cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A007226 a(n) = 2*det(M(n; -1))/det(M(n; 0)), where M(n; m) is the n X n matrix with (i,j)-th element equal to 1/binomial(n + i + j + m, n).

Original entry on oeis.org

2, 3, 10, 42, 198, 1001, 5304, 29070, 163438, 937365, 5462730, 32256120, 192565800, 1160346492, 7048030544, 43108428198, 265276342782, 1641229898525, 10202773534590, 63698396932170, 399223286267190, 2510857763851185, 15842014607109600
Offset: 0

Views

Author

Keywords

Comments

For n >= 1, a(n) is the number of distinct perforation patterns for deriving (v,b) = (n+1,n) punctured convolutional codes from (3,1). [Edited by Petros Hadjicostas, Jul 27 2020]
Apparently Bégin's (1992) paper was presented at a poster session at the conference and was never published.
a(n) is the total number of down steps between the first and second up steps in all 2-Dyck paths of length 3*(n+1). A 2-Dyck path is a nonnegative lattice path with steps (1,2), (1,-1) that starts and ends at y = 0. - Sarah Selkirk, May 07 2020
From Petros Hadjicostas, Jul 27 2020: (Start)
"A punctured convolutional code is a high-rate code obtained by the periodic elimination (i.e., puncturing) of specific code symbols from the output of a low-rate encoder. The resulting high-rate code depends on both the low-rate code, called the original code, and the number and specific positions of the punctured symbols." (The quote is from Haccoun and Bégin (1989).)
A high-rate code (v,b) (written as R = b/v) can be constructed from a low-rate code (v0,1) (written as R = 1/v0) by deleting from every v0*b code symbols a number of v0*b - v symbols (so that the resulting rate is R = b/v).
Even though my formulas below do not appear in the two published papers in the IEEE Transactions on Communications, from the theory in those two papers, it makes sense to replace "k|b" with "k|v0*b" (and "k|gcd(v,b)" with "k|gcd(v,v0*b)"). Pab Ter, however, uses "k|b" in the Maple programs in the related sequences A007223, A007224, A007225, A007227, and A007229. (End)
Conjecture: a(n) is odd iff n = A022341(k) for some k. - Peter Bala, Mar 13 2023

References

  • Guy Bégin, On the enumeration of perforation patterns for punctured convolutional codes, Séries Formelles et Combinatoire Algébrique, 4th colloquium, 15-19 Juin 1992, Montréal, Université du Québec à Montréal, pp. 1-10.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    [Binomial(3*n,n)/(2*n+1)+Binomial(3*n+1,n)/(n+1): n in [0..25]]; // Vincenzo Librandi, Aug 10 2014
    
  • Maple
    A007226:=n->2*binomial(3*n,n)-binomial(3*n,n+1): seq(A007226(n), n=0..30); # Wesley Ivan Hurt, Aug 11 2014
  • Mathematica
    Table[2*Binomial[3n,n]-Binomial[3n,n+1], {n,0,20}] (* Harvey P. Dale, Aug 10 2014 *)
  • PARI
    a(n) = {my(M1=matrix(n,n)); my(M0=matrix(n,n)); for(i=1, n, for(j=1, n, M1[i,j] = 1/binomial(n+i+j-1,n); M0[i,j] = 1/binomial(n+i+j,n);)); 2*matdet(M1)/matdet(M0);} \\ Petros Hadjicostas, Jul 27 2020

Formula

a(n) = (2/(n + 1))*binomial(3*n, n).
a(n) = (2n+1) * A000139(n). - F. Chapoton, Feb 23 2024
a(n) = 2*C(3*n, n) - C(3*n, n+1) for n >= 0. - David Callan, Oct 25 2004
a(n) = C(3*n, n)/(2*n + 1) + C(3*n + 1, n)/(n + 1) = C(3*n, n)/(2*n + 1) + 2*C(3*n + 1, n)/(2*n + 2) for n >= 0. - Paul Barry, Nov 05 2006
G.f.: g*(2 - g)/x, where g*(1 - g)^2 = x. - Mark van Hoeij, Nov 08 2011 [Thus, g = (4/3)*sin((1/3)*arcsin(sqrt(27*x/4)))^2. - Petros Hadjicostas, Jul 27 2020]
Recurrence: 2*(n+1)*(2*n-1)*a(n) - 3*(3*n-1)*(3*n-2)*a(n-1) = 0 for n >= 1. - R. J. Mathar, Nov 26 2012
G.f.: (1 - 1/B(x))/x, where B(x) is the g.f. of A006013. [Vladimir Kruchinin, Mar 05 2013]
G.f.: ( -16 * sin(asin((3^(3/2) * sqrt(x))/2)/3)^4 + 24 * sin(asin((3^(3/2) * sqrt(x))/2)/3)^2 ) / (9*x). [Vladimir Kruchinin, Nov 16 2013]
From Petros Hadjicostas, Jul 27 2020: (Start)
The number of perforation patterns to derive high-rate convolutional code (v,b) (written as R = b/v) from a given low-rate convolutional code (v0, 1) (written as R = 1/v0) is (1/b)*Sum_{k|gcd(v,b)} phi(k)*binomial(v0*b/k, v/k).
According to Pab Ter's Maple code in the related sequences (see above), this is the coefficient of z^v in the polynomial (1/b)*Sum_{k|b} phi(k)*(1 + z^k)^(v0*b/k).
Here (v,b) = (n+1,n) and (v0,1) = (3,1), so for n >= 1,
a(n) = (1/n)*Sum_{k|gcd(n+1,n)} phi(k)*binomial(3*n/k, (n+1)/k).
This simplifies to
a(n) = (1/n)*binomial(3*n, n+1) for n >= 1. (End)
G.f.: (-(p+r)*(4*r-12*p)^(1/3)+(p-r)*(4*r+12*p)^(1/3)+8)/(12*z), where p = i*sqrt(3*z), r = sqrt(4-27*z), and i = sqrt(-1) is the imaginary unit. - Karol A. Penson, Mar 20 2025

Extensions

Edited following a suggestion of Ralf Stephan, Feb 07 2004
Offset changed to 0 and all formulas checked by Petros Hadjicostas, Jul 27 2020

A007228 a(n) = 3*binomial(4*n,n)/(n+1).

Original entry on oeis.org

3, 6, 28, 165, 1092, 7752, 57684, 444015, 3506100, 28242984, 231180144, 1917334783, 16077354108, 136074334200, 1160946392760, 9973891723635, 86210635955220, 749191930237608, 6541908910355280, 57369142749576660, 505045163173167760, 4461713825057817120
Offset: 0

Views

Author

Keywords

Comments

For n >= 1, a(n) is the number of distinct perforation patterns for deriving (v,b) = (n+1,n) punctured convolutional codes from (4,1). [Edited by Petros Hadjicostas, Jul 27 2020]
Apparently Bégin's (1992) paper was presented at a poster session at the conference and was never published.
a(n) is the total number of down steps between the first and second up steps in all 3-Dyck paths of length 4*(n+1). A 3-Dyck path is a nonnegative lattice path with steps (1,3), (1,-1) that starts and ends at y = 0. - Sarah Selkirk, May 07 2020
From Petros Hadjicostas, Jul 27 2020: (Start)
"A punctured convolutional code is a high-rate code obtained by the periodic elimination (i.e., puncturing) of specific code symbols from the output of a low-rate encoder. The resulting high-rate code depends on both the low-rate code, called the original code, and the number and specific positions of the punctured symbols." (The quote is from Haccoun and Bégin (1989).)
A high-rate code (v,b) (written as R = b/v) can be constructed from a low-rate code (v0,1) (written as R = 1/v0) by deleting from every v0*b code symbols a number of v0*b - v symbols (so that the resulting rate is R = b/v).
Even though my formulas below do not appear in the two published papers in the IEEE Transactions on Communications, from the theory in those two papers, it makes sense to replace "k|b" with "k|v0*b" (and "k|gcd(v,b)" with "k|gcd(v,v0*b)"). Pab Ter, however, uses "k|b" in the Maple programs in the related sequences A007223, A007224, A007225, A007227, and A007229. (End)
Conjecture: for n >= 1, a(n) is odd iff n = 4*A263133(k) + 3 for some k. - Peter Bala, Mar 13 2023

Examples

			From _Petros Hadjicostas_, Jul 29 2020: (Start)
We give some examples to illustrate the comment by _Sarah Selkirk_ about the total number of downs between the 1st and 2nd ups in a 2-Dyck path of length 4*(n+1). We denote by (+3) an up movement by a vector of (1,3) and by (-1) a down movement by a vector of (1,-1). We use powers to denote repetition of the same movement.
(i) For n = 0, we have the following 2-Dyck path of length 4 that contributes to a(0) = 3: (+3)(-1)^3 (no 2nd up here) with a total of 3 downs after the 1st up.
(ii) For n = 1, we have the following 2-Dyck paths of length 8 that contribute to a(1) = 6: (+3)(-1)(+3)(-1)^5, (+3)(-1)^2(+3)(-1)^4, and (+3)(-1)^3(+3)(-1)^3 with a contribution of 1 + 2 + 3 = 6 downs between the 1st and 2nd ups.
(iii) For n = 2, we have the following 2-Dyck paths of length 12 that contribute to a(2) = 28: (+3)(-1)(+3)(-1)^i(+3)(-1)^(8-i) for i = 0..5, (+3)(-1)^2(+3)(-1)^i(+3)^(7-i) for i = 0..4, and (+3)(-1)^3(+3)(-1)^i(+3)(-1)^(6-i) for i = 0..3 with a contribution of 1 x 6 + 2 x 5 + 3 x 4 = 28 downs between the 1st and 2nd ups. (End)
		

References

  • Guy Bégin, On the enumeration of perforation patterns for punctured convolutional codes, Séries Formelles et Combinatoire Algébrique, 4th colloquium, 15-19 Juin 1992, Montréal, Université du Québec à Montréal, pp. 1-10.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    [3*Binomial(4*n,n)/(n+1) : n in [0..25]]; // Wesley Ivan Hurt, Jul 27 2020
  • Mathematica
    Table[3/(n+1) Binomial[4n,n],{n,0,30}] (* Harvey P. Dale, Nov 14 2013 *)
  • PARI
    a(n)={3*binomial(4*n,n)/(n+1)} \\ Andrew Howroyd, May 08 2020
    

Formula

a(n) = C(4*n,n)/(3*n+1) + 2*C(4*n+1,n)/(3*n+2) + 3*C(4*n+2,n)/(3*n+3). - Paul Barry, Nov 05 2006
G.f.: g + g^2 + g^3 where g = 1 + x*g^4 is the g.f. of A002293. - Mark van Hoeij, Nov 11 2011
3*(3*n-1)*(3*n-2)*(n+1)*a(n) - 8*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1) = 0. - R. J. Mathar, Nov 24 2012
From Petros Hadjicostas, Jul 27 2020: (Start)
The number of perforation patterns to derive high-rate convolutional code (v,b) (written as R = b/v) from a given low-rate convolutional code (v0, 1) (written as R = 1/v0) is (1/b)*Sum_{k|gcd(v,b)} phi(k)*binomial(v0*b/k, v/k).
According to Pab Ter's Maple code in the related sequences (see above), this is the coefficient of z^v in the polynomial (1/b)*Sum_{k|b} phi(k)*(1 + z^k)^(v0*b/k).
Here (v,b) = (n+1,n) and (v0,1) = (4,1), so for n >= 1,
a(n) = (1/n)*Sum_{k|gcd(n+1,n)} phi(k)*binomial(4*n/k, (n+1)/k).
This simplifies to
a(n) = (1/n)*binomial(4*n, n+1) for n >= 1. (End)

Extensions

Edited by N. J. A. Sloane, Feb 07 2004 following a suggestion of Ralf Stephan
Reedited by N. J. A. Sloane, May 31 2008 following a suggestion of R. J. Mathar

A275431 Triangle read by rows: T(n,k) = number of ways to insert n pairs of parentheses in k words.

Original entry on oeis.org

1, 2, 1, 5, 2, 1, 14, 8, 2, 1, 42, 24, 8, 2, 1, 132, 85, 28, 8, 2, 1, 429, 286, 100, 28, 8, 2, 1, 1430, 1008, 358, 105, 28, 8, 2, 1, 4862, 3536, 1309, 378, 105, 28, 8, 2, 1, 16796, 12618, 4772, 1410, 384, 105, 28, 8, 2, 1, 58786, 45220, 17556, 5220, 1435, 384, 105, 28, 8, 2, 1
Offset: 1

Views

Author

R. J. Mathar, Jul 27 2016

Keywords

Comments

Multiset transformation of A000108. Each word is dissected by a number of parentheses associated to its length.
Also the number of forests of exactly k (unlabeled) ordered rooted trees with a total of n non-root nodes where each tree has at least 1 non-root node. - Alois P. Heinz, Sep 20 2017

Examples

			       1
       2     1
       5     2     1
      14     8     2     1
      42    24     8     2     1
     132    85    28     8     2     1
     429   286   100    28     8     2     1
    1430  1008   358   105    28     8     2     1
    4862  3536  1309   378   105    28     8     2     1
   16796 12618  4772  1410   384   105    28     8     2     1
   58786 45220 17556  5220  1435   384   105    28     8     2     1
		

Crossrefs

Cf. A000108 (1st column), A007223 (2nd column), A056711 (3rd column), A088327 (row sums).
T(2n,n) gives A292668.

Programs

  • Maple
    C:= proc(n) option remember; binomial(2*n, n)/(n+1) end:
    b:= proc(n, i, p) option remember; `if`(p>n, 0, `if`(n=0, 1,
          `if`(min(i, p)<1, 0, add(b(n-i*j, i-1, p-j)*
           binomial(C(i)+j-1, j), j=0..min(n/i, p)))))
        end:
    T:= (n, k)-> b(n$2, k):
    seq(seq(T(n, k), k=1..n), n=1..12);  # Alois P. Heinz, Apr 13 2017
  • Mathematica
    c[n_] := c[n] = Binomial[2*n, n]/(n + 1);
    b[n_, i_, p_] := b[n, i, p] = If[p > n, 0, If[n == 0, 1, If[Min[i, p] < 1, 0, Sum[b[n - i*j, i - 1, p - j]*Binomial[c[i] + j - 1, j], {j, 0, Min[n/i, p]}]]]];
    T[n_, k_] := b[n, n, k];
    Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, May 18 2018, after Alois P. Heinz *)

Formula

T(n,1) = A000108(n).
T(n,k) = Sum_{c_i*N_i=n,i=1..k} binomial(T(N_i,1)+c_i-1,c_i) for 1
G.f.: Product_{j>=1} 1/(1-y*x^j)^A000108(j). - Alois P. Heinz, Apr 13 2017

A007224 Number of distinct perforation patterns for deriving (v,b) = (n+3,n) punctured convolutional codes from (2,1).

Original entry on oeis.org

2, 9, 38, 143, 546, 2066, 7752, 29070, 108968, 408595, 1533870, 5766243, 21710850, 81880920, 309328008, 1170524970, 4436618940, 16842720336, 64037794548, 243836217702, 929759970392, 3549992610584, 13571935767600, 51950354135888
Offset: 4

Author

Keywords

Comments

From Petros Hadjicostas, Jul 27 2020: (Start)
"A punctured convolutional code is a high-rate code obtained by the periodic elimination (i.e., puncturing) of specific code symbols from the output of a low-rate encoder. The resulting high-rate code depends on both the low-rate code, called the original code, and the number and specific positions of the punctured symbols." (The quote is from Haccoun and Bégin (1989).)
A high-rate code (v,b) (written as R = b/v) can be constructed from a low-rate code (v0,1) (written as R = 1/v0) by deleting from every v0*b code symbols a number of v0*b - v symbols (so that the resulting rate is R = b/v).
Even though the formulas below do not appear in the two published papers in the IEEE Transactions on Communications, from the theory in those two papers, it makes sense to replace "k|b" with "k|v0*b" (and "k|gcd(v,b)" with "k|gcd(v,v0*b)"). Pab Ter, however, uses "k|b" in the Maple program below. (End)

References

  • Guy Bégin, On the enumeration of perforation patterns for punctured convolutional codes, Séries Formelles et Combinatoire Algébrique, 4th colloquium, 15-19 Juin 1992, Montréal, Université du Québec à Montréal, pp. 1-10.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    with(numtheory):P:=proc(b,v0) local k: RETURN(add(phi(k)*(1+z^k)^(v0*(b/k)),k=divisors(b))/b): end; seq(coeff(P(b,2),z,b+3),b=4..40); (Pab Ter)
  • Mathematica
    a[n_] := DivisorSum[GCD[n+3, n], EulerPhi[#] Binomial[2n/#, (n+3)/#]&]/n;
    Table[a[n], {n, 4, 40}] (* Jean-François Alcover, Mar 09 2023 *)

Formula

From Petros Hadjicostas, Jul 27 2020: (Start)
The number of perforation patterns to derive high-rate convolutional code (v,b) (written as R = b/v) from a given low-rate convolutional code (v0, 1) (written as R = 1/v0) is (1/b)*Sum_{k|gcd(v,b)} phi(k)*binomial(v0*b/k, v/k).
According to Pab Ter's Maple code, this is the coefficient of z^v in the polynomial (1/b)*Sum_{k|b} phi(k)*(1 + z^k)^(v0*b/k).
Here (v,b) = (n+3,n) and (v0,1) = (2,1), so
a(n) = (1/n)*Sum_{k|gcd(n+3,n)} phi(k)*binomial(2*n/k, (n+3)/k).
This simplifies to
a(n) = (1/n)*(binomial(2*n, n+3) + 2*[(n mod 3) == 0]*binomial(2*n/3, (n/3) + 1)). (End)

Extensions

More terms from Pab Ter (pabrlos2(AT)yahoo.com), Nov 13 2005

A007225 Number of distinct perforation patterns for deriving (v,b) = (n+4,n) punctured convolutional codes from (2,1).

Original entry on oeis.org

2, 12, 52, 232, 952, 3888, 15504, 61333, 240350, 937508, 3641820, 14112560, 54587280, 210907168, 814278240, 3142611402, 12126758436, 46796872472, 180619420520, 697320058864, 2693097842512, 10405151052320, 40219629005920
Offset: 5

Author

Keywords

Comments

From Petros Hadjicostas, Jul 27 2020: (Start)
"A punctured convolutional code is a high-rate code obtained by the periodic elimination (i.e., puncturing) of specific code symbols from the output of a low-rate encoder. The resulting high-rate code depends on both the low-rate code, called the original code, and the number and specific positions of the punctured symbols." (The quote is from Haccoun and Bégin (1989).)
A high-rate code (v,b) (written as R = b/v) can be constructed from a low-rate code (v0,1) (written as R = 1/v0) by deleting from every v0*b code symbols a number of v0*b - v symbols (so that the resulting rate is R = b/v).
Even though the formulas below do not appear in the two published papers in the IEEE Transactions on Communications, from the theory in those two papers, it makes sense to replace "k|b" with "k|v0*b" (and "k|gcd(v,b)" with "k|gcd(v,v0*b)"). Pab Ter, however, uses "k|b" in the Maple program below. (End)

References

  • Guy Bégin, On the enumeration of perforation patterns for punctured convolutional codes, Séries Formelles et Combinatoire Algébrique, 4th colloquium, 15-19 Juin 1992, Montréal, Université du Québec à Montréal, pp. 1-10.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    with(numtheory):P:=proc(b,v0) local k: RETURN(add(phi(k)*(1+z^k)^(v0*(b/k)),k=divisors(b))/b): end; seq(coeff(P(b,2),z,b+4),b=5..40); (Pab Ter)
  • Mathematica
    a[n_] := DivisorSum[GCD[n+4, n], EulerPhi[#]*Binomial[2n/#, (n+4)/#]&]/n;
    Table[a[n], {n, 5, 27}] (* Jean-François Alcover, Mar 10 2023 *)

Formula

From Petros Hadjicostas, Jul 27 2020: (Start)
The number of perforation patterns to derive high-rate convolutional code (v,b) (written as R = b/v) from a given low-rate convolutional code (v0, 1) (written as R = 1/v0) is (1/b)*Sum_{k|gcd(v,b)} phi(k)*binomial(v0*b/k, v/k).
According to Pab Ter's Maple code, this is the coefficient of z^v in the polynomial (1/b)*Sum_{k|b} phi(k)*(1 + z^k)^(v0*b/k).
Here (v,b) = (n+4,n) and (v0,1) = (2,1), so
a(n) = (1/n)*Sum_{k|gcd(n+4,n)} phi(k)*binomial(2*n/k, (n+4)/k).
This simplifies to
a(n) = (1/n)*(binomial(2*n, n+4) + [(n mod 2) == 0]*binomial(n, (n/2) + 2) + 2*[(n mod 4) == 0]*binomial(n/2, (n/4) + 1)). (End)

Extensions

More terms from Pab Ter (pabrlos2(AT)yahoo.com), Nov 13 2005

A007227 Number of distinct perforation patterns for deriving (v,b) = (n+2,n) punctured convolutional codes from (3,1).

Original entry on oeis.org

9, 42, 236, 1287, 7314, 41990, 245256, 1448655, 8649823, 52106040, 316360752, 1933910820, 11893566078, 73537906926, 456864894288, 2850557192175, 17854854154215, 112230508880490, 707714010205020, 4475876883386895
Offset: 2

Author

Keywords

Comments

From Petros Hadjicostas, Jul 27 2020: (Start)
"A punctured convolutional code is a high-rate code obtained by the periodic elimination (i.e., puncturing) of specific code symbols from the output of a low-rate encoder. The resulting high-rate code depends on both the low-rate code, called the original code, and the number and specific positions of the punctured symbols." (The quote is from Haccoun and Bégin (1989).)
A high-rate code (v,b) (written as R = b/v) can be constructed from a low-rate code (v0,1) (written as R = 1/v0) by deleting from every v0*b code symbols a number of v0*b - v symbols (so that the resulting rate is R = b/v).
Even though the formulas below do not appear in the two published papers in the IEEE Transactions on Communications, from the theory in those two papers, it makes sense to replace "k|b" with "k|v0*b" (and "k|gcd(v,b)" with "k|gcd(v,v0*b)"). Pab Ter, however, uses "k|b" in the Maple program below. (End)

References

  • Guy Bégin, On the enumeration of perforation patterns for punctured convolutional codes, Séries Formelles et Combinatoire Algébrique, 4th colloquium, 15-19 Juin 1992, Montréal, Université du Québec à Montréal, pp. 1-10.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    with(numtheory):P:=proc(b,v0) local k: RETURN(add(phi(k)*(1+z^k)^(v0*(b/k)),k=divisors(b))/b): end; seq(coeff(P(b,3),z,b+2),b=2..40); (Pab Ter)

Formula

From Petros Hadjicostas, Jul 27 2020: (Start)
The number of perforation patterns to derive high-rate convolutional code (v,b) (written as R = b/v) from a given low-rate convolutional code (v0, 1) (written as R = 1/v0) is (1/b)*Sum_{k|gcd(v,b)} phi(k)*binomial(v0*b/k, v/k).
According to Pab Ter's Maple code, this is the coefficient of z^v in the polynomial (1/b)*Sum_{k|b} phi(k)*(1 + z^k)^(v0*b/k).
Here (v,b) = (n+2,n) and (v0,1) = (3,1), so
a(n) = (1/n)*Sum_{k|gcd(n+2,n)} phi(k)*binomial(3*n/k, (n+2)/k).
This simplifies to
a(n) = (1/n)*(binomial(3*n, n+2) + [(n mod 2) = 0]*binomial(3*n/2, (n/2) + 1)), where [ ] is the Iverson bracket. (End)

Extensions

More terms from Pab Ter (pabrlos2(AT)yahoo.com), Nov 13 2005

A275206 Expansion of (A(x)^2 - A(x^2))/2 where A(x) = A000108(x) - 1.

Original entry on oeis.org

0, 0, 0, 2, 6, 24, 80, 286, 994, 3536, 12576, 45220, 163372, 594320, 2172768, 7983990, 29464010, 109174560, 405990464, 1514797020, 5669004692, 21275014800, 80047213792, 301892460012, 1141068949396, 4321730134624, 16399422014400, 62340424959176, 237373155238104, 905251034394784
Offset: 0

Author

R. J. Mathar, Jul 19 2016

Keywords

Comments

Number of ways to distribute n pairs of parentheses into 2 distinct patterns where each pattern represents a Catalan ordering (A000108), and each pattern must contain at least one pair of parentheses.
If one of the groups is allowed to have no parentheses, we arrive at A000150 (with a different offset).
Analog of A216785 with Catalan number replacing connected graph counts.
From Petros Hadjicostas, Jul 27 2020: (Start)
It is proved in A050182 that A050182(n) = 1/(2*n + 4)*(binomial(2*n + 4, n) - [(n mod 2) == 0]*binomial(n + 2, n/2)).
Let C(x) = A(x) + 1 = Sum_{n >= 0} c(n)*x^n be the g.f. of the Catalan numbers A000108. Then C(x)^2 = (C(x) - 1)/x. Then (A(x) + 1)^2 = A(x)/x, and thus, A(x)^2 = -2*A(x) - 1 + A(x)/x. Thus, (A(x)^2 - A(x^2))/2 = (-2*A(x) - 1 + A(x)/x - A(x^2))/2.
Substituting A(x) = Sum_{n >= 1} c(n)*x^n in the above expression, we get (after some algebra) that a(n) = (-2*c(n) + c(n+1) - [(n mod 2) == 0]*c(n/2))/2 for n >= 1. It is then easy to prove that a(n) = 2*A050182(n-2) = (1/n)*(binomial(2*n, n-2) - [(n mod 2) == 0]*binomial(n, (n/2) - 1)) for n >= 2, thus proving the conjecture below. (End)

Crossrefs

Programs

  • Mathematica
    A[x_] = (1 - Sqrt[1 - 4x])/(2x) - 1;
    CoefficientList[(A[x]^2 - A[x^2])/2 + O[x]^30, x] (* Jean-François Alcover, Apr 30 2023 *)

Formula

a(2*n+1) = A007223(2*n+1).
Conjecture: a(n) = 2*A050182(n-2) for n >= 2.
From Petros Hadjicostas, Jul 27 2020: (Start)
a(n) = (-2*c(n) + c(n+1) - [(n mod 2) == 0]*c(n/2))/2 for n >= 1, where c = A000108.
a(n) = (1/n)*(binomial(2*n, n-2) - [(n mod 2) == 0]*binomial(n, (n/2) - 1)) for n >= 2. (End)
D-finite with recurrence: 5*(n-3)*(n+2)*(n+1)*a(n) + 32*(n-5)*(n-6)*(2*n-11)*a(n-6) + 80*(n-5)*(n-4)*(n-3)*a(n-5) - 40*(n-4)*(7*n^2-35*n+39)*a(n-4) + (120*n^3 - 600*n^2 + 660*n + 60)*a(n-3) + 2*(n-5)*(23*n^2-44*n+6)*a(n-2) - 5*(n+1)*(7*n^2-25*n+6)*a(n-1) = 0. - Georg Fischer, Feb 12 2025
Showing 1-7 of 7 results.