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.

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A185286 Triangle T(n,k) is the number of nonnegative walks of n steps with step sizes 1 and 2, starting at 0 and ending at k.

Original entry on oeis.org

1, 0, 1, 1, 2, 1, 1, 2, 1, 2, 5, 6, 3, 3, 3, 1, 11, 11, 13, 17, 13, 7, 6, 4, 1, 24, 41, 52, 44, 43, 40, 25, 14, 10, 5, 1, 93, 120, 152, 176, 161, 126, 107, 80, 45, 25, 15, 6, 1, 272, 421, 550, 559, 561, 524, 412, 303, 227, 146, 77, 41, 21, 7, 1, 971, 1381, 1813, 2056, 2045, 1835, 1615, 1309, 938, 648, 435, 251, 126, 63, 28, 8, 1
Offset: 0

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Author

Keywords

Comments

Equivalently, the number of paths from (0,0) to (n,k) using steps of the form (1,2),(1,1),(1,-1) or (1,-2) and staying on or above the x-axis.
It appears that A047002 gives the row sums of this triangle.

Examples

			The table starts:
1
0,1,1
2,1,1,2,1
2,5,6,3,3,3,1
		

Crossrefs

Columns k=0..2 are A187430, A055113, A296619.
Cf. A005408(row lengths), A047002(apparently row sums).

Programs

  • Maple
    T:= proc(n,k) option remember;
      if k < 0 or k > 2*n then return 0 fi;
      procname(n-1,k-2)+procname(n-1,k-1)+procname(n-1,k+1)+procname(n-1,k+2)
    end proc:
    T(0,0):= 1:
    for nn from 0 to 10 do
      seq(T(nn,k),k=0..2*nn)
    od; # Robert Israel, Dec 19 2017
  • Mathematica
    T[n_, k_] := T[n, k] = If[k < 0 || k > 2n, 0, T[n-1, k-2] + T[n-1, k-1] + T[n-1, k+1] + T[n-1, k+2]];
    T[0, 0] = 1;
    Table[T[n, k], {n, 0, 10}, {k, 0, 2n}] // Flatten (* Jean-François Alcover, Aug 19 2022, after Robert Israel *)
  • PARI
    flip(v)=vector(#v,i,v[#v+1-i])
    ar(n)={local(p);p=1;
    for(k=1,n,p*=1+x+x^3+x^4;p=(p-polcoeff(p,0)-polcoeff(p,1)*x)/x^2);
    flip(Vec(p))}

A278618 a(n) = Sum_{j=0..n/2} binomial(n-j-1,n-2*j)*binomial(2*n+1,j).

Original entry on oeis.org

1, 0, 5, 7, 45, 121, 533, 1800, 7157, 26239, 101640, 384583, 1483925, 5693247, 22013059, 85076183, 330014421, 1281349195, 4985766650, 19422653367, 75775163028, 295953650376, 1157212653030, 4529183513913, 17743019073381, 69565441895001
Offset: 0

Views

Author

Vladimir Kruchinin, Nov 23 2016

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n - j - 1, n - 2*j]*Binomial[2*n + 1, j], {j, 0, n/2}], {n,0,50}] (* G. C. Greubel, Jun 06 2017 *)
  • Maxima
    A(x):=(12-4/sqrt(1-4*x))/(8*sqrt(12*x+2*sqrt(1-4*x)+2))+1/(2*sqrt(1-4*x));
    B(x):=1/((x+1)*sqrt(-3*x^2-2*x+1));
    C(x):=sqrt(12*x+2*sqrt(1-4*x)+2)/4-sqrt(1-4*x)/4-1/4;
    taylor(x*A(x)/C(x)*B(C(x)),x,0,20);
    
  • PARI
    for(n=0,25, print1(sum(j=0,n, binomial(n-j-1,n-2*j)*binomial(2*n+1,j)), ", ")) \\ G. C. Greubel, Jun 06 2017

Formula

G.f.: x*A(x)/C(x)*B(C(x)), where
A(x) = (12-4/sqrt(1-4*x))/(8*sqrt(12*x+2*sqrt(1-4*x)+2))+1/(2*sqrt(1-4*x)),
B(x) = 1/((x+1)*sqrt(-3*x^2-2*x+1)),
C(x) = sqrt(12*x+2*sqrt(1-4*x)+2)/4-sqrt(1-4*x)/4-1/4.
a(n) ~ (1 - 1/sqrt(5)) * 4^n / sqrt(Pi*n). - Vaclav Kotesovec, Nov 24 2016
a(n) = (2*n + 1)*3F2(1-n/2,3/2-n/2,-2*n; 2,2-n; 4) for n>1. - Ilya Gutkovskiy, Nov 24 2016
Conjecture: 2*n*(5*n-8)*(2*n-1)*(n+1)*a(n) -n*(115*n^3-344*n^2+299*n-82)*a(n-1) -4*(2*n-1)*(5*n^3+27*n^2-74*n+30)*a(n-2) +36*(n-2)*(5*n-3)*(2*n-1)*(2*n-3)*a(n-3)=0. - R. J. Mathar, Dec 02 2016

A296619 The number of nonnegative walks of n steps with step sizes 1 and 2, starting at 0 and ending at 2.

Original entry on oeis.org

0, 1, 1, 6, 13, 52, 152, 550, 1813, 6453, 22427, 80330, 286895, 1038931, 3772801, 13807294, 50726893, 187332517, 694364517, 2583714636, 9644852364, 36115537269, 135607526865, 510496492338, 1926284451923, 7284476707597, 27602839227883, 104791979218326
Offset: 0

Views

Author

Feng Jishe, Dec 17 2017

Keywords

Comments

a(n) is the number of 2-D walks with n steps of type {(1,-2), (1,-1), (1,1), or (1,2)} starting at (0,0), ending at (n,2), and not dropping below the x-axis.
The sequence corresponds to element (1,3) of the matrix B(n)^n (see Maple script). Furthermore, element (1,1) of the matrix is A187430, the element (1,2) of these matrix is A055113.

Examples

			There are 6 walks of length 3:
        __
       |  |         __
     __|  |_     __|  |_     __    _
    |           |           |  |__|
   _|          _|          _|
    2+2-2=2     2+1-1=2     2-1+1=2
                    __
     __    _       |  |_           _
    |  |  |      __|         __   |
   _|  |__|    _|          _|  |__|
    2-2+2=2     1+2-1=2     1-1+2=2
		

Crossrefs

Programs

  • Maple
    B := n -> LinearAlgebra:-ToeplitzMatrix([0,1,1, seq(0, k=0..n-2)], symmetric):
    seq((B(n)^n)(1, 3), n=0..27);
    # alternative:
    T:= proc(n,k) option remember;
      if k < 0 or k > 2*n then return 0 fi;
      procname(n-1,k-2)+procname(n-1,k-1)+procname(n-1,k+1)+procname(n-1,k+2)
    end proc:
    T(0,0):= 1:
    seq(T(n,2),n=0..40); # Robert Israel, Dec 19 2017
  • Mathematica
    b[n_] := ToeplitzMatrix[Join[{0,1,1}, ConstantArray[0,n-1]]];
    Prepend[Table[MatrixPower[b[n],n][[1,3]], {n,20}], 0]
    (* Andrey Zabolotskiy, Dec 19 2017 *)
  • PARI
    Next(v)={vector(#v+2, i, if(i<3||i>#v-2, 0, v[i-2]+v[i-1]+v[i+1]+v[i+2]))}
    my(v=vector(7,i,i==3)); for(n=1, 50, print1(v[5],", "); v=Next(v)) \\ Andrew Howroyd, Dec 18 2017

Formula

a(n) = A185286(n,2). - Robert Israel, Dec 19 2017

A123880 Inverse of number triangle A123878.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 1, 3, 0, 1, 5, 3, 5, 0, 1, 11, 18, 5, 7, 0, 1, 41, 39, 35, 7, 9, 0, 1, 120, 157, 75, 56, 9, 11, 0, 1, 421, 459, 325, 119, 81, 11, 13, 0, 1, 1381, 1668, 950, 553, 171, 110, 13, 15, 0, 1, 4840
Offset: 0

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Author

Paul Barry, Oct 16 2006

Keywords

Comments

First column is A055113. Row sums are A000108.

Examples

			Number triangle begins
1,
0, 1,
1, 0, 1,
1, 3, 0, 1,
5, 3, 5, 0, 1,
11, 18, 5, 7, 0, 1,
41, 39, 35, 7, 9, 0, 1,
120, 157, 75, 56, 9, 11, 0, 1
		

A211867 a(n) = A097609(2*n-1,n), n>0; a(0)=1.

Original entry on oeis.org

1, 0, 2, 3, 18, 50, 215, 735, 2898, 10668, 41202, 156090, 601623, 2308878, 8923343, 34487453, 133749330, 519277512, 2020262660, 7869597840, 30699524018, 119894389380, 468768069882, 1834589752182, 7186572436887, 28175111736300, 110547143014050, 434049816801900
Offset: 0

Views

Author

Vladimir Kruchinin, Feb 12 2013

Keywords

Programs

  • Maple
    a := n -> (-1)^n*binomial(2*n-1,n-1)*hypergeom([-n,n/2,(n+1)/2], [n,n+1], 4):
    seq(simplify(a(n)), n=0..27); # Peter Luschny, Nov 02 2016
  • Mathematica
    a[n_] := ((-1)^(3*n)*(2*n)!*HypergeometricPFQ[{(n+1)/2, -n, n/2}, {n, n+1}, 4])/(2*n!^2); a[0]=1; Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Feb 13 2013, from A097609 *)
  • PARI
    a(n) = if(n==0, 1, sum(k=0, n/2, (binomial(2*n,k)*binomial(n-k-1,n-2*k))/2)); \\ Altug Alkan, Oct 05 2015

Formula

G.f.: x*G'(x)/G(x), where G(x) is the g.f. of A055113.
G.f.: x * d/dx (log(sqrt(12*x+2*sqrt(1-4*x)+2)/4-sqrt(1-4*x)/4-1/4)).
a(n) = sum(j=0..n, C(2*j+n-1,j)*(-1)^(n+j)*C(2*n,n-j))/2, n>0; a(0)=1.
a(n) = A097609(2*n-1,n), n>0; a(0)=1. (Corrected by M. F. Hasler, Feb 12 2013)
a(n) = Sum_{j=0..n/2} (binomial(2*n,j)*binomial(n-j-1,n-2*j))/2. - Vladimir Kruchinin, Oct 05 2015
a(n) ~ 2^(2*n-1) / sqrt(5*Pi*n). - Vaclav Kotesovec, Apr 27 2024

A239425 Expansion of -16/(sqrt(12*x+2*sqrt(1-4*x)+2)-sqrt(1-4*x)-1)^2+1/x^2-1.

Original entry on oeis.org

1, 2, 7, 16, 53, 156, 522, 1702, 5833, 19990, 70079, 247160, 882587, 3172196, 11492847, 41874864, 153452521, 564975570, 2089346157, 7756501690, 28898156364, 108010059036, 404890987653, 1521877280868, 5734545323859, 21657665796526
Offset: 0

Views

Author

Vladimir Kruchinin, Mar 17 2014

Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[-16/(Sqrt[12*x+2*Sqrt[1-4*x]+2]-Sqrt[1-4*x] -1)^2+1/x^2-1, {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 18 2014 *)
    Flatten[{1,Table[Sum[Binomial[n+2*j-1,j+n-1]*(-1)^(j+n)*Binomial[2*n+2,j+n],{j,0,n+2}]/(n+1),{n,1,20}]}] (* Vaclav Kotesovec, Mar 18 2014 *)
  • Maxima
    a(n):=(sum(binomial(n+2*j-1, j)*(-1)^(j+n)*binomial(2*n+2, j+n), j, 0, n+2))/(n+1)-kron_delta(n,0);
    
  • PARI
    my(x='x+O('x^50)); Vec(-16/(sqrt(12*x+2*sqrt(1-4*x)+2)-sqrt(1-4*x) -1)^2 + 1/x^2 -1) \\ G. C. Greubel, Jun 01 2017

Formula

a(n) = (Sum_{j=0..(n+2)} C(n+2*j-1,j)*(-1)^(j+n)*C(2*n+2,j+n))/(n+1) - delta(n,0).
a(n) ~ (5+3*sqrt(5)) * 2^(2*n+1) / (5*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 18 2014
Conjecture: 2*(2*n+1)*(n+2)*(n+1)*a(n) +(n+1)*(n^2-27*n+2)*a(n-1) +2*(-73*n^3+204*n^2-167*n+6)*a(n-2) +12*(n-3)*(2*n-3)*(4*n-7)*a(n-3) +216*(2*n-5)*(n-3)*(2*n-3)*a(n-4)=0. - R. J. Mathar, Apr 02 2014

A212696 Central coefficient of the triangle A097609.

Original entry on oeis.org

1, 0, 3, 4, 25, 66, 287, 960, 3789, 13810, 53240, 200652, 771641, 2952054, 11386065, 43910288, 170007429, 658979586, 2560258550, 9960335060, 38811668868, 151418146704, 591464244882, 2312774560296, 9052560751725, 35464735083726, 139054217427702, 545635715465596
Offset: 0

Views

Author

Vladimir Kruchinin, May 24 2012

Keywords

Crossrefs

Programs

  • Mathematica
    Table[((n + 1) Sum[Binomial[n + 2 j, n + j] (-1)^(n - j) Binomial[2 n + 1, n + j + 1], {j, 0, n}])/(2 n + 1), {n, 0, 27}] (* or *)
    CoefficientList[Series[(12 - 4/#)/(8 Sqrt[12 x + 2 # + 2]) + 1/(2 #) &@ Sqrt[1 - 4 x], {x, 0, 27}], x] (* Michael De Vlieger, Oct 08 2016 *)
    a[n_] := (-1)^n Binomial[2n, n] HypergeometricPFQ[{(n+1)/2, 1+n/2, -n}, {1+n, 2+n}, 4]; Table[a[n], {n, 0, 27}] (* Peter Luschny, Dec 26 2017 *)
  • PARI
    x='x+O('x^66);
    gf=(12-4/sqrt(1-4*x))/(8*sqrt(12*x+2*sqrt(1-4*x)+2))+1/(2*sqrt(1-4*x));
    Vec(Ser(gf))
    /* Joerg Arndt, Jun 09 2012 */

Formula

G.f.: (12-4/sqrt(1-4*x))/(8*sqrt(12*x+2*sqrt(1-4*x)+2))+1/(2*sqrt(1-4*x)).
a(n) = ((n+1)*Sum_{j=0..n} C(n+2*j, n+j)*(-1)^(n-j)*C(2*n+1, n+j+1)) / (2*n+1).
a(n) = (n+1)*A055113(n).
Conjecture: 2*n*(n-1)*(2*n+1)*(5*n-8)*a(n) -(n-1)*(115*n^3-344*n^2+299*n-82) *a(n-1) -4*(2*n-3)*(5*n^3+27*n^2-74*n+30)*a(n-2) +36*(n-1)*(5*n-3)*(2*n-3)*(2*n-5) *a(n-3)=0. - R. J. Mathar, Oct 08 2016
a(n) = (-1)^n*binomial(2*n, n)*hypergeom([(n+1)/2, 1+n/2, -n], [1+n, 2+n], 4). - Peter Luschny, Dec 26 2017
From Emanuele Munarini, Jul 14 2024: (Start)
a(n) = Sum_{k=0..floor(n/2)} binomial(2*n,k)*binomial(n-k-1,k-1)*(n+1)/(2*n-k+1).
a(n) = Sum_{k=0..n} (-1)^k*binomial(2*n,k)*binomial(3n-2k,2*n-k)*(n+1)/(2*n-k+1).
a(n) = (n+1)/(2n+1)*Sum_{k=0..n} binomial(2*n+i,2*n)*trinomial(2*n+1,n-k)*(-1)^{n-k}, where trinomial(n,k) are the trinomial coefficients (A027907).
a(n) = Sum_{k=0..n} (-1)^k*binomial(3*n-k,n-k)*trinomial(2*n,k)*(n+k+1)/(2*n+1). (End)

A239230 Expansion of -x*log'(-sqrt(12*x+2*sqrt(1-4*x)+2)/4+sqrt(1-4*x)/4+5/4).

Original entry on oeis.org

0, 1, 1, 4, 9, 36, 112, 428, 1505, 5692, 21026, 79806, 301488, 1151866, 4403778, 16929474, 65204353, 251947668, 975366094, 3784197606, 14705937794, 57242631464, 223121176224, 870805992278, 3402485053664, 13308485156086, 52104519751272, 204176144516818
Offset: 0

Views

Author

Vladimir Kruchinin, May 22 2014

Keywords

Crossrefs

Cf. A055113.

Programs

  • Maple
    a:= n-> add((-1)^(k+n)*binomial(2*n-k-1, n-1)*hypergeom([k-n, (n+1)/2, n/2], [n, n+1], 4), k=1..n);
    seq(round(evalf(a(n), 32)), n=0..24); # Peter Luschny, May 22 2014
  • Maxima
    a(n):=n*sum(sum(binomial(n+2*j-1,j+n-1)*(-1)^(k+j+n)*binomial(2*n-k,j+n), j,0,n-k)/(2*n-k), k,1,n);

Formula

G.f.: A(x) = x*F'(x)/(1-F(x)), where F(x) is g.f. of A055113.
a(n) = n * Sum_{k=1..n} (Sum_{j=0..n-k} C(n+2*j-1,j+n-1) * (-1)^(k+j+n) * C(2*n-k,j+n)) / (2*n-k).
a(n) = Sum_{k=1..n} (-1)^(k+n) * C(2*n-k-1,n-1) * hypergeom([k-n, n/2+1/2, n/2], [n, n+1], 4). - Peter Luschny, May 22 2014
Conjecture D-finite with recurrence +24*(365025561*n-1672569283)*(n-1)*(n-2)*(2*n-1)*a(n) -4*(n-2)*(27129169947*n^3-209577621466*n^2+463278020461*n-314084557758)*a(n-1) +2*(-28823853823*n^4+487259692534*n^3-3105214937957*n^2+8814274338098*n-9143920331436)*a(n-2) +4*(276083065830*n^4-4172118623320*n^3+24824880820695*n^2-69263721795041*n+75832154222148)*a(n-3) +(-587491214125*n^4+9941738070620*n^3-75070680472775*n^2+281912285021344*n-413197788157152)*a(n-4) +2*(-1186924847911*n^4+26108844767699*n^3-211936472383904*n^2+757584729548632*n-1009721693733312)*a(n-5) +4*(2*n-11)*(328530544924*n^3-5280431217363*n^2+28334632524947*n-50473913356356)*a(n-6) -72*(4143100547*n-18456753180)*(n-6)*(2*n-11)*(2*n-13)*a(n-7)=0. - R. J. Mathar, Jul 27 2022
Conjecture: a(n) = Sum_{i=0..n-1} A059260(2*(n-1),n+i-1)*A009766(n+i-1,i)*(-1)^(n+i-1) = n*Sum_{i=0..n-1} binomial(n+2*i, i)/(n+2*i)*(-1)^(n+i-1)*[x^(n+i-1)] (1+(x+1)^(2*n-1))/(x+2) for n >= 0. - Mikhail Kurkov, Feb 18 2025

A297074 Number of ways of inserting parentheses in x^x^...^x (with n x's) whose result is an integer where x = sqrt(2).

Original entry on oeis.org

0, 0, 1, 1, 2, 5, 10, 23, 55
Offset: 1

Views

Author

Jon E. Schoenfield, Dec 24 2017

Keywords

Comments

The largest value that can be obtained by inserting parentheses in x^x^x^x^x^x^x^x^x (9 x's), where x = sqrt(2), is x^(x^((((((x^x)^x)^x)^x)^x)^x)) = 2^128 = 340282366920938463463374607431768211456; this is one of the a(9) = 55 ways of inserting parentheses in x^x^x^x^x^x^x^x^x that yield an integer value.

Examples

			With x = sqrt(2),
x = 1.414213... is not an integer, so a(1) = 0;
x^x = 1.632526... is not an integer, so a(2) = 0.
(x^x)^x = 2 is an integer, but x^(x^x) = 1.760839... is not, so a(3) = 1;
((x^x)^x)^x, (x^x)^(x^x), (x^(x^x))^x, and x^(x^(x^x)) are noninteger values, but x^((x^x)^x) = 2, so a(4) = 1;
the only ways of inserting parentheses in x^x^x^x^x that yield integer values are x^(x^((x^x)^x)) = 2 and (((x^x)^x)^x)^x = 4, so a(5) = 2.
		

Crossrefs

Programs

A346380 Complement of A187430 in A000108.

Original entry on oeis.org

0, 0, 1, 0, 3, 3, 18, 39, 157, 459, 1668, 5503, 19638, 68325, 245144, 876438, 3177651, 11549939, 42307920, 155555733, 574881920, 2132231076, 7938771624, 29651189637, 111086480106, 417305224917, 1571633677078, 5932720163529, 22443721850064, 85075094996719, 323086777251300
Offset: 0

Views

Author

F. Chapoton, Jul 14 2021

Keywords

Comments

Related to the decomposition of A000108 as the sum of A055113 and A111160.

Crossrefs

Programs

  • Sage
    N = 30
    x = (PowerSeriesRing(QQ, 'x').0).O(N + 1)
    f = (x*(1-x^2)^2/(1+x^3)^2).reverse()
    g = sum(catalan_number(n)*x**n for n in range(N + 1)).O(N + 1)
    list(x*g-f)

Formula

G.f. y(x) satisfies x y^4 + 2 x^2 y^2 + x^3 + 3 x y^2 + y^3 - x y = 0.
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