A129890 -id:A129890 - cp's OEIS frontend

A180048 Coefficient triangle of the denominators of the (n-th convergents to) the continued fraction 1/(w+2/(w+3/(w+4/... . Conjectured to equal unsigned version of A137286.

1, 0, 1, 2, 0, 1, 0, 5, 0, 1, 8, 0, 9, 0, 1, 0, 33, 0, 14, 0, 1, 48, 0, 87, 0, 20, 0, 1, 0, 279, 0, 185, 0, 27, 0, 1, 384, 0, 975, 0, 345, 0, 35, 0, 1, 0, 2895, 0, 2640, 0, 588, 0, 44, 0, 1, 3840, 0, 12645, 0, 6090, 0, 938, 0, 54, 0, 1, 0, 35685, 0, 41685, 0, 12558, 0, 1422, 0, 65

Offset: 0

Wouter Meeussen, Aug 08 2010

Equivalence to the recurrence formula needs formal proof. This continued fraction converges to 0.525135276160981... for w=1. A conjecture by Ramanujan puts this equal to -1 + 1/(sqrt(e*Pi/2) - Sum_{k>=1} 1/(2k-1)!!).
From Alexander Kreinin, Oct 26 2015: (Start)
Let us denote the continued fraction by U(w).
Then it is easy to show that Mill's ratio, R(w) = (1 - Phi(w))/f(w), where Phi is the standard normal distribution function and f is the standard normal density function, satisfies R(w) = 1/(w + U(w)).
Indeed, R(w) = 1/(w+1/(w+2/(w+3/(w+... Then we find U(w) = 1/R(w) - w. It was proved in Alexander Kreinin (arXiv:1405.5852) that R(w+t) + Q(w, t) = exp(w*t + w^2/2)*R(t), where Q(w,t) = Sum_{k>=0} Sum_{m=0..k} q(k,m) * t^m * w^(k+1)/(k+1)!.
Substituting t=0, we obtain R(w) = exp(w^2/2)*sqrt(Pi/2) - Sum_{n>=0} w^(2n+1)/(2n+1)!!. If w=1 we obtain Ramanujan's formula. (End)

			The denominator of 1/(w+2/(w+3/(w+4/(w+5/(w+6/w))))) equals 48 + 87w^2 + 20w^4 + w^6.
From _Joerg Arndt_, Apr 20 2013: (Start)
Triangle begins
     1;
     0,     1;
     2,     0,     1;
     0,     5,     0,     1;
     8,     0,     9,     0,    1;
     0,    33,     0,    14,    0,   1;
    48,     0,    87,     0,   20,   0,   1;
     0,   279,     0,   185,    0,  27,   0,  1;
   384,     0,   975,     0,  345,   0,  35,  0,  1;
     0,  2895,     0,  2640,    0, 588,   0, 44,  0, 1;
  3840,     0, 12645,     0, 6090,   0, 938,  0, 54, 0, 1;
     0, 35685,     0, 41685,    0, ... (End)

G. C. Greubel, Table of n, a(n) for the first 75 rows, flattened
Authors?, Hungarian discussion forum
Alexander Kreinin, Combinatorial Properties of Mills' Ratio, arXiv:1405.5852 [math.CO], 2014. See Table 3.
Alexander Kreinin, Integer Sequences and Laplace Continued Fraction, preprint, 2016.
Alexander Kreinin, Integer Sequences Connected to the Laplace Continued Fraction and Ramanujan's Identity, Journal of Integer Sequences 19 (2016), Article 16.6.2.

Cf. A137286, A084950, A180047, A180049.

Cf. A111129.

Table[ CoefficientList[ Denominator[ Together[ Fold[ #2/(w+#1) &, Infinity, Reverse @ Table[ k, {k, 1, n} ] ] ] ], w ], {n, 16} ] (* or equivalently *) Clear[ p ];p[ 0 ]=1; p[ 1 ]=w; p[ n_ ]:=p[ n ]= w*p[ n-1 ] + n*p[ n-2 ]; Table[ CoefficientList[ p[ k ]//Expand, w ], {k,0,15} ]

p(0)=1; p(1)=w; p(n) = w*p(n-1) + n*p(n-2) (conjecture).
T(n,k) = T(n-1,k-1) + n*T(n-2,k), T(0,0) = 1, T(1,0) = 0, T(1,1) = 1. - Philippe Deléham, Oct 28 2013
sum_{k=0..n} T(n,k) = A000932(n). - Philippe Deléham, Oct 28 2013
T(2n,0) = A000165(n); T(2n+1,1) = A129890(n); T(2n+2,2) = A035101(n+2). - Philippe Deléham, Oct 28 2013

A035101 E.g.f. x(c(x/2)-1)/(1-2x), where c(x) = g.f. for Catalan numbers A000108.

Original entry on oeis.org

0, 1, 9, 87, 975, 12645, 187425, 3133935, 58437855, 1203216525, 27125492625, 664761133575, 17600023616175, 500706514833525, 15234653491682625, 493699195087473375, 16977671416936605375, 617528830880480644125, 23687738668934964248625

Offset: 1

Wolfdieter Lang

2nd column of triangular array A035342 whose first column is given by A001147(n), n >= 1. Recursion: a(n) = 2*n*a(n-1)+ A001147(n-1), n >= 2, a(1)=0.
a(n) gives the number of organically labeled forests (sets) with two rooted ordered trees with n non-root vertices. See the example a(3)=9 given in A035342. Organic labeling means that the vertex labels along the (unique) path from the root to any of the leaves (degree 1, non-root vertices) is increasing. - Wolfdieter Lang, Aug 07 2007
a(n), n>=2, enumerates unordered n-vertex forests composed of two plane (ordered) ternary (3-ary) trees with increasing vertex labeling. See A001147 (number of increasing ternary trees) and a D. Callan comment there. For a picture of some ternary trees see a W. Lang link under A001764.
a(n) is the number of linear chord diagrams on 2n vertices with one marked chord such that exactly 1 of the remaining n-1 chords are contained within the marked chord, see [Young]. - Donovan Young, Aug 11 2020

			a(2)=1 for the forest: {r1-1, r2-2} (with root labels r1 and r2). The order between the components of the forest is irrelevant (like for sets).
a(3)=9 increasing ternary 2-forest with n=3 vertices: there are three 2-forests (the one vertex tree together with any of the three different 2-vertex trees) each with three increasing labelings. - _Wolfdieter Lang_, Sep 14 2007

Robert Israel, Table of n, a(n) for n = 1..370
Selden Crary, Richard Diehl Martinez, Michael Saunders, The Nu Class of Low-Degree-Truncated Rational Multifunctions. Ib. Integrals of Matern-correlation functions for all odd-half-integer class parameters, arXiv:1707.00705 [stat.ME], 2017, Table 2.
Alexander Kreinin, Integer Sequences and Laplace Continued Fraction, Preprint 2016.
Alexander Kreinin, Integer Sequences Connected to the Laplace Continued Fraction and Ramanujan's Identity, Journal of Integer Sequences, 19 (2016), #16.6.2.
Donovan Young, A critical quartet for queuing couples, arXiv:2007.13868 [math.CO], 2020.

Cf. A000108, A002866, A008549, A336599.

Cf. A001147 (m=1 column of A035342). See a D. Callan comment there on the number of increasing ordered rooted trees on n+1 vertices.

I:=[0,1,9]; [n le 3 select I[n] else - 2*(n-1)*(2*n-3)*Self(n-2)+(4*n-3)*Self(n-1): n in [1..30]]; // Vincenzo Librandi, Sep 12 2015

F:= gfun:-rectoproc({(4*n^2+6*n+2)*a(n)+(-4*n-5)*a(n+1)+a(n+2),a(1)=0,a(2)=1,a(3)=9},a(n),remember):
map(f, [$1..30]); # Robert Israel, Sep 11 2015

Table[Round [n! (4^(n - 1) - Binomial[2 n, n]/2)/2^(n - 1)], {n, 1, 20}] (* Vincenzo Librandi, Sep 12 2015 *)

a(n) = n!*(4^(n-1)-binomial(2*n, n)/2)/2^(n-1);
vector(40, n, a(n)) \\ Altug Alkan, Oct 01 2015

a(n) = n!*A008549(n-1)/2^(n-1) = n!(4^(n-1)-binomial(2*n, n)/2)/2^(n-1).
a(n) = (2n-2)*a(n-1) + A129890(n-2). - Philippe Deléham, Oct 28 2013
a(n) = n!*2^(n-1) - A001147(n) = A002866(n) - A001147(n). - Peter Bala, Sep 11 2015
a(n) = -2*(n-1)*(2*n-3)*a(n-2)+(4*n-3)*a(n-1). - Robert Israel, Sep 11 2015

A336599 Triangle read by rows: T(n,k) is the number of linear chord diagrams on 2n vertices with one marked chord such that exactly k of the remaining n-1 chords are contained within the marked chord.

Original entry on oeis.org

1, 5, 1, 33, 9, 3, 279, 87, 39, 15, 2895, 975, 495, 255, 105, 35685, 12645, 6885, 4005, 2205, 945, 509985, 187425, 106785, 66465, 41265, 23625, 10395, 8294895, 3133935, 1843695, 1198575, 795375, 513135, 301455, 135135, 151335135, 58437855, 35213535, 23601375, 16343775, 11263455, 7453215, 4459455, 2027025

Offset: 1

Donovan Young, Jul 29 2020

			Triangle begins:
     1;
     5,    1;
    33,    9,    3;
   279,   87,   39,  15;
  2895,  975,  495, 255, 105;
...
For n = 2 and k = 1, let the four vertices be {1,2,3,4}. The marked chord can only be (1,4) and it contains one other chord, namely (2,3), hence T(2,1) = 1.

Donovan Young, Table of n, a(n) for n = 1..9870
Donovan Young, A critical quartet for queuing couples, arXiv:2007.13868 [math.CO], 2020.

Row sums are n*A001147(n) for n > 0.
Leading diagonal is A001147(n-1) for n > 0.
The first column is A129890(n-1) for n > 0.
The second column is A035101(n+1) for n > 0.
Cf. A336598, A336600, A336601.

CoefficientList[Normal[Series[(Sqrt[1-2*y*x]-Sqrt[1-2*x])/(1-2*x)/(1-y),{x,0,10}]]/.{x^n_.->x^n*n!},{x,y}]

E.g.f.: (sqrt(1 - 2*y*x) - sqrt(1 - 2*x))/(1 - 2*x)/(1 - y).

A135876 Square array, read by antidiagonals, where row n+1 is generated from row n by first removing terms at positions [(m+2)^2/4 - 1] for m>=0 and then taking partial sums, starting with all 1's in row 0.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 15, 8, 3, 1, 105, 48, 15, 4, 1, 945, 384, 105, 23, 5, 1, 10395, 3840, 945, 176, 33, 6, 1, 135135, 46080, 10395, 1689, 279, 44, 7, 1, 2027025, 645120, 135135, 19524, 2895, 400, 57, 8, 1, 34459425, 10321920, 2027025, 264207, 35685, 4384, 561

Offset: 0

Paul D. Hanna, Dec 14 2007

This is the double factorial analog of Moessner's factorial array (A125714). Compare to triangle A135877 which is generated by a complementary process. A very interesting variant is square array A135878.

			Square array begins:
(1),(1),1,(1),1,(1),1,1,(1),1,1,(1),1,1,1,(1),1,1,1,(1),1,1,1,1,(1),...;
(1),(2),3,(4),5,(6),7,8,(9),10,11,(12),13,14,15,(16),17,18,19,(20),...;
(3),(8),15,(23),33,(44),57,71,(86),103,121,(140),161,183,206,(230),..;
(15),(48),105,(176),279,(400),561,744,(950),1206,1489,(1800),2171,..;
(105),(384),945,(1689),2895,(4384),6555,9129,(12139),16161,20763,..;
(945),(3840),10395,(19524),35685,(56448),89055,129072,(177331),245778,...;
(10395),(46080),135135,(264207),509985,(836352),1381905,2071215,(2924172),.;
(135135),(645120),2027025,(4098240),8294895,(14026752),24137505,...; ...
where terms in parenthesis are removed before taking partial sums.
For example, to generate row 2 from row 1, remove terms at positions
{[(m+2)^2/4-1], m>=0} = [0,1,3,5,8,11,15,19,24,29,35,...] to obtain:
[3, 5, 7,8, 10,11, 13,14,15, 17,18,19, 21,22,23,24, 25,26,27,28, ...]
then take partial sums to get row 2:
[3, 8, 15,23, 33,44, 57,71,86, 103,121,140, 161,183,206,230, ...].
Repeating this process will generate all the rows of the triangle,
where column 0 will be the odd double factorials (A001147)
and column 1 will be the even double factorials (A000165).

Cf. columns: A001147, A000165, A004041, A129890; variants: A135878, A125714.

{T(n, k)=local(A=0, b=0, c=0, d=0); if(n==0, A=1, until(d>k, if(c==floor((b+2)^2/4)-1, b+=1, A+=T(n-1, c); d+=1); c+=1)); A}

T(n,0) = (2n)!/n!/2^n; T(n,1) = 2^n*n!; T(n,2) = (2n+1)!/n!/2^n; T(n,3) = A004041(n) = (2n+1)!/n!/2^n * Sum_{k=0..n} 1/(2k+1). T(n,4) = A129890(n) = 2^(n+1)*(n+1)! - (2n+1)!/n!/2^n = T(n+1,1)-T(n+1,0).

A232618 a(n) = (2n)!! mod (2n-1)!! where k!! = A006882(k).

Original entry on oeis.org

0, 2, 3, 69, 60, 4500, 104580, 186795, 13497435, 442245825, 13003053525, 64585694250, 3576632909850, 147580842959550, 5708173568847750, 27904470362393625, 2292043480058957625, 126842184377462428875, 6371504674680470700375, 312265748715684068930625

Offset: 1

Alex Ratushnyak, Nov 27 2013

nonn

(2n)!! is the product of first n even numbers, (2n-1)!! is the product of first n odd numbers.

			a(3) = A006882(6) mod A006882(5) = 2*4*6 mod 1*3*5 = 48 mod 15 = 3.

Cf. A006882, A122649, A129890, A232617.

Table[Mod[(2n)!!,(2n-1)!!],{n,20}] (* Harvey P. Dale, Sep 23 2020 *)

o=e=1
for n in range(1,99,2):
  o*=n
  e*=n+1
  print(str(e%o), end=',')

a(n) = A006882(2*n) mod A006882(2*n-1).

A160481 Row sums of the Beta triangle A160480.

Original entry on oeis.org

-1, -10, -264, -13392, -1111680, -137030400, -23500108800, -5351202662400, -1562069156659200, -568747270103040000, -252681700853514240000, -134539938778433126400000, -84573370199475510312960000, -61972704966344777143418880000, -52361960516341326660973363200000

Offset: 2

Johannes W. Meijer, May 24 2009, Sep 19 2012

It is conjectured that the row sums of the Beta triangle depend on three different sequences. Two Maple algorithms are given. The first one gives the row sums according to the Beta triangle A160480 and the second one gives the row sums according to our conjecture.

Christopher P. Herzog, Kuo-Wei Huang, and Kristan Jensen, Universal Entanglement and Boundary Geometry in Conformal Field Theory, arXiv preprint arXiv:1510.00021 [hep-th], 2015.
Kuo-Wei Huang, Resummation of Multi-Stress Tensors in Higher Dimensions, arXiv:2406.07458 [hep-th], 2024. See p. 10.

A160480 is the Beta triangle.
Row sum factors A120778, A000165 and A049606.
Cf. A002474, A009445, A129890.

nmax := 14; mmax := nmax: for n from 1 to nmax do BETA(n, n) := 0 end do: m := 1: for n from m+1 to nmax do BETA(n,m) := (2*n-3)^2*BETA(n-1, m)-(2*n-4)! od: for m from 2 to mmax do for n from m+1 to nmax do BETA(n, m) := (2*n-3)^2*BETA(n-1, m) - BETA(n-1, m-1) od: od: for n from 2 to nmax do s1(n) := 0: for m from 1 to n-1 do s1(n) := s1(n) + BETA(n, m) od: od: seq(s1(n), n=2..nmax);
# End first program
nmax := nmax; A120778 := proc(n): numer(sum(binomial(2*k1, k1)/(k1+1) / 4^k1, k1=0..n)) end proc: A000165 := proc(n): 2^n*n! end proc: A049606 := proc(n): denom(2^n/n!) end proc: for n from 2 to nmax do s2(n) := (-1)*A120778(n-2)*A000165(n-2)*A049606(n-1) end do: seq(s2(n), n=2..nmax);
# End second program

BETA[2, 1] = -1; BETA[n_, 1] := BETA[n, 1] = (2*n - 3)^2*BETA[n - 1, 1] - (2*n - 4)!; BETA[n_ /; n > 2, m_ /; m > 0] /; 1 <= m <= n := BETA[n, m] = (2*n - 3)^2*BETA[n - 1, m] - BETA[n - 1, m - 1]; BETA[, ] = 0;
Table[Sum[BETA[n, m], {m, 1, n - 1}], {n, 2, 14}] (* Jean-François Alcover, Dec 13 2017 *)

Rowsums(n) = (-1)*A120778(n-2)*A000165(n-2)*A049606(n-1) for n >= 2.
Conjecture: a(n) = (2*n-3)! - 2^(2*n-3)*(n-1)!*(n-2)!, for n >= 2 (gives the first 13 terms). - Christopher P. Herzog, Nov 25 2014
Meijer's and Herzog's conjectures can also be written as: a(n) = -A129890(n-2)*A000165(n-2) = A009445(n-2) - A002474(n-2). - Peter Luschny, Dec 01 2014

a(15)-a(16) from Stefano Spezia, Jun 28 2024

A232617 Product of first n odd numbers plus product of first n even numbers: (2n-1)!! + (2n)!!, where k!! = A006882(k).

Original entry on oeis.org

3, 11, 63, 489, 4785, 56475, 780255, 12348945, 220253985, 4370620275, 95498916975, 2278224696825, 58917607974225, 1641787169697675, 49040157044253375, 1563094742062478625, 52953322446161762625, 1899986948191060603875, 71977860935783603175375, 2870913642898706235455625

Offset: 1

Alex Ratushnyak, Nov 27 2013

nonn

			a(3) = 1*3*5 + 2*4*6 = 15 + 48 = 63.

Cf. A006882, A122649, A129890.

Table[n!!+(n+1)!!,{n,1,41,2}] (* Harvey P. Dale, Jan 22 2019 *)

a(n)=prod(i=1,n,2*i-1)+prod(i=1,n,2*i) \\ Ralf Stephan, Nov 28 2013

o=e=1
for n in range(1,99,2):
  o*=n
  e*=n+1
  print(str(e+o), end=',')

a(n) = A006882(2*n-1) + A006882(2*n).
a(n) = A001147(n) + A000165(n).
a(n) +(-4*n+3)*a(n-1) +2*(n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Nov 23 2014

A202212 Triangle read by rows: T(n,k) (1 <= k <= n-1, n >= 2) = d(2(n-k)-1)(d(2n-2)/d(2(n-k)-2) - d(2n-3)/d(2(n-k)-3)), where d = A006882 is the double factorial function.

Original entry on oeis.org

1, 3, 5, 15, 27, 33, 105, 195, 261, 279, 945, 1785, 2475, 2925, 2895, 10395, 19845, 28035, 34425, 37935, 35685, 135135, 259875, 371385, 465255, 533925, 562275, 509985, 2027025, 3918915, 5644485, 7158375, 8390025, 9218475, 9401805, 8294895, 34459425, 66891825, 96891795, 123898005, 147093975, 165209625, 176067675, 175313565, 151335135, 654729075, 1274998725, 1854727875, 2385808425, 2857013775, 3252014325, 3545408475, 3693650625, 3609649575, 3061162125

Offset: 2

N. J. A. Sloane, Dec 14 2011

			Triangle begins
1,
3, 5,
15, 27, 33,
105, 195, 261, 279,
945, 1785, 2475, 2925, 2895,
10395, 19845, 28035, 34425, 37935, 35685,
135135, 259875, 371385, 465255, 533925, 562275, 509985,
...

N. Ochiumi, On the total sum of number of nodes covering a given number of leaves in an unordered binary tree.

Edges of triangle are A006882 and A129890.

d:=doublefactorial;
a:=(n,k)-> d(2*(n-k)-1)*(d(2*n-2)/d(2*(n-k)-2) - d(2*n-3)/d(2*(n-k)-3));
f:=n->[seq(a(n,k),k=1..n-1)];
for n from 1 to 10 do lprint(f(n)); od:

a[n_, k_] := (2*(n-k)-1)!!*((2*n-2)!!/(2*(n-k)-2)!!-(2*n-3)!!/(2*(n-k)-3)!!); Table[a[n, k], {n, 2, 11}, {k, 1, n-1}] // Flatten (* Jean-François Alcover, Jan 08 2014 *)

A265649 Triangle of coefficients T(n,k) of polynomials p(n,x) = Sum_{k=0..n} T(n,k)x^k where T(0,0) = 1, and T(n,k) = 0 for k < 0 or k > n, and T(n,k) = T(n-1,k-1) + (2n-1+k)*T(n-1,k) for n > 0 and 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 3, 5, 1, 15, 33, 12, 1, 105, 279, 141, 22, 1, 945, 2895, 1830, 405, 35, 1, 10395, 35685, 26685, 7500, 930, 51, 1, 135135, 509985, 435960, 146685, 23310, 1848, 70, 1, 2027025, 8294895, 7921305, 3076290, 589575, 60270, 3318, 92, 1, 34459425, 151335135, 158799690, 69447105, 15457365, 1915515, 136584, 5526, 117, 1

Offset: 0

Werner Schulte, Dec 11 2015

The polynomials p(n,x) satisfy the differential equation: x*y''' + (3*x+1)*y'' + (2*x+2)*y' - 2*n*y = 0 where y' = dy/dx (first derivative).

Appears to be the exponential Riordan array [1/sqrt(1 - 2x), 1/(sqrt(1 - 2x) - 1)]. [Barry, Example 1] - Eric M. Schmidt, Sep 23 2017

			The triangle T(n,k) begins:
n\k:        0        1        2        3       4      5     6   7  8
  0:        1
  1:        1        1
  2:        3        5        1
  3:       15       33       12        1
  4:      105      279      141       22       1
  5:      945     2895     1830      405      35      1
  6:    10395    35685    26685     7500     930     51     1
  7:   135135   509985   435960   146685   23310   1848    70   1
  8:  2027025  8294895  7921305  3076290  589575  60270  3318  92  1
  etc.
The polynomial corresponding to row 3 is p(3,x) = 15 + 33*x + 12*x^2 + x^3.

Paul Barry, A Note on d-Hankel Transforms, Continued Fractions, and Riordan Arrays, arXiv:1702.04011 [math.CO], 2017.
Robert S. Maier, Boson Operator Ordering Identities from Generalized Stirling and Eulerian Numbers, arXiv:2308.10332 [math.CO], 2023. See pp. 21, 28.

Cf. A000007, A000326, A001147, A001497, A021009, A129890, A035342.

T := (n, k) -> local j; 2^n*add((-1)^(k-j)*binomial(k, j)*pochhammer((j+1)/2, n), j=0..k) / k!: for n from 0 to 6 do seq(T(n, k), k=0..n) od;  # Peter Luschny, Mar 04 2024

(* The function RiordanArray is defined in A256893. *)
rows = 10;
R = RiordanArray[1/Sqrt[1 - 2 #]&, 1/Sqrt[1 - 2 #] - 1&, rows, True];
R // Flatten (* Jean-François Alcover, Jul 20 2019 *)

Recurrence: p(0,x) = 1 and p(n+1,x) = (2*n+1+x)*p(n,x) + x*p'(n,x).
T(n,0) = A001147(n), T(n+1,1) = A129890(n), T(n+1,n) = A000326(n+1), and Sum_{k=0..n} (-1)^k*k!*T(n,k) = A000007(n).
Recurrence: k^2*(k+1)*T(n,k+1) = (2*n+2-2*k)*T(n,k-1)-k*(3*k-1)*T(n,k).
Conjecture: T(n,k) = 2^(n-k)*(n-k)!*binomial(n,k)*(Sum_{j=0..n-k} (-1/4)^j* binomial(2*j+k,j)*binomial(n,j+k)).
Conjecture: T(n,k) = (-1)^k*Sum_{j=0..n-1} A001497(n-1,j)*A021009(j+1,k).
T(n,k) = (Sum_{i=0..k} (-1)^(k-i) * binomial(k, i)*Product_{j=1..n} (2*j+i-1))/k!. - Werner Schulte, Mar 03 2024
T(n,k) = (2^n/k!)*(Sum_{j=0..k}(-1)^(k-j)*binomial(k,j)*Pochhammer((j + 1)/2, n)). - Peter Luschny, Mar 04 2024

A306184 a(n) = (2n+1)!! mod (2n)!! where k!! = A006882(k).

Original entry on oeis.org

1, 7, 9, 177, 2715, 42975, 91665, 3493665, 97345395, 2601636975, 70985324025, 57891366225, 9411029102475, 476966861546175, 20499289200014625, 847876038362978625, 35160445175104123875, 1487419121780448231375, 945654757149212735625, 357657177058846280240625

Offset: 1

Alex Ratushnyak, Jan 27 2019

nonn

a(n) is divisible by A049606(n). - Robert Israel, Jan 28 2019

			a(3) = A006882(7) mod A006882(6) = (7*5*3) mod (6*4*2) = 105 mod 48 = 9.

Robert Israel, Table of n, a(n) for n = 1..403

Cf. A006882, A049606, A122649, A129890, A232617, A232618, A306185.

f:= n -> doublefactorial(2*n+1) mod doublefactorial(2*n):
map(f, [$1..40]); # Robert Israel, Jan 28 2019

Mod[#[[2]],#[[1]]]&/@Partition[Range[2,42]!!,2] (* Harvey P. Dale, May 29 2025 *)

o=e=1
for n in range(2, 99, 2):
  o*=n+1
  e*=n
  print(o%e, end=', ')

a(n) = A006882(2*n+1) mod A006882(2*n).

cp's OEIS Frontend

A180048 Coefficient triangle of the denominators of the (n-th convergents to) the continued fraction 1/(w+2/(w+3/(w+4/... . Conjectured to equal unsigned version of A137286.

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A035101 E.g.f. x*(c(x/2)-1)/(1-2*x), where c(x) = g.f. for Catalan numbers A000108.

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A336599 Triangle read by rows: T(n,k) is the number of linear chord diagrams on 2n vertices with one marked chord such that exactly k of the remaining n-1 chords are contained within the marked chord.

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A135876 Square array, read by antidiagonals, where row n+1 is generated from row n by first removing terms at positions [(m+2)^2/4 - 1] for m>=0 and then taking partial sums, starting with all 1's in row 0.

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A232618 a(n) = (2n)!! mod (2n-1)!! where k!! = A006882(k).

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A160481 Row sums of the Beta triangle A160480.

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A232617 Product of first n odd numbers plus product of first n even numbers: (2n-1)!! + (2n)!!, where k!! = A006882(k).

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A035101 E.g.f. x(c(x/2)-1)/(1-2x), where c(x) = g.f. for Catalan numbers A000108.

A202212 Triangle read by rows: T(n,k) (1 <= k <= n-1, n >= 2) = d(2(n-k)-1)(d(2n-2)/d(2(n-k)-2) - d(2n-3)/d(2(n-k)-3)), where d = A006882 is the double factorial function.

A265649 Triangle of coefficients T(n,k) of polynomials p(n,x) = Sum_{k=0..n} T(n,k)x^k where T(0,0) = 1, and T(n,k) = 0 for k < 0 or k > n, and T(n,k) = T(n-1,k-1) + (2n-1+k)*T(n-1,k) for n > 0 and 0 <= k <= n.