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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A357960 a(n) = A005259(n-1)^5 * A005258(n)^6.

Original entry on oeis.org

729, 147018378125, 20917910914764786689697, 24148107115850058575342740485778125, 79477722547796770983047586179643766765851375729, 492664048531500749211923278756418311980637289373757041378125, 4671227340507161302417161873394448514470099313382652883508175438056640625
Offset: 1

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Author

Peter Bala, Oct 25 2022

Keywords

Comments

Conjectures:
1) a(p) == a(1) (mod p^5) for all primes p >= 3 (checked up to p = 271).
2) a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for r >= 2 and for all primes p >= 3. These are stronger supercongruences than those satisfied separately by the two types of Apéry numbers A005258 and A005259.

Crossrefs

Cf. A005258, A005259, A212334, A352655, A357567, A357956, A357957, A357958, A357959.

Programs

  • Maple
    seq( add(binomial(n-1,k)^2*binomial(n+k-1,k)^2, k = 0..n-1)^5 * add(binomial(n, k)^2*binomial(n+k,k), k = 0..n)^6, n = 1..20);

Formula

a(n) = ( Sum_{k = 0..n-1} binomial(n-1,k)^2*binomial(n+k-1,k)^2 )^5 * ( Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k) )^6.
a(n*p^r) == a(n*p^(r-1)) ( mod p^(3*r) ) for positive integers n and r and for all primes p >= 5.

A361713 a(n) = Sum_{k = 0..n-1} binomial(n,k)^2 * binomial(n+k-1,k)^2.

Original entry on oeis.org

0, 1, 17, 406, 10257, 268126, 7213166, 198978074, 5609330705, 161095277710, 4700175389142, 138986764820410, 4157185583199534, 125568602682092818, 3825026187780837266, 117376010145070696906, 3625095243230562818065, 112596592142021739522670, 3514965607470183733302470
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^(4*r+1)) holds for all primes p >= 7.
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 positive integers r and all primes p >= 5.

Links

Crossrefs

Cf. A005259, A060150, A177316, A212334, A361712, A361714, A361715, A361717.

Programs

  • Maple
    seq(add(binomial(n,k)^2*binomial(n+k-1,k)^2, k = 0..n-1), n = 0..25);
# Alternative:
A361713 := n -> hypergeom([-n, -n, n, n], [1, 1, 1], 1) - binomial(2*n - 1, n)^2:
seq(simplify(A361713(n)), n = 0..18); # Peter Luschny, Mar 27 2023
  • Mathematica
    A361713[n_] := HypergeometricPFQ[{-n, -n, n, n}, {1, 1, 1}, 1] - Binomial[2*n-1, n]^2; Array[A361713, 20, 0] (* Paolo Xausa, Jul 11 2024 *)

Formula

a(n) = (1/3)*(A005259(n) + A005259(n-1)) - (1/4)*binomial(2*n,n)^2 = A177316(n) - A060150(n).
a(n) ~ C*(12*sqrt(2) + 17)^n/n^(3/2), where C = 1/(2^(5/4)*Pi^(3/2)).
a(n) = hypergeom([-n, -n, n, n], [1, 1, 1], 1) - binomial(2*n-1, n)^2. This is another way to write the first formula. - Peter Luschny, Mar 27 2023

A364118 a(n) = 3*A364114(n) - 11*A364114(n-1).

Original entry on oeis.org

10, 412, 15076, 643900, 30440010, 1541377330, 81983235064, 4524150828092, 256902133600630, 14924997512212912, 883403610976880740, 53105747607145638706, 3234568078911042493578, 199234128948556264779390, 12391648147019445115584576, 777286417688953098495554620
Offset: 1

Views

Author

Peter Bala, Jul 12 2023

Keywords

Comments

It is conjectured that the sequence {A364114(n)} and the shifted sequence {A364114(n-1)} both satisfy the supercongruences A364114(p^r) == A364114(p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers r. Stronger supercongruences may hold for the present sequence, a linear combination of A364114(n) and A364114(n-1).
Conjectures: 1) the supercongruences a(p) == a(1) (mod p^5) hold for all primes p >= 7 (checked up to p = 101).
2) for r >= 2, the supercongruences a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) hold for all primes p >= 7. Cf. A212334.
There is also a multiplicative version of this sequence. Define a sequence of rational numbers {b(n) : n >= 1} by b(n) = A364114(n)^21 / A364114(n-1)^11. Then we conjecture that the above pair of supercongruences also hold for the sequence {b(n)}.

Links

Crossrefs

Cf. A212334, A357506, A357507, A357568, A364114, A364119.

Programs

  • Maple
    A364114 := n -> coeff(series( 1/(1-x)* LegendreP(n,(1+x)/(1-x))^3, x, 21), x, n):
seq(3*A364114(n) - 11*A364114(n-1), n = 1..20);

A364119 a(n) = 7*A364115(n) - 17*A364115(n-1).

Original entry on oeis.org

46, 1870, 95950, 6111054, 445850046, 35606390254, 3031075759870, 270542736416590, 25045919145436366, 2386963634176587870, 232926731552238831054, 23180020599857593886190, 2345286553765877009107710, 240670553547813070050900126, 25001383450621552178261089950
Offset: 1

Views

Author

Peter Bala, Jul 12 2023

Keywords

Comments

It is conjectured that the sequence {A364115(n)} and the shifted sequence {A364115(n-1)} both satisfy the supercongruences A364115(p^r) == A364115(p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers r. Stronger supercongruences may hold for the present sequence, a linear combination of A364115(n) and A364115(n-1).
Conjectures: 1) the supercongruences a(p) == a(1) (mod p^5) hold for all primes p >= 7 (checked up to p = 101).
2) for r >= 2, the supercongruences a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) hold for all primes p >= 7. Cf. A212334.
There is also a multiplicative version of this sequence. Define a sequence of rational numbers {b(n) : n >= 1} by b(n) = A364115(n)^63 / A364115(n-1)^17. Then we conjecture that the above pair of supercongruences also hold for the sequence {b(n)}.

Links

Crossrefs

Cf. A212334, A357506, A357507, A357568, A364115, A364118.

Programs

  • Maple
    A364115 := n -> coeff(series( 1/(1-x)* LegendreP(n,(1+x)/(1-x))^4, x, 21), x, n):
seq(7*A364115(n) - 17*A364115(n-1), n = 1..20);
Previous Showing 11-14 of 14 results.

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