A357560 -id:A357560 - cp's OEIS frontend

A357558 a(n) = Sum_{k = 0..n} (-1)^(n+k)kbinomial(n,k)*binomial(n+k,k)^2.

0, 4, 54, 648, 7500, 85440, 965202, 10849552, 121566744, 1359160020, 15172321890, 169175039616, 1884704860116, 20982512553912, 233474575117770, 2596777575029280, 28872014164369968, 320917108809011868, 3566175414049854306, 39620770883613043240, 440115513924937822020

Offset: 0

Peter Bala, Oct 03 2022

			Example of a supercongruence:
p = 17: a(17 - 1) = 28872014164369968 = (2^4)*3*(17^4)*107*251*268153 == 0 (mod 16*7^4)

A. Straub, Multivariate Apéry numbers and supercongruences of rational functions, arXiv:1401.0854 [math.NT] (2014).

Cf. A005258, A357510, A357511, A357512, A357513, A357559, A357560, A357561.

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

Conjecture: a(p-1) == 0 (mod (p - 1)*p^4) for all primes p >= 5 (checked up to p = 499).
Note: Let B(n) = A005258(n). It is known that B(n) = Sum_{k = 0..n} (-1)^(n+k)* binomial(n,k)*binomial(n+k,k)^2 and the supercongruences B(p-1) == 1 (mod p^3) hold for all primes p >= 5 (see, for example, Straub, Example 3.4).
Recurrence: a(0) = 0, a(1) = 4 and for n >= 2, (5*n - 2)*(n^2 - 1)*a(n) = (55*n^3 - 22*n^2 - 19*n + 10)*a(n-1) + n*(5*n + 3)*(n-1)*a(n-2).
a(n) ~ phi^(5*n + 7/2) / (2*Pi*5^(1/4)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Oct 05 2022

A357559 a(n) = Sum_{k = 0..n} (-1)^(n+k)k^3binomial(n,k)*binomial(n+k,k)^2.

Original entry on oeis.org

0, 4, 270, 8448, 192000, 3669300, 62952162, 1003770880, 15182515584, 220700443500, 3110529630450, 42769154678976, 576313309494000, 7636526099508852, 99765264496070250, 1287663145631539200, 16446680778536421888, 208154776511034178380, 2613380452317012835386

Offset: 0

Peter Bala, Oct 04 2022

Define S_m(n) = Sum_{k = 0..n} (-1)^(n+k)*k^m*binomial(n,k)*binomial(n+k,k)^2, so that S_0(n) = A005258(n), one type of Apéry numbers. The present sequence is the case m = 3. See A357558 for the case m = 1.
It is known that S_0(2*n) - 1 is divisible by (2*n + 1)^3 provided 2*n + 1 is a prime greater than 3 (see, for example, Straub, Example 3.4).
Let C_m = Product_{odd primes p <= m} p.
Conjectures:
1) S_2(2*n) is divisible by n^2*(2*n + 1)^2.
2) for even m >= 4, C_m * S_m(2*n) is divisible by n^3*(2*n + 1)^2.
3) for odd m >= 5, C_m * S_m(2*n) is divisible by n^2*(2*n + 1)^4.

A. Straub, Multivariate Apéry numbers and supercongruences of rational functions, arXiv:1401.0854 [math.NT] (2014).

Cf. A005258, A357510, A357512, A357558, A357560, A357561.

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

Conjecture: 3*a(2*n) == 0 ( mod n^2*(2*n + 1)^3 ).
Recurrence a(0) = 0, a(1) = 4, and for n >= 2, n*(n + 3)*(5*n - 6)*(n - 1)^4*a(n) = (n - 1)*(55*n^4 - 66*n^3 - 22*n^2 + 15*n + 6)*(n + 1)^2*a(n-1) + n^4*(5*n - 1)*(n + 1)^2*a(n-2).
a(n) ~ n^2 * phi^(5*n + 11/2) / (2*Pi*5^(1/4)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Oct 07 2022

A357561 a(n) = the numerator of ( Sum_{k = 1..n} (-1)^(n+k)(1/k^3)binomial(n,k)* binomial(n+k,k)^2 ).

Original entry on oeis.org

0, 4, -27, 1367, -15625, 3129353, -14749, 308477847, 14343020119, 80826490175689, 618729030402659, 6526775794564145231, 52975460244520902439, 965428117884339747694757, 8161435689582967449592663, 70159702295938799645630801, 4897311439674525483507166097, 212741477113936719632186271679919

Offset: 0

Peter Bala, Oct 04 2022

Define S_m(n) = the numerator of Sum_{k = 1..n} (-1)^(n+k)*(1/k^m)*binomial(n,k)* binomial(n+k,k)^2, so that S_0(n) = -1 + A005258(n), one type of Apéry numbers. The present sequence is the case m = 3. See A357560 for the case m = 1.

Conjecture: for odd m >= 1, S_m(p-1) == 0 (mod p^4) for all primes p > m+4.

			Example of a supercongruence:
p = 17: a(17 - 1) = 212741477113936719632186271679919 = (17^4)*4871421029* 12036670481533 == 0 (mod 17^4).

A. Straub, Multivariate Apéry numbers and supercongruences of rational functions, arXiv:1401.0854 [math.NT] (2014).

Cf. A005258, A357510, A357511, A357512, A357513, A357558, A357559, A357560.

seq( numer(add( (-1)^(n+k) * (1/k^3) * binomial(n,k) * binomial(n+k,k)^2, k = 1..n )), n = 0..20 );

Conjecture: a(p-1) == 0 (mod p^4) for p = 5 and all primes p >= 11 (checked up to p = 499).

Note: the Apéry numbers B(n) = A005258(n) = Sum_{k = 0..n} (-1)^(n+k)* binomial(n,k)*binomial(n+k,k)^2 satisfy the supercongruences B(p-1) == 1 (mod p^3) for all primes p >= 5 (see, for example, Straub, Example 3.4).

cp's OEIS Frontend

A357558 a(n) = Sum_{k = 0..n} (-1)^(n+k)kbinomial(n,k)*binomial(n+k,k)^2.

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A357559 a(n) = Sum_{k = 0..n} (-1)^(n+k)k^3binomial(n,k)*binomial(n+k,k)^2.

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A357561 a(n) = the numerator of ( Sum_{k = 1..n} (-1)^(n+k)(1/k^3)binomial(n,k)* binomial(n+k,k)^2 ).

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cp's OEIS Frontend

A357558 a(n) = Sum_{k = 0..n} (-1)^(n+k)*k*binomial(n,k)*binomial(n+k,k)^2.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Formula

A357559 a(n) = Sum_{k = 0..n} (-1)^(n+k)*k^3*binomial(n,k)*binomial(n+k,k)^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Formula

A357561 a(n) = the numerator of ( Sum_{k = 1..n} (-1)^(n+k)*(1/k^3)*binomial(n,k)* binomial(n+k,k)^2 ).

Views

Author

Keywords

Comments

Examples

Links

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Programs

Maple

Formula

A357558 a(n) = Sum_{k = 0..n} (-1)^(n+k)kbinomial(n,k)*binomial(n+k,k)^2.

A357559 a(n) = Sum_{k = 0..n} (-1)^(n+k)k^3binomial(n,k)*binomial(n+k,k)^2.

A357561 a(n) = the numerator of ( Sum_{k = 1..n} (-1)^(n+k)(1/k^3)binomial(n,k)* binomial(n+k,k)^2 ).