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 1-8 of 8 results.

A349362 G.f. A(x) satisfies: A(x) = 1 + x * A(x)^6 / (1 + x).

Original entry on oeis.org

1, 1, 5, 40, 370, 3740, 40006, 445231, 5102165, 59799505, 713496815, 8637432580, 105826926716, 1309793896431, 16351672606365, 205665994855320, 2603696877136060, 33151784577226295, 424258396639960591, 5454120586840761631, 70402732493668027775
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 15 2021

Keywords

Crossrefs

Programs

  • Maple
    a:= n-> coeff(series(RootOf(1+x*A^6/(1+x)-A, A), x, n+1), x, n):
    seq(a(n), n=0..20);  # Alois P. Heinz, Nov 15 2021
  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = 1 + x A[x]^6/(1 + x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    Table[Sum[(-1)^(n - k) Binomial[n - 1, k - 1] Binomial[6 k, k]/(5 k + 1), {k, 0, n}], {n, 0, 20}]

Formula

a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n-1,k-1) * binomial(6*k,k) / (5*k+1).
a(n) = (-1)^(n+1)* F([7/6, 4/3, 3/2, 5/3, 11/6, 1-n], [7/5, 8/5, 9/5, 2, 11/5], 6^6/5^5), where F is the generalized hypergeometric function. - Stefano Spezia, Nov 15 2021
a(n) ~ 43531^(n + 1/2) / (72 * sqrt(3*Pi) * n^(3/2) * 5^(5*n + 3/2)). - Vaclav Kotesovec, Nov 17 2021

A346664 a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * binomial(4*k,k) / (3*k + 1).

Original entry on oeis.org

1, 0, 3, 12, 73, 453, 2985, 20373, 142933, 1024302, 7466211, 55182240, 412586977, 3115105321, 23717115513, 181884676827, 1403719428485, 10894049061956, 84967420574247, 665643698649684, 5235570329071893, 41328838600501830, 327315349579739619, 2600034901186102182
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 27 2021

Keywords

Comments

Inverse binomial transform of A002293.

Crossrefs

Programs

  • Maple
    A346664 := proc(n)
        add( (-1)^(n-k)*binomial(n,k)*binomial(4*k,k)/(3*k+1),k=0..n) ;
    end proc:
    seq(A346664(n),n=0..80); # R. J. Mathar, Aug 17 2023
  • Mathematica
    Table[Sum[(-1)^(n - k) Binomial[n, k] Binomial[4 k, k]/(3 k + 1), {k, 0, n}], {n, 0, 23}]
    nmax = 23; A[] = 0; Do[A[x] = 1/(1 + x) + x (1 + x)^2 A[x]^4 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    nmax = 23; CoefficientList[Series[Sum[(Binomial[4 k, k]/(3 k + 1)) x^k/(1 + x)^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x]
    Table[(-1)^n HypergeometricPFQ[{1/4, 1/2, 3/4, -n}, {2/3, 1, 4/3}, 256/27], {n, 0, 23}]
  • PARI
    a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n,k)*binomial(4*k,k)/(3*k+1)); \\ Michel Marcus, Jul 28 2021

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 + x) + x * (1 + x)^2 * A(x)^4.
G.f.: Sum_{k>=0} ( binomial(4*k,k) / (3*k + 1) ) * x^k / (1 + x)^(k+1).
a(n) ~ 229^(n + 3/2) / (2048 * sqrt(2*Pi) * n^(3/2) * 3^(3*n + 3/2)). - Vaclav Kotesovec, Jul 30 2021
D-finite with recurrence +3*n*(3*n-1)*(3*n+1)*a(n) -74*n*(2*n-1) *(n-1)*a(n-1) -6*(n-1) *(101*n^2 -202*n +105)*a(n-2) -330*(n-1) *(n-2)*(2*n-3) *a(n-3) -229*(n-1)*(n-2) *(n-3)*a(n-4)=0. - R. J. Mathar, Aug 17 2023

A346665 a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * binomial(5*k,k) / (4*k + 1).

Original entry on oeis.org

1, 0, 4, 22, 172, 1409, 12216, 109904, 1016876, 9614584, 92490261, 902364918, 8907507708, 88802649446, 892833960460, 9042639746819, 92171773008828, 944819352291920, 9733592874215112, 100725697334689896, 1046535959932600141, 10913073121311627481, 114175868855824821752
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 27 2021

Keywords

Comments

Inverse binomial transform of A002294.

Crossrefs

Programs

  • Maple
    A346665 := proc(n)
        add((-1)^(n-k)*binomial(n,k)*binomial(5*k,k)/(4*k+1),k=0..n) ;
    end proc:
    seq(A346665(n),n=0..80); # R. J. Mathar, Aug 17 2023
  • Mathematica
    Table[Sum[(-1)^(n - k) Binomial[n, k] Binomial[5 k, k]/(4 k + 1), {k, 0, n}], {n, 0, 22}]
    nmax = 22; A[] = 0; Do[A[x] = 1/(1 + x) + x (1 + x)^3 A[x]^5 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    nmax = 22; CoefficientList[Series[Sum[(Binomial[5 k, k]/(4 k + 1)) x^k/(1 + x)^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x]
    Table[(-1)^n HypergeometricPFQ[{1/5, 2/5, 3/5, 4/5, -n}, {1/2, 3/4, 1, 5/4}, 3125/256], {n, 0, 22}]
  • PARI
    a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n,k)*binomial(5*k,k)/(4*k + 1)); \\ Michel Marcus, Jul 28 2021

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 + x) + x * (1 + x)^3 * A(x)^5.
G.f.: Sum_{k>=0} ( binomial(5*k,k) / (4*k + 1) ) * x^k / (1 + x)^(k+1).
a(n) ~ 2869^(n + 3/2) / (78125 * sqrt(Pi) * n^(3/2) * 2^(8*n + 7/2)). - Vaclav Kotesovec, Jul 30 2021
D-finite with recurrence 8*n*(4*n+1)*(2*n-1)*(4*n-1)*a(n) -(n-1) *(1845*n^3 -1333*n^2 -238*n +240)*a(n-1) -4*(n-1) *(2485*n^3 -7263*n^2 +7388*n -2580) *a(n-2) -2*(n-1) *(n-2) *(8095*n^2 -24029*n +18924) *a(n-3) -4*(n-1) *(n-2) *(n-3) *(2805*n -5578) *a(n-4) -2869*(n-1) *(n-2) *(n-3) *(n-4) *a(n-5)=0. - R. J. Mathar, Aug 17 2023

A346668 a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * binomial(8*k,k) / (7*k + 1).

Original entry on oeis.org

1, 0, 7, 70, 917, 12922, 192591, 2984156, 47594289, 776184997, 12884436285, 216981375849, 3698021707457, 63663537870121, 1105474964523293, 19339098305850757, 340519405008643561, 6030158137055187758, 107328892461895007043, 1918980244360791943044, 34450128513971163342013
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 27 2021

Keywords

Comments

Inverse binomial transform of A007556.
In general, for m > 1, Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * binomial(m*k,k) / ((m-1)*k + 1) ~ (m^m - (m-1)^(m-1))^(n + 3/2) / (sqrt(2*Pi) * m^((3*m-1)/2) * n^(3/2) * (m-1)^((m-1)*n + 3/2)). - Vaclav Kotesovec, Jul 30 2021

Crossrefs

Programs

  • Mathematica
    Table[Sum[(-1)^(n - k) Binomial[n, k] Binomial[8 k, k]/(7 k + 1), {k, 0, n}], {n, 0, 20}]
    nmax = 20; A[] = 0; Do[A[x] = 1/(1 + x) + x (1 + x)^6 A[x]^8 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    nmax = 20; CoefficientList[Series[Sum[(Binomial[8 k, k]/(7 k + 1)) x^k/(1 + x)^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x]
    Table[(-1)^n HypergeometricPFQ[{1/8, 1/4, 3/8, 1/2, 5/8, 3/4, 7/8, -n}, {2/7, 3/7, 4/7, 5/7, 6/7, 1, 8/7}, 16777216/823543], {n, 0, 20}]
  • PARI
    a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n,k)*binomial(8*k,k)/(7*k + 1)); \\ Michel Marcus, Jul 28 2021

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 + x) + x * (1 + x)^6 * A(x)^8.
G.f.: Sum_{k>=0} ( binomial(8*k,k) / (7*k + 1) ) * x^k / (1 + x)^(k+1).
a(n) ~ 15953673^(n + 3/2) / (34359738368 * sqrt(Pi) * n^(3/2) * 7^(7*n + 3/2)). - Vaclav Kotesovec, Jul 30 2021

A346628 G.f. A(x) satisfies: A(x) = 1 / (1 + x) + x * (1 + x) * A(x)^3.

Original entry on oeis.org

1, 0, 2, 5, 22, 92, 415, 1927, 9198, 44804, 221880, 1113730, 5653747, 28975962, 149725355, 779178092, 4080167790, 21483383992, 113670233848, 604070682354, 3222823434608, 17255628041720, 92689459311470, 499359484166994, 2697571066055611, 14608820993453132
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 25 2021

Keywords

Comments

Inverse binomial transform of A001764.

Crossrefs

Programs

  • Mathematica
    nmax = 25; A[] = 0; Do[A[x] = 1/(1 + x) + x (1 + x) A[x]^3 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    nmax = 25; CoefficientList[Series[Sum[(Binomial[3 k, k]/(2 k + 1)) x^k/(1 + x)^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x]
    Table[Sum[(-1)^(n - k) Binomial[n, k] Binomial[3 k, k]/(2 k + 1), {k, 0, n}], {n, 0, 25}]

Formula

G.f.: Sum_{k>=0} ( binomial(3*k,k) / (2*k + 1) ) * x^k / (1 + x)^(k+1).
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * binomial(3*k,k) / (2*k + 1).
a(n) ~ 23^(n + 3/2) / (81 * sqrt(Pi) * n^(3/2) * 2^(2*n+2)). - Vaclav Kotesovec, Jul 30 2021
D-finite with recurrence +2*n*(2*n+1)*a(n) -(15*n-4)*(n-1)*a(n-1) -2*(n-1)*(21*n-22)*a(n-2) -23*(n-1)*(n-2)*a(n-3)=0. - R. J. Mathar, Aug 05 2021

A346667 a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * binomial(7*k,k) / (6*k + 1).

Original entry on oeis.org

1, 0, 6, 51, 578, 7011, 89931, 1198798, 16445122, 230643888, 3292247673, 47672499727, 698569117499, 10339672571689, 154357100458366, 2321475460350492, 35140713973159266, 534971413383669580, 8185501429052369700, 125811555778930237392, 1941590759206061655069
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 27 2021

Keywords

Comments

Inverse binomial transform of A002296.

Crossrefs

Programs

  • Mathematica
    Table[Sum[(-1)^(n - k) Binomial[n, k] Binomial[7 k, k]/(6 k + 1), {k, 0, n}], {n, 0, 20}]
    nmax = 20; A[] = 0; Do[A[x] = 1/(1 + x) + x (1 + x)^5 A[x]^7 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    nmax = 20; CoefficientList[Series[Sum[(Binomial[7 k, k]/(6 k + 1)) x^k/(1 + x)^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x]
    Table[(-1)^n HypergeometricPFQ[{1/7, 2/7, 3/7, 4/7, 5/7, 6/7, -n}, {1/3, 1/2, 2/3, 5/6, 1, 7/6}, 823543/46656], {n, 0, 20}]
  • PARI
    a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n,k)*binomial(7*k,k)/(6*k + 1)); \\ Michel Marcus, Jul 28 2021

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 + x) + x * (1 + x)^5 * A(x)^7.
G.f.: Sum_{k>=0} ( binomial(7*k,k) / (6*k + 1) ) * x^k / (1 + x)^(k+1).
a(n) ~ 776887^(n + 3/2) / (282475249 * sqrt(Pi) * n^(3/2) * 2^(6*n + 2) * 3^(6*n + 3/2)). - Vaclav Kotesovec, Jul 30 2021

A349301 G.f. A(x) satisfies: A(x) = 1 / ((1 + x) * (1 - x * A(x)^5)).

Original entry on oeis.org

1, 0, 1, 5, 31, 200, 1351, 9430, 67531, 493505, 3665981, 27602081, 210179437, 1615820402, 12524590873, 97775503808, 768083233899, 6067097140799, 48159634951855, 383965003803985, 3073379977522321, 24688458872260007, 198968304164411309
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 13 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 22; A[] = 0; Do[A[x] = 1/((1 + x) (1 - x A[x]^5)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    Table[Sum[(-1)^(n - k) Binomial[n + 4 k, 5 k] Binomial[6 k, k]/(5 k + 1), {k, 0, n}], {n, 0, 22}]

Formula

a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n+4*k,5*k) * binomial(6*k,k) / (5*k+1).
a(n) ~ sqrt(1 - 4*r) / (2^(6/5) * 3^(7/10) * sqrt(5*Pi*(1+r)) * n^(3/2) * r^(n + 1/5)), where r = 0.11589193448796656683553561932282471511616945026903125324... is the real root of the equation 6^6 * r = 5^5 * (1+r)^5. - Vaclav Kotesovec, Nov 14 2021
From Peter Bala, Jun 02 2024: (Start)
A(x) = 1/(1 + x)*F(x/(1 + x)^5), where F(x) = Sum_{n >= 0} A002295(n)*x^n.
A(x) = 1/(1 + x) + x*A(x)^6. (End)

A346682 a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(6*k,k) / (5*k + 1).

Original entry on oeis.org

1, 0, 6, 45, 461, 5020, 57812, 691586, 8512048, 107095262, 1371219004, 17808830924, 234048288772, 3106795261083, 41593689788637, 560980967638479, 7614970691479315, 103957059568762775, 1426355910771621805, 19658792867492660060, 272046427837226505466
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 29 2021

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[(-1)^(n - k) Binomial[6 k, k]/(5 k + 1), {k, 0, n}], {n, 0, 20}]
    nmax = 20; A[] = 0; Do[A[x] = 1/(1 + x) + x (1 + x)^5 A[x]^6 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
  • PARI
    a(n) = sum(k=0, n, (-1)^(n-k)*binomial(6*k, k)/(5*k + 1)); \\ Michel Marcus, Jul 29 2021

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 + x) + x * (1 + x)^5 * A(x)^6.
a(n) ~ 2^(6*n + 6) * 3^(6*n + 13/2) / (49781 * sqrt(Pi) * n^(3/2) * 5^(5*n + 3/2)). - Vaclav Kotesovec, Jul 30 2021
Showing 1-8 of 8 results.