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 1-5 of 5 results.

A349254 G.f. A(x) satisfies A(x) = 1 / ((1 - x) * (1 - 3 * x * A(x)^2)).

Original entry on oeis.org

1, 4, 37, 478, 7159, 116497, 2000386, 35671756, 654218641, 12261271942, 233798163646, 4521194100541, 88458184054882, 1747850650032532, 34828329987024058, 699083528482636228, 14121906499195594537, 286877562430915732546, 5856866441794110926809
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 12 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 18; A[] = 0; Do[A[x] = 1/((1 - x) (1 - 3 x A[x]^2)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    a[n_] := a[n] = 1 + 3 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, 18}]
    Table[Sum[Binomial[n + k, n - k] 3^k Binomial[3 k, k]/(2 k + 1), {k, 0, n}], {n, 0, 18}]
    a[n_] := HypergeometricPFQ[{1/3, 2/3, -n, n + 1}, {1/2, 1, 3/2}, -81/16];
    Table[a[n], {n, 0, 18}] (* Peter Luschny, Nov 12 2021 *)

Formula

a(n) = 1 + 3 * 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+k,n-k) * 3^k * binomial(3*k,k) / (2*k+1).
a(n) = hypergeom([1/3, 2/3, -n, n + 1], [1/2, 1, 3/2], -(3/2)^4). - Peter Luschny, Nov 12 2021
a(n) ~ sqrt(873 + 89*sqrt(97)) * (89 + 9*sqrt(97))^n / (3^(5/2) * sqrt(Pi) * n^(3/2) * 2^(3*n + 5/2)). - Vaclav Kotesovec, Nov 13 2021

A349255 G.f. A(x) satisfies A(x) = 1 / ((1 + x) * (1 - 2 * x * A(x)^2)).

Original entry on oeis.org

1, 1, 7, 47, 369, 3113, 27631, 254239, 2403361, 23201393, 227771831, 2266983119, 22822484497, 231994748633, 2377894546783, 24548520253247, 255026759000897, 2664111200687969, 27967731861910759, 294900120348032623, 3121862973452544433, 33167268461833410569
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 12 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 21; A[] = 0; Do[A[x] = 1/((1 + x) (1 - 2 x A[x]^2)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    a[n_] := a[n] = (-1)^n + 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, 21}]
    Table[Sum[(-1)^(n - k) Binomial[n + k, n - k] 2^k Binomial[3 k, k]/(2 k + 1), {k, 0, n}], {n, 0, 21}]
    a[n_] := (-1)^n*HypergeometricPFQ[{1/3, 2/3, -n, n + 1}, {1/2, 1, 3/2}, (3/2)^3]; Table[a[n], {n, 0, 21}] (* Peter Luschny, Nov 12 2021 *)

Formula

a(n) = (-1)^n + 2 * 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} (-1)^(n-k) * binomial(n+k,n-k) * 2^k * binomial(3*k,k) / (2*k+1).
a(n) = (-1)^n*hypergeom([1/3, 2/3, -n, n + 1], [1/2, 1, 3/2], (3/2)^3). - Peter Luschny, Nov 12 2021
a(n) ~ sqrt(171 + 23*sqrt(57)) * (23 + 3*sqrt(57))^n / (9 * sqrt(Pi) * n^(3/2) * 2^(2*n + 5/2)). - Vaclav Kotesovec, Nov 13 2021

A349256 G.f. A(x) satisfies A(x) = 1 / ((1 + x) * (1 - 3 * x * A(x)^2)).

Original entry on oeis.org

1, 2, 19, 206, 2563, 34415, 486370, 7128488, 107364421, 1651615568, 25840137724, 409898503763, 6577319627506, 106571487893024, 1741193467526782, 28653852176675324, 474521786894159593, 7902112425718228064, 132243695376774536755, 2222925664652778182060
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 12 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 19; A[] = 0; Do[A[x] = 1/((1 + x) (1 - 3 x A[x]^2)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    a[n_] := a[n] = (-1)^n + 3 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, 19}]
    Table[Sum[(-1)^(n - k) Binomial[n + k, n - k] 3^k Binomial[3 k, k]/(2 k + 1), {k, 0, n}], {n, 0, 19}]
    a[n_] := (-1)^n*HypergeometricPFQ[{1/3, 2/3, -n, n + 1}, {1/2, 1, 3/2}, (3/2)^4]; Table[a[n], {n, 0, 19}] (* Peter Luschny, Nov 12 2021 *)

Formula

a(n) = (-1)^n + 3 * 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} (-1)^(n-k) * binomial(n+k,n-k) * 3^k * binomial(3*k,k) / (2*k+1).
a(n) = (-1)^n*hypergeom([1/3, 2/3, -n, n + 1], [1/2, 1, 3/2], (3/2)^4). - Peter Luschny, Nov 12 2021
a(n) ~ sqrt(585 + 73*sqrt(65)) * (73 + 9*sqrt(65))^n / (3^(5/2) * sqrt(Pi) * n^(3/2) * 2^(3*n + 5/2)). - Vaclav Kotesovec, Nov 13 2021

A349533 G.f. A(x) satisfies A(x) = 1 / ((1 - 2 * x) * (1 - x * A(x)^2)).

Original entry on oeis.org

1, 3, 13, 74, 499, 3719, 29494, 243888, 2078431, 18122369, 160885449, 1449268478, 13213370392, 121696581804, 1130565483472, 10581614352704, 99685591788687, 944490400760597, 8994266558594671, 86040075341770806, 826423263373253923, 7967095415955791687
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 21 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 21; A[] = 0; Do[A[x] = 1/((1 - 2 x) (1 - x A[x]^2)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    a[n_] := a[n] = 2^n + 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}]
    Table[Sum[Binomial[n + k, 2 k] Binomial[3 k, k] 2^(n - k)/(2 k + 1), {k, 0, n}], {n, 0, 21}]

Formula

a(n) = 2^n + 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+k,2*k) * binomial(3*k,k) * 2^(n-k) / (2*k+1).
a(n) = 2^n*F([1/3, 2/3, -n, 1+n], [1/2, 1, 3/2], -3^3/2^5), where F is the generalized hypergeometric function. - Stefano Spezia, Nov 21 2021
a(n) ~ 177^(1/4) * (43 + 3*sqrt(177))^(n + 1/2) / (9 * sqrt(Pi) * n^(3/2) * 2^(3*n + 5/2)). - Vaclav Kotesovec, Nov 22 2021

A364641 G.f. satisfies A(x) = 1/(1 - 2*x) - x*A(x)^3.

Original entry on oeis.org

1, 1, 1, 2, 3, 5, 10, 16, 31, 59, 101, 206, 376, 692, 1408, 2528, 4943, 9767, 17755, 35950, 68659, 129029, 262758, 490832, 958948, 1920580, 3581020, 7203080, 14054600, 26665160, 54195040, 103450560, 201749935, 406617695, 769870535, 1539785150, 3042812185
Offset: 0

Views

Author

Seiichi Manyama, Jul 31 2023

Keywords

Crossrefs

Programs

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

Formula

a(n) = Sum_{k=0..n} (-1)^k * 2^(n-k) * binomial(n+k,2*k) * binomial(3*k,k) / (2*k+1).
Showing 1-5 of 5 results.