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 1-10 of 11 results. Next

A362733 a(n) = [x^n] F(x)^n, where F(x) = exp( Sum_{k >= 1} A362732(k)*x^k/k ).

Original entry on oeis.org

1, 6, 234, 10428, 492522, 24033006, 1197423396, 60530725380, 3092592004074, 159295600885794, 8258018380659234, 430335300869496072, 22521831447746893092, 1182951246247357578348, 62325193477833011143260, 3292376206935392483917428, 174323297281680647978503146, 9248680725006429075147528150
Offset: 0

Views

Author

Peter Bala, May 06 2023

Keywords

Crossrefs

Programs

  • Maple
    E(n,x) := series(exp(n*add((3*k)!/k!^3*x^k/k, k = 1..20)), x, 21):
    A362732(n) := coeftayl(E(n,x), x = 0, n):
    F(n,x) := series(exp(n*add(A362732(k)*x^k/k, k = 1..20)), x, 21):
    seq(coeftayl(F(n,x), x = 0, n), n = 0..20);
    # alternative program
    G(n,x) := series(exp(n*add((3*k)!/k!^3*x^(2*k)/k, k = 1..40)), x, 41):
    seq((1/2)*coeftayl(G(2*n,x), x = 0, 2*n), n = 1..20); # Peter Bala, Oct 27 2024

Formula

a(n*p^r) == a(n*p^(r-1)) (mod p^r) (Gauss congruence) holds for all primes p and positive integers n and r.
Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 3 and positive integers n and r.
For n >= 1, a(n) = (1/2) * [x^n] G(x)^(2*n), where G(x) = exp(Sum_{k >= 1} A006480(k)*x^k/k) is the g.f. of A229451. - Peter Bala, Oct 27 2024

A362732 a(n) = [x^n] E(x)^n, where E(x) = exp( Sum_{k >= 1} A006480(k)*x^k/k ).

Original entry on oeis.org

1, 6, 162, 5082, 170274, 5920506, 210808494, 7631158674, 279617726754, 10341283241130, 385275082939662, 14439312879759378, 543815325940475694, 20565700004741265900, 780470358196543271622, 29708379800729905316832, 1133811403010621704628514, 43371319655978568356324868
Offset: 0

Views

Author

Peter Bala, May 06 2023

Keywords

Comments

The de Bruijn numbers A006480(n) = (3*n)!/n!^3 satisfy the supercongruences A006480(n*p^r) == A006480(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r (write a(n) as binomial(3*n,2*n)*binomial(2*n,n) and apply Mestrovic, equation 39, p. 12).
We inductively define a family of sequences {a(i,n) : n >= 0}, i >= 0, by setting a(0,n) = A006480(n) and, for i >= 1, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n. The present sequence is {a(1,n)}. See A362733 for {a(2,n)}.
We conjecture that the sequences {a(i,n) : n >= 0}, i >= 1, satisfy the same supercongruences as above.
For i >= 1, a(i, n) = 1/i *[x^n] E(x)^(i*n), where E(x) = exp(Sum_{k >= 1} (3*k)!/k!^3 * x^k/k) is the g.f. of A229451. - Peter Bala, Oct 30 2024

Crossrefs

Programs

  • Maple
    E(n,x) := series(exp(n*add(((3*k)!/k!^3*x^k)/k, k = 1..20)), x, 21):
    seq(coeftayl(E(n,x), x = 0, n), n = 0..20);
  • Mathematica
    nmax = 20; Table[SeriesCoefficient[E^(n*Sum[(3*k)!/k!^3*x^k/k, {k, 1, n}]), {x, 0, n}], {n, 0, nmax}] (* Vaclav Kotesovec, Nov 26 2024 *)
  • Python
    from sympy import symbols, factorial, exp, series
    x = symbols('x')
    nmax = 10
    result = []
    for n in range(0, nmax + 1):
        summation = sum(factorial(3 * k) / (factorial(k)**3 * k) * x**k for k in range(1, n + 1))
        exp_series = exp(n * summation)
        coefficient = exp_series.series(x, 0, n + 1).coeff(x, n)
        result.append(coefficient)
    print(result) # Robert C. Lyons, Jan 27 2025

Formula

a(n*p^r) == a(n*p^(r-1)) (mod p^r) (Gauss congruence) holds for all primes p and positive integers n and r.
Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 3 and positive integers n and r.
a(n) ~ c * d^n / sqrt(n), where d = 39.5963012687026... and c = 0.107979249748...

A362723 a(n) = [x^n] ( E(x)/E(-x) )^n where E(x)= exp( Sum_{k >= 1} A005259(k)*x^k/k ).

Original entry on oeis.org

1, 10, 200, 7390, 260800, 10263010, 407520920, 16758685030, 697767370240, 29525605934410, 1261570539980200, 54419751094210270, 2364396136291654720, 103393259758470870770, 4545671563318715532280, 200804420082143353690390, 8907295723280072012247040, 396570344897237949249382010
Offset: 0

Views

Author

Peter Bala, May 01 2023

Keywords

Comments

It is known that the sequence of Apéry numbers A005259 satisfies the Gauss congruences A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
One consequence is that the power series expansion of E(x) = exp( Sum_{k >= 1} A005259(k)*x^k/k ) = 1 + 5*x + 49*x^2 + 685*x^3 + 11807*x^4 + ... has integer coefficients. See A267220. For a proof see, for example, Beukers, Proposition, p 143. Therefore, the power series expansion of E(x)/E(-x) also has integer coefficients and so a(n) = [x^n] ( E(x)/E(-x) )^n is an integer.
In fact, the Apéry numbers satisfy stronger congruences than the Gauss congruences known as supercongruences: A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r (see Straub, Section 1).
We conjecture below that {a(n)} satisfies supercongruences similar to (but weaker than) the above supercongruences satisfied by the Apéry numbers.

Crossrefs

Programs

  • Maple
    A005259 := proc(n) add(binomial(n, k)^2*binomial(n+k,k)^2, k = 0..n) end;
    E(n,x) := series(exp(n*add(2*A005259(2*k+1)*x^(2*k+1)/(2*k+1), k = 0..10)), x, 21):
    seq(coeftayl(E(n,x), x = 0, n), n = 0..20);

Formula

a(n) = [x^n] exp( Sum_{k >= 1} n*( 2*A005259(2*k+1)*x^(2*k+1) )/(2*k+1) ).
Conjectures:
1) the supercongruence a(p) == a(1) (mod p^3) holds for all primes p >= 5 (checked up to p = 101).
2) for n >= 2, a(n*p) == a(n) (mod p^2) holds for all primes p >= 5.
3) for n >= 1, r >= 2, the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for all primes p >= 5.

A362724 a(n) = [x^n] E(x)^n, where E(x) = exp( Sum_{k >= 1} A005258(k)*x^k/k ).

Original entry on oeis.org

1, 3, 37, 525, 7925, 123878, 1980199, 32150030, 527984245, 8747075100, 145917510662, 2447835093498, 41253740275559, 697956867712705, 11847510103853090, 201678623730755525, 3441648250114203253, 58859380176953941937, 1008553120517397082420, 17311102730697482426850
Offset: 0

Views

Author

Peter Bala, May 02 2023

Keywords

Comments

Compare with A362722.
It is known that the sequence of Apéry numbers A005258 satisfies the Gauss congruences A005258(n*p^r) == A005258(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
One consequence is that the power series expansion of E(x) = exp( Sum_{k >= 1} A005258(k)*x^k/k ) = 1 + 3*x + 14*x^2 + 82*x^3 + 551*x^4 + ... has integer coefficients (see, for example, Beukers, Proposition, p. 143), and therefore a(n) = [x^n] E(x)^n is an integer.
In fact, the Apéry numbers satisfy stronger congruences than the Gauss congruences known as supercongruences: A005258(n*p^r) == A005258(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r (see Straub, Section 1).
We conjecture below that {a(n)} satisfies supercongruences similar to (but weaker than) the above supercongruences satisfied by the Apéry numbers.
More generally, we inductively define a family of sequences {a(i,n) : n >= 0}, i >= 0, by setting a(0,n) = A005258(n) and, for i >= 1, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n. In this notation the present sequence is {a(1,n)}.
We conjecture that the sequences {a(i,n) : n >= 0}, i >= 1, satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(2*r)) for all primes p >= 3, and positive integers n and r.

Crossrefs

Programs

  • Maple
    A005258 := proc(n) add(binomial(n,k)^2*binomial(n+k,k), k = 0..n) end proc:
    E(n,x) := series(exp(n*add(A005258(k)*x^k/k, k = 1..20)), x, 21):
    seq(coeftayl(E(n,x), x = 0, n), n = 0..20);

Formula

Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for all primes p >= 3 and positive integers n and r.

A362725 a(n) = [x^n] E(x)^n, where E(x) = exp( Sum_{k >= 1} A005259(k)*x^k/k ).

Original entry on oeis.org

1, 5, 123, 3650, 118059, 4015380, 141175410, 5082313276, 186243853995, 6920379988871, 260030830600748, 9860709859708350, 376821110248674594, 14494688046084958080, 560708803489098556632, 21797478402692370515400, 851057798310071946207915, 33356751162583463626417872
Offset: 0

Views

Author

Peter Bala, May 02 2023

Keywords

Comments

Compare with A362723.
It is known that the sequence of Apéry numbers A005259 satisfies the Gauss congruences A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
One consequence is that the power series expansion of E(x) = exp( Sum_{k
>= 1} A005259(k)*x^k/k ) = 1 + 5*x + 49*x^2 + 685*x^3 + 11807*x^4 + ... has integer coefficients (see, for example, Beukers, Proposition, p. 143), and therefore a(n) = [x^n] E(x)^n is an integer.
In fact, the Apéry numbers satisfy stronger congruences than the Gauss congruences known as supercongruences: A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r (see Straub, Section 1).
We conjecture below that the present sequence satisfies supercongruences similar to (but weaker than) the above supercongruences satisfied by the Apéry numbers.
More generally, we inductively define a family of sequences {a(i,n) : n >= 0}, i >= 0, by setting a(0,n) = A005259(n) and, for i >= 1, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n. In this notation the present sequence is {a(1,n)}.
We conjecture that the sequences {a(i,n) : n >= 0}, i >= 1, satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(2*r)) for all primes p >= 3, and positive integers n and r.

Crossrefs

Programs

  • Maple
    A005259 := proc(n) add(binomial(n,k)^2*binomial(n+k,k)^2, k = 0..n) end proc:
    E(n,x) := series(exp(n*add((A005259(k)*x^k)/k, k = 1..20)), x, 21):
    seq(coeftayl(E(n,x), x = 0, n), n = 0..20);

Formula

Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for all primes p >= 3 and positive integers n and r.

A362730 a(n) = [x^n] E(x)^n where E(x) = exp( Sum_{k >= 1} binomial(2*k,k)^2*x^k/k ).

Original entry on oeis.org

1, 4, 68, 1336, 27972, 607004, 13478072, 304083224, 6941422916, 159882680452, 3708781743068, 86526900550864, 2028273983776440, 47733938489878528, 1127187050415921304, 26694934151138897336, 633813403549444601156, 15082008687681962081088, 359592614152718921447108
Offset: 0

Views

Author

Peter Bala, May 05 2023

Keywords

Comments

Compare with A359108(n) = [x^n] F(x)^n where F(x) = exp( Sum_{k >= 1} binomial(2*k,k)*x^k/k ).
More generally, we inductively define a family of sequences {a(i,n) : n >= 0}, i >= 0, by setting a(0,n) = binomial(2*n,n)^2 and, for i >= 1, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n. In this notation the present sequence is {a(1,n)}.
We conjecture that the sequences {a(i,n) : n >= 0}, i >= 1, satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5, and positive integers n and r.

Crossrefs

Programs

  • Maple
    E(n,x) := series( exp(n*add(binomial(2*k,k)^2*x^k/k, k = 1..20)), x, 21 ):
    seq(coeftayl(E(n,x), x = 0, n), n = 0..20);

Formula

Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 5 and positive integers n and r.

A362726 a(n) = [x^n] E(x)^n where E(x) = exp( Sum_{k >= 1} A208675(k)*x^k/k ).

Original entry on oeis.org

1, 1, 7, 64, 647, 6901, 76120, 859216, 9863303, 114689746, 1347186307, 15954752903, 190235245976, 2281177393704, 27487043703672, 332588768198389, 4038905184944263, 49204502405466061, 601135759955624038, 7362647062772162397, 90380912127647103747
Offset: 0

Views

Author

Peter Bala, May 02 2023

Keywords

Comments

A208675(n) = B(n,n-1,n-1) in the notation of Straub, equation 24, where it is shown that the supercongruences A208675(n*p^k) == A208675(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k.
Inductively define a family of sequences {a(i,n) : n >= 0}, i >= 0, by setting a(0,n) = A208675(n) and, for i >= 1, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n. In this notation the present sequence is {a(1,n)}.
We conjecture that the sequences {a(i,n) : n >= 0}, i >= 1, satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 7, and positive integers n and r.

Crossrefs

Programs

  • Maple
    A208675 := proc(n) add( (-1)^k*binomial(n-1,k)*binomial(2*n-k-1,n-k)^2, k = 0..n-1) end:
    E(n,x) := series(exp(n*add(A208675(k)*x^k/k, k = 1..20)), x, 21):
    seq(coeftayl(E(n,x), x = 0, n), n = 0..20);

Formula

Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 7 and positive integers n and r.

A362727 a(n) = [x^n] ( E(x)/E(-x) )^n where E(x) = exp( Sum_{k >= 1} A208675(k)*x^k/k ).

Original entry on oeis.org

1, 2, 8, 110, 960, 12502, 136952, 1746558, 20951040, 267467294, 3347043208, 43051344074, 550991269824, 7146318966438, 92706899799480, 1211369977374310, 15857138035286016, 208493724775866726, 2747100161210031944, 36305149229744449050, 480750961929272288960
Offset: 0

Views

Author

Peter Bala, May 02 2023

Keywords

Comments

A208675(n) = B(n,n-1,n-1) in the notation of Straub, equation 24, where it is shown that the supercongruences A208675(n*p^k) == A208675(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k.

Crossrefs

Programs

  • Maple
    A208675 := proc(n) add( (-1)^k*binomial(n-1,k)*binomial(2*n-k-1,n-k)^2, k = 0..n-1) end:
    E(n,x) := series(exp(n*add(2*A208675(2*k+1)*x^(2*k+1)/(2*k+1), k = 0..10)), x, 21):
    seq(coeftayl(E(n,x), x = 0, n), n = 0..20);

Formula

Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 7 and positive integers n and r.

A362728 a(n) = [x^n] E(x)^n where E(x) = exp( Sum_{k >= 1} A108628(k-1)*x^k/k ).

Original entry on oeis.org

1, 1, 9, 91, 985, 11101, 128475, 1515032, 18116825, 218988046, 2669804209, 32776883899, 404733925435, 5022161428571, 62578069656776, 782560813918216, 9817011145746649, 123492956278927438, 1557295053170126994, 19681186581532094418
Offset: 0

Views

Author

Peter Bala, May 03 2023

Keywords

Comments

A108628(n) = B(n+1,n,n+1) in the notation of Straub, equation 24, where it is shown that the supercongruences A108628(n*p^k) == A108628(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k.
Inductively define a family of sequences {a(i,n) : n >= 0}, i >= 0, by setting a(0,n) = A108628(n) and, for i >= 1, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n. In this notation the present sequence is {a(1,n)}.
We conjecture that the sequences {a(i,n) : n >= 0}, i >= 1, satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 7, and positive integers n and r.

Crossrefs

Programs

  • Maple
    A108628 := proc(n) add(binomial(n,k)*binomial(n+1,k)*binomial(n+k+1,k), k = 0..n) end:
    E(n,x) := series( exp(n*add(A108628(k-1)*x^k/k, k = 1..20)), x, 21 ):
    seq(coeftayl(E(n,x), x = 0, n), n = 0..20);

Formula

Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 7 and positive integers n and r.

A362729 a(n) = [x^n] ( E(x)/E(-x) )^n where E(x) = exp( Sum_{k >= 1} A108628(k-1)*x^k/k ).

Original entry on oeis.org

1, 2, 8, 146, 1344, 18502, 214136, 2820834, 35377152, 465110894, 6038588808, 79936149174, 1056557893440, 14094461001558, 188319357861944, 2529143690991946, 34042038343081984, 459723572413090934, 6221522287903354568, 84397945280561045302, 1147007337762078241344
Offset: 0

Views

Author

Peter Bala, May 03 2023

Keywords

Comments

A108628(n) = B(n+1,n,n+1) in the notation of Straub, equation 24, where it is shown that the supercongruences A108628(n*p^k) == A108628(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k.

Crossrefs

Programs

  • Maple
    A108628 := proc(n) add(binomial(n,k)*binomial(n+1,k)*binomial(n+k+1,k) , k = 0..n) end:
    E(n,x) := series(exp(n*add(2*(A108628(2*k)*x^(2*k+1))/(2*k+1), k = 0..10)), x, 21):
    seq(coeftayl(E(n,x), x = 0, n), n = 0..20);

Formula

Conjecture:the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 7 and positive integers n and r.
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