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 1-8 of 8 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.

A357959 a(n) = 5*A005259(n-1) + 2*A005258(n).

Original entry on oeis.org

11, 63, 659, 9727, 187511, 4304943, 109312739, 2941124607, 82033399631, 2345394917563, 68306797052879, 2018580243252847, 60368874298729631, 1823588997226603663, 55558079041172790659, 1705174802761490321407, 52672634815976274443711, 1636296942340074307669443
Offset: 1

Views

Author

Peter Bala, Oct 25 2022

Keywords

Comments

Conjectures:
1) a(p) == a(1) (mod p^5) for all primes p >= 5 (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 >= 5.
These are stronger supercongruences than those satisfied separately by the two types of Apéry numbers A005258 and A005259. Cf. A357959.
There is also a product version of these conjectures:
3) the sequence {u(n): n >= 1} defined by u(n) = A005259(n-1)^5 * A005258(n)^6 also satisfies the congruences in 1) and 2) above. See A357960.

Examples

			Examples of supercongruences:
a(13) - a(1) = 60368874298729631 - 11 = (2^2)*3*5*(13^5)*131*20685869 == 0 (mod 13^5).
a(5^2) - a(5) = 51292638914356604042099497031437511 - 187511 = (2^4)*3*(5^10)* 37*72974432287*40526706713533 == 0 (mod 5^10).

Crossrefs

Cf. A005258, A005259, A212334, A352655, A357567, A357956, A356957, A357958, A357960.

Programs

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

Formula

a(n) = 5*Sum_{k = 0..n-1} binomial(n-1,k)^2*binomial(n+k-1,k)^2 + 2*Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k).
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.

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);

A357958 a(n) = 5*A005259(n) + 14*A005258(n-1).

Original entry on oeis.org

39, 407, 7491, 167063, 4112539, 107461667, 2923006251, 81853622423, 2343591359499, 68288538877907, 2018394003648391, 60366962358086243, 1823569260750104179, 55557874330437332267, 1705172670555862322491, 52672612525369663916183
Offset: 1

Views

Author

Peter Bala, Oct 25 2022

Keywords

Comments

Conjectures:
1) a(p) == a(1) (mod p^5) for all primes p >= 5 (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. Cf. A357959.
There is also a product version of these conjectures:
3) the sequence {u(n): n>= 1} defined by u(n) = A005259(n)^25 * A005258(n-1)^14 conjecturally satisfies the congruences in 1) and 2) above.

Examples

			Examples of supercongruences:
a(13) - a(1) = 1823569260750104179 - 39 = (2^2)*5*7*(13^5)*35081444357 == 0 (mod 13^5).
a(7^2) - a(7) = (2^3)*(7^9)* 10412078726049425470554760052126170543547100055154203726400782433 == 0 (mod 7^9).

Crossrefs

Cf. A005258, A005259, A212334, A352655, A357567, A357956, A357957, A357959, A357960.

Programs

  • Maple
    seq( add( 5*binomial(n,k)^2*binomial(n+k,k)^2 + 14*binomial(n-1,k)^2* binomial(n+k-1,k), k = 0..n ), n = 1..20);

Formula

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

A357957 a(n) = A005259(n)^5 - A005258(n)^2.

Original entry on oeis.org

0, 3116, 2073071232, 6299980938881516, 39141322964380888600000, 368495989505416178203682748116, 4552312485541626792249211584618373944, 68109360474242016374599574592870648425552876, 1174806832391451114413440151405736019461523615095744
Offset: 0

Views

Author

Peter Bala, Oct 24 2022

Keywords

Comments

Conjectures:
1) a(p - 1) == 0 (mod p^5) for all primes p >= 5 (checked up to p = 271).
2) a(p^r - 1) == a(p^(r-1) - 1) ( mod p^(3*r+3) ) for r >= 2 and all primes p >= 5.
These are stronger supercongruences than those satisfied separately by the two types of Apéry numbers A005258 and A005259.
3) Put u(n) = A005259(n)^5 / A005258(n)^2. Then u(p^r - 1) == u(p^(r-1) - 1) ( mod p^(3*r+3) ) for r >= 2 and all primes p >= 5.

Examples

			a(7) = 4552312485541626792249211584618373944 = (2^3)*(3^3)*(7^5)*29*107* 404116272977592231282158029 == 0 (mod 7^5).

Crossrefs

Cf. A005258, A005259, A212334, A352655, A357567, A357568, A357569, A357956, A357958, A357959, A357960.

Programs

  • Maple
    seq(add(binomial(n,k)^2*binomial(n+k,k)^2, k = 0..n)^5 - add(binomial(n,k)^2*binomial(n+k,k), k = 0..n)^2, n = 0..20);
# Alternatively:
a := n -> hypergeom([-n, -n, 1 + n, 1 + n], [1, 1, 1], 1)^5 - hypergeom([1 + n, -n, -n], [1, 1], 1)^2: seq(simplify(a(n)), n=0..8); # Peter Luschny, Nov 01 2022

Formula

a(n) = ( Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)^2 )^5 - ( Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k) )^2.
a(n*p^r - 1) == a(n*p^(r-1) - 1) ( mod p^(3*r) ) for positive integers n and r and for all primes p >= 5.
a(n) = hypergeom([-n, -n, 1 + n, 1 + n], [1, 1, 1], 1)^5 - hypergeom([1 + n, -n, -n], [1, 1], 1)^2. - Peter Luschny, Nov 01 2022

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

Original entry on oeis.org

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

Views

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.

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 1-8 of 8 results.

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