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

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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

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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

A334642 a(n) is the total number of down steps between the first and second up steps in all 2_1-Dyck paths of length 3*n. A 2_1-Dyck path is a lattice path with steps (1, 2), (1, -1) that starts and ends at y = 0 and stays above the line y = -1.

Original entry on oeis.org

0, 3, 9, 32, 139, 669, 3430, 18360, 101403, 573551, 3305445, 19340100, 114579348, 685962172, 4143459504, 25220816752, 154545611355, 952583230899, 5902090839715, 36738469359480, 229636903762035, 1440759023752125, 9070230371741490, 57278432955350880
Offset: 0

Views

Author

Benjamin Hackl, May 07 2020

Keywords

Comments

For n = 1, there is no 2nd up step, a(1) = 3 enumerates the total number of down steps between the 1st up step and the end of the path.

Examples

			For n = 1, the 2_1-Dyck paths are UDD, DUD. This corresponds to a(1) = 2 + 1 = 3 down steps between the 1st up step and the end of the path.
For n = 2, the 2_1-Dyck paths are UUDDDD, UDUDDD, UDDUDD, UDDDUD, DUDDUD, DUDUDD, DUUDDD. In total, there are a(2) = 0 + 1 + 2 + 3 + 2 + 1 + 0 = 9 down steps between the 1st and 2nd up step.
		

Crossrefs

Programs

  • Mathematica
    a[0] = 0; a[n_] := 2 * Binomial[3*n, n]/(n + 1) - Binomial[3*n + 1, n]/(n + 1) + 4 * Binomial[3*(n - 1), n - 1]/n - 2 * Boole[n == 1]; Array[a, 24, 0] (* Amiram Eldar, May 09 2020 *)
  • PARI
    a(n) = if (n==0, 0, 2*binomial(3*n, n)/(n+1) - binomial(3*n+1, n)/(n+1) + 4*binomial(3*(n-1), n-1)/n - 2*(n==1)); \\ Michel Marcus, May 09 2020

Formula

a(0) = 0 and a(n) = 2*binomial(3*n, n)/(n+1) - binomial(3*n+1, n)/(n+1) + 4*binomial(3*(n-1), n-1)/n - 2*[n=1] for n > 0, where [ ] is the Iverson bracket.

A334640 a(n) is the total number of down steps between the 2nd and 3rd up steps in all 2-Dyck paths of length 3*n. A 2-Dyck path is a nonnegative lattice path with steps (1, 2), (1, -1) that starts and ends at y = 0.

Original entry on oeis.org

0, 0, 9, 19, 72, 324, 1595, 8307, 44982, 250648, 1427679, 8274825, 48644310, 289334160, 1738043892, 10529070020, 64252519830, 394601627376, 2437058926871, 15126463230165, 94306717535940, 590318477063700, 3708527622652755, 23374587898663155, 147770791807427880
Offset: 0

Views

Author

Benjamin Hackl, May 07 2020

Keywords

Comments

For n = 2, there is no 3rd up step, a(2) = 9 enumerates the total number of down steps between the 2nd up step and the end of the path.

Examples

			For n = 2, there are the 2-Dyck paths UUDDDD, UDUDDD, UDDUDD. Between the 2nd up step and the end of the path there are a(2) = 4 + 3 + 2 = 9 down steps in total.
		

Crossrefs

Programs

  • Maple
    b:= proc(x, y, u, c) option remember; `if`(x=0, c,
         `if`(y+20, b(x-1, y-1, u, c+`if`(u=2, 1, 0)), 0))
        end:
    a:= n-> b(3*n, 0$3):
    seq(a(n), n=0..24);  # Alois P. Heinz, May 09 2020
    # second Maple program:
    a:= proc(n) option remember; `if`(n<3, [0$2, 9][n+1],
         (3*(n-1)*(3*n-8)*(3*n-7)*(13*n-20)*a(n-1))/
         (2*(13*n-33)*(n-2)*(2*n-3)*n))
        end:
    seq(a(n), n=0..24);  # Alois P. Heinz, May 09 2020
  • Mathematica
    a[0] = a[1] = 0; a[n_] := 2 * Sum[Binomial[3*j + 1, j] * Binomial[3*(n - j), n - j]/((3*j + 1)*(n - j + 1)), {j, 1, 2}]; Array[a, 25, 0] (* Amiram Eldar, May 09 2020 *)
  • PARI
    a(n) = if (n<=1, 0, 2*sum(j=1, 2, binomial(3*j+1,j) * binomial(3*(n-j),n-j)/((3*j+1)*(n-j+1)))); \\ Michel Marcus, May 09 2020

Formula

a(0) = a(1) = 0 and a(n) = 2*Sum_{j=1..2} binomial(3*j+1,j) * binomial(3*(n-j),n-j) / ((3*j+1)*(n-j+1)) for n > 1.

A334641 a(n) is the total number of down steps between the 3rd and 4th up steps in all 2-Dyck paths of length 3*n.

Original entry on oeis.org

0, 0, 0, 43, 108, 444, 2099, 10683, 56994, 314296, 1776519, 10236081, 59892690, 354886920, 2125117332, 12839859620, 78176677734, 479177993904, 2954360065247, 18309779343549, 114001476318240, 712751759478780, 4472908385838795, 28165267333869435
Offset: 0

Views

Author

Benjamin Hackl, May 07 2020

Keywords

Comments

A 2-Dyck path is a nonnegative lattice path with steps (1, 2), (1, -1) that starts and ends at y = 0.
For n = 3, there is no 4th up step, a(3) = 43 enumerates the total number of down steps between the 3rd up step and the end of the path.

Crossrefs

Programs

  • Mathematica
    a[0] = a[1] = a[2] = 0; a[n_] := 2 * Sum[Binomial[3*j + 1, j] * Binomial[3*(n - j), n - j]/((3*j + 1)*(n - j + 1)), {j, 1, 3}]; Array[a, 24, 0] (* Amiram Eldar, May 09 2020 *)
  • PARI
    a(n) = if (n<=2, 0, 2*sum(j=1, 3, binomial(3*j+1, j)*binomial(3*(n-j), n-j)/((3*j+1)*(n-j+1)))); \\ Michel Marcus, May 09 2020

Formula

a(0) = a(1) = a(2) = 0 and a(n) = 2*Sum_{j=1..3}binomial(3*j+1, j)*binomial(3*(n-j), n-j)/((3*j+1)*(n-j+1)) for n > 2.

A334651 a(n) is the total number of down steps between the first and second up steps in all 4_1-Dyck paths of length 5*n.

Original entry on oeis.org

0, 7, 25, 155, 1195, 10282, 94591, 910480, 9054965, 92310075, 959473878, 10129715890, 108327387675, 1170975480360, 12773887368040, 140445927510832, 1554748206904325, 17314584431331025, 193849445090545875, 2180550929942519685, 24632294533221865028
Offset: 0

Views

Author

Benjamin Hackl, May 13 2020

Keywords

Comments

A 4_1-Dyck path is a lattice path with steps (1, 4), (1, -1) that starts and ends at y = 0 and stays above the line y = -1.
For n = 1, there is no 2nd up step, a(1) = 7 enumerates the total number of down steps between the 1st up step and the end of the path.

Examples

			For n = 1, the 4_1-Dyck paths are DUDDD, UDDDD. This corresponds to a(1) = 3 + 4 = 7 down steps between the 1st up step and the end of the path.
		

Crossrefs

Programs

  • Mathematica
    a[0] = 0; a[n_] := 4 * Binomial[5*n, n]/(n + 1) - 3 * Binomial[5*n + 1, n]/(n + 1) + 8*Binomial[5*(n - 1), n - 1]/n - 2 * Boole[n == 1]; Array[a, 21, 0] (* Amiram Eldar, May 13 2020 *)
  • SageMath
    [4*binomial(5*n, n)/(n + 1) - 3*binomial(5*n + 1, n)/(n + 1) + 8*binomial(5*(n - 1), n - 1)/n - 2*(n==1) if n > 0 else 0 for n in srange(30)]

Formula

a(0) = 0 and a(n) = 4*binomial(5*n, n)/(n+1) - 3*binomial(5*n+1, n)/(n+1) + 8*binomial(5*(n-1), n-1)/n - 2*[n=1] for n > 0, where [ ] is the Iverson bracket.

A241262 Array t(n,k) = binomial(n*k, n+1)/n, where n >= 1 and k >= 2, read by ascending antidiagonals.

Original entry on oeis.org

1, 2, 3, 5, 10, 6, 14, 42, 28, 10, 42, 198, 165, 60, 15, 132, 1001, 1092, 455, 110, 21, 429, 5304, 7752, 3876, 1020, 182, 28, 1430, 29070, 57684, 35420, 10626, 1995, 280, 36, 4862, 163438, 444015, 339300, 118755, 24570, 3542, 408, 45, 16796, 937365, 3506100, 3362260, 1391280, 324632, 50344, 5850, 570, 55
Offset: 1

Views

Author

Jean-François Alcover, Apr 18 2014

Keywords

Comments

About the "root estimation" question asked in MathOverflow, one can check (at least numerically) that, for instance with k = 4 and a = 1/11, the series a^-1 + (k - 1) + Sum_{n>=} (-1)^n*binomial(n*k, n+1)/n*a^n evaluates to the positive solution of x^k = (x+1)^(k-1).
Row 1 is A000217 (triangular numbers),
Row 2 is A006331 (twice the square pyramidal numbers),
Row 3 is A067047(3n) = lcm(3n, 3n+1, 3n+2, 3n+3)/12 (from column r=4 of A067049),
Row 4 is A222715(2n) = (n-1)*n*(2n-1)*(4n-3)*(4n-1)/15,
Row 5 is not in the OEIS.
Column 1 is A000108 (Catalan numbers),
Column 2 is A007226 left shifted 1 place,
Column 4 is A007228 left shifted 1 place,
Column 5 is A124724 left shifted 1 place,
Column 6 is not in the OEIS.

Examples

			Array begins:
    1,    3,     6,     10,      15,      21, ...
    2,   10,    28,     60,     110,     182, ...
    5,   42,   165,    455,    1020,    1995, ...
   14,  198,  1092,   3876,   10626,   24570, ...
   42, 1001,  7752,  35420,  118755,  324632, ...
  132, 5304, 57684, 339300, 1391280, 4496388, ...
  etc.
		

References

  • N. S. S. Gu, H. Prodinger, S. Wagner, Bijections for a class of labeled plane trees, Eur. J. Combinat. 31 (2010) 720-732, doi|10.1016/j.ejc.2009.10.007, Theorem 2

Crossrefs

Programs

  • Mathematica
    t[n_, k_] := Binomial[n*k, n+1]/n; Table[t[n-k+2, k], {n, 1, 10}, {k, 2, n+1}] // Flatten
Showing 1-7 of 7 results.