = (2n+1)(n^2+n+25)a(n) - n^6a(n-1).

Original entry on oeis.org

0, 1, 81, 12491, 3176120, 1235165464, 697648230720, 550023729068736, 586201214122536960, 822460381655068717056, 1485544574481829982208000, 3389058487000919282503680000, 9606157364646714324010401792000

Offset: 0

Peter Bala, Jul 19 2008

This is the case m = 3 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+2*m^2+2*m+1 )*a(n) - n^6*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^3 for the constant zeta(3). For remarks on the general theory see A143003 (m=1). For other cases see A066989 (m=0), A143004 (m=2) and A143006 (m=4).

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

Seiichi Manyama, Table of n, a(n) for n = 0..180

Cf. A066989, A143003, A143004, A143006, A143007.

p := n -> (10*n^6+30*n^5+85*n^4+120*n^3+121*n^2+66*n+18)/18: a := n -> n!^3*p(n)*sum (1/(k^3*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..13);

RecurrenceTable[{a[0]==0,a[1]==1,a[n+1]==(2n+1)(n^2+n+25)a[n]-n^6 a[n-1]}, a, {n,15}] (* Harvey P. Dale, Dec 04 2011 *)

a(n) = n!^3*p(n)*sum {k = 1..n} 1/(k^3*p(k-1)*p(k)), where p(n) = (10*n^6 +30*n^5 +85*n^4 +120*n^3 +121*n^2 +66*n +18)/18. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+25)*a(n) - n^6*a(n-1). The sequence b(n):= n!^3*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 25. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(25- 1^6/(81- 2^6/(155- 3^6/(259-...- (n-1)^6/((2*n-1)*(n^2-n+25)))))), for n >=2. The behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = sum {k = 1..inf} 1/(k^3*p(k-1)*p(k)) = 1/(25- 1^6/(81- 2^6/(155- 3^6/(259-...- n^6/((2*n+1)*(n^2+n+25)-...))))) = zeta(3) - (1 + 1/2^3 + 1/3^3), where the final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii) at x = 3].

A143006 a(0) = 0, a(1) = 1, a(n+1) = (2n+1)(n^2+n+41)a(n) - n^6a(n-1).

Original entry on oeis.org

0, 1, 129, 30251, 11129080, 5985956824, 4501140404544, 4577449694922432, 6130634638657941504, 10576998168036159614976, 23068077837238986190848000, 62571876653848665596203008000, 208106990164246904792046895104000

Offset: 0

Peter Bala, Jul 19 2008

This is the case m = 4 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+2*m^2+2*m+1 )*a(n) - n^6*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^3 for the constant zeta(3). For remarks on the general theory see A143003 (m=1). For other cases see A066989 (m=0), A143004 (m=2) and A143005 (m=3).

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

Seiichi Manyama, Table of n, a(n) for n = 0..179

Cf. A066989, A143003, A143004, A143005, A143007.

p := n -> (35*n^8 +140*n^7 +630*n^6 +1400*n^5 +2595*n^4 +3020*n^3 +2500*n^2 +1200*n +288)/288: a := n -> n!^3*p(n)*sum (1/(k^3*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..13);

a(n) = n!^3*p(n)*sum {k = 1..n} 1/(k^3*p(k-1)*p(k)), where p(n) = (35*n^8 +140*n^7 +630*n^6 +1400*n^5 +2595*n^4 +3020*n^3 +2500*n^2 +1200*n +288)/288. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+41)*a(n) - n^6*a(n-1). The sequence b(n):= n!^3*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 41. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(41- 1^6/(129- 2^6/(235- 3^6/(371-...- (n-1)^6/((2*n-1)*(n^2-n+41)))))), for n >=2. The behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = sum {k = 1..inf} 1/(k^3*p(k-1)*p(k)) = 1/(41- 1^6/(129- 2^6/(235- 3^6/(371-...- n^6/((2*n+1)*(n^2+n+41)-...))))) = zeta(3) - (1 + 1/2^3 + 1/3^3 + 1/4^3), where the final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii) at x = 4].

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

cp's OEIS Frontend

A143003 a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+5)*a(n) - n^6*a(n-1).

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A001819 Central factorial numbers: second right-hand column of triangle A008955.

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A291556 Square array A(n,k), n>=0, k>=0, read by antidiagonals: A(n,k) = (n!)^k * Sum_{i=1..n} 1/i^k.

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A291456 a(n) = (n!)^6 * Sum_{i=1..n} 1/i^6.

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A143004 a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+13)*a(n) - n^6*a(n-1).

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A143005 a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+25)*a(n) - n^6*a(n-1).

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A143006 a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+41)*a(n) - n^6*a(n-1).

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A143003 a(0) = 0, a(1) = 1, a(n+1) = (2n+1)(n^2+n+5)a(n) - n^6a(n-1).

A143004 a(0) = 0, a(1) = 1, a(n+1) = (2n+1)(n^2+n+13)a(n) - n^6a(n-1).

A143005 a(0) = 0, a(1) = 1, a(n+1) = (2n+1)(n^2+n+25)a(n) - n^6a(n-1).

A143006 a(0) = 0, a(1) = 1, a(n+1) = (2n+1)(n^2+n+41)a(n) - n^6a(n-1).