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 1-7 of 7 results.

A357512 a(n) = Sum_{k = 0..n} k^5 * binomial(n,k)^2 * binomial(n+k,k)^2.

Original entry on oeis.org

0, 4, 1188, 126144, 10040000, 682492500, 41503541940, 2325305113600, 122429236976640, 6140504039242500, 296222848665342500, 13841644170257145792, 629814531655430506944, 28019919084086921883892, 1222770835880665252492500, 52476371578141941012480000, 2219374467089388085650636800
Offset: 0

Views

Author

Peter Bala, Oct 02 2022

Keywords

Comments

Let m >= 2 and set u(n) = Sum_{k = 0..n} k^(2*m+1) * binomial(n,k)^2 * binomial(n+k,k)^2. We conjecture that there is a finite set of primes P(m) such that u(n-1) == 0 (mod n^4) for all odd numbers n not divisible by an element of P(m). For example, calculation suggests that P(2) = {3} (the present sequence), while conjecturally P(3) = {3, 5}, P(4) = {3, 7}, P(5) = {3, 5}, P(6) = {3, 11}, P(7) = {3, 5, 7, 13}, P(8) = {3}, P(9) = {3} and P(10) = {3, 7, 19}.

Links

Crossrefs

Cf. A005259, A007310, A357510, A357511, A357513.

Programs

  • Maple
    seq( add( k^5 * binomial(n,k)^2 * binomial(n+k,k)^2, k = 0..n ), n = 0..20 );
  • PARI
    a(n) = sum(k = 0, n, k^5 * binomial(n,k)^2 * binomial(n+k,k)^2); \\ Michel Marcus, Oct 04 2022

Formula

Conjecture: a(n-1) == 0 (mod n^4) for all n congruent to 1 or 5 mod 6 (checked up to n = 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(1) = 4, a(2) = 1188, a(3) = 126144 and thereafter
(n - 3)*(2*n - 3)*(3*n^4 - 18*n^3 + 40*n^2 - 39*n + 13)*(n - 2)^3*(n - 1)^4*a(n) = (n - 3)*(2*n - 1)*(n + 1)*(105*n^7 - 879*n^6 + 2981*n^5 - 5124*n^4 + 4376*n^3 - 1236*n^2 - 442*n + 195)*(n - 2)^3*a(n-1) - n^3*(n - 3)*(2*n - 3)*(n + 1)*(105*n^7 - 591*n^6 + 1253*n^5 - 1346*n^4 + 784*n^3 - 124*n^2 - 90*n + 33)*a(n-2) + n^3*(2*n - 1)*(n + 1)*(3*n^4 - 6*n^3 + 4*n^2 - n - 1)*(n - 1)^4*a(n-3).
a(n) ~ n^(7/2) * (1 + sqrt(2))^(4*n + 2) / (2^(19/4) * Pi^(3/2)). - Vaclav Kotesovec, Oct 04 2022

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).

A357513 a(n) = numerator of Sum_{k = 1..n} (1/k^3) * binomial(n,k)^2 * binomial(n+k,k)^2 for n >= 1 with a(0) = 0.

Original entry on oeis.org

0, 4, 81, 14651, 956875, 1335793103, 697621869, 3929170277787, 573290332967211, 8235727724024089939, 172296487023049395523, 5032311952710217004416313, 114828404520381550476341513, 5947240175728534283432460589661, 144126887537331651710781931325261
Offset: 0

Views

Author

Peter Bala, Oct 02 2022

Keywords

Comments

Let m be a nonnegative integer and set u(n) = the numerator of Sum_{k = 0..n} 1/k^(2*m+1) * binomial(n,k)^2 * binomial(n+k,k)^2. We conjecture that u(p-1) == 0 (mod p^4) for all primes p, with a finite number of exceptions that depend on m.

Examples

			a(11 - 1) = 172296487023049395523 = (11^4)*(43^2)*6163*1032705769 == 0 (mod 11^4).

Links

Crossrefs

Cf. A005259, A357510, A357511, A357512.

Programs

  • Maple
    seq(numer(add( (1/k^3) * 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^3)), 0); \\ Michel Marcus, Oct 04 2022

Formula

Conjecture: a(p-1) == 0 (mod p^4) for all primes p >= 3 except 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).

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

Original entry on oeis.org

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

Views

Author

Peter Bala, Oct 03 2022

Keywords

Examples

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

Links

Crossrefs

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

Programs

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

Formula

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^3*binomial(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

Views

Author

Peter Bala, Oct 04 2022

Keywords

Comments

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.

Links

Crossrefs

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

Programs

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

Formula

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

A357560 a(n) = the numerator of ( Sum_{k = 1..n} (-1)^(n+k)*(1/k)*binomial(n,k)* binomial(n+k,k)^2 ).

Original entry on oeis.org

0, 4, 0, 94, 500, 19262, 50421, 2929583, 25197642, 2007045752, 3634262225, 368738402141, 6908530637021, 852421484283739, 1168833981781025, 56641833705924527, 276827636652242789, 46345946530867053437, 51051733540797155872, 9673584199611903429172
Offset: 0

Views

Author

Peter Bala, Oct 04 2022

Keywords

Comments

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 of the two types of Apéry numbers. The present sequence is the case m = 1. See A357561 for the case m = 3.
Conjectures:
1) for even m >= 2, S_m(p-1) == 0 (mod p^3) for all primes p > m + 3.
2) for odd m >= 1, S_m(p-1) == 0 (mod p^4) for all primes p > m + 4.

Examples

			Example of a supercongruence:
p = 19: a(19 - 1) = 51051733540797155872 = (2^5)*(19^4)*12241823444801 == 0 (mod 19^4).

Links

Crossrefs

Cf. A005258, A357510, A357511, A357512, A357513, A357558, A357559, A357561.

Programs

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

Formula

Conjecture: a(p-1) == 0 (mod p^4) for all primes p >= 7 (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).

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

Views

Author

Peter Bala, Oct 04 2022

Keywords

Comments

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.

Examples

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

Links

Crossrefs

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

Programs

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

Formula

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).
Showing 1-7 of 7 results.

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