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 13 results. Next

A362722 a(n) = [x^n] ( E(x)/E(-x) )^n where E(x) = exp( Sum_{k >= 1} A005258(k)*x^k/k ).

Original entry on oeis.org

1, 6, 72, 1266, 23232, 445506, 8740728, 174366114, 3519799296, 71696570010, 1470795168072, 30344633110710, 628994746308288, 13089254107521234, 273292588355096760, 5722454505166750266, 120119862431845048320, 2526922404360157374738, 53260275108329790626952
Offset: 0

Views

Author

Peter Bala, May 01 2023

Keywords

Comments

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). 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: 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.

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(2*A005258(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*A005258(2*k+1)*x^(2*k+1) )/(2*k+1) ).
Conjectures:
1) the supercongruence a(p^r) == a(p^(r-1)) (mod p^(2*r+1)) holds for all primes p >= 5.
2) for n >= 2, a(n*p) == a(n) (mod p^2) holds for all primes p >= 3.
3) for r >= 2, the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for all primes p >= 3 and n >= 1.

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.

A363418 Square array read by ascending antidiagonals: T(n,k) = [x^(n*k)] ((1 + x)/(1 - x))^k for n, k >= 1.

Original entry on oeis.org

2, 2, 8, 2, 16, 38, 2, 24, 146, 192, 2, 32, 326, 1408, 1002, 2, 40, 578, 4672, 14002, 5336, 2, 48, 902, 11008, 69002, 142000, 28814, 2, 56, 1298, 21440, 216002, 1038984, 1459810, 157184, 2, 64, 1766, 36992, 525002, 4320608, 15856206, 15158272, 864146
Offset: 1

Views

Author

Peter Bala, Jun 12 2023

Keywords

Comments

The n-th row sequence {T(n, k) : k >= 1} satisfies the Gauss congruences, that is, T(n, m*p^r) == T(n, m*p^(r-1)) ( mod p^r ) for all primes p and positive integers m and r.
We conjecture that each row sequence satisfies the stronger supercongruences T(n, m*p^r) == T(n, m*p^(r-1)) ( mod p^(3*r) ) for all primes p >= 5 and positive integers m and r.

Examples

			Square array begins
 n\k |  1   2     3      4        5          6           7
 - - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  1  |  2   8    38    192     1002       5336       28814   ...   (A002003)
  2  |  2  16   146   1408    14002     142000     1459810   ...   (A103885)
  3  |  2  24   326   4672    69002    1038984    15856206   ...   (A333715)
  4  |  2  32   578  11008   216002    4320608    87588482   ...
  5  |  2  40   902  21440   525002   13104184   331482062   ...
  6  |  2  48  1298  36992  1086002   32497680   985524066   ...
  7  |  2  56  1766  58688  2009002   70097384  2478629134   ...
  8  |  2  64  2306  87552  3424002  136485568  5513464322   ...
		

References

  • R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Crossrefs

A002003 (row 1), A103885 (row 2), A333715 (row 3). Cf. A035607, A362724 - A362733, A363419.

Programs

  • Maple
    # display as a square array
    T := (n,k) -> add( binomial(k, j)*binomial((n + 1)*k - j - 1, n*k - j) , j = 0..k): for n from 1 to 10 do seq(T(n, k), k = 1..10) end do;
    #alternative program
    seq(print(seq(simplify(2*k*hypergeom([1 - k, 1 - n*k], [2], 2)), k = 1..10)), n = 1..10);
    # display as a sequence
    seq(seq(T(n+1-i, i), i = 1..n), n = 1..10);
  • PARI
    T(n,k) = sum(j=0, k, binomial(k, j)*binomial((n + 1)*k - j - 1, n*k - j)) \\ Andrew Howroyd, Jan 05 2024

Formula

T(n,k) = Sum_{j = 0..k} binomial(k, j)*binomial((n + 1)*k - j - 1, n*k - j).
T(n,k) = 1/n * [x^k] ((1 + x)/(1 - x))*(n*k).
T(n,k) = (1/n)*Sum_{j = 0..k} binomial(n*k, j)*binomial((n + 1)*k - j - 1, k - j).
T(2*n,k) = [x^(n*k)] Chebyshev_T(k,(1 + x)/(1 - x)), where Chebyshev_T(n,x) denotes the n-th Chebyshev polynomial of the first kind. See A053120.
T(n,k) = Sum_{j = 1..k} (2^j)*binomial(k, j)*binomial(n*k - 1, n*k - j).
T(n,k) = (2*k) * hypergeom([1 - k, 1 - n*k], [2], 2).
Define E(n,x) = exp( Sum_{j >= 1} T(n,j)*x^j/j ). Then T(n+1,k) = [x^k] E(n,x)^k.
E(n,x) = (1/x) * the series reversion of x/E(n-1,x) for n >= 2.
E(n,x)^n = (1/x) * the series reversion of x*((1 - x)/(1 + x))^n.
E(m,x) appears to be the g.f. of the (m + 1)-Schroeder numbers. See A027307 (m = 2) and the cross references there.
The o.g.f. for row n is the diagonal of the bivariate rational function (1/n) * t*f(x)^n/(1 - t*f(x)^n), where f(x) = (1 + x)/(1 - x), and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.

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.

A363419 Square array read by ascending antidiagonals: T(n,k) = 1/n * [x^k] 1/((1 - x)*(1 - x^2))^(n*k) for n, k >= 1.

Original entry on oeis.org

1, 1, 5, 1, 7, 19, 1, 9, 46, 85, 1, 11, 82, 327, 376, 1, 13, 127, 793, 2376, 1715, 1, 15, 181, 1547, 7876, 17602, 7890, 1, 17, 244, 2653, 19376, 79686, 132056, 36693, 1, 19, 316, 4175, 40001, 247205, 816684, 1000263, 171820, 1, 21, 397, 6177, 73501, 614389, 3195046, 8450585, 7632433, 809380
Offset: 0

Views

Author

Peter Bala, Jun 13 2023

Keywords

Comments

The n-th row sequence {T(n, k) : k >= 1} satisfies the Gauss congruences, that is, T(n, m*p^r) == T(n, m*p^(r-1)) ( mod p^r ) for all primes p and positive integers m and r.
We conjecture that each row sequence satisfies the stronger supercongruences T(n, m*p^r) == T(n, m*p^(r-1)) ( mod p^(3*r) ) for all primes p >= 5 and positive integers m and r.
The table can be extended to negative values of n, and the row sequences also appear to satisfy the above supercongruences.

Examples

			The square array begins
 n\k |  1   2    3      4       5         6          7
 - - + - - - - - - - - - - - - - - - - - - - - - - - - - - -
  1  |  1   5   19     85     376      1715       7890   ...     (A348410)
  2  |  1   7   46    327    2376     17602     132056   ...
  3  |  1   9   82    793    7876     79686     816684   ...
  4  |  1  11  127   1547   19376    247205    3195046   ...
  5  |  1  13  181   2653   40001    614389    9560097   ...
  6  |  1  15  244   4175   73501   1318236   23952720   ...
  7  |  1  17  316   6177  124251   2546288   52867620   ...
  8  |  1  19  397   8723  197251   4544407  106076867   ...
  9  |  1  21  487  11877  298126   7624551  197571088   ...
 10  |  1  23  586  15703  433126  12172550  346618308   ...
Array extended to negative values of n:
 n\k |  1   2    3      4       5         6          7
 - - + - - - - - - - - - - - - - - - - - - - - - - - - - - -
 -5  |  1  -7   46   -247     626      8642    -194480   ...
 -4  |  1  -5   19     -5    -874     11569    -105300   ...
 -3  |  1  -3    1     77    -749      4641     -19893   ...
 -2  |  1  -1   -8     63    -249       440       1716   ...
 -1  |  1   1   -8     17       1      -116        344   ...     (-A234839)
		

References

  • R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Crossrefs

Programs

  • Maple
    # display as a square array
    T := (n,k) -> (1/n)*add( (-1)^(k+j) * binomial(-n*k,j)*binomial(-n*k, k-2*j) , j = 0..floor(k/2)): for n from 1 to 10 do seq(T(n, k), k = 1..10) end do;
    # display as a sequence
    seq(seq(T(n+1-i, i), i = 1..n), n = 1..10);

Formula

T(n,k) = (1/n)*Sum_{j = 0..floor(k/2)} binomial(n*k + j - 1, j)*binomial((n+1)*k - 2*j - 1, k - 2*j).
Define E(n,x) = exp( Sum_{j >= 1} T(n,j)*x^j/j ). Then T(n+1,k) = [x^k] E(n,x)^k.
E(n,x) = (1/x) * the series reversion of x/E(n-1,x) for n >= 2.
E(n,x)^n = (1/x) * the series reversion of x*((1 - x)(1 - x^2))^n.
T(n,k) = (1/n)*binomial(n*k+k-1,k) * hypergeom([n*k, -k/2, (1 - k)/2], [(1 - (n+1)*k)/2, (2 - (n+1)*k)/2], 1) except when n = 1 and k = 1 or 2.
The o.g.f. for row n is the diagonal of the bivariate rational function (1/n) * t*f(x)^n/(1 - t*f(x)^n), where f(x) = 1/((1 - x)*(1 - x^2)), and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.

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.
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