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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A361791 Expansion of 1/sqrt(1 - 4*x/(1+x)^5).

Original entry on oeis.org

1, 2, -4, -10, 30, 72, -238, -580, 1970, 4910, -16734, -42750, 144600, 379000, -1264700, -3402480, 11160730, 30828070, -99168820, -281279030, 885931600, 2580541580, -7948885910, -23779051760, 71572652480, 219906488302, -646332447086, -2039738985238, 5850898295170
Offset: 0

Views

Author

Seiichi Manyama, Mar 24 2023

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_]:=(-1)^(n+1)Pochhammer[n,4]HypergeometricPFQ[{3/2,1-n,1+n/4,(5+n)/4, (6+n)/4, (7+n)/4}, {6/5,7/5,8/5,9/5,2}, 2^10/5^5]/12; Join[{1},Array[a,28]] (* Stefano Spezia, Jul 11 2024 *)
  • PARI
    my(N=30, x='x+O('x^N)); Vec(1/sqrt(1-4*x/(1+x)^5))
    
  • PARI
    a(n) = sum(k=0, n, (-1)^(n-k) * binomial(2*k,k) * binomial(n+4*k-1,n-k)) \\ Winston de Greef, Mar 24 2023

Formula

a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(2*k,k) * binomial(n+4*k-1,n-k).
n*a(n) = -( (2*n-4)*a(n-1) + (11*n-14)*a(n-2) + 20*(n-3)*a(n-3) + 15*(n-4)*a(n-4) + 6*(n-5)*a(n-5) + (n-6)*a(n-6) ) for n > 5.
a(0) = 1; a(n) = (2/n) * Sum_{k=0..n-1} (-1)^(n-1-k) * (n+k) * binomial(n+3-k,4) * a(k).
a(n) = (-1)^(n+1)*Pochhammer(n,4)*hypergeom([3/2, 1-n, 1+n/4, (5+n)/4, (6+n)/4, (7+n)/4], [6/5, 7/5, 8/5, 9/5, 2], 2^10/5^5)/12 for n > 0. - Stefano Spezia, Jul 11 2024

A361792 Expansion of 1/sqrt(1 - 4*x/(1+x)^6).

Original entry on oeis.org

1, 2, -6, -10, 66, 60, -750, -236, 8682, -2098, -100792, 80286, 1162458, -1603412, -13225764, 26767020, 147428498, -409582818, -1596563202, 5941802122, 16587101544, -83014131140, -161717252990, 1126247965980, 1411774064970, -14905602076350
Offset: 0

Views

Author

Seiichi Manyama, Mar 24 2023

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_]:=(-1)^(n+1)Pochhammer[n,5]HypergeometricPFQ[{1-n,1+n/5,(6+n)/5, (7+n)/5, (8+n)/5, (9+n)/5}, {7/6,4/3,5/3,11/6,2}, 5^5/(2^4*3^6)]/60; Join[{1},Array[a,25]] (* Stefano Spezia, Jul 11 2024 *)
  • PARI
    my(N=30, x='x+O('x^N)); Vec(1/sqrt(1-4*x/(1+x)^6))

Formula

a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(2*k,k) * binomial(n+5*k-1,n-k).
n*a(n) = -( (3*n-5)*a(n-1) + (17*n-24)*a(n-2) + 35*(n-3)*a(n-3) + 35*(n-4)*a(n-4) + 21*(n-5)*a(n-5) + 7*(n-6)*a(n-6) + (n-7)*a(n-7) ) for n > 6.
a(0) = 1; a(n) = (2/n) * Sum_{k=0..n-1} (-1)^(n-1-k) * (n+k) * binomial(n+4-k,5) * a(k).
a(n) = (-1)^(n+1)*Pochhammer(n,5)*hypergeom([1-n, 1+n/5, (6+n)/5, (7+n)/5, (8+n)/5, (9+n)/5], [7/6, 4/3, 5/3, 11/6, 2], 5^5/(2^4*3^6))/60 for n > 0. - Stefano Spezia, Jul 11 2024

A084608 Triangle, read by rows, where the n-th row lists the (2n+1) coefficients of (1+2*x+3*x^2)^n.

Original entry on oeis.org

1, 1, 2, 3, 1, 4, 10, 12, 9, 1, 6, 21, 44, 63, 54, 27, 1, 8, 36, 104, 214, 312, 324, 216, 81, 1, 10, 55, 200, 530, 1052, 1590, 1800, 1485, 810, 243, 1, 12, 78, 340, 1095, 2712, 5284, 8136, 9855, 9180, 6318, 2916, 729, 1, 14, 105, 532, 2009, 5922, 13993, 26840, 41979
Offset: 0

Views

Author

Paul D. Hanna, Jun 01 2003

Keywords

Examples

			Triangle begins:
  1;
  1,  2,  3;
  1,  4, 10,  12,   9;
  1,  6, 21,  44,  63,   54,   27;
  1,  8, 36, 104, 214,  312,  324,  216,   81;
  1, 10, 55, 200, 530, 1052, 1590, 1800, 1485, 810, 243;
		

Crossrefs

Programs

  • Haskell
    a084608 n = a084608_list !! n
    a084608_list = concat $ iterate ([1,2,3] *) [1]
    instance Num a => Num [a] where
       fromInteger k = [fromInteger k]
       (p:ps) + (q:qs) = p + q : ps + qs
       ps + qs         = ps ++ qs
       (p:ps) * qs'@(q:qs) = p * q : ps * qs' + [p] * qs
        *                = []
    -- Reinhard Zumkeller, Apr 02 2011
    
  • Magma
    A084608:= func< n,k | (&+[Binomial(n, k-j)*Binomial(k-j, j)*2^(k-2*j)*3^j: j in [0..k]]) >;
    [A084608(n,k): k in [0..2*n], n in [0..13]]; // G. C. Greubel, Mar 27 2023
    
  • Maple
    f:= proc(n) option remember; expand((1+2*x+3*x^2)^n) end:
    T:= (n,k)-> coeff(f(n), x, k):
    seq(seq(T(n, k), k=0..2*n), n=0..10);  # Alois P. Heinz, Apr 03 2011
  • Mathematica
    row[n_] := (1+2x+3x^2)^n + O[x]^(2n+1) // CoefficientList[#, x]&; Table[row[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Feb 01 2017 *)
  • PARI
    for(n=0,10, for(k=0,2*n,t=polcoeff((1+2*x+3*x^2)^n,k,x); print1(t",")); print(" "))
    
  • SageMath
    def A084608(n,k): return sum(binomial(n,j)*binomial(n-j,k-2*j)*2^(k-2*j)*3^j for j in range(k//2+1))
    flatten([[A084608(n,k) for k in range(2*n+1)] for n in range(14)]) # G. C. Greubel, Mar 27 2023

Formula

From G. C. Greubel, Mar 27 2023: (Start)
T(n, k) = Sum_{j=0..k} binomial(n, k-j)*binomial(k-j, j)*2^(k-2*j)*3^j.
T(n, n) = A084609(n).
T(n, 2*n-1) = A212697(n), n >= 1.
T(n, 2*n) = A000244(n).
Sum_{j=0..2*n} T(n, k) = A000400(n).
Sum_{k=0..2*n} (-1)^k*T(n, k) = A000079(n).
Sum_{k=0..n} T(n-k, k) = A101822(n). (End)

A360266 a(n) = Sum_{k=0..floor(n/3)} binomial(n-2*k,k) * binomial(2*(n-2*k),n-2*k).

Original entry on oeis.org

1, 2, 6, 22, 82, 312, 1210, 4752, 18834, 75184, 301856, 1217604, 4930626, 20032052, 81615072, 333328532, 1364264250, 5594210292, 22977466864, 94517423444, 389316529512, 1605533230256, 6628467569292, 27393187077144, 113310732332274, 469101108803052
Offset: 0

Views

Author

Seiichi Manyama, Jan 31 2023

Keywords

Comments

Diagonal of rational function 1/(1 - (x + y + x^3*y^2)). - Seiichi Manyama, Mar 23 2023

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n\3, binomial(n-2*k, k)*binomial(2*(n-2*k), n-2*k));
    
  • PARI
    my(N=30, x='x+O('x^N)); Vec(1/sqrt(1-4*x*(1+x^2)))

Formula

G.f.: 1/sqrt(1 - 4*x*(1 + x^2)).
n*a(n) = 2*(2*n-1)*a(n-1) + 2*(2*n-3)*a(n-3).

A361812 Expansion of 1/sqrt(1 - 4*x*(1+x)^3).

Original entry on oeis.org

1, 2, 12, 62, 342, 1932, 11094, 64480, 378150, 2233304, 13263772, 79136844, 473969586, 2847911596, 17159547804, 103640073972, 627280131594, 3803643145596, 23102172930156, 140522319418164, 855880464524472, 5219168576004184, 31861229045809436
Offset: 0

Views

Author

Seiichi Manyama, Mar 25 2023

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_]:=Binomial[2*n, n]HypergeometricPFQ[{(1-3*n)/4, (2-3*n)/4, 3*(1-n)/4, -3*n/4}, {1/3-n, 1/2-n, 2/3-n}, -2^6/3^3]; Array[a,23,0] (* Stefano Spezia, Jul 11 2024 *)
  • PARI
    my(N=30, x='x+O('x^N)); Vec(1/sqrt(1-4*x*(1+x)^3))

Formula

a(n) = Sum_{k=0..n} binomial(2*k,k) * binomial(3*k,n-k).
n*a(n) = 2 * ( (2*n-1)*a(n-1) + 3*(2*n-2)*a(n-2) + 3*(2*n-3)*a(n-3) + (2*n-4)*a(n-4) ) for n > 3.
a(n) = binomial(2*n, n)*hypergeom([(1-3*n)/4, (2-3*n)/4, 3*(1-n)/4, -3*n/4], [1/3-n, 1/2-n, 2/3-n], -2^6/3^3). - Stefano Spezia, Jul 11 2024

A084606 Triangle, read by rows, where the n-th row lists the (2n+1) coefficients of (1+2x+2x^2)^n.

Original entry on oeis.org

1, 1, 2, 2, 1, 4, 8, 8, 4, 1, 6, 18, 32, 36, 24, 8, 1, 8, 32, 80, 136, 160, 128, 64, 16, 1, 10, 50, 160, 360, 592, 720, 640, 400, 160, 32, 1, 12, 72, 280, 780, 1632, 2624, 3264, 3120, 2240, 1152, 384, 64, 1, 14, 98, 448, 1484, 3752, 7448, 11776, 14896, 15008, 11872
Offset: 0

Views

Author

Paul D. Hanna, Jun 01 2003

Keywords

Examples

			Rows:
{1},
{1,2,2},
{1,4,8,8,4},
{1,6,18,32,36,24,8},
{1,8,32,80,136,160,128,64,16},
{1,10,50,160,360,592,720,640,400,160,32},
{1,12,72,280,780,1632,2624,3264,3120,2240,1152,384,64},
		

Crossrefs

Programs

  • PARI
    for(n=0,15, for(k=0,2*n,t=polcoeff((1+2*x+2*x^2)^n,k,x); print1(t",")); print(" "))

A106258 Expansion of 1/sqrt(1-8x-8x^2).

Original entry on oeis.org

1, 4, 28, 208, 1624, 13024, 106336, 879232, 7338592, 61699456, 521753728, 4433024512, 37812715264, 323603221504, 2777262164992, 23893731463168, 206005885076992, 1779480850438144, 15396895523989504, 133420304211238912
Offset: 0

Views

Author

Paul Barry, Apr 28 2005

Keywords

Comments

Central coefficient of (1+4x+6x^2)^n. Fourth binomial transform of 1/sqrt(1-24x^2). In general, 1/sqrt(1-4*r*x-4*r*x^2) has e.g.f. exp(2rx)BesselI(0,2r*sqrt((r+1)/r)x)), a(n)=sum{k=0..n, C(2k,k)C(k,n-k)r^k}, gives the central coefficient of (1+(2r)x+r(r+1)x^2) and is the (2r)-th binomial transform of 1/sqrt(1-8*C(n+1,2)x^2).
Also number of paths from (0,0) to (n,0) using steps U=(1,1), H=(1,0) and D=(1,-1), the H steps can have 4 colors and the U steps can have 6 colors. - N-E. Fahssi, Mar 31 2008

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/Sqrt[1-8*x-8*x^2], {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 17 2012 *)
    RecurrenceTable[{a[0]==1,a[1]==4,a[n]==(4(2n-1)a[n-1]+8(n-1)a[n-2])/n}, a,{n,20}] (* Harvey P. Dale, Mar 13 2013 *)

Formula

E.g.f.: exp(4*x)*BesselI(0, 4*sqrt(3/2)*x); a(n)=sum{k=0..n, C(2k, k)C(k, n-k)2^k}.
D-finite with recurrence: n*a(n) = 4*(2*n-1)*a(n-1) + 8*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 17 2012
a(n) ~ sqrt(18+6*sqrt(6))*(4+2*sqrt(6))^n/(6*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 17 2012

A106259 Expansion of 1/sqrt(1-12x-12x^2).

Original entry on oeis.org

1, 6, 60, 648, 7344, 85536, 1014336, 12182400, 147702528, 1803907584, 22159733760, 273508669440, 3389106769920, 42134712606720, 525323149885440, 6565657319866368, 82235651779657728, 1031956779869798400
Offset: 0

Views

Author

Paul Barry, Apr 28 2005

Keywords

Comments

Central coefficient of (1+6x+12x^2)^n. Sixth binomial transform of 1/sqrt(1-48x^2). In general, 1/sqrt(1-4*r*x-4*r*x^2) has e.g.f. exp(2rx)BesselI(0,2r*sqrt((r+1)/r)x)), a(n)=sum{k=0..n, C(2k,k)C(k,n-k)r^k}, gives the central coefficient of (1+(2r)x+r(r+1)x^2) and is the (2r)-th binomial transform of 1/sqrt(1-8*C(n+1,2)x^2).

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/Sqrt[1-12*x-12*x^2], {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 17 2012 *)

Formula

E.g.f.: exp(6*x)*BesselI(0, 6*sqrt(4/3)*x); a(n)=sum{k=0..n, C(2k, k)C(k, n-k)3^k}.
D-finite with recurrence: n*a(n) = 6*(2*n-1)*a(n-1) + 12*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 17 2012
a(n) ~ (1+sqrt(3))*(6+4*sqrt(3))^n/(2*sqrt(2*Pi*n)). - Vaclav Kotesovec, Oct 17 2012

A106260 Expansion of 1/sqrt(1-16x-16x^2).

Original entry on oeis.org

1, 8, 104, 1472, 21856, 333568, 5183744, 81590272, 1296426496, 20750839808, 334081306624, 5404163080192, 87763693060096, 1430025994108928, 23367175920287744, 382767375745810432, 6283401962864377856
Offset: 0

Views

Author

Paul Barry, Apr 28 2005

Keywords

Comments

Central coefficient of (1+8x+20x^2)^n. Eighth binomial transform of 1/sqrt(1-80x^2). In general, 1/sqrt(1-4*r*x-4*r*x^2) has e.g.f. exp(2rx)BesselI(0,2r*sqrt((r+1)/r)x)), a(n)=sum{k=0..n, C(2k,k)C(k,n-k)r^k}, gives the central coefficient of (1+(2r)x+r(r+1)x^2) and is the (2r)-th binomial transform of 1/sqrt(1-8*C(n+1,2)x^2).

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/Sqrt[1-16*x-16*x^2], {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 17 2012 *)

Formula

E.g.f.: exp(8*x)*BesselI(0, 8*sqrt(5/4)*x); a(n)=sum{k=0..n, C(2k, k)C(k, n-k)4^k}.
D-finite with recurrence: n*a(n) = 8*(2*n-1)*a(n-1) + 16*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 17 2012
a(n) ~ sqrt(50+20*sqrt(5))*(8+4*sqrt(5))^n/(10*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 17 2012

A106261 Expansion of 1/sqrt(1 - 20*x - 20*x^2).

Original entry on oeis.org

1, 10, 160, 2800, 51400, 970000, 18640000, 362800000, 7128700000, 141103000000, 2809273600000, 56197096000000, 1128614356000000, 22741607080000000, 459548117440000000, 9309106936000000000, 188980474087000000000
Offset: 0

Views

Author

Paul Barry, Apr 28 2005

Keywords

Comments

Central coefficient of (1 + 10x + 30x^2)^n. Tenth binomial transform of 1/sqrt(1 - 120x^2). In general, 1/sqrt(1 - 4*r*x - 4*r*x^2) has e.g.f. exp(2rx)*BesselI(0,2r*sqrt((r+1)/r)x)), and a(n) = Sum_{k=0..n} C(2k,k)*C(k,n-k)*r^k gives the central coefficient of (1 + (2r)*x + r(r+1)*x^2) and is the (2r)-th binomial transform of 1/sqrt(1 - 8*C(n+1,2)x^2).

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/Sqrt[1-20*x-20*x^2], {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 19 2013 *)
  • PARI
    for(n=0,25, print1(sum(k=0,n,binomial(2*k,k)*binomial(k,n-k)*5^k), ", ")) \\ G. C. Greubel, Jan 31 2017

Formula

E.g.f.: exp(10*x)*BesselI(0, 10*sqrt(6/5)*x).
a(n) = Sum_{k=0..n} C(2k, k)*C(k, n-k)*5^k.
D-finite with recurrence: n*a(n) + 10*(-2*n+1)*a(n-1) + 20*(-n+1)*a(n-2) = 0. - R. J. Mathar, Nov 26 2012
a(n) ~ sqrt((1+sqrt(5/6))/2) * (10+2*sqrt(30))^n / sqrt(Pi*n). - Vaclav Kotesovec, Oct 19 2013
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