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 1-7 of 7 results.

A363985 a(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(2*n+k,k)*binomial(2*k,k).

Original entry on oeis.org

1, 2, 26, 272, 3418, 44252, 597104, 8249152, 116158298, 1659335492, 23979247276, 349798313152, 5142733169776, 76108788764192, 1132729444052288, 16940944956246272, 254449319912898394, 3836162994088105172, 58028561918702719604
Offset: 0

Views

Author

Peter Bala, Jul 02 2023

Keywords

Comments

The sequence of Franel numbers A000172 satisfies the identity A000172(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(n+2*k,2*k)*binomial(2*k,k). The present sequence comes from a modification of the right-hand side of the identity.
The Franel numbers satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r. We conjecture that the present sequence satisfies the same supercongruences.
More generally, define two families of sequences {u_m(n): n >= 0} and {v_m(n): n >= 0}, depending on an integer parameter m, by u_m(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(m*n + k,k)*binomial(2*k,k) and v_m(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(m*n + 2*k,2*k)*binomial(2*k,k). In this notation, the Franel numbers A000172 = v_1. Note that u_0(n) = v_0(n) = (-1)^n*binomial(2*n,n). We conjecture that, for all m in Z, the sequences u_m and v_m satisfy the above supercongruences.

Links

Crossrefs

Cf. A000172, A000984, A362676, A363986 - A363990.

Programs

  • Maple
    seq(add((-4)^(n-k)*binomial(n,k)*binomial(2*n+k,k)*binomial(2*k,k), k = 0..n), n = 0..20);
# alternative faster program for large n
seq(simplify((-4)^n*hypergeom([-n, 2*n+1, 1/2], [1, 1], 1)), n = 0..20);
  • Mathematica
    Table[(-4)^n*HypergeometricPFQ[{-n, 2*n+1, 1/2}, {1, 1}, 1], {n, 0, 20}] (* Vaclav Kotesovec, Jul 17 2023 *)
Table[Sum[(-4)^(n-k) Binomial[n,k] Binomial[2n+k,k]Binomial[2k,k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Dec 26 2024 *)
  • Python
    from sympy import hyper, hyperexpand, S
def A363985(n): return int(hyperexpand(hyper((-n, (m:=n<<1)+1, S.Half), [1,1], 1))*(-(1<Chai Wah Wu, Jul 10 2023

Formula

a(n) = (-4)^n*hypergeom([-n, 2*n+1, 1/2], [1, 1], 1).
P-recursive: (20*n^2 - 55*n + 38)*n^2*(2*n - 1)^2*a(n) = (960*n^6 - 4560*n^5 + 8564*n^4 - 8107*n^3 + 4085*n^2 - 1044*n + 108)*a(n-1) + 64*(20*n^2 - 15*n + 3)*(n - 1)^2*(2*n - 3)^2*a(n-2) with a(0) = 1 and a(1) = 2.
a(n) ~ 2^(4*n - 1/2) / (Pi*n). - Vaclav Kotesovec, Jul 17 2023

A363990 a(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(4*n+2*k,2*k)* binomial(2*k,k).

Original entry on oeis.org

1, 26, 2266, 248912, 30319450, 3916519276, 525461758576, 72392116266048, 10170507622031194, 1450543675513428980, 209364056530208329516, 30512943166359499514816, 4482853776729105180939376, 663074836517763287008874912, 98645674256284705335605335360
Offset: 0

Views

Author

Peter Bala, Jul 02 2023

Keywords

Comments

The sequence of Franel numbers A000172 satisfies the identity A000172(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(n+2*k,2*k)*binomial(2*k,k). The present sequence comes from a modification of the right-hand side of the identity.
The Franel numbers satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r. We conjecture that the present sequence satisfies the same supercongruences.

Crossrefs

Cf. A000172, A362676, A363985 - A363989.

Programs

  • Maple
    seq(add((-4)^(n-k)*binomial(n,k)*binomial(4*n+2*k,2*k)*binomial(2*k,k), k = 0..n), n = 0..20);
# alternative faster program for large n
seq(simplify((-4)^n*hypergeom([-n, (4*n+1)/2, (4*n+2)/2], [1, 1], 1)), n = 0..20);
  • Mathematica
    Table[(-4)^n*HypergeometricPFQ[{-n, (4*n+1)/2, (4*n+2)/2}, {1, 1}, 1], {n, 0, 20}] (* Vaclav Kotesovec, Jul 17 2023 *)

Formula

a(n) = (-4)^n*hypergeom([-n, (4*n+1)/2, (4*n+2)/2], [1, 1], 1).
a(n) ~ 2^(n - 5/2) * (1 + sqrt(3))^(6*n + 3/2) / (Pi * n * 3^(3*n/2 + 1/4)). - Vaclav Kotesovec, Jul 17 2023

A363986 a(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(2*n+2*k,2*k)* binomial(2*k,k).

Original entry on oeis.org

1, 8, 196, 5984, 202276, 7247008, 269601424, 10297532288, 401058612004, 15856045354016, 634378106023696, 25627134232807808, 1043593599416897296, 42785184639021974144, 1764239681293889859136, 73111053557938599529984, 3042945108269382418885412, 127134885390090499643731744
Offset: 0

Views

Author

Peter Bala, Jul 02 2023

Keywords

Comments

The sequence of Franel numbers A000172 satisfies the identity A000172(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(n+2*k,2*k)*binomial(2*k,k). The present sequence comes from a modification of the right-hand side of the identity.
The Franel numbers satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r. We conjecture that the present sequence satisfies the same supercongruences.

Crossrefs

Cf. A000172, A362676, A363985 - A363990.

Programs

  • Maple
    seq(add((-4)^(n-k)*binomial(n,k)*binomial(2*n+2*k,2*k)*binomial(2*k,k), k = 0..n), n = 0..20);
# alternative faster program for large n
seq(simplify((-4)^n*hypergeom([-n, n+1/2, n+1], [1, 1], 1)), n = 0..20);
  • Mathematica
    Table[(-4)^n*HypergeometricPFQ[{-n, n+1/2, n+1}, {1, 1}, 1], {n, 0, 20}] (* Vaclav Kotesovec, Jul 17 2023 *)

Formula

a(n) = (-4)^n*hypergeom([-n, n+1/2, n+1], [1, 1], 1).
P-recursive: n^2*(10*n^2 - 26*n + 17)*(2*n - 1)^2*a(n) = 8*(220*n^6 - 1012*n^5 + 1834*n^4 - 1664*n^3 + 794*n^2 - 189*n + 18)*a(n-1) + 16*(10*n^2 - 6*n + 1)*(n - 1)^2*(2*n - 3)^2*a(n-2) with a(0) = 1 and a(1) = 8.
a(n) ~ 2^(2*n - 1) * phi^(5*n + 3/2) / (5^(1/4)*Pi*n), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Jul 17 2023

A363989 a(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(4*n+k,k)*binomial(2*k,k).

Original entry on oeis.org

1, 6, 142, 3732, 108750, 3345256, 106702756, 3491407296, 116430027470, 3939976259340, 134891827384392, 4662318475294992, 162421061711817252, 5696013237211704192, 200894768178332121280, 7120345964204221311232, 253453190156679801719502
Offset: 0

Views

Author

Peter Bala, Jul 02 2023

Keywords

Comments

The sequence of Franel numbers A000172 satisfies the identity A000172(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(n+2*k,2*k)*binomial(2*k,k). The present sequence comes from a modification of the right-hand side of the identity.
The Franel numbers satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r. We conjecture that the present sequence satisfies the same supercongruences.

Crossrefs

Cf. A000172, A362676, A363985 - A363990.

Programs

  • Maple
    seq(add((-4)^(n-k)*binomial(n,k)*binomial(4*n+k,k)*binomial(2*k,k), k = 0..n), n = 0..20);
# alternative faster program for large n
seq(simplify((-4)^n*hypergeom([-n, 4*n+1, 1/2], [1, 1], 1)), n = 0..20);
  • Mathematica
    Table[(-4)^n*HypergeometricPFQ[{-n, 4*n+1, 1/2}, {1, 1}, 1], {n, 0, 20}] (* Vaclav Kotesovec, Jul 17 2023 *)

Formula

a(n) = (-4)^n*hypergeom([-n, 4*n+1, 1/2], [1, 1], 1).
a(n) ~ 2^(10*n - 1/2) / (Pi * n * 3^(3*n)). - Vaclav Kotesovec, Jul 17 2023

A364111 a(n) = Sum_{k = 0..n} binomial(n+k-1,k)^2 * binomial(2*n-2*k,n-k) * binomial(2*k,k).

Original entry on oeis.org

1, 4, 76, 2560, 106060, 4864504, 237354880, 12079462560, 633885607500, 34050190896040, 1863047125801576, 103465470769890112, 5817117095161011328, 330450303019252600240, 18937657945720403830240, 1093557503049551583194560, 63566414131528881235953228, 3716526456851323626808570632
Offset: 0

Views

Author

Peter Bala, Jul 07 2023

Keywords

Comments

Compare with the Domb numbers A002895, which are defined by A002895(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(2*n-2*k,n-k) * binomial(2*k,k).
The supercongruences A002895(n*p^r) == A002895(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and r (see Osburn and Sahu).
We conjecture that the present sequence satisfies the same supercongruences.
More generally, let A >= 2, B and C be positive integers. Then we conjecture that the sequence whose terms are given by Sum_{k = 0..n} binomial(n+k-1,k)^A * binomial(2*n-2*k,n-k)^B * binomial(2*k,k)^C also satisfies the same supercongruences.

Links

Crossrefs

Cf. A002895, A362676.

Programs

  • Maple
    seq(add(binomial(n+k-1,k)^2 * binomial(2*n-2*k,n-k) * binomial(2*k,k)), n = 0..20);
# faster program for large n
seq(simplify(binomial(2*n,n)*hypergeom([-n, n, n, 1/2], [1, 1, 1/2 - n], 1)), n = 0..20);
  • Mathematica
    Table[Binomial[2*n,n] * HypergeometricPFQ[{-n, n, n, 1/2}, {1, 1, 1/2 - n}, 1], {n, 0, 20}] (* Vaclav Kotesovec, Jul 09 2023 *)

Formula

a(n) = Sum_{k = 0..n} binomial(-n,k)^2 * binomial(2*n-2*k,n-k) * binomial(2*k,k).
a(n) = binomial(2*n,n)*hypergeom([-n, n, n, 1/2], [1, 1, 1/2 - n], 1).
a(n) ~ 2^(6*n-1) / (sqrt(3) * Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Jul 09 2023

A363987 a(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(3*n+k,k)*binomial(2*k,k).

Original entry on oeis.org

1, 4, 72, 1336, 27816, 609504, 13849776, 322761632, 7665078696, 184706717728, 4502875483072, 110816709352848, 2748733193029488, 68633934354206784, 1723482744564382272, 43491464193991134336, 1102200555724995901864, 28038534962615758665120, 715655490507796997136960, 18321118155905647835092032
Offset: 0

Views

Author

Peter Bala, Jul 02 2023

Keywords

Comments

The sequence of Franel numbers A000172 satisfies the identity A000172(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(n+2*k,2*k)*binomial(2*k,k). The present sequence comes from a modification of the right-hand side of the identity.
The Franel numbers satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r. We conjecture that the present sequence satisfies the same supercongruences.

Crossrefs

Cf. A000172, A362676, A363985, A363986, A363987, A363988, A363989, A363990.

Programs

  • Maple
    seq(add((-4)^(n-k)*binomial(n,k)*binomial(3*n+k,k)*binomial(2*k,k), k = 0..n), n = 0..20);
# alternative faster program for large n
seq(simplify((-4)^n*hypergeom([-n, 3*n+1, 1/2], [1, 1], 1)), n = 0..20);
  • Mathematica
    Table[(-4)^n*HypergeometricPFQ[{-n, 3*n+1, 1/2}, {1, 1}, 1], {n, 0, 20}] (* Vaclav Kotesovec, Jul 17 2023 *)

Formula

a(n) = (-4)^n*hypergeom([-n, 3*n+1, 1/2], [1, 1], 1).
P-recursive: n^2*(4*n - 1)*(3*n - 1)*(3*n - 2)*(4*n - 3)*P(n-1)*a(n) = 16*(57753*n^10 - 500526*n^9 + 1886256*n^8 - 4059105*n^7 + 5508054*n^6 - 4910492*n^5 + 2904096*n^4 - 1121483*n^3 + 269766*n^2 - 36410*n + 2100)*a(n-1) + 192*(3*n - 5)^2*(n - 1)^2*(3*n - 4)^2*P(n)*a(n-2), where P(n) = 279*n^4 - 465*n^3 + 276*n^2 - 68*n + 6 and a(0) = 1, a(1) = 4.
a(n) ~ 3^(3*n) / (sqrt(2)*Pi*n). - Vaclav Kotesovec, Jul 17 2023

Extensions

a(15) corrected and more terms from Georg Fischer, Sep 19 2023

A363988 a(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(3*n+2*k,2*k)* binomial(2*k,k).

Original entry on oeis.org

1, 16, 828, 53836, 3879404, 296396016, 23517939996, 1916072466688, 159188357217516, 13425731614346272, 1145885291754711328, 98752981735587825288, 8579149369628730276860, 750365776730717473307920, 66009615160057048401092544, 5835864811138398925049262336
Offset: 0

Views

Author

Peter Bala, Jul 02 2023

Keywords

Comments

The sequence of Franel numbers A000172 satisfies the identity A000172(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(n+2*k,2*k)*binomial(2*k,k). The present sequence comes from a modification of the right-hand side of the identity.
The Franel numbers satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r. We conjecture that the present sequence satisfies the same supercongruences.

Crossrefs

Cf. A000172, A362676, A363985 - A363990.

Programs

  • Maple
    seq(add((-4)^(n-k)*binomial(n,k)*binomial(3*n+2*k,2*k)*binomial(2*k,k), k = 0..n), n = 0..20);
# alternative faster program for large n
seq(simplify((-4)^n*hypergeom([-n, (3*n+1)/2, (3*n+2)/2], [1, 1], 1)), n = 0..20);
  • Mathematica
    Table[(-4)^n*HypergeometricPFQ[{-n, (3*n+1)/2, (3*n+2)/2}, {1, 1}, 1], {n, 0, 20}] (* Vaclav Kotesovec, Jul 17 2023 *)

Formula

a(n) = (-4)^n*hypergeom([-n, (3*n+1)/2, (3*n+2)/2], [1, 1], 1).
a(n) ~ sqrt(5 + 9*sqrt(3/11)) * 3^(n - 1/2) * (63 + 11*sqrt(33))^n / (Pi * n * 2^(2*n + 3/2)). - Vaclav Kotesovec, Jul 17 2023
Showing 1-7 of 7 results.

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