A002391 -id:A002391 - cp's OEIS frontend

A004736 Triangle read by rows: row n lists the first n positive integers in decreasing order.

1, 2, 1, 3, 2, 1, 4, 3, 2, 1, 5, 4, 3, 2, 1, 6, 5, 4, 3, 2, 1, 7, 6, 5, 4, 3, 2, 1, 8, 7, 6, 5, 4, 3, 2, 1, 9, 8, 7, 6, 5, 4, 3, 2, 1, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1

Offset: 1

R. Muller

Old name: Triangle T(n,k) = n-k, n >= 1, 0 <= k < n. Fractal sequence formed by repeatedly appending strings m, m-1, ..., 2, 1.
The PARI functions t1 (this sequence), t2 (A002260) can be used to read a square array T(n,k) (n >= 1, k >= 1) by antidiagonals upwards: n -> T(t1(n), t2(n)). - Michael Somos, Aug 23 2002, edited by M. F. Hasler, Mar 31 2020
A004736 is the mirror of the self-fission of the polynomial sequence (q(n,x)) given by q(n,x) = x^n+  x^(n-1) + ... + x + 1. See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011
Seen as flattened list: a(A000217(n)) = 1; a(A000124(n)) = n and a(m) <> n for m < A000124(n). - Reinhard Zumkeller, Jul 22 2012
Sequence B is called a reverse reluctant sequence of sequence A, if B is triangle array read by rows: row number k lists first k elements of the sequence A in reverse order. Sequence A004736 is the reverse reluctant sequence of sequence 1,2,3,... (A000027). - Boris Putievskiy, Dec 13 2012
The row sums equal A000217(n). The alternating row sums equal A004526(n+1). The antidiagonal sums equal A002620(n+1) respectively A008805(n-1). - Johannes W. Meijer, Sep 28 2013
From Peter Bala, Jul 29 2014: (Start)
Riordan array (1/(1-x)^2,x). Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0  M/
having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the infinite matrix product M(0)*M(1)*M(2)*... is equal to A078812. (End)
T(n, k) gives the number of subsets of [n] := {1, 2, ..., n} with k consecutive numbers (consecutive k-subsets of [n]).  - Wolfdieter Lang, May 30 2018
a(n) gives the distance from (n-1) to the smallest triangular number > (n-1). - Ctibor O. Zizka, Apr 09 2020
To construct the sequence, start from 1,2,,3,,,4,,,,5,,,,,6... where there are n commas after each "n". Then fill the empty places by the sequence itself. - Benoit Cloitre, Aug 17 2021
T(n,k) is the number of cycles of length 2*(k+1) in the (n+1)-ladder graph. There are no cycles of odd length. - Mohammed Yaseen, Jan 14 2023
The first 77 entries are also the signature sequence of log(3)=A002391. Then the two sequences start to differ. - R. J. Mathar, May 27 2024

			The triangle T(n, k) starts:
   n\k  1  2  3  4  5  6  7  8  9 10 11 12 ...
   1:   1
   2:   2  1
   3:   3  2  1
   4:   4  3  2  1
   5:   5  4  3  2  1
   6:   6  5  4  3  2  1
   7:   7  6  5  4  3  2  1
   8:   8  7  6  5  4  3  2  1
   9:   9  8  7  6  5  4  3  2  1
  10:  10  9  8  7  6  5  4  3  2  1
  11:  11 10  9  8  7  6  5  4  3  2  1
  12:  12 11 10  9  8  7  6  5  4  3  2  1
  ... Reformatted. - _Wolfdieter Lang_, Feb 04 2015
T(6, 3) = 4 because the four consecutive 3-subsets of [6] = {1, 2, ..., 6} are {1, 2, 3}, {2, 3, 4}, {3, 4, 5} and {4, 5, 6}. - _Wolfdieter Lang_, May 30 2018

H. S. M. Coxeter, Regular Polytopes, 3rd ed., Dover, NY, 1973, pp 159-162.

Reinhard Zumkeller, Rows n = 1..100 of triangle, flattened
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and Graça Tomaz, Intrinsic Properties of a Non-Symmetric Number Triangle, J. Int. Seq., Vol. 26 (2023), Article 23.4.8.
Glen Joyce C. Dulatre, Jamilah V. Alarcon, Vhenedict M. Florida, and Daisy Ann A. Disu, On Fractal Sequences, DMMMSU-CAS Science Monitor (2016-2017) Vol. 15 No. 2, 109-113.
Joaquín Figueroa, Ivan Gonzalez, and Daniel Salinas-Arizmendi, A Novel Transfer Matrix Framework for Multiple Dirac Delta Potentials, arXiv:2503.23134 [quant-ph], 2025. See pp. 4, 9.
Clark Kimberling, Fractal sequences
Clark Kimberling, Numeration systems and fractal sequences, Acta Arithmetica 73 (1995) 103-117.
Boris Putievskiy, Transformations Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.
F. Smarandache, Sequences of Numbers Involved in Unsolved Problems.
Michael Somos, Sequences used for indexing triangular or square arrays
Eric Weisstein's World of Mathematics, Ladder Graph
Eric Weisstein's World of Mathematics, Smarandache Sequences

Cf. A000217, A002024, A002262, A003056, A025581.
Ordinal transform of A002260. See also A078812.
Cf. A141419 (partial sums per row).
Cf. A134546 (T * A051731, matrix product).
See A001511 for definition of ordinal transform.
Cf. A128174 (parity).

=if(row()>=column();row()-column()+1;"") [Mats Granvik, Jan 19 2009]

a004736 n k = n - k + 1
a004736_row n = a004736_tabl !! (n-1)
a004736_tabl = map reverse a002260_tabl
-- Reinhard Zumkeller, Aug 04 2014, Jul 22 2012

A004736 := proc(n,m) n-m+1 ; end:
T := (n, k) -> n-k+1: seq(seq(T(n,k), k=1..n), n=1..13); # Johannes W. Meijer, Sep 28 2013

Flatten[ Table[ Reverse[ Range[n]], {n, 12}]] (* Robert G. Wilson v, Apr 27 2004 *)
Table[Range[n,1,-1],{n,20}]//Flatten (* Harvey P. Dale, May 27 2020 *)

{a(n) = 1 + binomial(1 + floor(1/2 + sqrt(2*n)), 2) - n}

{t1(n) = binomial( floor(3/2 + sqrt(2*n)), 2) - n + 1} /* A004736 */

{t2(n) = n - binomial( floor(1/2 + sqrt(2*n)), 2)} /* A002260 */

apply( A004736(n)=1-n+(n=sqrtint(8*n)\/2)*(n+1)\2, [1..99]) \\ M. F. Hasler, Mar 31 2020

def agen(rows):
    for n in range(1, rows+1): yield from range(n, 0, -1)
print([an for an in agen(13)]) # Michael S. Branicky, Aug 17 2021

from math import comb, isqrt
def A004736(n): return comb((m:=isqrt(k:=n<<1))+(k>m*(m+1))+1,2)+1-n # Chai Wah Wu, Nov 08 2024

a(n+1) = 1 + A025581(n).
a(n) = (2 - 2*n + round(sqrt(2*n)) + round(sqrt(2*n))^2)/2. - Brian Tenneson, Oct 11 2003
G.f.: 1 / ((1-x)^2 * (1-x*y)). - Ralf Stephan, Jan 23 2005
Recursion: e(n,k) = (e(n - 1, k)*e(n, k - 1) + 1)/e(n - 1, k - 1). - Roger L. Bagula, Mar 25 2009
a(n) = (t*t+3*t+4)/2-n, where t = floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Dec 13 2012
From Johannes W. Meijer, Sep 28 2013: (Start)
T(n, k) = n - k + 1, n >= 1 and 1 <= k <= n.
T(n, k) = A002260(n+k-1, n-k+1). (End)
a(n) = A000217(A002024(n)) - n + 1. - Enrique Pérez Herrero, Aug 29 2016

New name from Omar E. Pol, Jul 15 2012

A002162 Decimal expansion of the natural logarithm of 2.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

Newton calculated the first 16 terms of this sequence.
Area bounded by y = tan x, y = cot x, y = 0. - Clark Kimberling, Jun 26 2020
Choose four values independently and uniformly at random from the unit interval [0,1]. Sort them, and label them a,b,c,d from least to greatest (so that a b^2+c^2. - Akiva Weinberger, Dec 02 2024
Define the trihyperboloid to be the intersection of the three solid hyperboloids x^2+y^2-z^2<1, x^2-y^2+z^2<1, and -x^2+y^2+z^2<1. This fits perfectly within the cube [-1,1]^3. Then this is the ratio of the volume of the trihyperboloid to its bounding cube. - Akiva Weinberger, Dec 02 2024

			0.693147180559945309417232121458176568075500134360255254120680009493393...

G. Boros and V. H. Moll, Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals, Cambridge University Press, 2004.
Calvin C. Clawson, Mathematical Mysteries: The Beauty and Magic of Numbers, Springer, 2013. See p. 227.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 24, 250.
Steven R. Finch, Mathematical Constants, Cambridge, 2003, Sections 1.3.3, 2.21, 6.2, and 7.2.
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).
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 25 and appendix A, equations 25:14:3 and A:7:3 at pages 232, 670.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 29.

Harry J. Smith, Table of n, a(n) for n = 0..20000
D. H. Bailey and J. M. Borwein, Experimental Mathematics: Examples, Methods and Implications, Notices of the AMS, May 2005, Volume 52, Issue 5.
Peter Bala, New series for old functions
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Paul Cooijmans, Odds.
X. Gourdon and P. Sebah, The logarithm constant:log(2), 2001.
X. Gourdon and P. Sebah, The logarithm constant:log(2), 2004. (Revised and extended version of above article.)
M. Kontsevich and D. Zagier, Periods, pp. 4-5.
Mathematical Reflections, Solution to Problem U376, Issue 4, 2016, p 17.
Isaac Newton, The method of fluxions and infinite series; with its application to the geometry of curve-lines, 1736; see p. 96.
Michael Penn, an alternating floor sum., YouTube video, 2020.
Michael Penn, A wonderful limit from the 2011 Virginia Tech Regional Math Competition, YouTube video, 2022.
Simon Plouffe, log(2), natural logarithm of 2 to 2000 places.
S. Ramanujan, Question 260, J. Ind. Math. Soc., III, p. 43.
Albert Stadler, Problem 3567, Crux Mathematicorum, Vol. 36 (Oct. 2010), p. 396; Oliver Geupel, Solution, Crux Mathematicorum, Vol. 37 (Oct. 2011), pp. 400-401.
D. W. Sweeney, On the computation of Euler's constant, Math. Comp., 17 (1963), 170-178.
Horace S. Uhler, Recalculation and extension of the modulus and of the logarithms of 2, 3, 5, 7 and 17, Proc. Nat. Acad. Sci. U. S. A. 26, (1940). 205-212.
Eric Weisstein's World of Mathematics, Natural Logarithm of 2, Masser-Gramain Constant, Logarithmic Integral, Dirichlet Eta Function.
Wikipedia, Natural logarithm of 2.
Wikipedia, Period (algebraic geometry).
Index entries for transcendental numbers.

Cf. A016730 (continued fraction), A002939, A008288, A142979, A142992.

RealDigits[N[Log[2],200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Feb 21 2011 *)
RealDigits[Log[2],10,120][[1]] (* Harvey P. Dale, Jan 25 2024 *)

{ default(realprecision, 20080); x=10*log(2); for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b002162.txt", n, " ", d)); } \\ Harry J. Smith, Apr 21 2009

log(2) = Sum_{k>=1} 1/(k*2^k) = Sum_{j>=1} (-1)^(j+1)/j.
log(2) = Integral_{t=0..1} dt/(1+t).
log(2) = (2/3) * (1 + Sum_{k>=1} 2/((4*k)^3-4*k)) (Ramanujan).
log(2) = 4*Sum_{k>=0} (3-2*sqrt(2))^(2*k+1)/(2*k+1) (Y. Luke). - R. J. Mathar, Jul 13 2006
log(2) = 1 - (1/2)*Sum_{k>=1} 1/(k*(2*k+1)). - Jaume Oliver Lafont, Jan 06 2009, Jan 08 2009
log(2) = 4*Sum_{k>=0} 1/((4*k+1)*(4*k+2)*(4*k+3)). - Jaume Oliver Lafont, Jan 08 2009
log(2) = 7/12 + 24*Sum_{k>=1} 1/(A052787(k+4)*A000079(k)). - R. J. Mathar, Jan 23 2009
From Alexander R. Povolotsky, Jul 04 2009: (Start)
log(2) = (1/4)*(3 - Sum_{n>=1} 1/(n*(n+1)*(2*n+1))).
log(2) = (230166911/9240 - Sum_{k>=1} (1/2)^k*(11/k + 10/(k+1) + 9/(k+2) + 8/(k+3) + 7/(k+4) + 6/(k+5) - 6/(k+7) - 7/(k+8) - 8/(k+9) - 9/(k+10) - 10/(k+11)))/35917. (End)
log(2) = A052882/A000670. - Mats Granvik, Aug 10 2009
From log(1-x-x^2) at x=1/2, log(2) = (1/2)*Sum_{k>=1} L(k)/(k*2^k), where L(n) is the n-th Lucas number (A000032). - Jaume Oliver Lafont, Oct 24 2009
log(2) = Sum_{k>=1} 1/(cos(k*Pi/3)*k*2^k) (cf. A176900). - Jaume Oliver Lafont, Apr 29 2010
log(2) = (Sum_{n>=1} 1/(n^2*(n+1)^2*(2*n+1)) + 11)/16. - Alexander R. Povolotsky, Jan 13 2011
log(2) = ((Sum_{n>=1} (2*n+1)/(Sum_{k=1..n} k^2)^2)+396)/576. - Alexander R. Povolotsky, Jan 14 2011
From Alexander R. Povolotsky, Dec 16 2008: (Start)
log(2) = 105*(319/44100 - Sum_{n>=1} 1/(2*n*(2*n+1)*(2*n+3)*(2*n+5)*(2*n+7))).
log(2) = 319/420 - (3/2)*Sum_{n>=1} 1/(6*n^2+39*n+63). (End)
log(2) = Sum_{k>=1} A191907(2,k)/k. - Mats Granvik, Jun 19 2011
log(2) = Integral_{x=0..oo} 1/(1 + e^x) dx. - Jean-François Alcover, Mar 21 2013
log(2) = lim_{s->1} zeta(s)*(1-1/2^(s-1)). - Mats Granvik, Jun 18 2013
From Peter Bala, Dec 10 2013: (Start)
log(2) = 2*Sum_{n>=1} 1/( n*A008288(n-1,n-1)*A008288(n,n) ), a result due to Burnside.
log(2) = (1/3)*Sum_{n >= 0} (5*n+4)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(1/2)^n = (1/12)*Sum_{n >= 0} (28*n+17)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(-1/4)^n.
log(2) = (3/16)*Sum_{n >= 0} (14*n+11)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(1/4)^n = (1/12)*Sum_{n >= 0} (34*n+25)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(-1/18)^n. For more series of this type see the Bala link.
See A142979 for series acceleration formulas for log(2) obtained from the Mercator series log(2) = Sum_{n >= 1} (-1)^(n+1)/n. See A142992 for series for log(2) related to the root lattice C_n. (End)
log(2) = lim_{n->oo} Sum_{k=2^n..2^(n+1)-1} 1/k. - Richard R. Forberg, Aug 16 2014
From Peter Bala, Feb 03 2015: (Start)
log(2) = (2/3)*Sum_{k >= 0} 1/((2*k + 1)*9^k).
Define a pair of integer sequences A(n) = 9^n*(2*n + 1)!/n! and B(n) = A(n)*Sum_{k = 0..n} 1/((2*k + 1)*9^k). Both satisfy the same second-order recurrence equation u(n) = (40*n + 16)*u(n-1) - 36*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion log(2) = (2/3)*(1 + 2/(54 - 36*3^2/(96 - 36*5^2/(136 - ... - 36*(2*n - 1)^2/((40*n + 16) - ... ))))). Cf. A002391, A073000 and A105531 for similar expansions. (End)
log(2) = Sum_{n>=1} (Zeta(2*n)-1)/n. - Vaclav Kotesovec, Dec 11 2015
From Peter Bala, Oct 30 2016: (Start)
Asymptotic expansions:
for N even, log(2) - Sum_{k = 1..N/2} (-1)^(k-1)/k ~ (-1)^(N/2)*(1/N - 1/N^2 + 2/N^4 - 16/N^6 + 272/N^8 - ...), where the sequence of unsigned coefficients [1, 1, 2, 16, 272, ...] is A000182 with an extra initial term of 1. See Borwein et al., Theorem 1 (b);
for N odd, log(2) - Sum_{k = 1..(N-1)/2} (-1)^(k-1)/k ~ (-1)^((N-1)/2)*(1/N - 1/N^3 + 5/N^5 - 61/N^7 + 1385/N^9 - ...), by Borwein et al., Lemma 2 with f(x) := 1/(x + 1/2), h := 1/2 and then set x = (N - 1)/2, where the sequence of unsigned coefficients [1, 1, 5, 61, 1385, ...] is A000364. (End)
log(2) = lim_{n->oo} Sum_{k=1..n} sin(1/(n+k)). See Mathematical Reflections link. - Michel Marcus, Jan 07 2017
log(2) = Sum_{n>=1} A006519(n) / ((1 + 2^A006519(n)) * A000265(n) * (1 + A000265(n))). - Nicolas Nagel, Mar 19 2018
From Amiram Eldar, Jul 02 2020: (Start)
Equals Sum_{k>=2} zeta(k)/2^k.
Equals -Sum_{k>=2} log(1 - 1/k^2).
Equals Sum_{k>=1} 1/A002939(k).
Equals Integral_{x=0..Pi/3} tan(x) dx. (End)
log(2) = Integral_{x=0..Pi/2} (sec(x) - tan(x)) dx. - Clark Kimberling, Jul 08 2020
From Peter Bala, Nov 14 2020: (Start)
log(2) = Integral_{x = 0..1} (x - 1)/log(x) dx (Boros and Moll, p. 97).
log(2) = (1/2)*Integral_{x = 0..1} (x + 2)*(x - 1)^2/log(x)^2 dx.
log(2) = (1/4)*Integral_{x = 0..1} (x^2 + 3*x + 4)*(x - 1)^3/log(x)^3 dx. (End)
log(2) = 2*arcsinh(sqrt(2)/4) = 2*sqrt(2)*Sum_{n >= 0} (-1)^n*C(2*n,n)/ ((8*n+4)*32^n) = 3*Sum_{n >= 0} (-1)^n/((8*n+4)*(2^n)*C(2*n,n)). - Peter Bala, Jan 14 2022
log(2) = Integral_{x=0..oo} ( e^(-x) * (1-e^(-2x)) * (1-e^(-4x)) * (1-e^(-6x)) ) / ( x * (1-e^(-14x)) ) dx (see Crux Mathematicorum link). - Bernard Schott, Jul 11 2022
From Peter Bala, Oct 22 2023: (Start)
log(2) = 23/32 + 2!^3/16 * Sum_{n >= 1} (-1)^n * (n + 1)/(n*(n + 1)*(n + 2))^2 = 707/1024 - 4!^3/(16^2 * 2!^2) * Sum_{n >= 1} (-1)^n * (n + 2)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4))^2 = 42611/61440 + 6!^3/(16^3 * 3!^2) * Sum_{n >= 1} (-1)^n * (n + 3)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n + 5)*(n + 6))^2.
More generally, it appears that for k >= 0, log(2) = c(k) + (2*k)!^3/(16^k * k!^2) * Sum_{n >= 1} (-1)^(n+k+1) * (n + k)/(n*(n + 1)*...*(n + 2*k))^2 , where c(k) is a rational approximation to log(2). The first few values of c(k) are [0, 23/32, 707/1024, 42611/61440, 38154331/55050240, 76317139/110100480, 26863086823/38755368960, ...].
Let P(n,k) = n*(n + 1)*...*(n + k).
Conjecture: for k >= 0 and r even with r - 1 <= k, the series Sum_{n >= 1} (-1)^n * (d/dn)^r (P(n,k)) / (P(n,k)^2  = A(r,k)*log(2) + B(r,k), where A(r,k) and B(r,k) are both rational numbers. (End)
From Peter Bala, Nov 13 2023: (Start)
log(2) = 5/8 + (1/8)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)^2 / ( k*(k + 1) )^4
  = 257/384 + (3!^5/2^9)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)*(2*k + 3)^2*(2*k + 5) / ( k*(k + 1)*(k + 2)*(k + 3) )^4
  = 267515/393216 + (5!^5/2^19)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)*(2*k + 3)*(2*k + 5)^2*(2*k + 7)*(2*k + 9) / ( k*(k + 1)*(k + 2)*(k + 3)*(k + 4)*(k + 5) )^4
log(2) = 3/4 - 1/128 * Sum_{k >= 0} (-1/16)^k * (10*k + 12)*binomial(2*k+2,k+1)/ ((k + 1)*(2*k + 3)). The terms of the series are O(1/(k^(3/2)*4^n)). (End)
log(2) = eta(1) is a period, where eta(x) is the Dirichlet eta function. - Andrea Pinos, Mar 19 2024
log(2) = K_{n>=0} (n^2 + [n=0])/1, where K is the Gauss notation for an infinite continued fraction. In the expanded form, log(2) = 1/(1 + 1/(1 + 4/(1 + 9/1 + 16/(1 + 25/(1 + ... (see Clawson at p. 227). - Stefano Spezia, Jul 01 2024
log(2) = lim_{n->oo} Sum_{k=1..n} 1/(n + k) = lim_{x->0} (2^x - 1)/x = lim_{x->0} (2^x - 2^(-x))/(2*x) (see Finch). - Stefano Spezia, Oct 19 2024
From Colin Linzer, Nov 08 2024: (Start)
log(2) = Integral_{t=0...oo} (1 - tanh(t)) dt.
log(2) = Integral_{t=0...1} arctanh(t) dt.
log(2) = (1/2) * Integral_{t=-1...1} |arctanh(t)| dt. (End)
log(2) = 1 + Sum_{n >= 1} (-1)^n/(n*(4*n^2 - 1)) = 1/2 + (1/2)*Sum_{n >= 1} 1/(n*(4*n^2 - 1)). - Peter Bala, Jan 07 2025
log(2) = Integral_{x=0..1} Integral_{y=0..1} 1/((1 - x*y)*(1 + x)*(1 + y)) dy dx. - Kritsada Moomuang, May 22 2025

A061347 Period 3: repeat [1, 1, -2].

Original entry on oeis.org

1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2

Offset: 1

Jason Earls, Jun 07 2001

WARNING: It is unclear whether this sequence should start at offset 1 (as written) or offset 0 (in analogy to many similar sequences, which seems to be assumed in many of the given formulas).
Inverse binomial transform of A057079. - Paul Barry, May 15 2003
The unsigned version, with g.f. (1 + x + 2*x^2)/(1 - x^3), has a(n) = 4/3 -cos(2*Pi*n/3)/3 - sqrt(3)*sin(2*Pi*n/3)/3 = gcd(Fib(n+4), Fib(n+1)). - Paul Barry, Apr 02 2004
a(n) = L(n-2,-1), where L is defined as in A108299; see also A010892 for L(n,+1). - Reinhard Zumkeller, Jun 01 2005
From the Taylor expansion of log(1 + x + x^2) at x = 1, Sum_{k > 0} a(k)/k = log(3) = A002391. This is case n = 3 of the general expression Sum_{k > 0} (1-n*!(k mod n))/k = log(n). - Jaume Oliver Lafont, Oct 16 2009
If used with offset zero, a non-simple continued fraction representation of 2+sqrt(2). - R. J. Mathar, Mar 08 2012
Periodic sequences of this type can be also calculated by a(n) = c + floor(q/(p^m-1)*p^n) mod p, where c is a constant, q is the number representing the periodic digit pattern and m is the period length. c, p and q can be calculated as follows: Let D be the array representing the number pattern to be repeated, m = size of D, max = maximum value of elements in D, min = minimum value of elements in D. Than c := min, p := max - min + 1 and q := p^m*Sum_{i=1..m} (D(i)-min)/p^i. Example: D = (1, 1, -2), c = -2, p = 4 and q = 60 for this sequence. - Hieronymus Fischer, Jan 04 2013
This is the Dirichlet inverse of A117997. - Petros Hadjicostas, Jul 25 2020

			G.f.: x + x^2 - 2*x^3 + x^4 + x^5 - 2*x^6 + x^7 + x^8 - 2*x^9 + ... - _Michael Somos_, Nov 27 2019

W. Florek, A class of generalized Tribonacci sequences applied to counting problems, Appl. Math. Comput., 338 (2018), 809-821.
Ralph E. Griswold, Shaft Sequences, 2001 (see also here).
Tanya Khovanova, Recursive Sequences.
W. O. J. Moser, Cyclic binary strings without long runs of like (alternating) bits, Fibonacci Quart. 31(1) (1993), 2-6.
Index entries for linear recurrences with constant coefficients, signature (-1,-1).

Apart from signs, same as A057079 (also bin. Transf), A100063. Cf. A000045, A010892 for the rules a(n) = a(n - 1) + a(n - 2), a(n) = a(n - 1) - a(n - 2). a(n) = - a(n - 1) + a(n - 2) gives a signed version of Fibonacci numbers.
Alternating row sums of A130777: repeat(1,-2,1).
Cf. A000032, A002391, A007040, A100886, A108299, A117997, A320769.

Flat(List([1..50],n->[1,1,-2])); # Muniru A Asiru, Aug 02 2018

&cat [[1, 1, -2]^^30]; // Wesley Ivan Hurt, Jul 01 2016

seq(op([1, 1, -2]), n=1..50); # Wesley Ivan Hurt, Jul 01 2016

a[n_] := {1, 1, -2}[[Mod[n - 1, 3] + 1]]; Table[a[n], {n, 108}] (* Jean-François Alcover, Jul 19 2013 *)
PadRight[{}, 90, {1, 1, -2}] (* After Harvey P. Dale, or *)
CoefficientList[ Series[(2x + 1)/(x^2 + x + 1), {x, 0, 89}], x]  (* or *)
LinearRecurrence[{-1, -1}, {1, 1}, 90] (* Robert G. Wilson v, Jul 30 2018 *)

a(n)=1-3*!(n%3) \\ Jaume Oliver Lafont, Oct 16 2009

def A061347():
    x, y = -1, -1
    while True:
        yield -x
        x, y = y, -x -y
a = A061347(); [next(a) for i in range(40)] # Peter Luschny, Jul 11 2013

With offset zero, a(n) = A057079(2n). a(n) = -a(n-1) - a(n-2) with a(0) = a(1) = 1.
From Mario Catalani (mario.catalani(AT)unito.it), Jan 07 2003: (Start)
G.f.: x*(1 + 2*x)/(1 + x + x^2).
a(n) = (-1)^floor(2n/3) + ((-1)^floor((2n-1)/3) + (-1)^floor((2n+1)/3))/2. (End)
a(n) = -2*cos(2*Pi*n/3). - Jaume Oliver Lafont, May 06 2008
Dirichlet g.f.: zeta(s)*(1-1/3^(s-1)). - R. J. Mathar, Feb 09 2011
a(n) = n * Sum_{k=1..n} binomial(k,n-k)/k*(-1)^(k+1). - Dmitry Kruchinin, Jun 03 2011
a(n) = -2 + floor(110/333*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 04 2013
a(n) = -2 + floor(20/21*4^(n+1)) mod 4. - Hieronymus Fischer, Jan 04 2013
a(n) = a(n-3) for n > 3. - Wesley Ivan Hurt, Jul 01 2016
E.g.f.: 2 - 2*cos(sqrt(3)*x/2)*exp(-x/2). - Ilya Gutkovskiy, Jul 01 2016
a(n) = (-1)^n*hypergeom([-n/2-1, -n/2-3/2], [-n-2], 4). - Peter Luschny, Dec 17 2016
a(n) = A000032(n) - A007040(n), for n > 1. - Wojciech Florek, Feb 20 2018

Better definition from M. F. Hasler, Jan 13 2013

A153020 Decimal expansion of log_17 (3).

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Oct 30 2009

			.38776193503849007543619936073441864683067362944387624234675...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for transcendental numbers

Cf. decimal expansion of log_17(m): A152782 (m=2), this sequence, A153109 (m=4), A153430 (m=5), A153607 (m=6), A153627 (m=7), A153858 (m=8), A154016 (m=9), A154167 (m=10), A154188 (m=11), A154209 (m=12), A154399 (m=13), A154489 (m=14), A154683 (m=15), A154827 (m=16), A154974 (m=18), A155080 (m=19), A155529 (m=20), A155684 (m=21), A155783 (m=22), A155880 (m=23), A155992 (m=24).

RealDigits[Log[17, 3], 10, 100][[1]] (* Vincenzo Librandi, Sep 03 2013 *)

The ratio A002391/A016640.

A153035 Decimal expansion of log_20 (3).

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Oct 30 2009

			.36672579134208462891280714265933265359576843093233206828898...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for transcendental numbers

Cf. decimal expansion of log_20(m): A152821 (m=2), this sequence, A153124 (m=4), A153454 (m=5), A153610 (m=6), A153630 (m=7), A153872 (m=8), A154019 (m=9), A154170 (m=10), A154191 (m=11), A154212 (m=12), A154433 (m=13), A154492 (m=14), A154705 (m=15), A154838 (m=16), A154900 (m=17), A154976 (m=18), A155115 (m=19), A155687 (m=21), A155789 (m=22), A155907 (m=23), A156015 (m=24).

RealDigits[Log[20, 3], 10, 100][[1]] (* Vincenzo Librandi, Sep 05 2013 *)

Equals A002391 / A016643 . - R. J. Mathar, Aug 13 2024

A058962 a(n) = 2^(2n)(2*n+1).

Original entry on oeis.org

1, 12, 80, 448, 2304, 11264, 53248, 245760, 1114112, 4980736, 22020096, 96468992, 419430400, 1811939328, 7784628224, 33285996544, 141733920768, 601295421440, 2542620639232, 10720238370816, 45079976738816, 189115999977472, 791648371998720

Offset: 0

N. J. A. Sloane, Jan 13 2001

Denominators in expansion of -1/2*i*Pi+i*arcsin((1+1/4*x^2)/(1-1/4*x^2)), where i=sqrt(-1); numerators are all 1.
Bisection of A001787. That is, a(n) = A001787(2n+1). - Graeme McRae, Jul 12 2006
Denominators of odd terms in expansion of 2*arctanh(s/2); numerators are all 1. - Gerry Martens, Jul 26 2015
Reciprocals of coefficients of Taylor series expansion of sinh(x/2) / (x/2). - Tom Copeland, Feb 03 2016

Harry J. Smith, Table of n, a(n) for n = 0..200
G. A. Campbell, Physical theory of the electric wave-filter, Bell Syst. Tech. J., 1 (1922), 1-32, see Eq. (15d). Also reprinted in M. E. Van Valkebburg, ed., Circuit Theory, Dowden, Hutchinson and Ross, 1974.
Index entries for linear recurrences with constant coefficients, signature (8,-16).

Cf. A001787, A002391, A073000, A118413, A118415.
Cf. A154920. - Jaume Oliver Lafont, Jan 29 2009
Factor of the LS1[-2,n] matrix coefficients in A160487. - Johannes W. Meijer, May 24 2009

[2^(2*n)*(2*n+1) : n in [0..30]]; // Wesley Ivan Hurt, Aug 07 2015

a[n_] := 1/SeriesCoefficient[2 ArcTanh[s/2],{s,0,n}]
Table[a[n], {n, 1, 40, 2}] (* Gerry Martens, Jul 26 2015 *)
Table[2^(2 n) (2 n + 1), {n, 0, 40}] (* Vincenzo Librandi, Aug 08 2015 *)
a[ n_] := With[{m = 2 n + 2}, If[ n < 0, -a[-1 - n] 4^(m - 1), m! SeriesCoefficient[ x^2 D[x Sinc[I x]^2, x]/2, {x, 0, m}]]]; (* Michael Somos, Jun 18 2017 *)

A058962(n)=2^(2*n)*(2*n+1) \\ M. F. Hasler, Aug 11 2015

{a(n) = my(m = 2*n + 2); if( n<0, -a(-1 - n) * 4^(m - 1), m! * polcoeff( x^2 * deriv(x * sinc(I*x + x * O(x^m))^2, x) / 2, m))}; /* Michael Somos, Jun 18 2017 */

Central terms of the triangle in A118416: a(n) = A118416(2*n+1, n+1) - Reinhard Zumkeller, Apr 27 2006
Sum_{n>=0} 1/a(n) = log(3). - Jaume Oliver Lafont, May 22 2007; corrected by Jaume Oliver Lafont, Jan 26 2009
a(n) = 4((2n+1)/(2n-1))*a(n-1) = 4*a(n-1)+2^(2n+1) = 8*a(n-1)-16*a(n-2). - Jaume Oliver Lafont, Dec 09 2008
G.f.: (1+4*x)/(1-4*x)^2. - Jaume Oliver Lafont, Jan 29 2009
E.g.f.: exp(4*x)*(1+8*x). - Robert Israel, Aug 10 2015
a(n) = -a(-1-n) * 4^(2*n+1) for all n in Z. - Michael Somos, Jun 18 2017
a(n) = Sum_{k = 0..n} (2*k + 1)^2*binomial(2*n + 1, n - k). - Peter Bala, Feb 25 2019
Sum_{n>=0} (-1)^n/a(n) = 2 * arctan(1/2) = 2 * A073000. - Amiram Eldar, Jul 03 2020

A256425 Decimal expansion of the generalized Euler constant gamma(1,3).

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Apr 09 2015

			0.67780716378423221053372461245491438316931257963255620415268...

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 1.5.3, p. 32.

G. C. Greubel, Table of n, a(n) for n = 0..5000
D. H. Lehmer, Euler constants for arithmetic progressions, Acta Arith. 27 (1975), 125-142. [gamma=A001620, gamma(0,2)=A239097, gamma(1,2)=A228725, N_k=A007947, gamma(1,3)=A256425, the present sequence]

Cf. A001620, A002391, A007947, A047787, A228725, A239097.

RealDigits[EulerGamma/3 + Pi*Sqrt[3]/18 + Log[3]/6, 10, 50][[1]] (* G. C. Greubel, Jun 08 2017 *)

Euler/3 + Pi*sqrt(3)/18 + log(3)/6 \\ Charles R Greathouse IV, Sep 08 2015

-(psi(1/3) + log(3))/3 \\ Charles R Greathouse IV, Feb 03 2025

Equals gamma/3+Pi*sqrt(3)/18+log(3)/6.

Equals -(psi(1/3) + log(3))/3 = (A047787 - A002391)/3. - Amiram Eldar, Jan 07 2024

A073000 Decimal expansion of arctangent of 1/2.

Original entry on oeis.org

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

Offset: 0

Robert G. Wilson v, Aug 03 2002

The angle at which you must shoot a cue ball on a standard pool table so that it will strike all four sides and return to its origin. [Barrow] - Robert G. Wilson v, Nov 29 2015

			Arctan(1/2)
=0.463647609000806116214256231461214402028537054286120263810933088720197864165... radians
=26°.56505117707798935157219372045329467120421429964522102798601631528806582148474...
=26°33'.9030706246793610943316232271976802722528579787132616791609789172839492890...
=26°33'54".184237480761665659897393631860816335171478722795700749658735037036957...
complement = 63°.43494882292201064842780627954670532879578570035477897201398368471...
supplement = 153°.4349488229220106484278062795467053287957857003547789720139836847...

John D. Barrow, One Hundred Essential Things You Didn't Know You Didn't Know, W. W. Norton & Co., NY & London, 2008.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 242.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Peter Bala, New series for old functions.
R. J. Mathar, Hierarchical Subdivision of the Simple Cubic Lattice, arXiv preprint arXiv:1309.3705 [math.MG], 2013.
Simon Plouffe, arctan(1/2).
Index entries for transcendental numbers

Cf. A001622, A002162, A002391, A105531, A254619.

evalf(arctan(0.5)) ; # R. J. Mathar, Aug 22 2013

Mathematica
```
RealDigits[ ArcTan[1/2], 10, 110] [[1]]
```

default(realprecision,2000); atan(1/2) \\ Anders Hellström, Nov 30 2015

Equals Pi/2 - A105199 = A019669-A105199. - R. J. Mathar, Aug 21 2013
From Peter Bala, Feb 04 2015: (Start)
Arctan(1/2) = 1/2*Sum_{k >= 0} (-1)^k/((2*k + 1)*4^k).
Define a pair of integer sequences A(n) = 4^n*(2*n + 1)!/n! and B(n) = A(n)*Sum_{k = 0..n} (-1)^k/((2*k + 1)*4^k). Both sequences satisfy the same second order recurrence equation u(n) = (12*n + 10)*u(n-1) + 16*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion 2*arctan(1/2) = 1 - 2/(24 + 16*3^2/(34 + 16*5^2/(46 + ... + 16*(2*n - 1)^2/((12*n + 10) + ...)))). See A002391, A105531 and A002162 for similar expansions.
Arctan(1/2) = 2/5 * Sum_{k >= 0} (4/5)^k/((2*k + 1)*binomial(2*k,k)).
Define a pair of integer sequences C(n) = 5^n*(2*n + 1)!/n! and D(n) = C(n)*Sum_{k = 0..n} (4/5)^k/((2*k + 1)*binomial(2*k,k)). Both sequences satisfy the same second order recurrence equation u(n) = (24*n + 10)*u(n-1) - 40*n*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion 5/2*arctan(1/2) = 1 + 4/(30 - 240/(58 - 600/(82 - ... - 40*n*(2*n - 1)/((24*n + 10) - ... )))).
Arctan(1/2) = 2/25 * Sum_{k >= 0} (24*k + 17)*(4/5)^(2*k)/( (4*k + 1)*(4*k + 3)*binomial(4*k,2*k) ).
Arctan(1/2) = 2/125 * Sum_{k >= 0} (1116*k^2 + 1446*k + 433)*(4/5)^(3*k)/( (6*k + 1)*(6*k + 3)*(6*k + 5)*binomial(6*k,3*k) ). (End)
Equals Integral_{x = 0..oo} exp(-2*x)*sin(x)/x dx. - Peter Bala, Nov 05 2019
Equals 2 * arccot(phi^3), where phi is the golden ratio (A001622). - Amiram Eldar, Jul 06 2023
Equals Sum_{n >= 1} i/(n*P(n, 2*i)*P(n-1, 2*i)) = (1/2)*Sum_{n >= 1} (-1)^(n+1)*4^n/(n*A098443(n)*A098443(n-1)), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial. The n-th summand of the series is O( 1/(3 + 2*sqrt(2))^n ). - Peter Bala, Mar 16 2024

A016629 Decimal expansion of log(6).

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane

			1.791759469228055000812477358380702272722990692183004705855374343130887...

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].
Matt Parker, What's the story with log(1 + 2 + 3)?, standupmaths video (2019)
Index entries for transcendental numbers

Cf. A016734 (continued fraction).

First[RealDigits[Log[6], 10, 100]] (* Paolo Xausa, Mar 21 2024 *)

default(realprecision, 20080); x=log(6); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b016629.txt", n, " ", d)); \\ Harry J. Smith, May 16 2009

log(6) = 2*Sum_{n >= 1} 1/(n*P(n, 7/5)*P(n-1, 7/5)), where P(n, x) denotes the n-th Legendre polynomial. The first 20 terms of the series gives the approximation log(6) = 1.79175946922805(27...), correct to 14 decimal places. - Peter Bala, Mar 19 2024

Equals A002162 + A002391. - R. J. Mathar, Jun 10 2024

A034910 One quarter of octo-factorial numbers.

Original entry on oeis.org

1, 12, 240, 6720, 241920, 10644480, 553512960, 33210777600, 2258332876800, 171633298636800, 14417197085491200, 1326382131865190400, 132638213186519040000, 14324927024144056320000, 1661691534800710533120000, 206049750315288106106880000

Offset: 1

Wolfdieter Lang

A034910 occurs in connection with the Vandermonde permanent of (1,3,5,7,9,...); see the Mathematica section of A203516. - Clark Kimberling, Jan 03 2012

			G.f. = x + 12*x^2 + 240*x^3 + 6720*x^4 + 241920*x^5 + 10644480*x^6 + ...

G. C. Greubel, Table of n, a(n) for n = 1..325
Index entries for sequences related to factorial numbers.

Cf. A001147, A045755, A034908, A034909, A034911, A034912, A203516.

[n le 2 select 12^(n-1) else (7*n-3)*Self(n-1) +4*(n-1)*(2*n-3)*Self(n-2): n in [1..30]]; // G. C. Greubel, Oct 20 2022

[seq((2*n)!/(n)!*2^(n-2), n=1..14)]; # Zerinvary Lajos, Sep 25 2006

s=1;lst={s};Do[s+=n*s;AppendTo[lst, s], {n, 11, 5!, 8}];lst (* Vladimir Joseph Stephan Orlovsky, Nov 08 2008 *)
a[ n_] := Pochhammer[ 1/2, n] 8^n / 4; (* Michael Somos, Feb 04 2015 *)

{a(n) = if( n==1, 1, n>1, a(n-1) * (8*n - 4), a(n+1) / (8*n + 4))}; /* Michael Somos, Feb 04 2015 */

[2^(3*n-2)*rising_factorial(1/2, n) for n in range(1,40)] # G. C. Greubel, Oct 20 2022

4*a(n) = (8*n-4)(!^8) = Product_{j=1..n} (8*j-4) = 4^n*A001147(n) = 2^n*(2*n)!/n!, A001147(n) = (2*n-1)!!.
E.g.f. (-1+(1-8*x)^(-1/2))/4.
a(n) = A090802(2n-1, n). - Ross La Haye, Oct 18 2005
a(n) = ((2*n)!/n!)*2^(n-2). - Zerinvary Lajos, Sep 25 2006
G.f.: x/(1-12*x/(1-8*x/(1-20*x/(1-16*x/(1-28*x/(1-24*x/(1-36*x/(1-32*x/(1-... (continued fraction). - Philippe Deléham, Jan 07 2011
From Peter Bala, Feb 01 2015: (Start)
Recurrence equation: a(n) = (7*n - 3)*a(n-1) + 4*(n - 1)*(2*n - 3)*a(n-2).
The sequence b(n) := a(n)* Sum_{k = 0..n-1} (-1)^k/( 2^k*(2*k + 1)*binomial(2*k,k) ) beginning [1, 11, 222, 6210, 223584, ...] satisfies the same recurrence. This leads to the finite continued fraction expansion b(n)/a(n) = 1/(1 + 1/(11 + 24/(18 + 60/(25 + ... + 4*(n - 1)*(2*n - 3)/(7*n - 3) )))) for n >= 3.
Letting n tend to infinity gives the continued fraction expansion Sum_{k>=0} (-1)^k/( 2^k*(2*k + 1)*binomial(2*k,k) ) = (4/3)*log(2) = 1/(1 + 1/(11 + 24/(18 + 60/(25 + ... + 4*(n - 1)*(2*n - 3)/((7*n - 3) + ... ))))). (End)
From Peter Bala, Feb 03 2015: (Start)
This sequence satisfies several other second order recurrence equations leading to some continued fraction expansions.
1) a(n) = (9*n + 4)*a(n-1) - 4*n*(2*n - 1)*a(n-2).
This recurrence is also satisfied by the (integer) sequence c(n) := a(n)*Sum_{k = 0..n} 1/( 2^k*(2*k + 1)*binomial(2*k,k) ). From this we can obtain the continued fraction expansion Sum_{k >= 0} 1/( 2^k*(2*k + 1)*binomial(2*k,k) ) = (8/sqrt(7))*arctan(sqrt(7)/7) = (8/sqrt(7))*A195699 = 1 + 1/(12 - 24/(22 - 60/(31 - ... - 4*n*(2*n - 1)/((9*n + 4) - ... )))).
2) a(n) = (12*n + 2)*a(n-1) - 8*(2*n - 1)^2*a(n-2).
This recurrence is also satisfied by the (integer) sequence d(n) := a(n)*Sum_{k = 0..n} 1/( (2*k + 1)*2^k ). From this we can obtain the continued fraction expansion Sum_{k >= 0} 1/( (2*k + 1)*2^k ) = (1/sqrt(2))*log(3 + 2*sqrt(2)) = 1 + 2/(12 - 8*3^2/(26 - 8*5^2/(38 - ... - 8*(2*n - 1)^2/((12*n + 2) - ... )))). Cf. A002391.
3) a(n) = (4*n + 6)*a(n-1) + 8*(2*n - 1)^2*a(n-2).
This recurrence is also satisfied by the (integer) sequence e(n) := a(n)*Sum_{k = 0..n} (-1)^k/( (2*k + 1)*2^k ). From this we can obtain the continued fraction expansion Sum_{k >= 0} (-1)^k/( (2*k + 1)*2^k ) = (1/sqrt(2))*arctan(sqrt(2)/2) = 1 - 2/(12 + 8*3^2/(14 + 8*5^2/(18 + ... + 8*(2*n - 1)^2/((4*n + 6) + ... )))). Cf. A073000. (End)
a(n) = (-1)^n / (16*a(-n)) for all n in Z. - Michael Somos, Feb 04 2015
From Amiram Eldar, Jan 08 2022: (Start)
Sum_{n>=1} 1/a(n) = e^(1/8)*sqrt(2*Pi)*erf(1/(2*sqrt(2))), where erf is the error function.
Sum_{n>=1} (-1)^(n+1)/a(n) = e^(-1/8)*sqrt(2*Pi)*erfi(1/(2*sqrt(2))), where erfi is the imaginary error function. (End)

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A004736 Triangle read by rows: row n lists the first n positive integers in decreasing order.

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A002162 Decimal expansion of the natural logarithm of 2.

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A061347 Period 3: repeat [1, 1, -2].

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A153020 Decimal expansion of log_17 (3).

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A153035 Decimal expansion of log_20 (3).

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A058962 a(n) = 2^(2*n)*(2*n+1).

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A058962 a(n) = 2^(2n)(2*n+1).