A142983 -id:A142983 - cp's OEIS frontend

A005900 Octahedral numbers: a(n) = n(2n^2 + 1)/3.

0, 1, 6, 19, 44, 85, 146, 231, 344, 489, 670, 891, 1156, 1469, 1834, 2255, 2736, 3281, 3894, 4579, 5340, 6181, 7106, 8119, 9224, 10425, 11726, 13131, 14644, 16269, 18010, 19871, 21856, 23969, 26214, 28595, 31116, 33781, 36594, 39559, 42680

Offset: 0

N. J. A. Sloane

Series reversion of g.f.: A(x) is Sum_{n>0} - A066357(n)(-x)^n.
Partial sums of centered square numbers A001844. - Paul Barry, Jun 26 2003
Also as a(n) = (1/6)*(4n^3 + 2n), n>0: structured tetragonal diamond numbers (vertex structure 5) (cf. A000447 - structured diamonds); and structured trigonal anti-prism numbers (vertex structure 5) (cf. A100185 - structured anti-prisms). Cf. A100145 for more on structured polyhedral numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Schlaefli symbol for this polyhedron: {3,4}.
If X is an n-set and Y and Z are disjoint 2-subsets of X then a(n-4) is equal to the number of 5-subsets of X intersecting both Y and Z. - Milan Janjic, Aug 26 2007
Starting with 1 = binomial transform of [1, 5, 8, 4, 0, 0, 0, ...] where (1, 5, 8, 4) = row 3 of the Chebyshev triangle A081277. - Gary W. Adamson, Jul 19 2008
a(n) = largest coefficient of (1 + ... + x^(n-1))^4. - R. H. Hardin, Jul 23 2009
Convolution square root of (1 + 6x + 19x^3 + ...) = (1 + 3x + 5x^2 + 7x^3 + ...) = A005408(x). - Gary W. Adamson, Jul 27 2009
Starting with offset 1 = the triangular series convolved with [1, 3, 4, 4, 4, ...]. - Gary W. Adamson, Jul 28 2009
One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral, and icosahedral) numbers (cf. A053012). - Daniel Forgues, May 14 2010
Let b be any product of four different primes. Then the divisor lattice of b^n is of width a(n+1). - Jean Drabbe, Oct 13 2010
Arises in Bezdek's proof on contact numbers for congruent sphere packings (see preprint). - Jonathan Vos Post, Feb 08 2011
Euler transform of length 2 sequence [6, -2]. - Michael Somos, Mar 27 2011
a(n+1) is the number of 2 X 2 matrices with all terms in {0,1,...,n} and (sum of terms) = 2n. - Clark Kimberling, Mar 19 2012
a(n) is the number of semistandard Young tableaux over all partitions of 3 with maximal element <= n. - Alois P. Heinz, Mar 22 2012
Self convolution of the odd numbers. - Reinhard Zumkeller, Apr 04 2012
a(n) is the number of (w,x,y,z) with all terms in {1,...,n} and w+x=y+z; also the number of (w,x,y,z) with all terms in {0,...,n} and |w-x|<=y. - Clark Kimberling, Jun 02 2012
The sequence is the third partial sum of (0, 1, 3, 4, 4, 4, ...). - Gary W. Adamson, Sep 11 2015
a(n) is the number of join-irreducible elements in the Weyl group of type B_n with respect to the strong Bruhat order. - Rafael Mrden, Aug 26 2020
Number of unit octahedra contained in an n-scale octahedron composed of a tetrahedral-octahedral honeycomb. The number of unit tetrahedra in it is 8*A000292(n-1) = 4*(n^3 - n)/3. Also, the number of unit tetrahedra and unit octahedra contained in an n-scale tetrahedron composed of a tetrahedral-octahedral honeycomb is respectively A006527(n) = (n^3 + 2*n)/3 and A000292(n-1) = (n^3 - n)/6. - Jianing Song, Feb 24 2025

			G.f. = x + 6*x^2 + 19*x^3 + 44*x^4 + 85*x^5 + 146*x^6 + 231*x^7 + ...

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 50.
H. S. M. Coxeter, Polyhedral numbers, pp. 25-35 of R. S. Cohen, J. J. Stachel and M. W. Wartofsky, eds., For Dirk Struik: Scientific, historical and political essays in honor of Dirk J. Struik, Reidel, Dordrecht, 1974.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
X. Acloque, Polynexus Numbers and other mathematical wonders [broken link]
Karoly Bezdek, Contact numbers for congruent sphere packings, arXiv:1102.1198 [math.MG], 2011.
Matteo Cavaleri and Alfredo Donno, Some degree and distance-based invariants of wreath products of graphs, arXiv:1805.08989 [math.CO], 2018.
Y-h. Guo, Some n-Color Compositions, J. Int. Seq. 15 (2012) 12.1.2, eq (5), m=2.
Milan Janjic, Two Enumerative Functions
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
Muhammad Fatih Killik and Bünyamin Şahi̇n, Further Results on Level Matrix, Int'l J. Math. Combin. (2024) Vol 4, 68-74. See p. 71.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.
Hankyung Ko, Volodymyr Mazorchuk and Rafael Mrđen, Join operation for the Bruhat order and Verma modules, arXiv:2109.01067 [math.RT], 2021. See Remark 5.10 p. 19.
A. Lascoux and M.-P. Schützenberger, Treillis et bases des groupes de Coxeter, Electron. J. Combin. 3 (1996), #R27.
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).
J. K. Merikoski, R. Kumar and R. A. Rajput, Upper bounds for the largest eigenvalue of a bipartite graph, Electronic Journal of Linear Algebra ISSN 1081-3810, A publication of the International Linear Algebra Society, Volume 26, pp. 168-176, April 2013.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
B. K. Teo and N. J. A. Sloane, Magic numbers in polygonal and polyhedral clusters, Inorgan. Chem. 24 (1985), 4545-4558.
Eric Weisstein's World of Mathematics, Octahedral Number.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Sums of 2 consecutive terms give A001845. Cf. A001844.
1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Cf. A022521.
Cf. A081277.
Row n=3 of A210391. - Alois P. Heinz, Mar 22 2012
Cf. A005408.
Cf. A053676, A053677, A053678.
Cf. A002061.
Cf. A000292 (tetrahedral numbers), A000578 (cubes), A006566 (dodecahedral numbers), A006564 (icosahedral numbers).
Similar sequence: A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099193(m=7), A099195 (m=8), A099196 (m=9), A099197 (m=10).

a005900 n = sum $ zipWith (*) odds $ reverse odds
            where odds = take n a005408_list
a005900_list = scanl (+) 0 a001844_list
-- Reinhard Zumkeller, Jun 16 2013, Apr 04 2012

[n*(2*n^2+1)/3: n in [0..50]]; // Wesley Ivan Hurt, Sep 11 2015

I:=[0,1,6,19]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Sep 12 2015

al:=proc(s,n) binomial(n+s-1,s); end; be:=proc(d,n) local r; add( (-1)^r*binomial(d-1,r)*2^(d-1-r)*al(d-r,n), r=0..d-1); end; [seq(be(3,n), n=0..100)];
A005900:=(z+1)**2/(z-1)**4; # Simon Plouffe in his 1992 dissertation
with(combinat): seq(fibonacci(4,2*n)/12, n=0..40); # Zerinvary Lajos, Apr 21 2008

Table[(2n^3+n)/3, {n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1}, {0,1,6,19},50] (* Harvey P. Dale, Oct 10 2013 *)
CoefficientList[Series[x (1 + x)^2/(1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Sep 12 2015 *)

makelist(n*(2*n^2+1)/3, n, 0, 20); /* Martin Ettl, Jan 07 2013 */

PARI
```
{a(n) = n*(2*n^2+1)/3};
```

concat([0],Vec(x*(1 + x)^2/(1 - x)^4 + O(x^50))) \\ Indranil Ghosh, Mar 16 2017

def a(n): return n*(2*n*n + 1)//3
print([a(n) for n in range(41)]) # Michael S. Branicky, Sep 03 2021

a(n) = 1^2 + 2^2 + ... + (n-1)^2 + n^2 + (n-1)^2 + ... + 2^2 + 1^2. - Amarnath Murthy, May 28 2001
G.f.: x * (1 + x)^2 / (1 - x)^4. a(n) = -a(-n) = (2*n^3 + n) / 3.
a(n) = ( ((n+1)^5-n^5) - (n^5-(n-1)^5) )/30. - Xavier Acloque, Oct 17 2003
a(n) is the sum of the products pq, where p and q are both positive and odd and p + q = 2n, e.g., a(4) = 7*1 + 5*3 + 3*5 + 1*7 = 44. - Jon Perry, May 17 2005
a(n) = 4*binomial(n,3) + 4*binomial(n,2) + binomial(n,1). - Mitch Harris, Jul 06 2006
a(n) = binomial(n+2,3) + 2*binomial(n+1,3) + binomial(n,3), (this pair generalizes; see A014820, the 4-cross polytope numbers).
Sum_{n>=1} 1/a(n) = 3*gamma + 3*Psi((I*(1/2))*sqrt(2)) - (1/2)*(3*I)*Pi*coth((1/2)*Pi*sqrt(2)) - (1/2)*(3*I)*sqrt(2) = A175577, where I=sqrt(-1). - Stephen Crowley, Jul 14 2009
a(n) = A035597(n)/2. - J. M. Bergot, Jun 11 2012
a(n) = A000578(n) - 2*A000292(n-1) for n>0. - J. M. Bergot, Apr 05 2014
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3. - Wesley Ivan Hurt, Sep 11 2015
E.g.f.: (1/3)*x*(3 + 6*x + 2*x^2)*exp(x). - Ilya Gutkovskiy, Mar 16 2017
a(n) = (A002061(A002061(n+1)) - A002061(A002061(n)))/6. - Daniel Poveda Parrilla, Jun 10 2017
a(n) = 6*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018
Sum_{n >= 1} (-1)^(n+1)/(a(n)*a(n+1)) = 6*log(2) - 4 = 1/(6 + 2/(6 + 6/(6 + ... + n*(n-1)/(6 + ...)))). See A142983. - Peter Bala, Mar 06 2024

A014820 a(n) = (1/3)(n^2 + 2n + 3)*(n+1)^2.

Original entry on oeis.org

1, 8, 33, 96, 225, 456, 833, 1408, 2241, 3400, 4961, 7008, 9633, 12936, 17025, 22016, 28033, 35208, 43681, 53600, 65121, 78408, 93633, 110976, 130625, 152776, 177633, 205408, 236321, 270600, 308481

Offset: 0

N. J. A. Sloane

a(n) is the number of 4 X 4 pandiagonal magic squares with sum 2n. - Sharon Sela (sharonsela(AT)hotmail.com), May 10 2002
Figurate numbers based on the 4-dimensional regular convex polytope called the 16-cell, hexadecachoron, 4-cross polytope or 4-hyperoctahedron with Schlaefli symbol {3,3,4}. a(n)=(n^2*(n^2+2))/3 if the offset were 1. - Michael J. Welch (mjw1(AT)ntlworld.com), Apr 01 2004, R. J. Mathar, Jul 18 2009
If X is an n-set and Y_i (i=1,2,3) mutually disjoint 2-subsets of X then a(n-6) is equal to the number of 7-subsets of X intersecting each Y_i (i=1,2,3). - Milan Janjic, Aug 26 2007
Equals binomial transform of [1, 7, 18, 20, 8, 0, 0, 0, ...], where (1, 7, 18, 20, 8) = row 4 of the Chebyshev triangle A081277. Also = row 4 of the array in A142978. - Gary W. Adamson, Jul 19 2008

T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Ahmed, J. De Loera and R. Hemmecke, Polyhedral Cones of Magic Cubes and Squares, arXiv:math/0201108 [math.CO], 2002.
Maya Ahmed, Jesús De Loera and Raymond Hemmecke, Polyhedral cones of magic cubes and squares, in Discrete and Computational Geometry, Springer, Berlin, 2003, pp. 25-41.
Milan Janjic, Two Enumerative Functions
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
M. Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.
Eric Weisstein's World of Mathematics, 16-Cell
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1)

Cf. A000332, A000583, A005900, A069038, A069039, A070212, A081277, A092181, A092182, A092183, A099175, A099193, A099195, A099196, A099197, A142978.

Cf. A061927, A002415, A000290, A039623.

List([0..40], n -> (n+1)^2*((n+1)^2 +2)/3); # G. C. Greubel, Feb 10 2019

[(1/3)*(n^2+2*n+3)*(n+1)^2: n in [0..40]]; // Vincenzo Librandi, May 22 2011

al:=proc(s,n) binomial(n+s-1,s); end; be:=proc(d,n) local r; add( (-1)^r*binomial(d-1,r)*2^(d-1-r)*al(d-r,n),r=0..d-1); end; [seq(be(4,n),n=0..100)];

LinearRecurrence[{5, -10, 10, -5, 1}, {1, 8, 33, 96, 225}, 31] (* Jean-François Alcover, Jan 17 2018 *)

a(n)=(n+1)^2*(n^2+2*n+3)/3 \\ Charles R Greathouse IV, Apr 17 2012

a <- c(1, 8, 33, 96,225)
for(n in (length(a)+1):30) a[n] <- 5*a[n-1]-10*a[n-2]+10*a[n-3]-5*a[n-4]+a[n-5]
a # Yosu Yurramendi, Sep 03 2013

[((n+1)^2+2)*(n+1)^2/3 for n in range(40)] # G. C. Greubel, Feb 10 2019

Or, a(n-1) = n^2*(n^2+2)/3. - Corrected by R. J. Mathar, Jul 18 2009
From Vladeta Jovovic, Apr 03 2002: (Start)
G.f.: (1+x)^3/(1-x)^5.
Recurrence: a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). (End)
a(n-1) = C(n+3,4) + 3 C(n+2,4) + 3 C(n+1,4) + C(n,4).
Sum_{n>=0} 1/((1/3*(n^2 + 2*n + 3))*(n+1)^2) = (1/4)*Pi^2 - 3*sqrt(2)*Pi*coth(Pi*sqrt(2))*(1/8) + 3/8 = 1.1758589... - Stephen Crowley, Jul 14 2009
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), with n > 4, a(0)=1, a(1)=8, a(2)=33, a(3)=96, a(4)=225. - Yosu Yurramendi, Sep 03 2013
From Bruce J. Nicholson, Jan 23 2019: (Start)
Sum_{i=0..n} a(i) = A061927(n+1).
a(n) = 4*A002415(n+1) + A000290(n+1) = A039623(n+1) + A002415(n+1). (End)
E.g.f.: (3 + 21*x + 27*x^2 + 10*x^3 + x^4)*exp(x)/3. - G. C. Greubel, Feb 10 2019
Sum_{n >= 0} (-1)^n/(a(n)*a(n+1)) = 17/3 - 8*log(2) = 1/(8 + 2/(8 + 6/(8 + ... + n*(n-1)/(8 + ...)))). See A142983. - Peter Bala, Mar 06 2024

Formula index corrected by R. J. Mathar, Jul 18 2009

A142978 Table of figurate numbers for the n-dimensional cross polytopes.

Original entry on oeis.org

1, 1, 2, 1, 4, 3, 1, 6, 9, 4, 1, 8, 19, 16, 5, 1, 10, 33, 44, 25, 6, 1, 12, 51, 96, 85, 36, 7, 1, 14, 73, 180, 225, 146, 49, 8, 1, 16, 99, 304, 501, 456, 231, 64, 9, 1, 18, 129, 476, 985, 1182, 833, 344, 81, 10

Offset: 1

Peter Bala, Jul 15 2008

The n-th row entries for this array are the regular polytope numbers for the n-dimensional cross polytope as defined by [Kim]. The rows are the partial sums of the rows of the square array of Delannoy numbers A008288.
The odd numbered rows of this array form A142977. For a triangular version of this table see A104698. Cf. also A101603.
The n-th row of the array is the binomial transform of n-th row of triangle A081277, followed by zeros. Example: row 4 (1, 6, 19, 44, 85, ...) = binomial transform of row 3 of A081277: (1, 5, 8, 4, 0, 0, 0, ...). - Gary W. Adamson, Jul 17 2008
The main diagonal of the array T(n,k) is A047781 Sum_{k=0..n-1} binomial(n-1,k)*binomial(n+k,k). Also a(n) = T(n,n), array T as in A049600. The link from A099193 to J. V. Post, Table of polytope numbers, Sorted, Through 1,000,000, includes all n-D Hyperoctahedron (n-Cross Polytope) Numbers through 10-Cross(20) = 1669752016. - Jonathan Vos Post, Jul 16 2008

			The square array A(n, k) begins:
  n\k| 1   2    3     4     5       6
  ---+-------------------------------
   1 | 1   2    3     4      5      6    A000027
   2 | 1   4    9    16     25     36    A000290
   3 | 1   6   19    44     85    146    A005900
   4 | 1   8   33    96    225    456    A014820
   5 | 1  10   51   180    501   1182    A069038
   6 | 1  12   73   304    985   2668    A069039
   7 | 1  14   99   476   1765   5418    A099193

Reinhard Zumkeller, Rows n = 1..100 of triangle, flattened
Peter Bala, A142978 and a family of continued fraction expansions for log(2)
Valentin Bonzom and Etera R. Livine, Self-duality of the 6j-symbol and Fisher zeros for the Tetrahedron, arXiv:1905.00348 [math-ph], 2019.
Steven Edwards and William Griffiths, Generalizations of Delannoy and cross polytope numbers, Fib. Q., Vol. 55, No. 4 (2017), pp. 356-366.
Steven Edwards and William Griffiths, On Generalized Delannoy Numbers, J. Int. Seq., Vol. 23 (2020), Article 20.3.6.
Milan Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Milan Janjic and B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.
Eric W. Weisstein, Meixner polynomial of the first kind
Eric W. Weisstein, Mittag-Leffler polynomial
Yutong Yang, From Simplest Recursion to the Recursion of Generalizations of Cross Polytope Numbers, (2017), Honors College Capstone and Theses, 13.

Cf. A008288 (Delannoy numbers), A005900 (row 3), A014820 (row 4), A069038 (row 5), A069039 (row 6), A099193 (row 7), A099195 (row 8), A099196 (row 9), A099197 (row 10), A101603, A104698 (triangle version), A142977, A142983.

Cf. A049600, A047781, A081277.

a142978 n k = a142978_tabl !! (n-1) !! (k-1)
a142978_row n = a142978_tabl !! (n-1)
a142978_tabl = map reverse a104698_tabl
-- Reinhard Zumkeller, Jul 17 2015

with(combinat): T:=(n,k) -> add(binomial(n-1,i)*binomial(k+i,n),i = 0..n-1); for n from 1 to 10 do seq(T(n,k),k = 1..10) end do; # Program restored by Peter Bala, Oct 02 2008
A := (n, k) -> k*hypergeom([1 - n, 1 - k], [2], 2):
seq(print(seq(simplify(A(n, k)), k = 1..9)), n=1..7); # Peter Luschny, Mar 23 2023

t[n_, k_] := Sum[ Binomial[n-1, i]*Binomial[k+i, n], {i, 0, n-1}]; Table[t[n-k, k], {n, 1, 11}, {k, 1, n-1}] // Flatten (* Jean-François Alcover, Mar 06 2013 *)

T(n,k) = Sum_{i = 0..n-1} C(n-1,i)*C(k+i,n).
Reciprocity law: n*T(n,k) = k*T(k,n).
Recurrence relation: T(n,1) = 1, T(1,k) = k, T(n,k) = T(n,k-1) + T(n-1,k-1) + T(n-1,k), n,k > 1.
O.g.f. row n: x*(1 + x)^(n-1)/(1 - x)^(n+1).
O.g.f. for array: Sum_{n >= 1, k >= 1} T(n, k)*x^k*y^n = x*y/((1 - x)*(1 - x - y - x*y)).
The n-th row entries are the values [p_n(k)], k >= 1, of the polynomial function p_n(x) = Sum_{k = 1..n} 2^(k-1)*C(n-1,k-1)*C(x,k). The first few values are p_1(x) = x, p_2(x) = x^2, p_3(x) = (2*x^3 + x)/3 and p_4(x) = (x^4 + 2*x^2)/3.
The polynomial p_n(x) is the unique polynomial solution of the difference equation x*( f(x+1) - f(x-1) ) = 2*n*f(x), normalized so that f(1) = 1.
The o.g.f. for the p_n(x) is 1/2*((1 + t)/(1 - t))^x = 1/2 + x*t + x^2*t^2 + (2*x^3 + x)/3*t^3 + .... Thus p_n(x) is, apart from a constant factor, the Meixner polynomial of the first kind M_n(x;b,c) at b = 0, c = -1, also known as a Mittag-Leffler polynomial.
The entries in the n-th row appear in the series acceleration formula for the constant log(2): Sum_{k >= 1} (-1)^(k+1)/(T(n,k)*T(n,k+1)) = 1 + (-1)^(n+1) * (2*n)*(log(2) - (1 - 1/2 + 1/3 - ... + (-1)^(n+1)/n)). For example, n = 3 gives log(2) = 4/6 + (1/6)*(1/(1*6) - 1/(6*19) + 1/(19*44) - 1/(44*85) + ...). See A142983 for further details.
From Peter Bala, Oct 02 2008: (Start)
The odd-indexed columns of this array form the array A142992 of crystal ball sequences for lattices of type C_n.
Conjectural congruences for main diagonal entries: Put A(n) = T(n,n). Calculation suggests the following congruences: for prime p > 3 and m, r >= 1, A(m*p^r) == A(m*p^(r-1)) (mod p^(3*r));
Sum_{k = 0..p-1} A(k)^2 == 0 (mod p) if p is a prime of the form 8*n+1 or 8*n+7;
Sum_{k = 0..p-1} A(k)^2 == -1 (mod p) if p is a prime of the form 8*n+3 or 8*n+5.
(End)
From Peter Bala, Sep 27 2021: (Start)
T(n,k) = (1/2)*Sum_{i = 0..k} binomial(k,i)*binomial(n+k-1-i,k-1).
T(n,k) = (1/2)*[x^n] ((1+x)/(1-x))^k = (1/2)*(k/n)*[x^k] ((1+x)/(1-x))^n.
n*T(n,k) = 2*k*T(n-1,k) + (n - 2)*T(n-2,k). (End)
A(n,k) = k*hypergeom([1 - n, 1 - k], [2], 2). - Peter Luschny, Mar 23 2023
T(n,k) = 2*(Sum_{j=1..k-1} T(n-1,j)) + T(n-1,k) for n > 1. - Robert FERREOL, Jun 25 2024

A142979 a(1) = 1, a(2) = 3, a(n+2) = 3a(n+1) + (n+1)^2a(n).

Original entry on oeis.org

1, 3, 13, 66, 406, 2868, 23220, 210192, 2116656, 23375520, 281792160, 3673814400, 51599514240, 775673176320, 12440524320000, 211848037632000, 3820318338816000, 72685037892096000, 1455838255452672000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 1 of the more general recurrence a(1) = 1, a(2) = 2*m + 1, a(n+2) = (2*m + 1)*a(n+1) + (n + 1)^2*a(n) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of Mercator's series for the constant log(2). For other cases see A024167 (m = 0), A142980 (m = 2), A142981 (m = 3) and A142982 (m = 4).
The solution to the general recurrence may be expressed as a sum: a(n) = n!*p_m(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p_m(k-1)*p_m(k)), where p_m(x) = Sum_{k = 0..m} 2^k*C(m,k)*C(x,k) = Sum_{k = 0..m} C(m,k)*C(x+k,m), is the Ehrhart polynomial of the m-dimensional cross polytope (the hyperoctahedron).
The first few values are p_0(x) = 1, p_1(x) = 2*x + 1, p_2(x) = 2*x^2 + 2*x + 1 and p_3(x) = (4*x^3 + 6*x^2 + 8*x + 3)/3.
The sequence {p_m(k)},k>=0 is the crystal ball sequence for the product lattice A_1 x... x A_1 (m copies). The table of values [p_m(k)]m,k>=0 is the array of Delannoy numbers A008288.
The polynomial p_m(x) is the unique polynomial solution of the difference equation (x + 1)*f(x+1) - x*f(x-1) = (2*m + 1)*f(x), normalized so that f(0) = 1. These polynomials 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 [BUMP et al., Theorems 4 and 6]. The o.g.f. for the p_m(x) is (1 + t)^x/(1 - t)^(x+1) = 1 + (2*x + 1)*t + (2*x^2 + 2*x + 1)*t^2 + ....
The general recurrence in the first paragraph above also has a second solution b(n) = n!*p_m(n) with initial conditions b(1) = 2*m + 1, b(2) = (2*m + 1)^2 + 1.
Hence the behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k*p_m(k-1)*p_m(k)) = 1/((2*m + 1) + 1^2/((2*m + 1) + 2^2/((2*m + 1) + 3^2/((2*m + 1) + ... + n^2/((2*m + 1) + ...))))) = (-1)^m * (log(2) - (1 - 1/2 + 1/3 - ... + (-1)^(m+1)/m)), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 29]).
For other sequences defined by similar recurrences and related to log(2) see A142983 and A142988. See also A142992 for the connection between log(2) and the C_n lattices. For corresponding results for the constants e, zeta(2) and zeta(3) see A000522, A142995 and A143003 respectively.

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

Seiichi Manyama, Table of n, a(n) for n = 1..448
D. Bump, K. Choi, P. Kurlberg and J. Vaaler, A local Riemann hypothesis, I, Math. Zeit. 233, (2000), 1-19.

Cf. A008288, A024167, A142980, A142981, A142982, A142983, A142988, A142992, A317057.

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

RecurrenceTable[{a[1]==1,a[2]==3,a[n+2]==3a[n+1]+(n+1)^2 a[n]},a,{n,20}] (* Harvey P. Dale, May 20 2012 *)

a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = 2*n + 1.
Recurrence: a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1) + (n + 1)^2*a(n).
The sequence b(n):= n!*p(n) satisfies the same recurrence with b(1) = 3, b(2) = 10.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(3 + 1^2/(3 + 2^2/(3 + 3^2/(3 + ... + (n-1)^2/3)))), for n >= 2.
Limit_{n -> oo} a(n)/b(n) = 1/(3 + 1^2/(3 + 2^2/(3 + 3^2/(3 + ... + (n-1)^2/(3 + ...))))) = Sum_{k >= 1} (-1)^(k+1)/(k*(4k^2 - 1)) = 1 - log(2).
Thus a(n) ~ c*n*n! as n -> oo, where c = 2*(1 - log(2)).
From Peter Bala, Dec 09 2024: (Start)
E.g.f.: A(x) = (2*x - (1 + x)*log(1 + x))/(1 - x)^2 satisfies the differential equation 1 + (x + 3)*A(x) + (x^2 - 1)*A'(x) = 0 with A(0) = 0.
Sum_{k = 1..n} Stirling_2(n, k) * a(k) = A317057(n+1). (End)

A069038 Expansion of g.f. x*(1+x)^4/(1-x)^6.

Original entry on oeis.org

0, 1, 10, 51, 180, 501, 1182, 2471, 4712, 8361, 14002, 22363, 34332, 50973, 73542, 103503, 142544, 192593, 255834, 334723, 432004, 550725, 694254, 866295, 1070904, 1312505, 1595906, 1926315, 2309356, 2751085, 3258006, 3837087, 4495776, 5242017, 6084266

Offset: 0

Vladeta Jovovic, Apr 03 2002

Hyun Kwang Kim asserts that every nonnegative integer can be represented by the sum of no more than 14 of these numbers. - Jonathan Vos Post, Nov 16 2004
If Y_i (i=1,2,3,4) are 2-blocks of a (n+4)-set X then a(n-4) is the number of 9-subsets of X intersecting each Y_i (i=1,2,3,4). - Milan Janjic, Oct 28 2007
Starting with 1 = binomial transform of [1, 9, 32, 56, 48, 16, 0, 0, 0, ...], where (1, 9, 32, 56, 48, 16) = row 5 of the Chebyshev triangle A081277. Also = row 5 of the array in A142978. - Gary W. Adamson, Jul 19 2008
Starting with the term 1 this is the self-convolution of A001844(n). - Anton Zakharov, Sep 02 2016

H. S. M. Coxeter, Regular Polytopes, New York: Dover Publications, 1973.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Milan Janjic, Two Enumerative Functions
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000332, A014820, A005900.
Cf. A142978, A081277.
Cf. A001844.

[n*(2*n^4 + 10*n^2 + 3)/15: n in [0..40]]; // Vincenzo Librandi, May 22 2011

al:=proc(s,n) binomial(n+s-1,s); end; be:=proc(d,n) local r; add( (-1)^r*binomial(d-1,r)*2^(d-1-r)*al(d-r,n), r=0..d-1); end; [seq(be(5,n),n=0..100)];

CoefficientList[Series[x (1 + x)^4/(1 - x)^6, {x, 0, 32}], x] (* Michael De Vlieger, Sep 02 2016 *)
LinearRecurrence[{6,-15,20,-15,6,-1},{0,1,10,51,180,501},40] (* Harvey P. Dale, Jun 19 2021 *)

concat(0, Vec(x*(1+x)^4/(1-x)^6 + O(x^99))) \\ Altug Alkan, Sep 02 2016

{a(n) = n * (2*n^4 + 10*n^2 + 3) / 15}; /* Michael Somos, Jun 17 2018 */

Recurrence: a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
a(n) = n*(2*n^4 + 10*n^2 + 3)/15. - Jonathan Vos Post, Nov 16 2004
a(n) = C(n+4,5) + 4*C(n+3,5) + 6*C(n+2,5) + 4*C(n+1,5) + C(n,5).
Sum_{n>=1} 1/((1/15)*n*(2*n^4 + 10*n^2 + 3)*n!) = hypergeom([1, 1, 1+i*sqrt(10-2*sqrt(19))*(1/2), 1-i*sqrt(10-2*sqrt(19))*(1/2), 1+i*sqrt(10+2*sqrt(19))*(1/2), 1-i*sqrt(10+2*sqrt(19))*(1/2)], [2, 2, 2+i*sqrt(10-2*sqrt(19))*(1/2), 2-i*sqrt(10-2*sqrt(19))*(1/2), 2+i*sqrt(10+2*sqrt(19))*(1/2), 2-i*sqrt(10+2*sqrt(19))*(1/2)], 1) = 1.05351734968093116819345664995829700099916... - Stephen Crowley, Jul 14 2009
a(n) = a(n-1) + A014820(n-1) + A014820(n-2). - Bruce J. Nicholson, Apr 18 2018
a(n) = 10*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018
Euler transform of length 2 sequence [10, -4]. - Michael Somos, Jun 19 2018
Sum_{k >= 1} (-1)^k/(a(k)*a(k+1)) = 10*log(2) - 41/6 = 1/(10 + 2/(10 + 6/(10 + ... + n*(n-1)/(10 + ...)))). See A142983. Cf. A005900 and A014820. - Peter Bala, Mar 08 2024
E.g.f.: exp(x)*x*(15 + 60*x + 60*x^2 + 20*x^3 + 2*x^4)/15. - Stefano Spezia, Mar 10 2024

A060524 Triangle read by rows: T(n,k) = number of degree-n permutations with k odd cycles, k=0..n, n >= 0.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 5, 0, 1, 9, 0, 14, 0, 1, 0, 89, 0, 30, 0, 1, 225, 0, 439, 0, 55, 0, 1, 0, 3429, 0, 1519, 0, 91, 0, 1, 11025, 0, 24940, 0, 4214, 0, 140, 0, 1, 0, 230481, 0, 122156, 0, 10038, 0, 204, 0, 1, 893025, 0, 2250621, 0, 463490, 0, 21378, 0, 285, 0, 1, 0

Offset: 0

Vladeta Jovovic, Apr 01 2001

The row polynomials t(n,x):=Sum_{k=0..n} T(n,k)*x^k satisfy the recurrence relation t(n,x) = x*t(n-1,x) + ((n-1)^2)*t(n-2,x); t(-1,x)=0, t(0,x)=1. - Wolfdieter Lang, see above.
This is an example of a Sheffer triangle (coefficient triangle for Sheffer polynomials). In the umbral calculus (see the Roman reference given under A048854) s(n,x) := Sum_{k=0..n} T(n,k)*x^k would be called Sheffer polynomials for (1/cosh(t),tanh(t)), which translates to the e.g.f. for column number k>=0 given by (1/sqrt(1-x^2))*((arctanh(x))^k)/k!. The e.g.f. given below is rewritten in this Sheffer context as (1/sqrt(1-x^2))*exp(y*log(sqrt((1+x)/(1-x))))= (1/sqrt(1-x^2))*exp(y*arctanh(x)). The rows of the Jabotinsky type triangle |A049218| provide the coefficients of the associated polynomials. - Wolfdieter Lang, Feb 24 2005
The solution of the differential-difference relation f(n+1,x)= (d/dx)f(n,x) + (n^2)*f(n-1,x), n >= 1, with inputs f(0,x) and f(1,x) = (d/dx)f(0,x) is f(n,x) = t(n,d_x)*f(0,x), with the differential operator d_x:=d/dx and the row polynomials t(n,x) defined above. This problem appears in a computation of thermo field dynamics where f(0,x)=1/cosh(x). See the triangle A060081. - Wolfdieter Lang, Feb 24 2005
The inverse of the Sheffer matrix T with elements T(n,k) is the Sheffer matrix A060081. - Wolfdieter Lang, Jul 22 2005
T(n,k)=0 if n-k= 1(mod 2), else T(n,k) = sum of M2(n,p), p from {1,...,A000041(n)} restricted to partitions with exactly k odd parts and any nonnegative number of even parts. For the M2-multinomial numbers in A-St order see A036039(n,p). - Wolfdieter Lang, Aug 07 2007

			Triangle begins:
  [1],
  [0, 1],
  [1, 0, 1],
  [0, 5, 0, 1],
  [9, 0, 14, 0, 1],
  [0, 89, 0, 30, 0, 1],
  [225, 0, 439, 0, 55, 0, 1],
  [0, 3429, 0, 1519, 0, 91, 0, 1],
  [11025, 0, 24940, 0, 4214, 0, 140, 0, 1],
  [0, 230481, 0, 122156, 0, 10038, 0, 204, 0, 1],
  [893025, 0, 2250621, 0, 463490, 0, 21378, 0, 285, 0, 1],
  [0, 23941125, 0, 14466221, 0, 1467290, 0, 41778, 0, 385, 0, 1],
  ...
Signed version begins:
  [1],
  [0, 1],
  [-1, 0, 1],
  [0, -5, 0, 1],
  [9, 0, -14, 0, 1],
  [0, 89, 0, -30, 0, 1],
  [-225, 0, 439, 0, -55, 0, 1],
  [0, -3429, 0, 1519, 0, -91, 0, 1],
  ...
From _Peter Bala_, Feb 23 2024: (Start)
Maple can verify the following series for Pi:
Row 1 polynomial R(1, x) = x:
Pi = 3 + 4*Sum_{n >= 1} (-1)^(n+1)/((2*n + 1)*R(1, 2*n)*R(1, 2*n+2)).
Row 3 polynomial R(3, x) = 5*x + x^3:
(3/2)^2 * Pi = 7 + 4*(3^4)*Sum_{n >= 1} (-1)^(n+1)/((2*n + 1)*R(3, 2*n)*R(3, 2*n+2)).
Row 5 polynomial R(5, x) = 89*x + 30*x^3 + x^5:
((3*5)/(2*4))^2 * Pi = 11 + 4*(3*5)^4*Sum_{n >= 1} (-1)^(n+1)/((2*n + 1)*R(5, 2*n)*R(5, 2*n+2)). (End)

Alois P. Heinz, Rows n = 0..140, flattened
Marcelo Aguiar, Sarah Brauner, and Vic Reiner, Configuration spaces and peak representations, Sém. Lotharingien Comb., Proc. 36th Conf. Formal Power Series Alg. Comb. (2024) Vol. 91B, Art. No. 13. See p. 12.
Peter Bala, Meixner polynomials and Brouncker's continued fraction for Pi
Paul L. Butzer and Tom H. Koornwinder, Josef Meixner: His life and his orthogonal polynomials, Indagationes Mathematicae, Volume 30, Issue 1, January 2019, Pages 250-264.
Adel Hamdi and Jiang Zeng, Orthogonal polynomials and operator orderings, J. Math. Phys., 51:043506, (2010); arXiv:1006.0808 [math.CO], 2010.
Eric Weisstein's World of Mathematics, Meixner polynomial of the first kind.

Cf. A060338, A060523, A094368, A028353 (col 1), A103916 (col 2), A103917 (col 3), A103918 (col 4).

Cf. A111594 (associated Sheffer polynomials), A142979, A142983.

with(combinat):
b:= proc(n, i) option remember; expand(`if`(n=0, 1, `if`(i<1, 0,
      add(multinomial(n, n-i*j, i$j)*(i-1)!^j/j!*b(n-i*j, i-1)*
      `if`(irem(i, 2)=1, x^j, 1), j=0..n/i))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n$2)):
seq(T(n), n=0..12);  # Alois P. Heinz, Mar 09 2015
# alternative
A060524 := proc(n,k)
    option remember;
    if nR. J. Mathar, Jul 06 2023

nn = 6; Range[0, nn]! CoefficientList[
   Series[(1 - x^2)^(-1/2) ((1 + x)/(1 - x))^(y/2), {x, 0, nn}], {x, y}] // Grid  (* Geoffrey Critzer, Aug 28 2012 *)

E.g.f.: (1+x)^((y-1)/2)/(1-x)^((y+1)/2).
T(n, k) = T(n-1, k-1) + ((n-1)^2)*T(n-2, k); T(-1, k):=0, T(n, -1):=0, T(0, 0)=1, T(n, k)=0 if nWolfdieter Lang, see above.
The Meixner polynomials defined by S_0(x)=1, S_1(x) = x; S_{n+1}(x) = x*S_n(x) - n^2*S_{n-1}(x) give a signed version of this triangle (cf. A060338). - N. J. A. Sloane, May 30 2013
From Peter Bala, Apr 10 2024: (Start)
The n-th row polynomial R(n, x) satisfies
(4*n + 2)*R(n, x) = (x + 1)*R(n, x+2) - (x - 1)*R(n, x-2).
Series for Pi involving the row polynomials R(n, x): for n >= 0 there holds
((2*n + 1)!!/(2^n*n!))^2 * Pi = (4*n + 3) + 4*((2*n + 1)!!^4) * Sum_{k >= 1} (-1)^(k+1)/((2*k + 1)*R(2*n+1, 2*k)*R(2*n+1, 2*k+2)). Cf. A142979 and A142983.
R(2*n, 0) = A001147(n)^2 = A001818(n); R(2*n+1, 0) = 0.
R(n, 1) = n! = A000142(n).
R(2*n, 2) = (4*n + 1)*A001147(n)^2 = (4*n + 1)*((2*n)!/(2^n*n!))^2;
R(2*n+1, 2) = 2*A001447(n+1)^2 = 2*(2*n + 1)!^2/(n!^2*4^n).
R(n, 3) = (2*n + 1)*n! = A007680(n). (End)

A016687 Decimal expansion of log(64) = 6*log(2).

Original entry on oeis.org

4, 1, 5, 8, 8, 8, 3, 0, 8, 3, 3, 5, 9, 6, 7, 1, 8, 5, 6, 5, 0, 3, 3, 9, 2, 7, 2, 8, 7, 4, 9, 0, 5, 9, 4, 0, 8, 4, 5, 3, 0, 0, 0, 8, 0, 6, 1, 6, 1, 5, 3, 1, 5, 2, 4, 7, 2, 4, 0, 8, 0, 0, 5, 6, 9, 6, 0, 3, 6, 1, 7, 3, 1, 8, 1, 8, 1, 6, 8, 2, 9, 3, 6, 3, 5, 1, 7, 9, 9, 6, 1, 9, 7, 8, 5, 1, 2, 1, 2

Offset: 1

N. J. A. Sloane

			4.158883083359671856503392728749059408453000806161531524724080056960361...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 2.

Harry J. Smith, Table of n, a(n) for n = 1..20000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Index entries for transcendental numbers

Cf. A002162, A005900, A016492 (continued fraction), A016627, A016631.

RealDigits[Log[64],10,120][[1]] (* Harvey P. Dale, May 06 2022 *)

default(realprecision, 20080); x=log(64); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b016687.txt", n, " ", d)); \\ Harry J. Smith, May 22 2009

Equals 2*A016631 = 3*A016627 = 6*A002162. - Alois P. Heinz, Aug 07 2023
From Peter Bala, Mar 05 2024: (Start)
log(64) = 4 + Sum_{n >= 1} (-1)^(n+1)/(p(n)*p(n+1)), where p(n) = n*(2*n^2 + 1)/3 = A005900.
Continued fraction: log(64) = 4 + 1/(6 + (1*2)/(6 + (2*3)/(6 + (3*4)/(6 + (4*5)/(6 + ... ))))). See A142983. Cf. A016627. (End)

A142988 a(1) = 1, a(2) = 3, a(n+2) = 3a(n+1) + (n + 1)(n + 3)*a(n).

Original entry on oeis.org

1, 3, 17, 96, 696, 5448, 49752, 492480, 5457600, 65128320, 850296960, 11864240640, 178442611200, 2848854758400, 48517709184000, 872011090944000, 16589133517824000, 331426982928384000, 6966369015201792000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 0 of the general recurrence a(1) = 1, a(2) = 2*m + 3, a(n+2) = (2*m + 3)*a(n+1) + (n + 1)*(n + 3)*a(n) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series Sum_{k >= 1} (-1)^(k+1)/(k*(k + 1)*(k + 2)) = 2*(log(2) - 5/8). For other cases see A142989 (m = 1), A142990 (m = 2) and A142991 (m = 3).
The solution to the general recurrence may be expressed as a series: a(n) = (n+2)!*p_m(n+2)*Sum_{k = 1..n} (-1)^(k+1)/(k*(k + 1)*(k + 2)*p_m(k+1)*p_m(k+2)), where p_m(x) = 1/(2*x*(x - 1))*Sum_{k = 2..m+2} 2^k*C(m,k-2)*C(x,k).
The first few values are p_0(x) = 1, p_1(x) = (2*x - 1)/3, p_2(x) = (x^2 - x + 1)/3 and p_3(x) = (2*x^3 - 3*x^2 + 7*x - 3)/15.
The polynomial p_m(x) is the unique polynomial solution (up to multiplication by a constant) of the difference equation (x + 1)*f(x+1) - (x - 2)*f(x-1) = (2*m + 3)*f(x).
O.g.f. for the p_m(x): Sum_{k >= 0} p_m(x)*t^m = 1/(2*x*(x - 1)*t^2) * ((1 + t)^x/(1 - t)^(x-1) - 1 + t - 2*x*t) = 1 + (2*x - 1)/3*t + (x^2 - x + 1)/3*t^2 + ....
These polynomials satisfy a Riemann hypothesis: their zeros all lie on the vertical line Re x = 1/2 in the complex plane (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) = (1/2)*(n + 2)!*p_m(n+2), with b(1) = 2*m + 3, b(2) = 4*(m^2 + 3*m + 3).
Hence the behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = 2*Sum_{k >= 1} (-1)^(k+1)/(k*(k + 1)*(k + 2)*p_m(k+1)*p_m(k+2)) = 1/((2*m + 3) + 1*3/((2*m + 3) + 2*4/((2*m + 3) + 3*5/((2*m + 3) +...+ (n - 1)*(n + 1)/((2*m + 3) + ...))))).
The infinite continued fraction has the value (-1)^m*2*(m + 1)*(m + 2)*(log(2) - 5/8 - (1/2)*(1/(1*2*3) - 1/(2*3*4) + 1/(3*4*5) - ... + (-1)^(m+1)/(m*(m + 1)*(m + 2)))) for m > 0. This evaluation follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 34] (set l = n in Entry 34 and then let n tend to 1).
For related results see A142979 and A142983.

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

Seiichi Manyama, Table of n, a(n) for n = 1..448
D. Bump, K. Choi, P. Kurlberg and J. Vaaler, A local Riemann hypothesis, I, Math. Zeit. 233, (2000), 1-19.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2003), 65-75.

Cf. A142979, A142983, A142989, A142990, A142991.

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

a(n) = (n + 2)!*Sum_{k = 1..n} (-1)^(k+1)/(k*(k + 1)*(k + 2)).
Recurrence: a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1) + (n + 1)*(n + 3)*a(n).
The sequence b(n) := (1/2)*(n + 2)!*p(n+2) satisfies the same recurrence with b(1) = 3, b(2) = 12.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(3 + 1*3/(3 + 2*4/(3 + 3*5/(3 + ... + (n - 1)*(n + 1)/3)))), for n >= 2.
Limit_{n -> oo} a(n)/b(n) = 1/(3 + 1*3/(3 + 2*4/(3 + 3*5/(3 + ... + (n - 1)*(n + 1)/(3 + ...))))) = 2*Sum_{k >= 1} (-1)^(k+1)/(k*(k + 1)*(k + 2)) = 4*log(2) - 5/2.

A142984 a(1) = 1, a(2) = 4, a(n+2) = 4a(n+1) + (n + 1)(n + 2)*a(n).

Original entry on oeis.org

1, 4, 22, 136, 984, 8016, 73392, 742464, 8254080, 99838080, 1307301120, 18407831040, 277570298880, 4460506444800, 76131788544000, 1375048700928000, 26208041287680000, 525597067634688000, 11065538390925312000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 2 of the general recurrence a(1) = 1, a(2) = 2*m, a(n+2) = 2*m*a(n+1) + (n + 1)*(n + 2)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). See A142983 for remarks on the general case.

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

Reinhard Zumkeller, Table of n, a(n) for n = 1..250

Cf. A142983, A142985, A142986, A142987.

Cf. A002378.

a142984 n = a142984_list !! (n-1)
a142984_list = 1 : 4 : zipWith (+)
                       (map (* 4) $ tail a142984_list)
                       (zipWith (*) (drop 2 a002378_list) a142984_list)
-- Reinhard Zumkeller, Jul 17 2015

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

a(n) = n!*p(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = n^2.
Recurrence: a(1) = 1, a(2) = 4, a(n+2) = 4*a(n+1) + (n + 1)*(n + 2)*a(n).
The sequence b(n) := n!*p(n+1) satisfies the same recurrence with b(1) = 4 and b(2) = 18.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(4 + 1*2/(4 + 2*3/(4 + 3*4/(4 + ... + n*(n - 1)/(4))))), for n >= 2.
The behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k^2*(k + 1)^2) = 1/(4 + 1*2/(4 + 2*3/(4 + 3*4/(4 + ... + n*(n - 1)/(4 + ...))))) = 3 - 4*log(2), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).
a(n) = n! * (((-1)^n * (2 * Psi(n/2 + 1) - 2 * Psi(n/2 + 3/2)) - 4 * log(2) + 3) * (n+1)^2 + (-1)^n * (2 * n + 1)). - Robert Israel, Mar 07 2024

A142985 a(1) = 1, a(2) = 6, a(n+2) = 6a(n+1) + (n + 1)(n + 2)*a(n).

Original entry on oeis.org

1, 6, 42, 324, 2784, 26424, 275472, 3132576, 38629440, 513708480, 7331489280, 111798455040, 1814503057920, 31234337164800, 568451665152000, 10906950910464000, 220060558384128000, 4657890328906752000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 3 of the general recurrence a(1) = 1, a(2) = 2*m, a(n+2) = 2*m*a(n+1) + (n + 1)*(n + 2)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). See A142983 for remarks on the general case.

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

Reinhard Zumkeller, Table of n, a(n) for n = 1..250

Cf. A005900, A142983, A142984, A142986, A142987.

Cf. A002378.

a142985 n = a142985_list !! (n-1)
a142985_list = 1 : 6 : zipWith (+)
                       (map (* 6) $ tail a142985_list)
                       (zipWith (*) (drop 2 a002378_list) a142985_list)
-- Reinhard Zumkeller, Jul 17 2015

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

RecurrenceTable[{a[1]==1,a[2]==6,a[n]==6a[n-1]+(n-1)n*a[n-2]},a,{n,20}] (* Harvey P. Dale, Sep 20 2013 *)

a(n) = n!*p(n+1)*Sum {k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = (2*n^3 + n)/3 = A005900(n).
Recurrence: a(1) = 1, a(2) = 6, a(n+2) = 6*a(n+1) + (n + 1)*(n + 2)*a(n).
The sequence b(n):= n!*p(n+1) satisfies the same recurrence with b(1) = 6, b(2) = 38. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(6 + 1*2/(6 + 2*3/(6 + 3*4/(6 + ... + n*(n - 1)/6)))), for n >= 2.
The behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = Sum_{k = 1..oo} (-1)^(k+1)/(p(k)*p(k+1)) = 1/(6 + 1*2/(6 + 2*3/(6 + 3*4/(6 + ... + n*(n - 1)/(6 + ...))))) = 6*log(2) - 4, where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).

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A005900 Octahedral numbers: a(n) = n*(2*n^2 + 1)/3.

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A014820 a(n) = (1/3)*(n^2 + 2*n + 3)*(n+1)^2.

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A142978 Table of figurate numbers for the n-dimensional cross polytopes.

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A142979 a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1) + (n+1)^2*a(n).

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A069038 Expansion of g.f. x*(1+x)^4/(1-x)^6.

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A060524 Triangle read by rows: T(n,k) = number of degree-n permutations with k odd cycles, k=0..n, n >= 0.

A005900 Octahedral numbers: a(n) = n(2n^2 + 1)/3.

A014820 a(n) = (1/3)(n^2 + 2n + 3)*(n+1)^2.

A142979 a(1) = 1, a(2) = 3, a(n+2) = 3a(n+1) + (n+1)^2a(n).

A142988 a(1) = 1, a(2) = 3, a(n+2) = 3a(n+1) + (n + 1)(n + 3)*a(n).

A142984 a(1) = 1, a(2) = 4, a(n+2) = 4a(n+1) + (n + 1)(n + 2)*a(n).

A142985 a(1) = 1, a(2) = 6, a(n+2) = 6a(n+1) + (n + 1)(n + 2)*a(n).