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 13 results. Next

A082143 First subdiagonal of number array A082137.

Original entry on oeis.org

1, 3, 20, 140, 1008, 7392, 54912, 411840, 3111680, 23648768, 180590592, 1384527872, 10650214400, 82158796800, 635361361920, 4924050554880, 38233804308480, 297374033510400, 2316387208396800, 18067820225495040, 141101072237199360, 1103153837490831360
Offset: 0

Views

Author

Paul Barry, Apr 06 2003

Keywords

Examples

			a(0)=(2^(-1)+(0^0)/2)C(1,0)=2*(1/2)=1 (use 0^0=1).
		

Crossrefs

Programs

  • Haskell
    a082143 0 = 1
    a082143 n = (a000079 $ n - 1) * (a001700 n)
    -- Reinhard Zumkeller, Jan 15 2015
    
  • Magma
    [(2^(n-1) + 0^n/2)*Binomial(2*n+1,n): n in [0..30]]; // G. C. Greubel, Feb 05 2018
  • Mathematica
    Join[{1}, Table[2^(n-1)* Binomial[2*n+1,n], {n,1,30}]] (* G. C. Greubel, Feb 05 2018 *)
  • PARI
    for(n=0,30, print1((2^(n-1) + 0^n/2)*Binomial(2*n+1,n), ", ")) \\ G. C. Greubel, Feb 05 2018
    

Formula

a(n) = (2^(n-1) + 0^n/2)*C(2n+1, n).
Conjecture: (n+1)*a(n) +4*(-2*n-1)*a(n-1)=0. - R. J. Mathar, Oct 19 2014
From Reinhard Zumkeller, Jan 15 2015: (Start)
a(n) = A000079(n-1) * A001700(n), for n > 0.
a(n) = A069720(n+1)/2. (End)
From Amiram Eldar, Jan 16 2024: (Start)
Sum_{n>=0} 1/a(n) = 64*arcsin(1/(2*sqrt(2)))/(7*sqrt(7)) + 1/7.
Sum_{n>=0} (-1)^n/a(n) = 32*log(2)/27 - 1/9. (End)

A082144 A subdiagonal of number array A082137.

Original entry on oeis.org

1, 4, 30, 224, 1680, 12672, 96096, 732160, 5601024, 42997760, 331082752, 2556051456, 19778969600, 153363087360, 1191302553600, 9268801044480, 72219408138240, 563445537177600, 4401135695953920, 34414895667609600, 269374774271016960, 2110381254330286080
Offset: 0

Views

Author

Paul Barry, Apr 06 2003

Keywords

Examples

			a(0)=(2^(-1)+(0^0)/2)C(2,0)=2*(1/2)=1 (use 0^0=1).
		

Crossrefs

Programs

  • Magma
    [(2^(n-1) + 0^n/2)*Binomial(2*n+2,n): n in [0..30]]; // G. C. Greubel, Feb 05 2018
  • Mathematica
    Join[{1}, Table[2^(n-1)*Binomial[2*n+2, n], {n,1,50}]] (* G. C. Greubel, Feb 05 2018 *)
  • PARI
    for(n=0,30, print1((2^(n-1) + 0^n/2)*Binomial(2*n+2,n), ", ")) \\ G. C. Greubel, Feb 05 2018
    

Formula

a(n) = (2^(n-1) + 0^n/2)*C(2*n+2, n).
(n+2)*a(n) +12*(-n-1)*a(n-1) +16*(2*n-1)*a(n-2)=0. - R. J. Mathar, Oct 29 2014
From Amiram Eldar, Jan 16 2024: (Start)
Sum_{n>=0} 1/a(n) = 88*arccot(sqrt(7))/(7*sqrt(7)) - 3/7.
Sum_{n>=0} (-1)^n/a(n) = 52*log(2)/27 - 5/9. (End)

A082145 A subdiagonal of number array A082137.

Original entry on oeis.org

1, 5, 42, 336, 2640, 20592, 160160, 1244672, 9674496, 75246080, 585761792, 4564377600, 35602145280, 277970595840, 2172375244800, 16992801914880, 133035751833600, 1042374243778560, 8173537721057280, 64136851016908800, 503613708419727360, 3956964851869286400
Offset: 0

Views

Author

Paul Barry, Apr 06 2003

Keywords

Examples

			a(0) = ( 2^(-1)+(0^0)/2 )*C(3,0) = ( 1/2+1/2 )*1 = 1 (use 0^0 = 1). - clarified by _Jon Perry_, Oct 29 2014
		

Crossrefs

Programs

  • Magma
    [(2^(n-1)+(0^n)/2)*Binomial(2*n+3, n): n in [0..30]]; // Vincenzo Librandi, Oct 30 2014
    
  • Maple
    Z:=(1-3*z-sqrt(1-4*z))/sqrt(1-4*z)/64: Zser:=series(Z, z=0, 32): seq(coeff(Zser*2^(n+1), z, n), n=4..23); # Zerinvary Lajos, Jan 01 2007
  • Mathematica
    Join[{1}, Table[2^(n-1)* Binomial[2*n+3,n], {n,1,30}]] (* G. C. Greubel, Feb 05 2018 *)
  • PARI
    for(n=0,30, print1((2^(n-1) + 0^n/2)*Binomial(2*n+3,n), ", ")) \\ G. C. Greubel, Feb 05 2018

Formula

a(n) = ( 2^(n-1) + (0^n)/2 )*binomial(2*n+3, n).
(n+3)*a(n) +2*(-7*n-13)*a(n-1) +24*(2*n+1)*a(n-2)=0. - R. J. Mathar, Oct 29 2014
From Amiram Eldar, Jan 16 2024: (Start)
Sum_{n>=0} 1/a(n) = 37/7 - 208*arcsin(1/(2*sqrt(2)))/(7*sqrt(7)).
Sum_{n>=0} (-1)^n/a(n) = 296*log(2)/27 - 61/9. (End)

A134308 A007318 * A082137.

Original entry on oeis.org

1, 2, 1, 4, 4, 2, 8, 12, 12, 4, 16, 32, 48, 32, 8, 32, 80, 160, 160, 80, 16, 64, 192, 480, 640, 480, 192, 32, 128, 448, 1344, 2240, 2240, 1344, 448, 64, 256, 1024, 3584, 7168, 8960, 7168, 3584, 1024, 128, 512, 2304, 9216, 21504, 32256, 32256, 21504, 9216, 2304, 256
Offset: 0

Views

Author

Gary W. Adamson, Oct 19 2007

Keywords

Comments

row sums = A007582: (1, 3, 10, 36, 136, 528, ...).

Examples

			First few rows of the triangle:
   1;
   2,  1;
   4,  4,   2;
   8, 12,  12,   4;
  16, 32,  48,  32,  8;
  32, 80, 160, 160, 80, 16;
  ...
		

Crossrefs

Formula

Binomial transform of A082137, as infinite lower triangular matrices.

Extensions

a(5), a(46) corrected, a(27), a(34) split by Georg Fischer, Jun 01 2023

A069720 a(n) = 2^(n-1)*binomial(2*n-1, n).

Original entry on oeis.org

1, 6, 40, 280, 2016, 14784, 109824, 823680, 6223360, 47297536, 361181184, 2769055744, 21300428800, 164317593600, 1270722723840, 9848101109760, 76467608616960, 594748067020800, 4632774416793600, 36135640450990080, 282202144474398720, 2206307674981662720, 17266755717247795200
Offset: 1

Views

Author

Valery A. Liskovets, Apr 07 2002

Keywords

Comments

Number of rooted unicursal planar maps with n edges (unicursal means that exactly two nodes are of odd valency; there is an Eulerian path).

Crossrefs

First superdiagonal of number array A082137.

Programs

  • Haskell
    a069720 n = (a000079 $ n - 1) * (a001700 $ n - 1)
    -- Reinhard Zumkeller, Jan 15 2015
    
  • Magma
    [2^(n-2)*Binomial(2*n, n): n in [1..25]]; // Vincenzo Librandi, Apr 14 2018
    
  • Maple
    Z:=(1-sqrt(1-2*z))*4^(n-1)/sqrt(1-2*z): Zser:=series(Z, z=0, 32): seq(coeff(Zser, z, n), n=1..20); # Zerinvary Lajos, Jan 01 2007
  • Mathematica
    Table[2^(n-1) Binomial[2n-1,n],{n,20}] (* Harvey P. Dale, Jan 20 2013 *)
  • PARI
    a(n) = binomial(2*n-1,n)<<(n-1) \\ Charles R Greathouse IV, Feb 06 2017
    
  • SageMath
    def A069720(n): return 2^(n-2)*binomial(2*n, n)
    print([A069720(n) for n in range(1,31)]) # G. C. Greubel, Jan 18 2025

Formula

a(n) = 2^(n-2)*binomial(2*n, n).
G.f.: (1-sqrt(1-8*x))/(4*x*sqrt(1-8*x)) = 2/(sqrt(1-8*x)*(1-sqrt(1-8*x))) - 1/(2*x). - Paul Barry, Sep 06 2004
D-finite with recurrence: n*a(n) - 4*(2*n-1)*a(n-1) = 0. - R. J. Mathar, Apr 01 2012
E.g.f.: a(n) = n! * [x^n] (exp(4*x)*BesselI(0, 4*x) - 1)/4. - Peter Luschny, Aug 25 2012
From Reinhard Zumkeller, Jan 15 2015: (Start)
a(n) = A000079(n-1) * A001700(n-1); for n > 1:
a(n) = 2*A082143(n-1). (End)
From Amiram Eldar, Jan 16 2024: (Start)
Sum_{n>=1} 1/a(n) = 4/7 + 32*arcsin(1/(2*sqrt(2)))/(7*sqrt(7)).
Sum_{n>=1} (-1)^(n+1)/a(n) = 4/9 + 16*log(2)/27. (End)
a(n) = ((2*n)!/4) * [x^n] (BesselI(0, 2*sqrt(2*x)) - 1). - G. C. Greubel, Jan 18 2025

A069723 a(n) = 2^(n-1)*binomial(2*n-3, n-1).

Original entry on oeis.org

1, 2, 12, 80, 560, 4032, 29568, 219648, 1647360, 12446720, 94595072, 722362368, 5538111488, 42600857600, 328635187200, 2541445447680, 19696202219520, 152935217233920, 1189496134041600, 9265548833587200, 72271280901980160, 564404288948797440, 4412615349963325440
Offset: 1

Views

Author

Valery A. Liskovets, Apr 07 2002

Keywords

Comments

Number of rooted unicursal planar maps with n edges and two vertices of valency 1 (unicursal means that exactly two vertices are of odd valency; there is an Eulerian path).

Crossrefs

Main diagonal of array A082137.

Programs

  • Maple
    Z:=(1-sqrt(1-z))*8^n/sqrt(1-z)/2: Zser:=series(Z, z=0, 33): seq(coeff(Zser, z, n), n=0..20); # Zerinvary Lajos, Jan 16 2007
  • Mathematica
    Table[2^(n - 1) * Binomial[2*n - 3, n - 1], {n, 1,50}] (* G. C. Greubel, Jan 15 2017 *)
  • Sage
    # Assuming offset 0:
    A069723  = lambda n: (rising_factorial(n, n)/factorial(n)) << n
    [A069723(n) for n in (0..20)] # Peter Luschny, Nov 30 2014

Formula

a(n) = A069722(n)/2, n>1.
G.f.: 4*x/(sqrt(1-8*x) * (1-sqrt(1-8*x))). - Paul Barry, Sep 06 2004
With offset 0: a(n) = (0^n + 2^n*binomial(2n, n))/2. - Paul Barry, Sep 24 2004
D-finite with recurrence (-n+1)*a(n) + 4*(2*n-3)*a(n-1) = 0. - R. J. Mathar, Dec 03 2012
With offset 0: a(n) = 2^n*rf(n,n)/n! = 2^n*A088218(n), where rf denotes the rising factorial. - Peter Luschny, Nov 30 2014
a(n) = Sum_{k=0..n} binomial(n+k-1,k)*binomial(2*n-1, n-k). - Vladimir Kruchinin, Nov 11 2016
a(n) ~ 2^(3*n-4)/sqrt(Pi*n). - Ilya Gutkovskiy, Nov 11 2016
From Amiram Eldar, Jan 16 2024: (Start)
Sum_{n>=1} 1/a(n) = 9/7 + 16*arcsin(1/(2*sqrt(2)))/(7*sqrt(7)).
Sum_{n>=1} (-1)^(n+1)/a(n) = 7/9 - 8*log(2)/27. (End)

A134309 Triangle read by rows, where row n consists of n zeros followed by 2^(n-1).

Original entry on oeis.org

1, 0, 1, 0, 0, 2, 0, 0, 0, 4, 0, 0, 0, 0, 8, 0, 0, 0, 0, 0, 16, 0, 0, 0, 0, 0, 0, 32, 0, 0, 0, 0, 0, 0, 0, 64, 0, 0, 0, 0, 0, 0, 0, 0, 128, 0, 0, 0, 0, 0, 0, 0, 0, 0, 256, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 512, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1024, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2048, 0, 0, 0, 0, 0, 0, 0
Offset: 0

Views

Author

Gary W. Adamson, Oct 19 2007

Keywords

Comments

As infinite lower triangular matrices, binomial transform of A134309 = A082137. A134309 * A007318 = A055372. A134309 * [1,2,3,...] = A057711: (1, 2, 6, 16, 40, 96, 224,...).
Triangle read by rows given by [0,0,0,0,0,0,0,0,...] DELTA [1,1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 20 2007

Examples

			Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
  1;
  0, 1;
  0, 0, 2;
  0, 0, 0, 4;
  0, 0, 0, 0, 8;
  0, 0, 0, 0, 0, 16;
  ...
		

Crossrefs

Cf. A011782 (diagonal elements: 1 followed by 1, 2, 4, 8, ... = A000079: 2^n).

Programs

  • Mathematica
    Join[{1},Flatten[Table[Join[{PadRight[{},n],2^(n-1)}],{n,20}]]] (* Harvey P. Dale, Jan 04 2024 *)
  • PARI
    A134309(r,c)=if(r==c,2^max(r-1,0),0) \\ M. F. Hasler, Mar 29 2022

Formula

Triangle, T(0,0) = 1, then for n > 0, n zeros followed by 2^(n-1). Infinite lower triangular matrix with (1, 1, 2, 4, 8, 16, ...) in the main diagonal and the rest zeros.
G.f.: (1 - y*x)/(1 - 2*y*x). - Philippe Deléham, Feb 04 2012
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A011782(n), A081294(n), A081341(n), A092811(n), A093143(n), A067419(n) for x = 0, 1, 2, 3, 4, 5, 6 respectively. - Philippe Deléham, Feb 04 2012
Diagonal is A011782, other elements are 0. - M. F. Hasler, Mar 29 2022

A082139 A transform of binomial(n,5).

Original entry on oeis.org

1, 6, 42, 224, 1008, 4032, 14784, 50688, 164736, 512512, 1537536, 4472832, 12673024, 35094528, 95256576, 254017536, 666796032, 1725825024, 4410441728, 11142168576, 27855421440, 68975329280, 169303080960, 412216197120
Offset: 0

Views

Author

Paul Barry, Apr 06 2003

Keywords

Comments

Sixth row of number array A082137. C(n,5) has e.g.f. (x^5/5!)exp(x). The transform averages the binomial and inverse binomial transforms.

Examples

			a(0) = (2^(-1) + 0^0/2)*binomial(5,0) = 2*(1/2) = 1 (use 0^0 = 1).
		

Crossrefs

Programs

  • Magma
    [(Ceiling(Binomial(n+5, 5)*2^(n-1))) : n in [0..30]]; // Vincenzo Librandi, Sep 22 2011
    
  • Maple
    [seq (ceil(binomial(n+5,5)*2^(n-1)),n=0..23)]; # Zerinvary Lajos, Nov 01 2006
  • Mathematica
    Drop[With[{nmax = 56}, CoefficientList[Series[x^5*Exp[x]*Cosh[x]/5!, {x, 0, nmax}], x]*Range[0, nmax]!], 5] (* or *) Join[{1}, Table[2^(n-1)* Binomial[n+5,n], {n,1,30}]] (* G. C. Greubel, Feb 05 2018 *)
  • PARI
    my(x='x+O('x^30)); Vec(serlaplace(x^5*exp(x)*cosh(x)/5!)) \\ G. C. Greubel, Feb 05 2018

Formula

Equals 2 * A080952.
a(n) = (2^(n-1) + 0^n/2)*C(n+5, n).
a(n) = Sum_{j=0..n} C(n+5, j+5)*C(j+5, 5)*(1+(-1)^j)/2.
G.f.: (1 -6*x +30*x^2 -80*x^3 +120*x^4 -96*x^5 +32*x^6)/(1-2*x)^6.
E.g.f.: x^5*exp(x)*cosh(x)/5! (preceded by 5 zeros).
a(n) = ceiling(binomial(n+5,5)*2^(n-1)). - Zerinvary Lajos, Nov 01 2006
From Amiram Eldar, Jan 07 2022: (Start)
Sum_{n>=0} 1/a(n) = 20*log(2) - 38/3.
Sum_{n>=0} (-1)^n/a(n) = 1620*log(3/2) - 656. (End)

A082140 A transform of binomial(n,6).

Original entry on oeis.org

1, 7, 56, 336, 1680, 7392, 29568, 109824, 384384, 1281280, 4100096, 12673024, 38019072, 111132672, 317521920, 889061376, 2444918784, 6615662592, 17641766912, 46425702400, 120706826240, 310388981760, 790081044480
Offset: 0

Views

Author

Paul Barry, Apr 06 2003

Keywords

Comments

Seventh row of number array A082137. C(n,6) has e.g.f. (x^6/6!)exp(x). The transform averages the binomial and inverse binomial transforms.

Examples

			a(0) = (2^(-1) + 0^0/2)*binomial(6,0) = 2*(1/2) = 1 (use 0^0 = 1).
		

Crossrefs

For n>0, a(n) = 1/2 * A002409(n).

Programs

  • Magma
    [(2^(n-1) + 0^n/2)*Binomial(n+6,n): n in [0..30]]; // G. C. Greubel, Feb 05 2018
  • Maple
    [seq (ceil(binomial(n+6,6)*2^(n-1)),n=0..22)]; # Zerinvary Lajos, Nov 01 2006
  • Mathematica
    Drop[With[{nmax = 56}, CoefficientList[Series[x^6*Exp[x]*Cosh[x]/6!, {x, 0, nmax}], x]*Range[0, nmax]!], 5] (* or *) Join[{1}, Table[2^(n-1)* Binomial[n+6,n], {n,1,30}]] (* G. C. Greubel, Feb 05 2018 *)
    LinearRecurrence[{14,-84,280,-560,672,-448,128},{1,7,56,336,1680,7392,29568,109824},30] (* Harvey P. Dale, Jul 18 2023 *)
  • PARI
    my(x='x+O('x^30)); Vec(serlaplace(x^6*exp(x)*cosh(x)/6!)) \\ G. C. Greubel, Feb 05 2018
    

Formula

a(n) = (2^(n-1) + 0^n/2)*C(n+6,n).
a(n) = Sum_{j=0..n} C(n+6, j+6)*C(j+6, 6)*(1+(-1)^j)/2.
G.f.: (1 - 7*x + 42*x^2 - 140*x^3 + 280*x^4 - 336*x^5 + 224*x^6 - 64*x^7)/ (1-2*x)^7.
E.g.f.: (x^6/6!)*exp(x)*cosh(x) (with 6 leading zeros).
a(n) = ceiling(binomial(n+6,6)*2^(n-1)). - Zerinvary Lajos, Nov 01 2006
From Amiram Eldar, Jan 07 2022: (Start)
Sum_{n>=0} 1/a(n) = 89/5 - 24*log(2).
Sum_{n>=0} (-1)^n/a(n) = 5832*log(3/2) - 11819/5. (End)

A082138 A transform of C(n,3).

Original entry on oeis.org

1, 4, 20, 80, 280, 896, 2688, 7680, 21120, 56320, 146432, 372736, 931840, 2293760, 5570560, 13369344, 31752192, 74711040, 174325760, 403701760, 928514048, 2122317824, 4823449600, 10905190400, 24536678400, 54962159616, 122607894528
Offset: 0

Views

Author

Paul Barry, Apr 06 2003

Keywords

Comments

Fourth row of number array A082137. C(n,3) has e.g.f. (x^3/3!)exp(x). The transform averages the binomial and inverse binomial transforms.

Examples

			a(0) = (2^(-1) + 0^0/2)*C(3,0) = 2*(1/2) = 1 (using 0^0=1).
		

Crossrefs

Programs

  • GAP
    a:=[4,20,80,280];; for n in [5..30] do a[n]:=8*a[n-1]-24*a[n-2] +32*a[n-3]-16*a[n-4]; od; Concatenation([1], a);
  • Magma
    [(Ceiling(Binomial(n+3, 3)*2^(n-1))) : n in [0..30]]; // Vincenzo Librandi, Sep 22 2011
    
  • Maple
    [seq (ceil(binomial(n+3,3)*2^(n-1)),n=0..30)]; # Zerinvary Lajos, Nov 01 2006
  • Mathematica
    Join[{1}, LinearRecurrence[{8,-24,32,-16}, {4,20,80,280}, 30]] (* G. C. Greubel, Jul 23 2019 *)
  • PARI
    my(x='x+O('x^30)); Vec((1-4*x+12*x^2-16*x^3 + 8*x^4)/(1-2*x)^4) \\ G. C. Greubel, Jul 23 2019
    
  • Sage
    ((1-4*x+12*x^2-16*x^3+8*x^4)/(1-2*x)^4).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jul 23 2019
    

Formula

a(n) = (2^(n-1) + 0^n/2)*C(n+3, n).
a(n) = Sum_{j=0..n} C(n+3, j+3)*C(j+3, 3)*(1 + (-1)^j)/2.
G.f.: (1 - 4*x + 12*x^2 - 16*x^3 + 8*x^4)/(1-2*x)^4.
E.g.f.: (x^3/3!)*exp(x)*cosh(x) (preceded by 3 zeros).
a(n) = ceiling(binomial(n+3,3)*2^(n-1)). - Zerinvary Lajos, Nov 01 2006
From Amiram Eldar, Jan 07 2022: (Start)
Sum_{n>=0} 1/a(n) = 12*log(2) - 7.
Sum_{n>=0} (-1)^n/a(n) = 108*log(3/2) - 43. (End)
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