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 1-5 of 5 results.

A168593 G.f.: exp( Sum_{n>=1} A132303(n)*x^n/n ), where A132303(n) = sum of the cubes of the trinomial coefficients in row n of triangle A027907.

Original entry on oeis.org

1, 3, 27, 349, 5484, 96408, 1824758, 36393090, 754696998, 16130052394, 353134333470, 7884110379006, 178908263232959, 4115917059924057, 95806493175049929, 2252809457441037107, 53443567449376649304
Offset: 0

Views

Author

Paul D. Hanna, Dec 01 2009

Keywords

Comments

Self-convolution cube-root yields the integer sequence A251686.

Examples

			G.f.: A(x) = 1 + 3*x + 27*x^2 + 349*x^3 + 5484*x^4 + 96408*x^5 +...
log(A(x)) = 3*x + 45*x^2/2 + 831*x^3/3 + 17181*x^4/4 + 375903*x^5/5 +...+ A132303(n)*x^n/n +...

Crossrefs

Cf. A168590, A168592, A132303, A027907, A251686.

Programs

  • PARI
    {A027907(n,k) = polcoeff((1+x+x^2)^n, k)}
{A132303(n) = sum(k=0, 2*n, A027907(n,k)^3)}
{a(n) = local(A); A = exp(sum(m=1, n+1, A132303(m)*x^m/m) +x*O(x^n)); polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

A251686 G.f.: exp( Sum_{n>=1} A132303(n)/3 * x^n/n ), where A132303(n) = sum of the cubes of the trinomial coefficients in row n of triangle A027907.

Original entry on oeis.org

1, 1, 8, 100, 1556, 27260, 515510, 10284094, 213433728, 4566363088, 100082133066, 2236952393302, 50817223209451, 1170319824912699, 27268900054818390, 641812268110993694, 15239341125950643462, 364655982858022960206, 8785745372509009963892, 212976842702489760621536
Offset: 0

Views

Author

Paul D. Hanna, Feb 28 2015

Keywords

Comments

Self-convolution cube yields A168593.

Examples

			G.f.: A(x) = 1 + x + 8*x^2 + 100*x^3 + 1556*x^4 + 27260*x^5 +...
where
log(A(x)) = 1*x + 15*x^2/2 + 277*x^3/3 + 5727*x^4/4 + 125301*x^5/5 + 2843643*x^6/6 + 66214485*x^7/7 + 1571497119*x^8/8 +...+ A132303(n)/3*x^n/n +...

Crossrefs

Cf. A168593, A132303, A027907.

Programs

  • PARI
    {A027907(n,k) = polcoeff((1+x+x^2)^n, k)}
{A132303(n) = sum(k=0, 2*n, A027907(n,k)^3)}
{a(n) = if(n==0, 1, polcoeff(exp(sum(m=1, n, A132303(m)/3 * x^m/m) +x*O(x^n)), n))}
for(n=0,30,print1(a(n),", "))

A082758 Sum of the squares of the trinomial coefficients (A027907).

Original entry on oeis.org

1, 3, 19, 141, 1107, 8953, 73789, 616227, 5196627, 44152809, 377379369, 3241135527, 27948336381, 241813226151, 2098240353907, 18252025766941, 159114492071763, 1389754816243449, 12159131877715993, 106542797484006471, 934837217271732457, 8212609533895771131
Offset: 0

Views

Author

Emanuele Munarini, May 21 2003

Keywords

Comments

a(n) = T(2*n, 2*n), the coefficient of x^(2*n) in (1+x+x^2)^(2*n), where T is the trinomial triangle A027907; Integral representation: a(n) = (1/Pi) * Integral_{x=-1..1} ((1+2*x)^(2*n)/sqrt(1-x^2)), i.e., a(n) is the moment of order 2n of the random variable 1+2X, where the distribution of X is an arcsin law on the interval (-1,1). - N-E. Fahssi, Jan 22 2008

Examples

			G.f. = 1 + 3*x + 19*x^2 + 141*x^3 + 1107*x^4 + 8953*x^5 + 73789*x^6 + ...

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 77. (In the integral formula a left bracket is missing for the cosine argument.)

Links

Crossrefs

Cf. A002426, A027907, A132303 - A132305, A168592, A273055, A337389.

Programs

  • Maple
    a := n -> simplify(GegenbauerC(2*n,-2*n,-1/2)):
seq(a(n), n=0..19); # Peter Luschny, May 07 2016
  • Mathematica
    Table[Sum[(-1)^(i)*Binomial[2*n,i]*Binomial[4*n-3*i-1,2*n-3*i],{i,0,2*n/3}],{n, 0,25}] (* Adi Dani, Jul 03 2011 *)
Table[Hypergeometric2F1[1/2-n,-n,1,4], {n,0,19}] (* Peter Luschny, May 15 2016 *)
a[ n_] := SeriesCoefficient[ (1 - 2 x - 3 x^2)^(-1/2), {x, 0, 2 n}]; (* Michael Somos, Jan 08 2017 *)
  • Maxima
    makelist(sum(binomial(2*n-k, k)*binomial(2*n, k),k,0,n),n,0,40);
  • PARI
    a(n)={local(v=Vec((1+x+x^2)^n));sum(k=1,#v,v[k]^2);}
  • PARI
    a(n)=sum(k=0,n,binomial(2*n-k,k)*binomial(2*n,k));
  • PARI
    {a(n)=sum(k=0,n,binomial(n,k)^2*binomial(2*n,n)/binomial(2*k,k))} \\ Paul D. Hanna, Sep 29 2012
  • Sage
    def A():
    a, b, n = 1, 3, 1
    yield a
    while True:
        yield b
        n += 1
        a, b = b, ((9-9*n)*(4*n-1)*(2*n-3)*a+(4*n-3)*(20*n^2-30*n+7)*b)//(n*(2*n-1)*(4*n-5))
A082758 = A()
print([next(A082758) for  in range(20)]) # _Peter Luschny, May 16 2016

Formula

a(n) = Sum_{k=0..2n} T(n, k)^2, where T(n, k) are trinomial coefficients (A027907).
a(n) = Sum_{k=0..n} binomial(2*n-k, k)*binomial(2*n, k). - Benoit Cloitre, Jul 30 2003
G.f.: (1/sqrt(1+2*x-3*x^2) + 1/sqrt(1-2*x-3*x^2))/2 (with interpolated zeros). - Paul Barry, Jan 04 2005
a(n) = Sum_{k=0..n} binomial(2*n,2*k)*binomial(2*k,k) = Sum_{k=0..n} binomial(n+k,2k)*binomial(2*n,n+k). - Paul Barry, Dec 16 2008
a(n) = Sum_{k=0..n} binomial(n,k)^2*binomial(2*n,n)/ binomial(2*k,k). - Paul D. Hanna, Sep 29 2012
Recurrence: n*(2*n-1)*a(n) = (14*n^2+n-12)*a(n-1) + 3*(14*n^2-71*n+78)*a(n-2) - 27*(n-2)*(2*n-5)*a(n-3). - Vaclav Kotesovec, Oct 14 2012
a(n) ~ 3^(2*n+1/2)/(2*sqrt(2*Pi*n)). - Vaclav Kotesovec, Oct 14 2012
a(n) = GegenbauerC(2*n, -2*n, -1/2). - Peter Luschny, May 07 2016
From Peter Luschny, May 15 2016: (Start)
a(n) = ((9-9*n)*(4*n-1)*(2*n-3)*a(n-2)+(4*n-3)*(20*n^2-30*n+7)*a(n-1))/(n*(2*n-1)*(4*n-5)) for n>=2.
a(n) = hypergeom([1/2-n, -n], [1], 4). (End)
a(n) = A002426(2*n). - Michael Somos, Jan 08 2017
From Peter Bala, Mar 16 2018: (Start)
a(n) = sqrt(-3)^(2*n)*P(2*n,-1/sqrt(-3)), where P(n,x) is the Legendre polynomial of degree n.
a(n) = 1/C(2*n,n)*Sum_{k = 0..n} C(n,k)*C(n+k,k)*C(2*n+2*k,n+k)*(-3)^(n-k). Cf. A273055. (End)
From Wolfdieter Lang, Apr 19 2018 : (Start)
a(n) = (2/Pi)*Integral_{phi=0..Pi/2} (sin(3*phi)/sin(phi))^(2*n) [Comtet, p. 77, q=3, n=k -> 2*n] = (2/Pi)*Integral_{x=0..2} (x^2 - 1)^(2*n)/sqrt(4-x^2) (with x = 2*cos(phi)). See also the integral of the above comment.
a(n) = 3^(2*n)*Sum_{k=0..2*n} binomial(2*n, k)*binomial(2*k, k)*(-1/3)^k = 3^(2*n)*hypergeometric([-2*n, 1/2], [1], 4/3) = (-3)^n*LegendreP(2*n, 1/sqrt(-3)). (End)
From Peter Bala, Apr 03 2022: (Start)
Conjecture: a(n) = [x^n] ( (1 + x + x^3 + x^4)/(1 - x)^2 )^n.
If the conjecture is true then the Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. [added Aug 29 2025: The conjecture is true. The conjecture leads to the double sum representation a(n) = Sum_{j, k} binomial(n, k)*binomial(n, j)*binomial(3*n-3*j-k-1, n-3*j-k), which satisfies the second-order recurrence given above by Peter Luschny, as can be verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger Maple package.]
Calculation suggests that the supercongruence a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) holds for primes p >= 5 and positive integers n and k.
Column 1 of A337389. (End)

A132304 Sum of fourth powers of trinomial coefficients: a(n) = Sum_{k=0..2n} trinomial(n,k)^4 where trinomial(n,k) = [x^k] (1 + x + x^2)^n.

Original entry on oeis.org

1, 3, 115, 5157, 281907, 16688953, 1043460469, 67769148555, 4527813298227, 309207348577017, 21487099512128265, 1514508365575327455, 108015601963868232885, 7780760215720406500095, 565264860408377433558523, 41369186303433709127364757, 3047140179357736909766753331
Offset: 0

Views

Author

Paul D. Hanna, Aug 18 2007

Keywords

Comments

Conjecture: the supercongruence a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) holds for all primes p >= 5 and positive integers n and k. - Peter Bala, Aug 29 2025

Crossrefs

Cf. A027907, A082758, A132303, A132305.

Programs

  • PARI
    a(n)=sum(k=0,2*n,polcoeff((1+x+x^2)^n,k)^4)

A132305 Sum of fifth powers of trinomial coefficients: a(n) = Sum_{k=0..2n} trinomial(n,k)^5 where trinomial(n,k) = [x^k] (1 + x + x^2)^n.

Original entry on oeis.org

1, 3, 309, 32847, 4775301, 764206503, 131689759209, 23857704965727, 4487745064421061, 869024373004460823, 172218243516497425809, 34778752398142969125627, 7134304921516864247956281, 1482965360342923597534988883, 311760492785929879483633778049
Offset: 0

Views

Author

Paul D. Hanna, Aug 18 2007

Keywords

Comments

Conjecture: the supercongruence a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) holds for all primes p >= 5 and positive integers n and k. - Peter Bala, Aug 29 2025

Crossrefs

Cf. A027907, A082758, A132303, A132304.

Programs

  • PARI
    a(n)=sum(k=0,2*n,polcoeff((1+x+x^2)^n,k)^5)
Showing 1-5 of 5 results.

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