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.

Previous Showing 11-16 of 16 results.

A214668 G.f. A(x) satisfies A(x) = 1 + 9*x*A(x)^(4/3).

Original entry on oeis.org

1, 9, 108, 1458, 21060, 318087, 4960116, 79227720, 1289516436, 21308126895, 356506456680, 6027199821864, 102804351279084, 1766931074710515, 30570993847594800, 532022685332573016, 9306598678048938420, 163549467160708850910, 2886019647490699098588
Offset: 0

Views

Author

Paul D. Hanna, Jul 24 2012

Keywords

Comments

Radius of convergence of g.f. A(x) is r = 1/(3*4^(4/3)) where A(r) = 4.
Self-convolution cube of A078532.

Examples

			G.f.: A(x) = 1 + 9*x + 108*x^2 + 1458*x^3 + 21060*x^4 + 318087*x^5 + ...
where A(x) = 1 + 9*x*A(x)^(4/3).
Radius of convergence: r = 1/(3*4^(4/3)) = 0.052496710...
Related expansions:
A(x)^(4/3) = 1 + 12*x + 162*x^2 + 2340*x^3 + 35343*x^4 + 551124*x^5 + ... + a(n+1)/9*x^n + ...
A(x)^(1/3) = 1 + 3*x + 27*x^2 + 315*x^3 + 4158*x^4 + 59049*x^5 + 880308*x^6 + 13586859*x^7 + 215233605*x^8 + ... + A078532(n)*x^n + ...
		

Crossrefs

Programs

  • Mathematica
    a[n_] := 9^n * Binomial[4*n/3, n] / (n/3 + 1); Array[a, 20, 0] (* Amiram Eldar, Sep 02 2025 *)
  • PARI
    {a(n)=9^n*binomial(4*n/3, n)/(n/3+1)}
    
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A =1+9*x*(A+x*O(x^n))^(4/3));polcoeff(A,n)}
    for(n=0,30,print1(a(n),", "))

Formula

a(n) = 9^n * binomial(4*n/3, n) / (n/3 + 1).
From Karol A. Penson, Mar 24 2024: (Start)
G.f. = 4F3([1/4, 1/2, 3/4, 1], [1/3, 2/3, 2], 6912*z^3) + 9*z*3F2([7/12, 5/6, 13/12], [2/3, 7/3], 6912*z^3) + 108*z^2*3F2([11/12, 7/6, 17/12], [4/3, 8/3], 6912*z^3);
a(n) = Integral_{x=0..6912^(1/3)} x^n*W(x), where
W(x) = h1(x) + h2(x) + h3(x), with
h1(x) = sqrt(6)*3F2([-3/4, 7/12, 11/12], [1/2, 3/4], x^3/6912)/(18*Pi*x^(1/4)),
h2(x) = sqrt(x)*3F2([-1/2, 5/6, 7/6], [3/4, 5/4], x^3/6912)/(36*Pi),
h3(x) = (5*sqrt(6)*x^(5/4)*3F2([-1/4, 13/12, 17/12], [5/4, 3/2], x^3/6912))/(5184*Pi).
This integral representation is unique as W(x) is the solution of the Hausdorff power moment problem on x = (0, 6912^(1/3)). Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0, with singularity x^(-1/4), and for x > 0 is first monotonically decreasing up to a local minimum at x around x = 2, then it is monotonically increasing up to a local maximum at x around x = 10.8, and then finally is monotonically decreasing up to zero at x = 6912^(1/3). For x -> 6912^(1/3), W'(x) tends to -infinity. (End)
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^(5/3)). - Seiichi Manyama, Jun 18 2025
D-finite with recurrence (n-1)*(n-2)*(n+3)*a(n) - 216*(4*n-9)*(4*n-3)*(2*n-3)*a(n-3) = 0. - R. J. Mathar, Jul 30 2025
a(n) ~ 2^(8*n/3 + 1/2) * 3^(n+1) / (sqrt(Pi) * n^(3/2)). - Amiram Eldar, Sep 02 2025

A386415 G.f. A(x) satisfies A(x) = (1 + 9*x*A(x)^4)^(2/3).

Original entry on oeis.org

1, 6, 135, 4140, 146475, 5629338, 228355281, 9622693080, 417122726490, 18480617374050, 833136935399208, 38094723501749460, 1762459398803643930, 82353342267057244950, 3880848811889775489300, 184228926273804535479216, 8801795826996054546077865, 422898288144162288398536860
Offset: 0

Views

Author

Seiichi Manyama, Jul 21 2025

Keywords

Crossrefs

Programs

  • Maple
    A386415 := proc(n)
        9^n*binomial((8*n+2)/3,n)/(4*n+1) ;
    end proc:
    seq(A386415(n),n=0..80) ; # R. J. Mathar, Jul 30 2025
  • Mathematica
    A386415[n_] := 9^n * Binomial[(8*n + 2)/3, n]/(4*n + 1);
    Array[A386415, 20, 0] (* Paolo Xausa, Aug 01 2025 *)
  • PARI
    apr(n, p, r) = r*binomial(n*p+r, n)/(n*p+r);
    a(n) = 9^n*apr(n, 8/3, 2/3);

Formula

a(n) = 9^n * binomial((8*n+2)/3,n)/(4*n+1).
G.f.: B(x)^2, where B(x) is the g.f. of A386416.
D-finite with recurrence 5*n*(n-1)*(n-2)*(5*n-4)*(5*n+2)*(5*n-7)*(5*n-1)*a(n) -3456*(4*n-11)*(8*n-19)*(8*n-13)*(4*n-5)*(8*n-7)*(2*n-1)*(8*n-1)*a(n-3)=0. - R. J. Mathar, Jul 30 2025

A386413 G.f. A(x) satisfies A(x) = (1 + 9*x*A(x)^2)^(2/3).

Original entry on oeis.org

1, 6, 63, 792, 10935, 160056, 2438667, 38263752, 614014830, 10029572280, 166203389781, 2787232297680, 47213065271268, 806618756189736, 13883029872725475, 240491818267745760, 4189678646994012501, 73357895462268102840, 1290223574267814268290, 22784365638084466567800
Offset: 0

Views

Author

Seiichi Manyama, Jul 21 2025

Keywords

Crossrefs

Programs

  • Mathematica
    A386413[n_] := 9^n*Binomial[(4*n + 2)/3, n]/(2*n + 1);
    Array[A386413, 25, 0] (* Paolo Xausa, Aug 01 2025 *)
  • PARI
    apr(n, p, r) = r*binomial(n*p+r, n)/(n*p+r);
    a(n) = 9^n*apr(n, 4/3, 2/3);

Formula

a(n) = 9^n * binomial((4*n+2)/3,n)/(2*n+1).
G.f.: B(x)^2, where B(x) is the g.f. of A078532.
D-finite with recurrence n*(n-2)*(n+2)*a(n) -216*(2*n-5)*(4*n-7)*(4*n-1)*a(n-3)=0. - R. J. Mathar, Jul 30 2025

A385208 G.f. A(x) satisfies A(x) = ( 1 + 49*x*A(x)^8 )^(1/7).

Original entry on oeis.org

1, 7, 245, 11319, 593047, 33429123, 1977326743, 121034349975, 7601257418678, 487008549508481, 31705597390195820, 2091361378163375955, 139468121325692304390, 9387480337647754305649, 636914947847207765431080, 43512658997082838985965655, 2990750175103769856729417627
Offset: 0

Views

Author

Seiichi Manyama, Jun 21 2025

Keywords

Crossrefs

Programs

  • PARI
    a(n) = 49^n*binomial(8*n/7+1/7, n)/(8*n+1);

Formula

a(n) = 49^n * binomial(8*n/7+1/7,n)/(8*n+1).
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^9).
G.f.: ( (1/x) * Series_Reversion(x/(1+49*x)^(8/7)) )^(1/8).

A386416 G.f. A(x) satisfies A(x) = (1 + 9*x*A(x)^8)^(1/3).

Original entry on oeis.org

1, 3, 63, 1881, 65610, 2499336, 100777122, 4228144596, 182674383705, 8072369224920, 363154406671485, 16576444298006658, 765806677899249168, 35739548618003938440, 1682429522012566325460, 79793991407758199002740, 3809208342822290233767522, 182890356905449116974950200
Offset: 0

Views

Author

Seiichi Manyama, Jul 21 2025

Keywords

Crossrefs

Programs

  • Mathematica
    A386416[n_] := 9^n*Binomial[(8*n + 1)/3, n]/(8*n + 1);
    Array[A386416, 20, 0] (* Paolo Xausa, Aug 01 2025 *)
  • PARI
    apr(n, p, r) = r*binomial(n*p+r, n)/(n*p+r);
    a(n) = 9^n*apr(n, 8/3, 1/3);

Formula

a(n) = 9^n * binomial((8*n+1)/3,n)/(8*n+1).
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^13).
D-finite with recurrence 5*n*(n-1)*(n-2)*(5*n-8)*(5*n-11)*(5*n+1)*(5*n-2)*a(n) -3456*(8*n-11)*(8*n-5)*(4*n-1)*(8*n-23)*(2*n-5)*(8*n-17)*(4*n-7)*a(n-3)=0. - R. J. Mathar, Jul 30 2025

A377267 G.f. A(x) satisfies A(x) = 1/(1 - 9*x*A(x))^(2/3).

Original entry on oeis.org

1, 6, 81, 1386, 26676, 551124, 11939967, 267647490, 6155681103, 144442697256, 3444310087155, 83222570083068, 2033143391152440, 50136558534943776, 1246312401751305306, 31197886757177379570, 785732617740674763684, 19896044539519213482690
Offset: 0

Views

Author

Seiichi Manyama, Oct 22 2024

Keywords

Crossrefs

Programs

  • Mathematica
    A377267[n_] := 9^n*Binomial[(5*n - 1)/3, n]/(n + 1);
    Array[A377267, 20, 0] (* Paolo Xausa, Aug 05 2025 *)
  • PARI
    a(n) = 9^n*binomial(5*n/3-1/3, n)/(n+1);

Formula

G.f.: (1/x) * Series_Reversion( x*(1-9*x)^(2/3) ).
a(n) = 9^n * binomial(5*n/3 - 1/3,n)/(n+1).
G.f.: B(x)^2, where B(x) is the g.f. of A245114.
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^(7/2)). - Seiichi Manyama, Jun 22 2025
E.g.f.: (2*hypergeom([2/15, 8/15, 11/15, 14/15], [1/3, 2/3, 2/3, 5/6, 1, 4/3], 5^5*x^3/4) + 12*x*hypergeom([7/15, 13/15, 16/15, 19/15], [2/3, 1, 7/6, 4/3, 4/3, 5/3], 5^5*x^3/4) + 81*x^2*hypergeom([4/5, 6/5, 7/5, 8/5], [4/3, 4/3, 3/2, 5/3, 5/3, 2], 5^5*x^3/4))/2. - Stefano Spezia, Jun 22 2025
From Karol A. Penson, Aug 23 2025: (Start)
a(n) = Integral_{x=0..15*5^(2/3)*2^(1/3)/2} x^n*W(x)*dx, where W(x) can be represented in terms of 4 different generalized hypergeometric functions of type 4F3 of argument 4*x^3/84375, W(x) = W1(x) + W2(x) + W3(x) + W4(x):
W1(x) = sqrt(15)*3^(7/10)*csc((2*Pi)/5)*sin((3*Pi)/10)*hypergeom([-1/5, 2/15, 3/10, 7/15], [1/5, 2/5, 3/5], (4*x^3)/84375)/(18*Pi*x^(3/5)),
W2(x) = -sqrt(15)*3^(3/10)*sec(Pi/10)*sin((3*Pi)/10)*x^(3/5)*hypergeom([1/5, 8/15, 7/10, 13/15], [3/5, 4/5, 7/5], (4*x^3)/84375)/(1350*Pi),
W3(x) = -sqrt(15)*3^(1/10)*sec((3*Pi)/10)*sin(Pi/10)*x^(6/5)*hypergeom([2/5, 11/15, 9/10, 16/15], [4/5, 6/5, 8/5], (4*x^3)/84375)/(30375*Pi),
W4(x) = -sqrt(15)*3^(9/10)*csc(Pi/5)*sin(Pi/10)*x^(9/5)*hypergeom([3/5, 14/15, 11/10, 19/15], [6/5, 7/5, 9/5], (4*x^3)/84375)/(911250*Pi).
W(x) is a positive function on x = (0,15*5^(2/3)*2^(1/3)/2), is singular at x=0 with the singularity x^(-3/5), and tends to zero at x = 15*5^(2/3)*2^(1/3)/2.
Thus a(n) is a positive definite sequence. This representation is unique as W(x) is the solution of the Hausdorff moment problem. (End)
a(n) ~ 3^n * 5^(5*n/3+1/6) / (sqrt(Pi) * 2^(2*(n+1)/3) * n^(3/2)). - Amiram Eldar, Sep 02 2025
Previous Showing 11-16 of 16 results.