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.

A357509 a(n) = 2*binomial(3*n,n) - 9*binomial(2*n,n).

Original entry on oeis.org

-7, -12, -24, -12, 360, 3738, 28812, 201672, 1355112, 8936070, 58427226, 380724552, 2479017996, 16151245488, 105359408760, 688338793488, 4504288103784, 29521135717470, 193771020939510, 1273649831269200, 8382448392851610, 55234026483856110, 364347399072847320
Offset: 0

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Author

Peter Bala, Oct 01 2022

Keywords

Comments

For integers j and k, not necessarily positive, define u(n) = k^2*(k - 1)*binomial(j*n,n) - j^2*(j - 1)*binomial(k*n,n). We conjecture that u(p) == u(1) (mod p^5) for all primes p >= 7. This is essentially the case (j, k) = (3, 2). [Follows from Helou and Terjanian (2008), Section 3, Proposition 2.]
Conjecture: for r >= 2, u(p^r) == u(p^(r-1)) ( mod p^(3*r+3) ) for all primes p >= 5. - Peter Bala, Oct 13 2022

Links

Crossrefs

Cf. A000984, A005809, A268589, A357508.

Programs

  • Maple
    seq(2*binomial(3*n,n) - 9*binomial(2*n,n), n = 0..20);

Formula

a(n) = 2*A005809(n) - 9*A000984(n).
a(p) == a(1) (mod p^4) for all primes p >= 5 by Meštrović, Section 3, equation 15.
Conjecture: the stronger supercongruence a(p) == a(1) (mod p^5) holds for all primes p >= 7.
The conjecture is true: apply Helou and Terjanian, Section 3, Proposition 2. - Peter Bala, Oct 22 2022
The conjecture was proved by Aidagulov and Alekseyev; see the remarks following Corollary 2. - Peter Bala, Oct 29 2022

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.

A357506 a(n) = A005258(n)^3 * A005258(n-1).

Original entry on oeis.org

27, 20577, 60353937, 287798988897, 1782634331587527, 13011500170881726987, 106321024671550496694837, 943479109706472533832704097, 8916177779855571182824077866307, 88547154924474394601268826256953077, 915376390434997094066775480671975209017
Offset: 1

Views

Author

Peter Bala, Oct 01 2022

Keywords

Comments

The Apéry numbers B(n) = A005258(n) satisfy the supercongruences B(p) == 3 (mod p^3) and B(p-1) == 1 (mod p^3) for all primes p >= 5 (see, for example, Straub, Example 3.4). It follows that a(p) == 27 (mod p^3) for all primes p >= 5. We conjecture that, in fact, the stronger congruence a(p) == 27 (mod p^5) holds for all primes p >= 3 (checked up to p = 251). Compare with the congruence B(p) + B(p-1) == 4 (mod p^5) conjectured to hold for all primes p >= 5. See A352655.
Conjecture: for r >= 2, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for all primes p >= 5. - Peter Bala, Oct 13 2022

Examples

			Example of a supercongruence:
a(7) - a(1) = 106321024671550496694837 - 27 = 2*(3^3)*5*(7^5)* 11*18143* 117398731273 == 0 (mod 7^5)

Links

Crossrefs

Cf. A005258, A212334, A339946, A352655, A357507, A357508, A357509.

Programs

  • Maple
    A005258 := n -> add(binomial(n,k)^2*binomial(n+k,k), k = 0..n):
seq(A005258(n)^3*A005258(n-1), n = 1..20);

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);
Showing 1-4 of 4 results.

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