A193534 -id:A193534 - cp's OEIS frontend

A008544 Triple factorial numbers: Product_{k=0..n-1} (3*k+2).

1, 2, 10, 80, 880, 12320, 209440, 4188800, 96342400, 2504902400, 72642169600, 2324549427200, 81359229952000, 3091650738176000, 126757680265216000, 5577337931669504000, 262134882788466688000

Offset: 0

Joe Keane (jgk(AT)jgk.org)

a(n-1), n >= 1, enumerates increasing plane (aka ordered) trees with n vertices (one of them a root labeled 1) where each vertex with outdegree r >= 0 comes in r+1 types (like an (r+1)-ary vertex). See the increasing tree comments under A004747. - Wolfdieter Lang, Oct 12 2007
An example for the case of 3 vertices is shown below. For the enumeration of non-plane trees of this type see A029768. - Peter Bala, Aug 30 2011
a(n) is the product of the positive integers k <= 3*n that have k modulo 3 = 2. - Peter Luschny, Jun 23 2011
See A094638 for connections to differential operators. - Tom Copeland, Sep 20 2011
Partial products of A016789. - Reinhard Zumkeller, Sep 20 2013
The Mathar conjecture is true. Generally from the factorial form, the last term is the "extra" product beyond the prior term, from k=n-1 and 3k+2 evaluates to 3*(n-1)+2 = 3n-1, yielding a(n) = a(n-1)*(3n-1) (eqn1). Similarly, a(n) = a(n-2)*(3n-1)*(3(n-2)+2) = a(n-2)*(3n-1)*(3n-4) (eqn2) and a(n) = a(n-3)*(3n-1)*(3n-4)*(3*(n-2)+2) = a(n-3)*(3n-1)*(3n-4)*(3n-7) (eqn3). We equate (eqn2) and (eqn3) to get a(n-2)*(3n-1)*(3n-4) = a(n-3)*(3n-1)*(3n-4)*(3n-7) or a(n-2)+(7-3n)*a(n-3) = 0 (eqn4). From (eqn1) we have a(n)+(1-3n)*a(n-1) = 0 (eqn5). Combining (eqn4) and (eqn5) yields a(n)+(1-3n)*a(n-1)+a(n-2)+(7-3n)*a(n-3) = 0. - Bill McEachen, Jan 01 2016
a(n-1), n>=1, is the dimension of the n-th component of the operad encoding the multilinearization of the following identity in nonassociative algebras: s*(a,a,b)-(s+t)*(a,b,a)+t*(b,a,a)=0, for any given pair of scalars (s,t). Here (a,b,c) is the associator (ab)c-a(bc). This is proved in the referenced article on associator dependent algebras by Bremner and me. - Vladimir Dotsenko, Mar 22 2022

			a(2) = 10 from the described trees with 3 vertices: there are three trees with a root vertex (label 1) with outdegree r=2 (like the three 3-stars each with one different ray missing) and the four trees with a root (r=1 and label 1) a vertex with (r=1) and a leaf (r=0). Assigning labels 2 and 3 yields 2*3+4=10 such trees.
a(2) = 10. The 10 possible plane increasing trees on 3 vertices, where vertices of outdegree 1 come in 2 colors (denoted a or b) and vertices of outdegree 2 come in 3 colors (a, b or c), are:
.
   1a    1b    1a    1b        1a       1b       1c
   |     |     |     |        / \      / \      / \
   2a    2b    2b    2a      2   3    2   3    2   3
   |     |     |     |
   3     3     3     3         1a       1b       1c
                              / \      / \      / \
                             3   2    3   2    3   2

T. D. Noe, Table of n, a(n) for n = 0..100
Murray Bremner and Vladimir Dotsenko, Associator dependent algebras and Koszul duality, arXiv:2203.11142 [math.RA], 2022.
Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seq., Vol. 3 (2000), Article 00.2.4.
Keiichi Shigechi, On the lattice of weighted partitions, arXiv:2212.14666 [math.CO], 2022. See p. 27.
Garrett Southwood and Hua Wang, The Statistics and Combinatorics of Increasing Trees and Colored Increasing Trees, J. Combin. Math. Combin. Comput. (2024) Vol. 123, 475-487. See p. 485.

a(n) = A004747(n+1, 1) (first column of triangle). Cf. A051141.
Cf. A000165, A001813, A047055, A047657, A084947, A084948, A084949.
Cf. A029768, A034724, A049308.
Cf. A024396, A193534.
Cf. A225470, A290596 (first columns).
Subsequence of A007661.

a008544 n = a008544_list !! n
a008544_list = scanl (*) 1 a016789_list
-- Reinhard Zumkeller, Sep 20 2013

[Round((Gamma(2*n-5/3)/Gamma(n-5/6)*Gamma(2/3)/Gamma(5/6) )/ Sqrt(3)*3^n/4^(n-1)): n in [1..20]]; // Vincenzo Librandi, Feb 21 2015

[Round(3^n*Gamma(n+2/3)/Gamma(2/3)): n in [0..20]]; // G. C. Greubel, Mar 31 2019

a := n -> mul(3*k-1, k = 1..n);
A008544 := n -> mul(k, k = select(k-> k mod 3 = 2, [$1 .. 3*n])): seq(A008544(n), n = 0 .. 16); # Peter Luschny, Jun 23 2011

k = 3; b[1]=2; b[n_]:= b[n] = b[n-1]+k; a[0]=1; a[1]=2; a[n_]:= a[n] = a[n-1]*b[n]; Table[a[n], {n,0,20}] (* Roger L. Bagula, Sep 17 2008 *)
Product[3 k + 2, {k, 0, # - 1}] & /@ Range[0, 16] (* Michael De Vlieger, Jan 02 2016 *)
Table[3^n*Pochhammer[2/3, n], {n,0,20}] (* G. C. Greubel, Mar 31 2019 *)

a(n):=((n)!*sum(binomial(k,n-k)*binomial(n+k,k)*3^(-n+k)*(-1)^(n-k),k,floor(n/2),n)); /* Vladimir Kruchinin, Sep 28 2013 */

PARI
```
a(n) = prod(k=0,n-1, 3*k+2 );
```

vector(20, n, n--; round(3^n*gamma(n+2/3)/gamma(2/3))) \\ G. C. Greubel, Mar 31 2019

@CachedFunction
def A008544(n): return 1 if n == 0 else (3*n-1)*A008544(n-1)
[A008544(n) for n in (0..16)]  # Peter Luschny, May 20 2013

[3^n*rising_factorial(2/3, n) for n in (0..20)] # G. C. Greubel, Mar 31 2019

a(n) = Product_{k=0..n-1} (3*k+2) = A007661(3*n-1) (with A007661(-1) = 1).
E.g.f.: (1-3*x)^(-2/3).
a(n) = 2*A034000(n), n >= 1, a(0) = 1.
a(n) ~ 2^(1/2)*Pi^(1/2)*Gamma(2/3)^-1*n^(1/6)*3^n*e^-n*n^n*{1 - 1/36*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 22 2001
a(n) = (Gamma(2*n-5/3)/Gamma(n-5/6)*Gamma(2/3)/Gamma(5/6))/sqrt(3)*3^n/4^(n-1). - Jeremy L. Martin, Mar 31 2002 (typo fixed by Vincenzo Librandi, Feb 21 2015)
From Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003: (Start)
a(n) = A084939(n)/A000142(n)*A000079(n).
a(n) = 3^n*Pochhammer(2/3, n) = 3^n*Gamma(n+2/3)/Gamma(2/3). (End)
Let T = A094638 and c(t) = column vector(1, t, t^2, t^3, t^4, t^5,...), then A008544 = unsigned [ T * c(-3) ] and the list partition transform A133314 of [1,T * c(-3)] gives [1,T * c(3)] with all odd terms negated, which equals a signed version of A007559; i.e., LPT[(1,signed A008544)] = signed A007559. Also LPT[A007559] = (1,-A008544) and e.g.f. [1,T * c(t)] = (1-x*t)^(-1/t) for t = 3 or -3. Analogous results hold for the double factorial, quadruple factorial and so on. - Tom Copeland, Dec 22 2007
G.f.: 1/(1-2x/(1-3x/(1-5x/(1-6x/(1-8x/(1-9x/(1-11x/(1-12x/(1-...))))))))) (continued fraction). - Philippe Deléham, Jan 08 2012
a(n) = (-1)^n*Sum_{k=0..n} 3^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
G.f.: 1/Q(0) where Q(k) = 1 - x*(3*k+2)/(1 - x*(3*k+3)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 20 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(3*k+2)/(x*(3*k+2) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 25 2013
D-finite with recurrence: a(n) = (9*(n-2)*(n-1)+2)*a(n-2) + 4*a(n-1), n>=2. - Ivan N. Ianakiev, Aug 09 2013
a(n) = n!*Sum_{k=floor(n/2)..n} binomial(k,n-k)*binomial(n+k,k)*3^(-n+k)*(-1)^(n-k). - Vladimir Kruchinin, Sep 28 2013
Recurrence equation: a(n) = 3*a(n-1) + (3*n - 4)^2*a(n-2) with a(0) = 1 and a(1) = 2. A024396 satisfies the same recurrence (but with different initial conditions). This observation leads to a continued fraction expansion for the constant A193534 due to Euler. - Peter Bala, Feb 20 2015
a(n) = A225470(n, 0), n >= 0. - Wolfdieter Lang, May 29 2017
G.f.: Hypergeometric2F0(1, 2/3; -; 3*x). - G. C. Greubel, Mar 31 2019
D-finite with recurrence: a(n) + (-3*n+1)*a(n-1)=0. - R. J. Mathar, Jan 17 2020
G.f.: 1/(1-2*x-6*x^2/(1-8*x-30*x^2/(1-14*x-72*x^2/(1-20*x-132*x^2/(1-...))))) (Jacobi continued fraction). - Nikolaos Pantelidis, Feb 28 2020
G.f.: 1/G(0), where G(k) = 1 - (6*k+2)*x - 3*(k+1)*(3*k+2)*x^2/G(k+1). - Nikolaos Pantelidis, Feb 28 2020
Sum_{n>=0} 1/a(n) = 1 + (e/3)^(1/3) * (Gamma(2/3) - Gamma(2/3, 1/3)). - Amiram Eldar, Mar 01 2022

A181048 Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+1).

Original entry on oeis.org

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

Offset: 0

Jonathan D. B. Hodgson, Oct 01 2010, Oct 06 2010

			0.86697298733991103757399516388287071365217536734524490433....
At N = 100000 the truncated series 2*Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 1) = 1.7339(3)5974(5)7982(5)075(25)79(846)27(404)7... to 32 digits. The bracketed numbers show where this decimal expansion differs from that of 2*A181048. The numbers 1, 1, -3, -11, 57, 361 must be added to the bracketed numbers to give the correct decimal expansion to 32 digits: 2*( (log(1 + sqrt(2)) + Pi/2)/(2*sqrt(2)) ) = 1.7339(4)5974(6)7982(2)075(14)79(903)27(765)7.... - _Peter Bala_, Sep 23 2016

Jolley, Summation of Series, Dover (1961) eq 82 page 16.
Murray R. Spiegel, Seymour Lipschutz, John Liu. Mathematical Handbook of Formulas and Tables, 3rd Ed. Schaum's Outline Series. New York: McGraw-Hill (2009): p. 135, equation 21.17

Ivan Panchenko, Table of n, a(n) for n = 0..1000
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Eric W. Weisstein, Euler's Series Transformation.
Herbert S. Wilf, Accelerated series for universal constants, by the WZ method, Discrete Mathematics & Theoretical Computer Science, Vol 3, No 4 (1999).

Cf. A001586, A003881, A024396, A091648, A093954, A113476, A181049, A181122, A188458, A193534, A262246.

RealDigits[(Pi Sqrt[2])/8 + (Sqrt[2] Log[1 + Sqrt[2]])/4, 10, 100][[1]] (* Alonso del Arte, Aug 11 2011 *)

(log(1+sqrt(2))+Pi/2)/(2*sqrt(2)) \\ G. C. Greubel, Jul 05 2017

(asinh(1)+Pi/2)/sqrt(8) \\ Charles R Greathouse IV, Jul 06 2017

Equals (A093954 + A091648/sqrt(2))/2.
Integral_{x = 0..1} 1/(1+x^4) = Sum_{k >= 0} (-1)^k/(4*k+1) = (log(1+sqrt(2)) + Pi/2)/(2*sqrt(2)).
1 - 1/5 + 1/9 - 1/13 + 1/17 - ... = (Pi*sqrt(2))/8 + (sqrt(2)*log(1 + sqrt(2)))/4 = (Pi + 2*log(1 + sqrt(2)))/(4 sqrt(2)). The first two are the formulas as given in Spiegel et al., the third is how Mathematica rewrites the infinite sum. - Alonso del Arte, Aug 11 2011
Let N be a positive integer divisible by 4. We have the asymptotic expansion 2*( (log(1 + sqrt(2)) + Pi/2)/(2*sqrt(2)) - Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 1) ) ~ 1/N + 1/N^2 - 3/N^3 - 11/N^4 + 57/N^5 + 361/N^6 - ..., where the sequence of coefficients [1, 1, -3, -11, 57, 361, ...] is A188458. This follows from Borwein et al., Lemma 2 with f(x) = 1/x and then set x = N/4 and h = 1/4. An example is given below. Cf. A181049. - Peter Bala, Sep 23 2016
Equals Sum_{n >= 0} 2^(n-1)*n!/(Product_{k = 0..n} 4*k + 1) = Sum_{n >= 0} 2^(n-1)*n!/A007696(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(4*k + 1)). - Peter Bala, Dec 01 2021
From Peter Bala, Oct 23 2023: (Start)
The slowly converging series representation Sum_{n >= 0} (-1)^n/(4*n + 1) for the constant can be accelerated to give the following faster converging series:
1/2 + 2*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5));
7/10 + 8*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9));
71/90 + 48*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9)*(4*n + 13));
971/1170 + 384*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9)*(4*n + 13)*(4*n + 17)).
These results may be easily verified by taking the partial fraction expansions of the summands. The general result appears to be that for r >= 0, the constant equals
C(r) + (2^r)*r!*Sum_{n >= 0} (-1)^n/((4*n + 1)*(4*n + 5)*...*(4*n + 4*r + 1)), where C(r) is the rational number Sum_{k = 0..r-1} 2^(k-1)*k!/(1*5*9*...*(4*k + 1)). [added 19 Feb 2024: the general result can be proved by the WZ method as described in Wilf.]
In the limit as r -> oo we find that the constant equals Sum_{k >= 0} 2^(k-1)*k!/(Product_{i = 0..k} 4*i + 1) as noted above. (End)
From Peter Bala, Mar 03 2024: (Start)
Continued fraction: 1/(1 + 1^2/(4 + 5^2/(4 + 9^2/(4 + 13^2/(4 + ... ))))) due to Euler.
Equals hypergeom([1/4, 1], [5/4], -1).
Gauss's continued fraction: 1/(1 + 1^2/(5 + 4^2/(9 + 5^2/(13 + 8^2/(17 + 9^2/(21 + 12^2/(25 + 13^2/(29 + 16^2/(33 + 17^2/(37 + ... )))))))))). (End)

A024396 a(n) = ( Product {k = 1..n} 3k - 1 ) ( Sum {k = 1..n} (-1)^(k+1)/(3*k - 1) ).

Original entry on oeis.org

1, 3, 34, 294, 4996, 72612, 1661680, 34029840, 981118240, 25947526560, 902963019520, 29279156256000, 1193967167680000, 45861003136704000, 2144641818280192000, 95220827527499520000, 5023176259163442688000, 253121597596239128064000, 14869466904778827894784000

Offset: 1

Clark Kimberling

Original name: s(1)*s(2)*...*s(n)*(1/s(1) - 1/s(2) + ... + c/s(n)), where c = (-1)^(n+1) and s(k) = 3k-1 for k = 1,2,3,...

Harvey P. Dale, Table of n, a(n) for n = 1..381
L. Euler, De fractionibus continuis observationes, The Euler Archive, Index Number 123, Section 5

Cf. A008544, A024217, A193534.

I:=[1,3]; [n le 2 select I[n] else 3*Self(n-1)+(3*n-4)^2*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 21 2015

a[1] := 1: a[2] := 3: for n from 3 to 20 do a[n] := 3*a[n-1]+(3*n-4)^2*a[n-2] end do: seq(a[n], n = 1 .. 20); # Peter Bala, Feb 20 2015

Table[Product[3*k-1,{k,1,n}] * Sum[(-1)^(k+1)/(3*k-1),{k,1,n}],{n,1,20}] (* Vaclav Kotesovec, Feb 21 2015 *)
nxt[{n_,a_,b_}]:={n+1,b,3b+a*(3n-1)^2}; NestList[nxt,{2,1,3},20][[;;,2]] (* Harvey P. Dale, Jun 07 2023 *)

From Peter Bala, Feb 20 2015: (Start)
Recurrence: a(n+1) = 3*a(n) + (3*n - 1)^2*a(n-1) with a(1) = 1 and a(2) = 3.
The triple factorial numbers A008544 satisfy the same second-order recurrence equation. This leads to the continued fraction representation a(n)/A008544(n) = 1/(2 + 2^2/(3 + 5^2/(3 + 8^2/(3 + ... + (3*n - 1)^2/(3 )))).
Taking the limit as n -> infinity gives the generalized continued fraction: Sum {k >= 1} (-1)^(k+1)/(3*k - 1) = 1/(2 + 2^2/(3 + 5^2/(3 + 8^2/(3 + 11^2/(3 + ... ))))) due to Euler. The alternating sum has the value 1/3*( Pi/sqrt(3) - log(2) ) = A193534. Cf. A024217. (End)
a(n) ~ GAMMA(1/3) * 3^(n-1) * n^(n+1/6) * (Pi - sqrt(3)*log(2)) / (sqrt(2*Pi) * exp(n)). - Vaclav Kotesovec, Feb 21 2015

New name from Peter Bala, Feb 20 2015

a(18)-a(19) from Vincenzo Librandi, Feb 21 2015

A113476 Decimal expansion of (log(2) + Pi/sqrt(3))/3.

Original entry on oeis.org

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

Offset: 0

Benoit Cloitre, Jan 08 2006

This number is transcendental - this follows from a result of Baker (1968) on linear forms of algebraic numbers.

			0.835648848264721053337... = A073010 + A193535.

Jolley, Summation of Series, Dover (1961), eq (79) page 16.
Murray R. Spiegel, Seymour Lipschutz, John Liu. Mathematical Handbook of Formulas and Tables, 3rd Ed. Schaum's Outline Series. New York: McGraw-Hill (2009): p. 135, equation 21.16

Ivan Panchenko, Table of n, a(n) for n = 0..1000
A. Baker, Linear forms in the logarithms of algebraic numbers (IV). Mathematika, 15 (1968) pp. 204-216
L. Euler, De fractionibus continuis observationes, The Euler Archive, Index Number 123, Section 5
Eric W. Weisstein, Euler's Series Transformation.
Index entries for transcendental numbers

Cf. A007559, A016777, A024217, A073010, A193534, A193535.

RealDigits[(Log[2]+\[Pi]/Sqrt[3])/3,10,120][[1]]  (* Harvey P. Dale, Mar 26 2011 *)

PARI
```
1/3*(log(2)+Pi/sqrt(3))
```

Equals Integral_{x = 0..1} dx/(1+x^3) = Sum_{k >= 0} (-1)^k/(3*k+1) = 1 - 1/4 + 1/7 - 1/10 + 1/13 - 1/16 + ... (see A016777). - Benoit Cloitre, Alonso del Arte, Jul 29 2011
Generalized continued fraction: 1/(1 + 1^2/(3 + 4^2/(3 + 7^2/(3 + 10^2/(3 + ... ))))) due to Euler. For a sketch proof see A024217. - Peter Bala, Feb 22 2015
Equals (1/2)*Sum_{n >= 0} n!*(3/2)^n/(Product_{k = 0..n} 3*k + 1) = (1/2)*Sum_{n >= 0} n!*(3/2)^n/A007559(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(3*k + 1)). - Peter Bala, Dec 01 2021
From Peter Bala, Mar 03 2024: (Start)
Equals hypergeom([1/3,  1], [4/3], -1).
Gauss's continued fraction: 1/(1 + 1/(4 + 3^2/(7 + 4^2/(10 + 6^2/(13 + 7^2/(16 + 9^2/(19 + 10^2/(22 + 12^2/(25 + 13^2/(28 + ... )))))))))). (End)
Equals (1/12) * Sum_{n >= 0} (-1/2)^n * (9*n + 7)/((3*n + 2)*(n + 1)*binomial(2*n+1/3, n+1)). - Peter Bala, Mar 05 2025

A258969 E.g.f.: A'(x) = 1 + A(x)^3, with A(0)=1.

Original entry on oeis.org

1, 2, 6, 42, 390, 4698, 69174, 1203498, 24163110, 549811962, 13982486166, 393026414922, 12099527531910, 404881353252378, 14632253175107574, 567974815524008298, 23567351945550373350, 1040985881615266375482, 48767788927611416600406, 2415210691383917131432842

Offset: 0

Vaclav Kotesovec, Jun 15 2015

nonn

Conjecture: A227250(n+1) = a(n).

			A(x) = 1 + 2*x + 6*x^2/2! + 42*x^3/3! + 390*x^4/4! + 4698*x^5/5! + ...
A'(x) = 2 + 6*x + 21*x^2 + 65*x^3 + 783*x^4/4 + 11529*x^5/20 + ...
1 + A(x)^3 = 2 + 6*x + 21*x^2 + 65*x^3 + 783*x^4/4 + 11529*x^5/20 + ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..200

Cf. A000831, A024396, A193534, A227250, A258970, A258971, A258880, A258994.

nmax=20; Subscript[a,0]=1; egf=Sum[Subscript[a,k]*x^k, {k,0,nmax+1}]; Table[Subscript[a,k]*k!, {k,0,nmax}] /.Solve[Take[CoefficientList[Expand[1+egf^3-D[egf,x]],x],nmax]==ConstantArray[0,nmax]][[1]]

{a(n) = local(A=1); A = 1 + serreverse( intformal( 1/((2+x)*(1+x+x^2) +x*O(x^n)) )); n!*polcoeff(A,n)}
for(n=0,25,print1(a(n),", ")) \\ Paul D. Hanna, Jun 16 2015

a(n) ~ (3/(Pi/sqrt(3)-log(2)))^(n+1/2) * n^n / exp(n).

E.g.f.: 1 + Series_Reversion( Integral 1/((2+x)*(1+x+x^2)) dx ). - Paul D. Hanna, Jun 16 2015

A181122 Decimal expansion of Sum_{k>=0} (-1)^k/(5k+1).

Original entry on oeis.org

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

Offset: 0

Jonathan D. B. Hodgson, Oct 05 2010

			0.88831357265178863804075522702037934627811083077545817120597...

Gheorghe Coserea, Table of n, a(n) for n = 0..1000
H. Wilf, Accelerated series for universal constants, by the WZ method, Discrete Mathematics and Theoretical Computer Science 3(4) (1999), 189-192.

Cf. A002162, A003881, A024396, A113476, A181048, A181049, A181122, A193534, A262246.

(int(1/(1+x^5),x=0..1));
evalf(LerchPhi(-1,1,1/5)/5) ; # R. J. Mathar, Oct 16 2011

(Sqrt[8 + 8/Sqrt[5]]*Pi + 2*Sqrt[5]*ArcCoth[3/Sqrt[5]] + Log[16])/20 // RealDigits[#, 10, 105]& // First (* Jean-François Alcover, Feb 13 2013 *)

default(realprecision, 106);
eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(5*n+1)))), "3..-2")) \\ Gheorghe Coserea, Oct 06 2015

Sum_{k>=0} (-1)^k/(5k+1) = Integral_{x=0..1}dx/(1+x^5) = (1/10)*sqrt(10-2*sqrt(5))*arctan((3/4)*sqrt(10-2*sqrt(5)) + (1/4)*sqrt(10-2*sqrt(5))*sqrt(5)) + (1/20)*sqrt(10-2*sqrt(5))*arctan(-(1/4)*sqrt(10-2*sqrt(5)) + (1/4)*sqrt(10-2*sqrt(5))*sqrt(5)) + (1/20)*sqrt(10-2*sqrt(5))*sqrt(5)*arctan(-(1/4)*sqrt(10-2*sqrt(5)) + (1/4)*sqrt(10-2*sqrt(5))*sqrt(5)) + (1/20)*log(2)*sqrt(5) + (1/5)*log(2) - (1/20)*log(7-3*sqrt(5))*sqrt(5).
Equals Pi*sqrt(phi)/5^(5/4) + log(phi)/sqrt(5) + log(2)/5, where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Nov 01 2015
From Peter Bala, Feb 19 2024: (Start)
Equals (1/2)*Sum_{n >= 0} n!*(5/2)^n/(Product_{k = 0..n} 5*k + 1) = (1/2)*Sum_{n >= 0} n!*(5/2)^n/A008548(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(5*k + 1)).
Continued fraction: 1/(1 + 1^2/(5 + 6^2/(5 + 11^2/(5 + ... + (5*n + 1)^2/(5 + ... ))))).
The slowly converging series representation Sum_{n >= 0} (-1)^n/(5*n + 1) for the constant can be accelerated to give the following faster converging series:
1/2 + (5/2)*Sum_{n >= 0} (-1)^n/((5*n + 1)(5*n + 6)) and
17/24 + (25/2)*Sum_{n >= 0} (-1)^n/((5*n + 1)(5*n + 6)*(5*n + 11)).
These two series are the cases r = 1 and r = 2 of the general result: for r >= 0, the constant equals
C(r) + ((5/2)^r)*r!*Sum_{n >= 0} (-1)^n/((5*n + 1)*(5*n + 6)*...*(5*n + 5*r + 1)), where C(r) is the rational number (1/2)*Sum_{k = 0..r-1} (5/2)^k*k!/(1*6*11*...*(5*k + 1)). The general result can be proved by the WZ method as described in Wilf. (End)
From Peter Bala, Mar 03 2024: (Start)
Equals hypergeom([1/5, 1], [6/5], -1).
Gauss's continued fraction: 1/(1 + 1^2/(6 + 5^2/(11 + 6^2/(16 + 10^2/(21 + 11^2/(26 + 15^2/(31 + 16^2/(36 + 20^2/(41 + 21^2/(46 + ... )))))))))). (End)

A262246 Decimal expansion of Sum_{k>=0} (-1)^k/(5k+2).

Original entry on oeis.org

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

Offset: 0

Gheorghe Coserea, Oct 06 2015

			0.4069016342...

Gheorghe Coserea, Table of n, a(n) for n = 0..2015
H. Wilf, Accelerated series for universal constants, by the WZ method, Discrete Mathematics and Theoretical Computer Science 3(4) (1999), 189-192.

Cf. A003881, A024396, A113476, A181048, A181049, A181122, A193534.

N[(1/5)*((Sqrt[5]-1)*Log[2] + Sqrt[5]*Log[Sin[3*Pi/10]] + (Pi/2)*Sec[Pi/10]), 100] (* G. C. Greubel, Oct 07 2015 *) (* fixed by Vaclav Kotesovec, Dec 11 2017 *)

default(realprecision, 87);
eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(5*n+2)))), "3..-2"))

Sum_{n>=0} (-1)^n/(5n+2) = Integral_{x=0..1} x/(1+x^5)dx.
From G. C. Greubel, Oct 07 2015: (Start)
Sum_{n>=0} (-1)^n/(5n+2) = (1/5)*(sqrt(5)*log(phi) - log(2) + Pi*(5*phi^2)^(-1/4)), where 2*phi=1+sqrt(5).
Sum_{n>=0} (-1)^n/(5n+2) = (1/5)*(sqrt(5)*log(2*sin(3*Pi/10)) - log(2) + (Pi/2)*sec(Pi/10)).
(End)
Sum_{n>=0} (-1)^n/(5n+2) = (Psi(1/5) - Psi(7/10))/10 , see A200135 and A354643. - Robert Israel, Oct 08 2015
From Peter Bala, Feb 19 2024: (Start)
Equals (1/2)*Sum_{n >= 0} n!*(5/2)^n/(Product_{k = 0..n} 5*k + 2) = (1/2)*Sum_{n >= 0} n!*(5/2)^n/A047055(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(5*k + 2)).
Continued fraction: 1/(2 + 2^2/(5 + 7^2/(5 + 12^2/(5 + ... + (5*n + 2)^2/(5 + ... ))))).
The slowly converging series representation Sum_{n >= 0} (-1)^n/(5*n + 2) for the constant can be accelerated to give the following faster converging series
1/4 + (5/2)*Sum_{n >= 0} (-1)^n/((5*n + 2)*(5*n + 7)) and
19/56 + (5^2/2)*Sum_{n >= 0} (-1)^n/((5*n + 2)*(5*n + 7)*(5*n + 12)).
These two series are the cases r = 1 and r = 2 of the general result:
for r >= 0, the constant equals C(r) + ((5/2)^r)*r!*Sum_{n >= 0} (-1)^n/((5*n + 2)*(5*n + 7)*...*(5*n + 5*r + 2)), where C(r) is the rational number (1/2)*Sum_{k = 0..r-1} (5/2)^k*k!/(2*7*12*...*(5*k + 2)). The general result can be proved by the WZ method as described in Wilf. (End)
From Peter Bala, Mar 03 2024: (Start)
Equals (1/2)*hypergeom([2/5, 1], [7/5], -1).
Gauss's continued fraction: 1/(2 + 2^2/(7 + 5^2/(12 + 7^2/(17 + 10^2/(22 + 12^2/(27 + 15^2/(32 + 17^2/(37 + 20^2/(42 + 22^2/(47 + ... )))))))))). (End)

cp's OEIS Frontend

A008544 Triple factorial numbers: Product_{k=0..n-1} (3*k+2).

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A181048 Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2*sqrt(2)) = Sum_{k>=0} (-1)^k/(4*k+1).

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A024396 a(n) = ( Product {k = 1..n} 3*k - 1 ) * ( Sum {k = 1..n} (-1)^(k+1)/(3*k - 1) ).

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A113476 Decimal expansion of (log(2) + Pi/sqrt(3))/3.

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A258969 E.g.f.: A'(x) = 1 + A(x)^3, with A(0)=1.

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A181122 Decimal expansion of Sum_{k>=0} (-1)^k/(5k+1).

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A181048 Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+1).

A024396 a(n) = ( Product {k = 1..n} 3k - 1 ) ( Sum {k = 1..n} (-1)^(k+1)/(3*k - 1) ).

A196548 Decimal expansion of Sum_{i>=0} 1/((6i+2)(6*i+5)).