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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A366366 G.f. satisfies A(x) = (1 + x/A(x)^4)/(1 - x).

Original entry on oeis.org

1, 2, -6, 58, -574, 6402, -75878, 939290, -12000318, 157050178, -2094657926, 28368411194, -389079656446, 5393118559938, -75431624084838, 1063251390845338, -15088643098754942, 215396586102923138, -3091050571516120582, 44566089825496186170
Offset: 0

Views

Author

Seiichi Manyama, Oct 08 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = (-1)^(n-1)*sum(k=0, n, binomial(5*k-1, k)*binomial(4*k-1, n-k)/(5*k-1));

Formula

a(n) = (-1)^(n-1) * Sum_{k=0..n} binomial(5*k-1,k) * binomial(4*k-1,n-k)/(5*k-1).

A107708 Number of paths from (0,0) to (3n,0) that stay in the first quadrant (x,y >= 0) and where each step is (3,0), (2,1), (1,2), or (1,-1).

Original entry on oeis.org

1, 3, 18, 144, 1323, 13176, 138348, 1507977, 16900650, 193536864, 2254630788, 26635735440, 318350663748, 3842488208997, 46770206742342, 573435609537600, 7075551692662875, 87794803094586336, 1094807464312435344
Offset: 0

Views

Author

Emeric Deutsch, Jun 10 2005

Keywords

Examples

			a(1)=3 because we have H, uD and Udd, where H=(3,0), u=(2,1), U=(1,2) and D=(1,-1).
		

Crossrefs

Programs

  • Maple
    a:=n->(1/n)*sum(3^j*binomial(n,j)*binomial(n+j,2*n+1-j),j=ceil((n+1)/2)..n): 1,seq(a(n),n=1..22);
  • Mathematica
    Flatten[{1,Table[1/n*Sum[3^j*Binomial[n, j]*Binomial[n+j, 2n+1-j], {j,Floor[(n+1)/2],n}],{n,1,20}]}] (* Vaclav Kotesovec, Mar 17 2014 *)
  • PARI
    concat([1], for(n=1,50, print1((1/n)*sum(j=floor((n+1)/2),n, 3^j*binomial(n,j)*binomial(n+j,2*n+1-j)), ", "))) \\ G. C. Greubel, Mar 16 2017

Formula

a(n) = (1/n)*Sum(3^j*binomial(n, j)*binomial(n+j, 2n+1-j), j=ceiling((n+1)/2)..n) for n >= 1; a(0)=1.
G.f. = (2/3)w*sin((1/3)*arcsin((36-7z)/2/(3-2z)/w))-1/3, where w=sqrt((3-2z)/z).
Recurrence: 2*n*(2*n+1)*(17*n-25)*a(n) = 4*(238*n^3 - 588*n^2 + 395*n - 72)*a(n-1) - 12*(n-2)*(34*n^2 - 67*n + 21)*a(n-2) + 3*(n-3)*(n-2)*(17*n - 8)*a(n-3). - Vaclav Kotesovec, Mar 17 2014
a(n) ~ (1/204)*sqrt(102)*sqrt((134963 + 21573*sqrt(17))^(1/3) * ((134963 + 21573*sqrt(17))^(2/3) + 2176 + 68*(134963 + 21573*sqrt(17))^(1/3))) / ((134963 + 21573*sqrt(17))^(1/3)*sqrt(Pi)) * 6^(-n) * ((19009 + 153*sqrt(17))^(2/3) + 712 + 28*(19009 + 153*sqrt(17))^(1/3))^n * (19009 + 153*sqrt(17))^(-n/3)*(1/n)^(3/2). - Vaclav Kotesovec, Mar 17 2014
D-finite with recurrence 8*n*(2*n+1)*a(n) +2*(-106*n^2+97*n-18)*a(n-1) +36*(-2*n^2+12*n-15)*a(n-2) +12*(5*n-14)*(n-3)*a(n-3) -9*(n-3)*(n-4)*a(n-4)=0. - R. J. Mathar, Jul 26 2022
G.f. satisfies A(x) = 1 + x * A(x) * (1 + A(x) + A(x)^2). - Seiichi Manyama, Apr 01 2024

A348793 G.f. A(x) satisfies A(x) = (1 + x * A(x)^3) / (1 - 2 * x).

Original entry on oeis.org

1, 3, 15, 102, 807, 6951, 63240, 597864, 5815167, 57815553, 584919951, 6002197914, 62321630100, 653553174756, 6912106219176, 73642451396160, 789642274208271, 8515008918555573, 92281921130853213, 1004600177464845450, 10980406558088695599, 120454756647900759543
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 03 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 21; A[] = 0; Do[A[x] = (1 + x A[x]^3)/(1 - 2 x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    a[0] = 1; a[n_] := a[n] = 2 a[n - 1] + Sum[Sum[a[i] a[j] a[n - i - j - 1], {j, 0, n - i - 1}], {i, 0, n - 1}]; Table[a[n], {n, 0, 21}]
  • PARI
    a(n) = sum(k=0, n, 2^(n-k)*binomial(n, k)*binomial(n+2*k+1, n)/(n+2*k+1)); \\ Seiichi Manyama, Jul 24 2023

Formula

a(0) = 1; a(n) = 2 * a(n-1) + Sum_{i=0..n-1} Sum_{j=0..n-i-1} a(i) * a(j) * a(n-i-j-1).
a(n) ~ sqrt((2 + s^3)/(3*Pi*s*(1 - 2*r))) / (2*n^(3/2)*r^n), where r = (2 + (3*(-2 + sqrt(6))^(1/3))/2^(2/3) - 3/(2*(-2 + sqrt(6)))^(1/3)) / 4 = 0.084819663336750180604484695162155813902734598764355... and s = 1/2 + (-sqrt(2) + sqrt(3))/(2^(5/6)*(-2 + sqrt(6))^(1/3)) + 1/(2*(-2 + sqrt(6)))^(2/3) = 1.8064439323587723772036249693148814564378856424032... - Vaclav Kotesovec, Nov 04 2021
a(n) = Sum_{k=0..n} 2^(n-k) * binomial(n,k) * binomial(n+2*k+1,n) / (n+2*k+1). - Seiichi Manyama, Jul 24 2023

A348912 G.f. A(x) satisfies A(x) = (1 + 2 * x * A(x)^3) / (1 - x).

Original entry on oeis.org

1, 3, 21, 201, 2217, 26535, 335001, 4391553, 59203137, 815580507, 11430639165, 162470033625, 2336381642649, 33930648153615, 496935405133617, 7331179445170689, 108846406625097729, 1625145134034548019, 24385673680861258533, 367546405595389076649, 5561980053932228243529
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 03 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = (1 + 2 x A[x]^3)/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    a[0] = 1; a[n_] := a[n] = a[n - 1] + 2 Sum[Sum[a[i] a[j] a[n - i - j - 1], {j, 0, n - i - 1}], {i, 0, n - 1}]; Table[a[n], {n, 0, 20}]
  • PARI
    a(n) = sum(k=0, n, 2^k*binomial(n, k)*binomial(n+2*k+1, n)/(n+2*k+1)); \\ Seiichi Manyama, Jul 24 2023

Formula

a(0) = 1; a(n) = a(n-1) + 2 * Sum_{i=0..n-1} Sum_{j=0..n-i-1} a(i) * a(j) * a(n-i-j-1).
a(n) ~ sqrt(-50 + 30*sqrt(3) + (22 - 12*sqrt(3))*(2*(sqrt(3) - 1))^(1/3) + (2*(sqrt(3) - 1))^(2/3)*(-11 + 7*sqrt(3)))/(4*sqrt(3*Pi)*(-1 + sqrt(3))^(3/2) * n^(3/2) * (1 + (3*(-1 + sqrt(3))^(1/3))/2^(2/3) - 3/(2*(-1 + sqrt(3)))^(1/3))^n). - Vaclav Kotesovec, Nov 04 2021
a(n) = Sum_{k=0..n} 2^k * binomial(n,k) * binomial(n+2*k+1,n) / (n+2*k+1). - Seiichi Manyama, Jul 24 2023

A363380 G.f. satisfies A(x) = 1 + x * A(x)^4 * (1 + A(x)^2).

Original entry on oeis.org

1, 2, 20, 284, 4712, 85392, 1638112, 32699472, 672188768, 14133399744, 302535052160, 6570819330688, 144442463464704, 3207564324825600, 71848240540852224, 1621452789508328704, 36831997860270007808, 841470878382566444032
Offset: 0

Views

Author

Seiichi Manyama, May 29 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(n, k)*binomial(4*n+2*k+1, n)/(4*n+2*k+1));

Formula

a(n) = Sum_{k=0..n} binomial(n,k) * binomial(4*n+2*k+1,n)/(4*n+2*k+1).

A378236 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,0) = 0^n and T(n,k) = k * Sum_{r=0..n} binomial(n,r) * binomial(n+2*r+k,n)/(n+2*r+k) for k > 0.

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 4, 8, 0, 1, 6, 20, 44, 0, 1, 8, 36, 120, 280, 0, 1, 10, 56, 236, 800, 1936, 0, 1, 12, 80, 400, 1656, 5696, 14128, 0, 1, 14, 108, 620, 2960, 12192, 42416, 107088, 0, 1, 16, 140, 904, 4840, 22592, 92960, 326304, 834912, 0, 1, 18, 176, 1260, 7440, 38352, 176800, 727824, 2572992, 6652608, 0
Offset: 0

Views

Author

Seiichi Manyama, Nov 20 2024

Keywords

Examples

			Square array begins:
   1,     1,     1,     1,      1,      1,      1, ...
   0,     2,     4,     6,      8,     10,     12, ...
   0,     8,    20,    36,     56,     80,    108, ...
   0,    44,   120,   236,    400,    620,    904, ...
   0,   280,   800,  1656,   2960,   4840,   7440, ...
   0,  1936,  5696, 12192,  22592,  38352,  61248, ...
   0, 14128, 42416, 92960, 176800, 308560, 507152, ...
		

Crossrefs

Columns k=0..1 give A000007, A346626.

Programs

  • PARI
    T(n, k, t=1, u=2) = if(k==0, 0^n, k*sum(r=0, n, binomial(n, r)*binomial(t*n+u*r+k, n)/(t*n+u*r+k)));
    matrix(7, 7, n, k, T(n-1, k-1))

Formula

G.f. A_k(x) of column k satisfies A_k(x) = ( 1 + x * A_k(x)^(1/k) * (1 + A_k(x)^(2/k)) )^k for k > 0.
G.f. of column k: B(x)^k where B(x) is the g.f. of A346626.
B(x)^k = B(x)^(k-1) + x * B(x)^k + x * B(x)^(k+2). So T(n,k) = T(n,k-1) + T(n-1,k) + T(n-1,k+2) for n > 0.

A364552 G.f. satisfies A(x) = 1 + x*A(x) + x^4*A(x)^3.

Original entry on oeis.org

1, 1, 1, 1, 2, 5, 11, 21, 39, 78, 169, 373, 808, 1727, 3719, 8153, 18100, 40315, 89770, 200250, 448755, 1010685, 2284295, 5173961, 11740697, 26699780, 60863291, 139045991, 318247190, 729572315, 1675085099, 3851795549, 8869990949, 20453679944, 47223844863
Offset: 0

Views

Author

Seiichi Manyama, Jul 28 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n\4, binomial(n-k, 3*k)*binomial(3*k, k)/(2*k+1));

Formula

a(n) = Sum_{k=0..floor(n/4)} binomial(n-k,3*k) * binomial(3*k,k) / (2*k+1).

A366178 G.f. A(x) satisfies A(x) = 1/(1 - x) + x*A(x)^3/(1 - x)^3.

Original entry on oeis.org

1, 2, 10, 67, 502, 4045, 34279, 301232, 2720266, 25091431, 235394601, 2239139980, 21546299491, 209361514219, 2051379996574, 20245794958408, 201079938971546, 2008276118393320, 20157131084034349, 203215717750220949, 2056913539436637829
Offset: 0

Views

Author

Seiichi Manyama, Oct 03 2023

Keywords

Crossrefs

Partial sums give A366179.

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(n+4*k, n-k)*binomial(3*k, k)/(2*k+1));

Formula

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

A371655 G.f. satisfies A(x) = 1 + x * A(x) * (1 + A(x))^2.

Original entry on oeis.org

1, 4, 32, 336, 4032, 52352, 716032, 10161408, 148229120, 2208921600, 33482670080, 514630230016, 8001860567040, 125640146354176, 1989285578473472, 31725578742464512, 509178657425326080, 8217766225008656384, 133287551280741351424, 2171450128344786403328
Offset: 0

Views

Author

Seiichi Manyama, Apr 01 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = if(n==0, 1, sum(k=0, (n-1)\2, 4^(n-k)*binomial(n, k)*binomial(2*n-k, n-1-2*k))/n);

Formula

a(n) = (1/n) * Sum_{k=0..floor((n-1)/2)} 4^(n-k) * binomial(n,k) * binomial(2*n-k,n-1-2*k) for n > 0.
a(n) = 2^n * A100327(n). - Seiichi Manyama, Dec 26 2024

A349514 G.f. A(x) satisfies: A(x) = (1 + x * A(x)^3) / (1 - 3 * x).

Original entry on oeis.org

1, 4, 24, 192, 1792, 18240, 196224, 2194176, 25247232, 296979456, 3555010560, 43165900800, 530362220544, 6581594275840, 82373440339968, 1038579580796928, 13179023462498304, 168183976239562752, 2157085003249876992, 27790652486543474688, 359485965093121818624
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 20 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = (1 + x A[x]^3)/(1 - 3 x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    a[0] = 1; a[n_] := a[n] = 3 a[n - 1] + Sum[Sum[a[i] a[j] a[n - i - j - 1], {j, 0, n - i - 1}], {i, 0, n - 1}]; Table[a[n], {n, 0, 20}]
    Table[Sum[Binomial[n + 2 k, 3 k] Binomial[3 k, k] 3^(n - k)/(2 k + 1), {k, 0, n}], {n, 0, 20}]
  • PARI
    a(n) = sum(k=0, n, binomial(n+2*k,3*k) * binomial(3*k,k) * 3^(n-k) / (2*k+1)) \\ Andrew Howroyd, Nov 20 2021

Formula

a(0) = 1; a(n) = 3 * a(n-1) + Sum_{i=0..n-1} Sum_{j=0..n-i-1} a(i) * a(j) * a(n-i-j-1).
a(n) = Sum_{k=0..n} binomial(n+2*k,3*k) * binomial(3*k,k) * 3^(n-k) / (2*k+1).
a(n) ~ (3/4*(7 + (3*(69 + 16*sqrt(3)))^(1/3) + (3*(69 - 16*sqrt(3)))^(1/3)))^n / (sqrt((4 - (2 + sqrt(3))^(1/3) - (2 - sqrt(3))^(1/3)) * Pi) * n^(3/2)). - Vaclav Kotesovec, Nov 21 2021
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