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 1-4 of 4 results.

A361712 a(n) = Sum_{k = 0..n-1} binomial(n,k)^2*binomial(n+k,k)*binomial(n+k-1,k).

Original entry on oeis.org

0, 1, 25, 649, 16921, 448751, 12160177, 336745053, 9513822745, 273585035755, 7988828082775, 236367018090017, 7072779699975601, 213701611408357567, 6511338458568750853, 199850727914988936149, 6173376842290368719385, 191776434791965521115235, 5987554996434696230487955
Offset: 0

Views

Author

Peter Bala, Mar 21 2023

Keywords

Comments

Conjecture 1: the supercongruence a(p) == a(1) (mod p^5) holds for all primes p >= 7 (checked up to p = 199).
Conjecture 2: for r >= 2, the supercongruence a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) holds for all primes p >= 5.
Compare with the Apéry numbers A005259(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(n+k,k)^2, which satisfy the weaker supercongruences A005259(p^r) == A005259(p^(r-1)) (mod p^(3*r)) for all primes p >= 5.

Examples

			a(7) - a(1) = (2^2)*(7^5)*5009 == 0 (mod 7^5)
a(11) - a(1) = (2^5)*(11^5)*45864163 == 0 (mod 11^5)
a(7^2) - a(7) = (2*3)*(7^9)*377052719*240136524699189343838527* 17965610580703155723668147409587 == 0 (mod 7^9)

Links

Crossrefs

Cf. A005259, A212334, A361713, A361714, A361715, A361717.
Cf. A361877, A361878.

Programs

  • Maple
    seq(add(binomial(n,k)^2*binomial(n+k,k)*binomial(n+k-1,k), k = 0..n-1), n = 0..25);
# Alternative:
A361712 := n -> hypergeom([-n, -n, n, n + 1], [1, 1, 1], 1) - binomial(2*n, n)*binomial(2*n-1, n): seq(simplify(A361712(n)), n = 0..18); # Peter Luschny, Mar 27 2023
  • Mathematica
    A361712[n_] := HypergeometricPFQ[{-n, -n, n, n+1}, {1, 1, 1}, 1] - Binomial[2*n, n]*Binomial[2*n-1, n]; Array[A361712, 20, 0] (* Paolo Xausa, Jul 10 2024 *)

Formula

a(n) = (1/12)*(7*A005259(n) + A005259(n-1)) - (1/2)*binomial(2*n,n)^2.
a(n) ~ 2^(1/4)*(1 + sqrt(2))^(4*n+1)/(4*Pi^(3/2)*n^(3/2)).
a(n) = hypergeom([-n, -n, n, n + 1], [1, 1, 1], 1) - binomial(2*n, n)*binomial(2*n - 1, n) = A361878(n) - A361877(n). - Peter Luschny, Mar 27 2023

A361717 a(n) = Sum_{k = 0..n-1} binomial(n-1,k)^2*binomial(n+k,k).

Original entry on oeis.org

0, 1, 4, 27, 216, 1875, 17088, 160867, 1549936, 15195843, 151017780, 1517232189, 15379549056, 157058738343, 1614039427224, 16676755365555, 173118505001952, 1804500885273123, 18877476988765404, 198120856336103017, 2085303730716475960
Offset: 0

Views

Author

Peter Bala, Mar 26 2023

Keywords

Comments

Compare with the Apery numbers A005258(n) = Sum_{k = 0..n} binomial(n,k)^2* binomial(n+k,k).
Conjecture 1: the supercongruence a(p) == 0 (mod p^4) holds for all primes p >= 5 (checked up to p = 199).
Conjecture 2: the supercongruence a(p-1) == 1 - 2*p - p^2 (mod p^3) holds for all primes except p = 3 (checked up to p = 199).

Examples

			a(5) = 3*(5^4); a(7) = (7^4)*67; a(11) = 3*(11^4)*34543; a(13) = (3^3)*(13^4)*203669.

Links

Crossrefs

Cf. A005258, A006480, A060542, A361712, A361713, A361714, A361715.

Programs

  • Maple
    seq( add(binomial(n-1,k)^2*binomial(n+k,k), k = 0..n), n = 0..20);
  • Mathematica
    A361717[n_]:=Sum[Binomial[n-1,k]^2Binomial[n+k,k],{k,0,n-1}];Array[A361717,30,0] (* Paolo Xausa, Oct 06 2023 *)
  • PARI
    a(n) = sum(k=0, n-1, binomial(n-1,k)^2*binomial(n+k,k)) \\ Winston de Greef, Mar 27 2023

Formula

a(n) = hypergeom([1 + n, 1 - n, 1 - n], [1, 1], 1) for n >= 1.
P-recursive:
n*(n-1)*(5*n-7)*a(n) = (55*n^3-187*n^2+190*n-48)*a(n-1) + (n-1)*(n-3)*(5*n-2)* a(n-2) with a(0) = 0 and a(1) = 1.
a(n) ~ phi^(5*n - 3/2) / (2*5^(1/4)*Pi*n), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Mar 27 2023
a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1) * binomial(n-1, k) * binomial(n+k-1, k) * binomial(n+k, k+1) = (-1)^(n+1) * n * hypergeom([n, n + 1, 1 - n], [1, 2], 1). - Peter Bala, Sep 08 2023
a(n) = Sum_{k = 0..n-1} (-1)^k * binomial(n-2, k) * binomial(2*n-2-k, n-1-k)^2. - Peter Bala, Oct 09 2024
From Peter Bala, Jul 31 2025: (Start)
a(n) = n * Sum_{k = 0..n} 1/(k+1) * binomial(n-1, k)^2 * binomial(n+k-1,k).
a(n) = n * hypergeom([n, 1 - n, 1 - n], [1, 2], 1). (End)

A361714 a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(n,k)*binomial(n+k-1,k)^2.

Original entry on oeis.org

0, 1, 7, 82, 1063, 14376, 199204, 2806770, 40053031, 577468684, 8397778882, 123029274666, 1814016998116, 26898142793068, 400836647993292, 5999796281063082, 90162110212198695, 1359731143731297396, 20571691450059355174, 312134224830052880826, 4748435338386591995938
Offset: 0

Views

Author

Peter Bala, Mar 21 2023

Keywords

Comments

Conjecture 1: the supercongruence a(p) == a(1) (mod p^5) holds for all primes p >= 7 (checked up to p = 199).
Conjecture 2: for r >= 2, the supercongruence a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) holds for all primes p >= 7.
Compare with the Apéry numbers A005258(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n,k) * binomial(n+k,k)^2, which satisfy the weaker supercongruences A005258(p^r) == A005258(p^(r-1)) (mod p^(3*r)) for positive integer r and all primes p >= 5.

Examples

			Examples of supercongruence:
a(11) - a(1) = 23029274666 - 1 = 5*(11^5)*152783 == 0 (mod 11^5).
a(13) - a(1) = 26898142793068 - 1 = (3^2)*7*(13^5)*1149913 == 0 (mod 13^5).
a(5^2) - a(5) = 3994642669575050040375014376 - 14376 = (2^6)*(3^6)*(5^9)*103* 425601520324429 == 0 (mod 5^9).

Links

Crossrefs

Cf. A005258, A352654, A361712, A361713, A361715, A361717.

Programs

  • Maple
    seq(add((-1)^(n+k+1)*binomial(n,k)*binomial(n+k-1,k)^2, k = 0..n-1), n = 0..20);
# Alternative:
A361714 := n -> binomial(2*n-1, n)^2 - (-1)^n*hypergeom([-n, n, n], [1, 1], 1):
seq(simplify(A361714(n)), n = 0..20); # Peter Luschny, Mar 27 2023
  • Mathematica
    A361714[n_] := Binomial[2*n-1, n]^2 - (-1)^n*HypergeometricPFQ[{-n, n, n}, {1, 1}, 1]; Array[A361714, 20, 0] (* Paolo Xausa, Jul 11 2024 *)

Formula

a(n) = binomial(2*n-1,n)^2 - (1/5)*(A005258(n) - 3*A005258(n-1)) for n >= 1.
P-recursive:
(395*n^10 - 6083*n^9 + 39816*n^8 - 144606*n^7 + 318639*n^6 - 436307*n^5 + 362870*n^4 - 167820*n^3 + 33096*n^2)*a(n) = (10665*n^10 - 174906*n^9 + 1243697*n^8 - 5033114*n^7 + 12789951*n^6 - 21235254*n^5 + 23221451*n^4 - 16437246*n^3 + 7182940*n^2 - 1753656*n + 185472)*a(n-1) - (69125*n^10 - 1202775*n^9 + 9159576*n^8 - 40005738*n^7 + 110271201*n^6 - 198723383*n^5 + 234346978*n^4 - 175661976*n^3 + 78402944*n^2 - 18529392*n + 1901088)*a(n-2) - 4*(n - 3)^2*(1580*n^8 - 19592*n^7 + 101515*n^6 - 284307*n^5 + 464411*n^4 - 444309*n^3 + 236490*n^2 - 62500*n + 7000)*a(n-3) with a(0) = 0, a(1) = 1 and a(2) = 7.
a(n) = binomial(2*n-1, n)^2 - (-1)^n*hypergeom([-n, n, n], [1, 1], 1). This is another way to write the first formula. - Peter Luschny, Mar 27 2023

A361715 a(n) = Sum_{k = 0..n-1} binomial(n,k)^2*binomial(n+k-1,k).

Original entry on oeis.org

0, 1, 9, 82, 745, 6876, 64764, 621860, 6070761, 60085720, 601493134, 6078225792, 61907445340, 634751002718, 6545478537810, 67830084149832, 705950951578089, 7375212511115184, 77310175072063914, 812839577957617640, 8569327793354169870, 90562666708303706642, 959212007563384494522, 10180245921386807485152
Offset: 0

Views

Author

Peter Bala, Mar 23 2023

Keywords

Comments

Conjecture 1: the supercongruence a(p) == a(1) (mod p^5) holds for all primes p >= 7 (checked up to p = 199).
Conjecture 2: for r >= 2, the supercongruence a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) holds for all primes p >= 5.
Compare with the Apéry numbers A005258(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(n+k,k), which satisfy the weaker supercongruences A005258(p^r) == A005258(p^(r-1)) (mod p^(3*r)) for all primes p >= 5.

Crossrefs

Cf. A005258, A103882, A361712, A361713, A361714, A361717.

Programs

  • Maple
    seq( add( binomial(n,k)^2*binomial(n+k-1,k), k = 0..n-1), n = 0..25);
#faster alternative program
P(n) := 145*n^4 - 1217*n^3 + 3763*n^2 - 5079*n + 2532:
Q(n) := (n - 1)*(n - 2)*(2175*n^6 - 20140*n^5 + 73132*n^4 - 131786*n^3 + 122789*n^2 - 55626*n + 9936):
R(n) := (n - 2)*(6235*n^7 - 67846*n^6 + 304860*n^5 - 731294*n^4 + 1008701*n^3 - 798060*n^2 + 335340*n - 58320):
a := proc(n) option remember; if n = 0 then 0 elif n = 1 then 1 elif n = 2 then 9 else (Q(n)*a(n-1) - R(n)*a(n-2) - 2*(n - 1)*(n - 3)^2*(2*n - 5)*P(n+1)*a(n-3))/((n - 1)*(n - 2)*n^2*P(n)) end if; end:
seq(a(n), n = 0..25);
# Alternative:
A361715 := n -> hypergeom([-n, -n, n], [1, 1], 1) - binomial(2*n-1, n):
seq(simplify(A361715(n)), n = 0..23); # Peter Luschny, Mar 27 2023
  • Mathematica
    Table[Sum[Binomial[n,k]^2 Binomial[n+k-1,k],{k,0,n-1}],{n,0,30}] (* Harvey P. Dale, Nov 01 2023 *)

Formula

a(n) = A103882(n) - binomial(2*n-1,n) = (3*A005258(n) + A005258(n-1))/5 - binomial(2*n-1,n) for n >= 1.
a(n) ~ sqrt(sqrt(5)/10 + 1/4)*(5*sqrt(5)/2 + 11/2)^n/(Pi*n)
P-recursive:
(n - 1)*(n - 2)*n^2*P(n)*a(n) = Q(n)*a(n - 1) - R(n)*a(n-2) - 2*(n - 1)*(n - 3)^2*(2*n - 5)*P(n+1)*a(n-3) with a(0) = 0, a(1) = 1 and a(2) = 9 and where
P(n) = 145*n^4 - 1217*n^3 + 3763*n^2 - 5079*n + 2532,
Q(n) = (n - 1)*(n - 2)*(2175*n^6 - 20140*n^5 + 73132*n^4 - 131786*n^3 + 122789*n^2 - 55626*n + 9936) and
R(n) = (n - 2)*(6235*n^7 - 67846*n^6 + 304860*n^5 - 731294*n^4 + 1008701*n^3 - 798060*n^2 + 335340*n - 58320).
a(n) = hypergeom([-n, -n, n], [1, 1], 1) - binomial(2*n-1, n). This is another way to write the first formula. - Peter Luschny, Mar 27 2023
Showing 1-4 of 4 results.

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