A333579 -id:A333579 - cp's OEIS frontend

A351858 a(n) = [x^n] (1 + x + x^2)^(3n)/(1 + x)^(2n).

1, 1, 7, 19, 103, 376, 1825, 7547, 35175, 153838, 708132, 3181091, 14616481, 66582283, 306501377, 1407473269, 6497464679, 29991098982, 138844558150, 643215119214, 2985368996228, 13868212710623, 64508509024241, 300324344452479, 1399598738196897, 6527698842078501

Offset: 0

Peter Bala, Feb 27 2022

Given an integer sequence (g(n))n>=1, there exists a formal power series G(x), with rational coefficients, such that g(n) = [x^n] G(x)^n. The power series G(x) has integer coefficients iff the Gauss congruences g(n*p^r) == g(n*p^(r-1)) (mod p^r) hold for all primes p and positive integers n and r.
The central binomial coefficient binomial(2*n,n) = A000984(n) may be defined using the coefficient extraction operator as binomial(2*n,n) = [x^n] ((1 + x)^2)^n and hence the Gauss congruences hold for A000984. Moreover, it is known that the stronger supercongruences A000984(n*p^r) == A000984(n*p^(r-1)) (mod p^(3*r)) hold for primes p >= 5 and positive integers n and r. See Meštrović, equation 39.
We define an infinite family of sequences as follows. Let k be a positive integer. Define the rational function G_k(x) = (1 + x + ... + x^k)^(k+1)/(1 + x + ... + x^(k-1))^k and define the sequence u_k by u_k(n) = [x^n] G_k(x)^n. In particular, G_1(x) = (1 + x)^2 and the sequence u_1 is the sequence of central binomial coefficients. The present sequence is the case k = 2. See A351859 for the case k = 3.
Conjecture: for k >= 2, each sequence u_k satisfies the same supercongruences as the central binomial coefficients.
More generally, if r is a positive integer and s an integer then the sequence defined by  u_k(r,s;n) = [x^(r*n)] G_k(x)^(s*n) may satisfy the same supercongruences.

			Examples of supercongruences:
a(5) - a(1) = 376 - 1 = 3*(5^3) == 0 (mod 5^3)
a(2*7)- a(2) = 306501377 - 7 = 2*5*(7^3)*193*463 == 0 (mod 7^3)
A(5^2) - a(5) = 6527698842078501 - 376 = (5^6)*17*107*229671647 == 0 (mod 5^6)

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

R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.

Cf. A000984, A103885, A333090, A333564, A333565, A333579, A333715, A351856, A351857, A351859, A372211, A372212, A372213.

seq(add(add((-1)^(n-k-j)*binomial(n,k)*binomial(3*n,j)* binomial(4*n-2*j-k-1,n-k-j), j = 0..n-k), k = 0..n), n = 0..25);

A351858[n_]:=Sum[(-1)^(n-k-j)Binomial[n,k]Binomial[3n,j]Binomial[4n-2j-k-1,n-k-j],{k,0,n},{j,0,n-k}];Array[A351858,25,0] (* Paolo Xausa, Oct 04 2023 *)
a[n_]:=SeriesCoefficient[(1 + x + x^2)^(3*n)/(1 + x)^(2*n),{x,0,n}]; Array[a,26,0] (* Stefano Spezia, Apr 30 2024 *)

a(n) = Sum_{k = 0..n} Sum_{j = 0..n-k} (-1)^(n-k-j)*C(n,k)*C(3*n,j)*C(4*n-2*j-k-1,n-k-j).
Conjecture: a(n) = Sum_{k = 0..floor(n/2)} C(3*n,k)*C(n-k,k).
The o.g.f. A(x) = 1 + x + 7*x^2 + 19*x^3 + ... is the diagonal of the bivariate rational function 1/(1 - t*(1 + x + x^2)^3/(1 + x)^2) and hence is an algebraic function over the field of rational functions Q(x) by Stanley, Theorem 6.33, p. 197.
Let F(x) = (1/x)*Series_Reversion( x*(1 + x)^2/(1 + x + x^2)^3 ) = 1 + x + 4*x^2 + 10*x^3 + 40*x^4 + 133*x^5 + 536*x^6 + .... Then A(x) = 1 + x*F'(x)/F(x).
a(n) ~ sqrt(2/9 + 2*sqrt(53/47)*cos(arccos(1259*sqrt(47/53)/1696)/3)/9) * (2*sqrt(164581)*cos(arccos(-90631279/(1316648*sqrt(164581)))/3)/81 - 293/81)^n / sqrt(Pi*n). - Vaclav Kotesovec, Jun 05 2022

A362408 a(n) = [x^n] (F(x)/F(-x))^n where F(x) = (1 + x)*(1 + x^3).

Original entry on oeis.org

1, 2, 8, 44, 256, 1502, 8912, 53510, 324352, 1980332, 12160008, 75015162, 464566144, 2886488906, 17985045464, 112333392044, 703119387648, 4409231140086, 27696141476336, 174229516043630, 1097501783152256, 6921721148337452, 43701895245221848

Offset: 0

Peter Bala, Apr 18 2023

Compare with A228960(n) = [x^n] F(x)^n.
Let k and m be positive integers and let f(x) be a finite product of cyclotomic polynomials. Define b(n) = [x^(k*n)] (f(x)/f(-x))^(m*n). Then we conjecture that the supercongruences a(p) == a(1) (mod p^3) and, for n >= 2, a(n*p) == a(n) (mod p^2) hold for all primes p, with a finite number of exceptions depending on f(x).
The present sequence is the case k = m = 1 and f(x) = (1 + x)*(1 + x^3) = C(2,x)^2 * C(6,x), where C(n,x) denotes the n-th cyclotomic polynomial. See A002003 for the case k = m = 1 and f(x) = (1 + x).

Wikipedia, Cyclotomic polynomial

Cf. A002003, A228960, A333579.

F(x) := (1 + x)*(1 + x^3): G(x) := taylor(F(x)/F(-x),x = 0, 50); seq(coeftayl(G(x)^n, x = 0, n), n = 0..50);

Conjectures: 1) the supercongruence a(p) == 2 (mod p^3) holds for all primes p >= 5 (checked up to p = 47).

2) for n >= 2, the supercongruence a(n*p) == a(n) (mod p^2) holds for all primes p >= 5.

A351859 a(n) = [x^n] (1 + x + x^2 + x^3)^(4n)/(1 + x + x^2)^(3n).

Original entry on oeis.org

1, 1, 3, 19, 67, 251, 1137, 4803, 20035, 87013, 377753, 1634469, 7134385, 31261114, 137121113, 603206144, 2660097603, 11749336328, 51981371895, 230336544210, 1021976441817, 4539784391763, 20188837618799, 89871081815631, 400427435522737, 1785639575031501

Offset: 0

Peter Bala, Mar 01 2022

This sequence is the third in an infinite family of sequences defined as follows. Let k be a positive integer. Define the rational function G_k(x) = (1 + x + ... + x^k)^(k+1)/(1 + x + ... + x^(k-1))^k, so that G_1(x) = (1 + x)^2, and define the sequence u_k by u_k(n) = [x^n] G_k(x)^n. See A000984, the sequence of central binomial coefficients, for the case k = 1 and A351858 for the case k = 2. The present sequence is the case k = 3.
Given a power series G(x) with integer coefficients it is known that the sequence (g(n))n>=1 defined by g(n) := [x^n] G(x)^n satisfies the Gauss congruences g(n*p^r) == g(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
Thus a(n) satisfies the Gauss congruences. Calculation suggests that, in fact, the stronger supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for primes p >= 5 and positive integers n and r. These supercongruences are known to hold for the central binomial coefficients A000984(n) = [x^n] ((1 + x)^2)^n (Meštrović, equation 39).
More generally, if r is a positive integer and s an integer then the sequence defined by a(r,s;n) = [x^(r*n)] G_3(x)^(s*n) may satisfy the same supercongruences.

			Examples of supercongruences:
a(5) - a(1) = 251 - 1 = 2*(5^3) == 0 (mod 5^3)
a(2*7) - a(2) = 137121113 - 3 = 2*5*(7^4)*5711 == 0 (mod 7^4)
a(5^2) - a(5) = 1785639575031501 - 251 = 2*(5^6)*1373*3989*10433 == 0 (mod 5^6)

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

Paolo Xausa, Table of n, a(n) for n = 0..425
R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.

Cf. A000984, A103885, A333090, A333564, A333565, A333579, A333715, A351856, A351857, A351858.

seq(add(add(add((-1)^j*binomial(4*n,n-2*i-j-k)*binomial(4*n,i)* binomial(3*n+j-1,j)*binomial(j,k), k = 0..j), j = 0..n), i = 0..n), n = 0..25);

A351859[n_] := Sum[(-1)^j*Binomial[4*n, n-2*i-j-k]*Binomial[4*n, i]*Binomial[3*n+j-1, j]*Binomial[j, k], {i, 0, n}, {j, 0, n}, {k, 0, j}];
Array[A351859, 25, 0] (* Paolo Xausa, May 30 2025 *)

a(n)=sum(i=0,n,sum(j=0,n,sum(k=0,j,(-1)^j*binomial(4*n,n-2*i-j-k)*binomial(4*n,i)*binomial(3*n+j-1,j)*binomial(j,k))));
vector(25,n,a(n-1)) \\ Paolo Xausa, May 04 2022

a(n) = Sum_{i = 0..n} Sum_{j = 0..n} Sum_{k = 0..j} (-1)^j* C(4n,n-2*i-j-k) *C(4n,i)*C(3n+j-1,j)*C(j,k).
The o.g.f. A(x) = 1 + x + 3*x^2 + 19*x^3 + 67*x^4 + ... is the diagonal of the bivariate rational function 1/(1 - t*(1 + x + x^2 + x^3)^4/(1 + x + x^2)^3) and hence is an algebraic function over the field of rational functions Q(x) by Stanley, Theorem 6.33, p. 197.
Let F(x) = (1/x)*Series_Reversion( x*(1 + x + x^2)^3/(1 + x + x^2 + x^3)^4 ) = 1 + x + 2*x^2 + 8*x^3 + 25*x^4 + 81*x^5 + 305*x^6 + .... Then A(x) = 1 + x*F'(x)/F(x).

cp's OEIS Frontend

A351858 a(n) = [x^n] (1 + x + x^2)^(3n)/(1 + x)^(2n).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A362408 a(n) = [x^n] (F(x)/F(-x))^n where F(x) = (1 + x)*(1 + x^3).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Formula

A351859 a(n) = [x^n] (1 + x + x^2 + x^3)^(4n)/(1 + x + x^2)^(3n).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

cp's OEIS Frontend

A351858 a(n) = [x^n] (1 + x + x^2)^(3*n)/(1 + x)^(2*n).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A362408 a(n) = [x^n] (F(x)/F(-x))^n where F(x) = (1 + x)*(1 + x^3).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Formula

A351859 a(n) = [x^n] (1 + x + x^2 + x^3)^(4*n)/(1 + x + x^2)^(3*n).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A351858 a(n) = [x^n] (1 + x + x^2)^(3n)/(1 + x)^(2n).

A351859 a(n) = [x^n] (1 + x + x^2 + x^3)^(4n)/(1 + x + x^2)^(3n).