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 1-6 of 6 results.

A357567 a(n) = 5*A005259(n) - 14*A005258(n).

Original entry on oeis.org

-9, -17, 99, 5167, 147491, 3937483, 105834699, 2907476527, 81702447651, 2342097382483, 68273597307599, 2018243113678027, 60365426282638091, 1823553517258576723, 55557712038989195099, 1705170989220937925167, 52672595030914982754851, 1636296525812843554700323
Offset: 0

Views

Author

Peter Bala, Oct 19 2022

Keywords

Comments

Conjectures:
1) a(p) == a(1) (mod p^5) for all primes p >= 5.
2) For r >= 2, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for all primes p >= 5.
These are stronger supercongruences than those satisfied separately by the two types of Apéry numbers A005258 and A005259.
From Peter Bala, Oct 25 2022: (Start)
Additional conjectures:
3) the sequence {u(n): n>= 1} defined by u(n) = (3^42)*A005259(n)^25 - (5^25)* A005258(n)^42 also satisfies the congruences in 1) and 2) above.
4) u(n) == 0 (mod n^5) for integer n of the form 3^i*5^j (see A003593). (End)

Examples

			a(11) - a(1) = 2018243113678027 + 17 = (2^2)*(3^2)*(11^5)*17*20476637 == 0 (mod 11^5).

Crossrefs

Cf. A005258, A005259, A212334, A352655, A357506, A357507, A357508, A357509, A357568, A357569, A357956, A357957, A357958, A357959, A357960.

Programs

  • Maple
    seq(add(5*binomial(n,k)^2*binomial(n+k,k)^2 - 14*binomial(n,k)^2*binomial(n+k,k), k = 0..n), n = 0..20);

Formula

a(n) = 5*Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)^2 - 14*Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k).
For positive integers n and r, a(n*p^r) == a(n*p^(r-1)) ( mod p^(3*r) ) for all primes p >= 5.

A357510 a(n) = Sum_{k = 0..n} k * binomial(n,k)^2 * binomial(n+k,k)^2.

Original entry on oeis.org

0, 4, 108, 3144, 95000, 2935020, 92054340, 2918972560, 93330811440, 3003683380020, 97177865060540, 3157623679795992, 102973952434618824, 3368460743291372092, 110480459392323735540, 3631941224582026770720, 119637879389041977365600, 3947968300820696313987780
Offset: 0

Views

Author

Peter Bala, Oct 01 2022

Keywords

Examples

			a(11 - 1) = 97177865060540 = (2^2)*5*(11^4)*37*239*37529 == 0 (mod 11^4).

Links

Crossrefs

Cf. A005259, A357506, A357507, A357511, A357512, A357513.

Programs

  • Maple
    seq( add( k*binomial(n,k)^2 * binomial(n+k,k)^2, k = 0..n ), n = 0..20 );
  • PARI
    a(n) = sum(k = 0, n, k * binomial(n,k)^2 * binomial(n+k,k)^2); \\ Michel Marcus, Oct 04 2022

Formula

Conjecture: a(p-1) == 0 (mod p^4) for all primes p >= 5 (checked up to p = 499).
Note: the Apery numbers A(n) = A005259(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(n+k,k)^2 satisfy the supercongruence A(p-1) == 1 (mod p^3) for all primes p >= 5 (see, for example, Straub, Introduction).
Recurrence: a(0) = 0, a(1) = 4, a(2) = 108 and thereafter n^2*(n-1)*(2*n-3)*(3*n^2-9*n+7)*a(n) = (2*n-1)*(105*n^5 - 576*n^4 + 1208*n^3 - 1195*n^2 + 556*n - 104)*a(n-1) - (2*n-3)*(105*n^5 - 474*n^4 + 800*n^3 - 629*n^2 + 240*n - 36)*a(n-2) + (n-1)*(2*n-1)*(3*n^2-3*n+1)*(n-2)^2*a(n-3).
a(n) ~ (1 + sqrt(2))^(4*n + 2) / (2^(11/4) * Pi^(3/2) * sqrt(n)). - Vaclav Kotesovec, Oct 04 2022

A357507 a(n) = A005259(n)^5 * (A005259(n-1))^7.

Original entry on oeis.org

3125, 161958718203125, 69598400094777710760545478125, 514885225734532980507136994998009584838203125, 15708056924221066705174364772957342407662356116035885781253125, 1125221282019374727979322420623179115437017599670596496532725068048858642578125
Offset: 1

Views

Author

Peter Bala, Oct 01 2022

Keywords

Comments

The Apéry numbers A(n) = A005259(n) satisfy the supercongruences A(p) == 5 (mod p^3) and A(p-1) == 1 (mod p^3) for all primes p >= 5 (see, for example, Straub, Introduction). It follows that a(p) == 3125 (mod p^3) for all primes p >= 5. We conjecture that, in fact, the stronger congruence a(p) == 3125 (mod p^5) holds for all primes p >= 3 (checked up to p = 251). Compare with the congruence A(p) + 7*A(p-1) == 12 (mod p^5) conjectured to hold for all primes p >= 5. See A212334.
Conjecture: a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for r >= 2 and all primes p >= 5. - Peter Bala, Oct 26 2022

Links

Crossrefs

Cf. A005259, A212334, A339946, A352655, A357506, A357508, A357509, A357567, A357568, A357569, A357956, A357957, A357958, A357959.

Programs

  • Maple
    A005259 := n -> add(binomial(n,k)^2*binomial(n+k,k)^2, k = 0..n):
seq(A005259(n)^5 * A005259(n-1)^7, n = 1..10);

A357511 a(n) = numerator of Sum_{k = 1..n} (1/k) * binomial(n,k)^2 * binomial(n+k,k)^2 for n >= 1 with a(0) = 0.

Original entry on oeis.org

0, 4, 54, 2182, 36625, 3591137, 25952409, 4220121443, 206216140401, 47128096330129, 1233722785504429, 364131107601152519, 9971452750252847789, 3611140187389794708497, 102077670374035974509597, 2922063451137950165057717, 169140610796591477659644439
Offset: 0

Views

Author

Peter Bala, Oct 01 2022

Keywords

Examples

			a(13 - 1) = 9971452750252847789 = (13^4)*37*2477*24197*157433 == 0 (mod 13^4).

Links

Crossrefs

Cf. A005259, A357506, A357507, A357510, A357512, A357513.

Programs

  • Maple
    seq(numer(add( (1/k) * binomial(n,k)^2 * binomial(n+k,k)^2, k = 1..n )), n = 0..20);
  • PARI
    a(n) = if (n, numerator(sum(k=1, n, binomial(n,k)^2*binomial(n+k,k)^2/k)), 0); \\ Michel Marcus, Oct 04 2022

Formula

Conjecture: a(p-1) == 0 (mod p^4) for all primes p >= 7 (checked up to p = 499).
Note: the Apery numbers A(n) = A005259(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(n+k,k)^2 satisfy the supercongruence A(p-1) == 1 (mod p^3) for all primes p >= 5 (see, for example, Straub, Introduction).

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);
Showing 1-6 of 6 results.

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