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 1-10 of 15 results. Next

A024497 Duplicate of A007228.

Original entry on oeis.org

3, 6, 28, 165, 1092, 7752, 57684, 444015, 3506100, 28242984, 231180144, 1917334783
Offset: 1

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Author

Keywords

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

A124724 a(n) = (4/(n + 1)) * C(5*n, n).

Original entry on oeis.org

4, 10, 60, 455, 3876, 35420, 339300, 3362260, 34179860, 354465254, 3735373880, 39884521950, 430571952300, 4691735290080, 51534335175776, 570003171679020, 6343110854237300, 70968228417131850, 797820661622862900
Offset: 0

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Author

Paul Barry, Nov 05 2006

Keywords

Comments

a(n) is the total number of down steps between the first and second up steps in all 4-Dyck paths of length 5*(n+1). A 4-Dyck path is a nonnegative lattice path with steps (1,4), (1,-1) that starts and ends at y = 0. - Sarah Selkirk, May 07 2020

Crossrefs

Programs

  • Mathematica
    Array[(4/(# + 1))*Binomial[5 #, #] &, 28, 0] (* Michael De Vlieger, Apr 12 2023 *)
  • PARI
    a(n) = (4/(n+1)) * binomial(5*n,n); \\ Michel Marcus, May 08 2020

Formula

a(n) = C(5*n, n)/(4*n + 1) + 2*C(5*n + 1, n)/(4*n + 2) + 3*C(5*n + 2, n)/(4*n + 3) + 4*C(5*n + 3, n)/(4*n + 4).

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

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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.

A334645 a(n) is the total number of down steps between the 2nd and 3rd up steps in all 3-Dyck paths of length 4*n. A 3-Dyck path is a nonnegative lattice path with steps (1, 3), (1, -1) that starts and ends at y = 0.

Original entry on oeis.org

0, 0, 18, 52, 277, 1752, 12120, 88692, 674751, 5282160, 42267384, 344152080, 2842055359, 23746693240, 200383750632, 1705243729560, 14617677294675, 126106202849760, 1094034474058488, 9538676631305712, 83536778390997780, 734521734171474400, 6481894477750488160
Offset: 0

Views

Author

Benjamin Hackl, May 12 2020

Keywords

Comments

For n = 2, there is no 3rd up step, a(2) = 18 enumerates the total number of down steps between the 2nd up step and the end of the path.

Examples

			For n = 2, the 3-Dyck paths are UDDDUDDD, UDDUDDDD, UDUDDDDD, UUDDDDDD. In total, there are a(2) = 3 + 4 + 5 + 6 = 18 down steps between the 2nd up step and the end of the path.
		

Crossrefs

Programs

  • SageMath
    [3*sum([binomial(4*j + 1, j)*binomial(4*(n - j), n - j)/(4*j + 1)/(n - j + 1) for j in srange(1, 3)]) if n > 1 else 0 for n in srange(30)] # Benjamin Hackl, May 12 2020

Formula

a(0) = a(1) = 0 and a(n) = 3*Sum_{j=0..2} binomial(4*j+1, j) * binomial(4*(n-j), n-j)/((4*j+1) * (n-j+1)) for n > 1.

A334647 a(n) is the total number of down steps between the first and second up steps in all 3_1-Dyck paths of length 4*n.

Original entry on oeis.org

0, 5, 16, 78, 470, 3153, 22588, 169188, 1308762, 10374460, 83829856, 687929086, 5717602930, 48030047206, 407142435000, 3478286028840, 29917720938690, 258866494630164, 2251694583485824, 19677972159742360, 172694287830500440, 1521328368800877065
Offset: 0

Views

Author

Benjamin Hackl, May 12 2020

Keywords

Comments

A 3_1-Dyck path is a lattice path with steps (1, 3), (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) = 5 enumerates the total number of down steps between the 1st up step and the end of the path.

Examples

			For n = 1, the 3_1-Dyck paths are UDDD, DUDD. This corresponds to a(1) = 3 + 2 = 5 down steps between the 1st up step and the end of the path.
For n = 2, the 3_1-Dyck paths are DUDDDUDD, DUDDUDDD, DUDUDDDD, DUUDDDDD, UDDDDUDD, UDDDUDDD, UDDUDDDD, UDUDDDDD, UUDDDDDD. In total, there are a(2) = 3 + 2 + 1 + 0 + 4 + 3 + 2 + 1 + 0 = 16 down steps between the 1st and 2nd up step.
		

Crossrefs

Programs

  • Mathematica
    a[0] = 0; a[n_] := 3 * Binomial[4*n, n]/(n + 1) - 2 * Binomial[4*n + 1, n]/(n + 1) + 6 * Binomial[4*(n - 1), n - 1]/n - 2 * Boole[n == 1]; Array[a, 22, 0] (* Amiram Eldar, May 12 2020 *)
  • SageMath
    [3*binomial(4*n, n)/(n+1) - 2*binomial(4*n+1, n)/(n+1) + 6*binomial(4*(n-1), n-1)/n - 2*(n==1) if n > 0 else 0 for n in srange(30)] # Benjamin Hackl, May 12 2020

Formula

a(0) = 0 and a(n) = 3*binomial(4*n, n)/(n+1) - 2*binomial(4*n+1, n)/(n+1) + 6*binomial(4*(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.

A334646 a(n) is the total number of down steps between the 3rd and 4th up steps in all 3-Dyck paths of length 4*n.

Original entry on oeis.org

0, 0, 0, 118, 409, 2368, 15750, 112716, 845295, 6551208, 52035714, 421286280, 3463401007, 28832656408, 242565115858, 2058945519936, 17611312647075, 151647023490480, 1313460091978458, 11435310622320552, 100019000856225156, 878443730199290560
Offset: 0

Views

Author

Benjamin Hackl, May 12 2020

Keywords

Comments

A 3-Dyck path is a nonnegative lattice path with steps (1, 3), (1, -1) that starts and ends at y = 0.

Examples

			For n = 3, there is no 4th up step, a(3) = 118 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_] := 3 * Sum[Binomial[4*j + 1, j] * Binomial[4*(n - j), n - j]/((4*j + 1)*(n - j + 1)), {j, 1, 3}]; Array[a, 22, 0] (* Amiram Eldar, May 12 2020 *)
  • SageMath
    [3*sum([binomial(4*j + 1, j)*binomial(4*(n - j), n - j)/(4*j + 1)/(n - j + 1) for j in srange(1, 4)]) if n > 2 else 0 for n in srange(30)] # Benjamin Hackl, May 12 2020

Formula

a(0) = a(1) = a(2) = 0 and a(n) = 3*Sum_{j=1..3} binomial(4*j+1, j)*binomial(4*(n-j), n-j)/((4*j+1)*(n-j+1)) for n > 2.

A334648 a(n) is the total number of down steps between the second and third up steps in all 3_1-Dyck paths of length 4*n.

Original entry on oeis.org

0, 0, 34, 132, 722, 4638, 32416, 238956, 1827918, 14370595, 115384756, 942115942, 7798224226, 65286060253, 551838621972, 4702955036640, 40366238473530, 348631520142879, 3027590307082804, 26420699531880832, 231571468023697960, 2037650653547067005
Offset: 0

Views

Author

Benjamin Hackl, May 12 2020

Keywords

Comments

A 3_1-Dyck path is a lattice path with steps (1, 3), (1, -1) that starts and ends at y = 0 and stays above the line y = -1.
For n = 2, there is no 3rd up step, a(2) = 34 enumerates the total number of down steps between the 2nd up step and the end of the path.

Examples

			For n = 2, the 3_1-Dyck paths are DUDDDUDD, DUDDUDDD, DUDUDDDD, DUUDDDDD, UDDDDUDD, UDDDUDDD, UDDUDDDD, UDUDDDDD, UUDDDDDD. In total, there are a(2) = 2 + 3 + 4 + 5 + 2 + 3 + 4 + 5 + 6 = 34 down steps between the 2nd up step and the end of the path.
		

Crossrefs

Programs

  • Mathematica
    a[0] = a[1] = 0; a[n_] := Binomial[4*n + 1, n]/(4*n + 1) + 6 * Sum[Binomial[4*j + 2, j] * Binomial[4*(n - j), n - j]/((4*j + 2)*(n - j + 1)), {j, 1, 2}] - 9 * Boole[n == 2]; Array[a, 22, 0] (* Amiram Eldar, May 12 2020 *)
  • SageMath
    [binomial(4*n + 1, n)/(4*n + 1) + 6*sum([binomial(4*j + 2, j)*binomial(4*(n - j), n - j)/(4*j + 2)/(n - j + 1) for j in srange(1, 3)]) - 9*(n==2) if n > 1 else 0 for n in srange(30)] # Benjamin Hackl, May 12 2020

Formula

a(0) = a(1) = 0 and a(n) = binomial(4*n+1, n)/(4*n+1) + 6*Sum_{j=1..2} binomial(4*j+2, j)*binomial(4*(n-j), n-j)/((4*j+2)*(n-j+1)) - 9*[n=2] for n > 1, where [ ] is the Iverson bracket.

A334649 a(n) is the total number of down steps between the third and fourth up steps in all 3_1-Dyck paths of length 4*n.

Original entry on oeis.org

0, 0, 0, 236, 1034, 6094, 40996, 295740, 2231022, 17370163, 138473536, 1124433142, 9266859394, 77307427741, 651540030688, 5538977450256, 47442103851930, 409000732566399, 3546232676711824, 30903652601552272, 270529448396053576, 2377829916885541565
Offset: 0

Views

Author

Benjamin Hackl, May 12 2020

Keywords

Comments

A 3_1-Dyck path is a lattice path with steps (1, 3), (1, -1) that starts and ends at y = 0 and stays above the line y = -1.
For n = 3, there is no 4th up step, a(3) = 236 enumerates the total number of down steps between the 3rd up step and the end of the path.

Crossrefs

Programs

  • SageMath
    [binomial(4*n + 1, n)/(4*n + 1) + 6*sum([binomial(4*j + 2, j)*binomial(4*(n - j), n - j)/(4*j + 2)/(n - j + 1) for j in srange(1, 4)]) - 52*(n==3) if n > 2 else 0 for n in srange(30)] # Benjamin Hackl, May 12 2020

Formula

a(0) = a(1) = a(2) = 0 and a(n) = binomial(4*n+1, n)/(4*n+1) + 6*Sum_{j=1..3} binomial(4*j+2, j)*binomial(4*(n-j), n-j)/((4*j+2)*(n-j+1)) - 52*[n=3] for n > 2, where [ ] is the Iverson bracket.
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