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 1-4 of 4 results.

A006442 Expansion of 1/sqrt(1 - 10*x + x^2).

Original entry on oeis.org

1, 5, 37, 305, 2641, 23525, 213445, 1961825, 18205345, 170195525, 1600472677, 15122515985, 143457011569, 1365435096485, 13033485491077, 124715953657025, 1195966908404545, 11490534389896325, 110584004488276645, 1065853221648055025
Offset: 0

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Author

Keywords

Comments

Number of Delannoy paths from (0,0) to (n,n) with steps U(0,1), H(1,0) and D(1,1) where H can choose from two colors. - Paul Barry, May 25 2005
Number of lattice paths from (0,0) to (n,n) using steps (0,1), (1,1), and two kinds of steps (1,0). - Joerg Arndt, Jul 01 2011
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 09 2022

Crossrefs

Column k=2 of A335333.
Sequences of the form LegendreP(n, 2*m+1): A000012 (m=0), A001850 (m=1), this sequence (m=2), A084768 (m=3), A084769 (m=4).
Cf. A098270, A243943 (a(n)^2).

Programs

  • Magma
    [Evaluate(LegendrePolynomial(n), 5): n in [0..40]]; // G. C. Greubel, May 21 2023
    
  • Maple
    seq(orthopoly[P](n,5), n = 0 .. 20); # Robert Israel, Aug 18 2014
  • Mathematica
    Table[LegendreP[n, 5], {n, 0, 19}] (* Arkadiusz Wesolowski, Aug 13 2012 *)
    CoefficientList[Series[1 / Sqrt[1 - 10 x + x^2], {x, 0, 20}], x] (* Vincenzo Librandi, Nov 23 2014 *)
  • PARI
    a(n)=subst(pollegendre(n),x,5)
    
  • PARI
    /* as lattice paths: same as in A092566 but use */
    steps=[[1,0], [1,0], [0,1], [1,1]]; /* note the double [1,0] */
    /* Joerg Arndt, Jul 01 2011 */
    
  • PARI
    {a(n)=sum(k=0,n,binomial(n,k)^2*2^k*3^(n-k))} /* Paul D. Hanna */
    
  • PARI
    {a(n) = sum(k=0, n, 2^k * binomial(2*k, k) * binomial(n+k, n-k) )}
    for(n=0, 25, print1(a(n), ", ")) \\ Paul D. Hanna, Aug 17 2014
    
  • SageMath
    [gen_legendre_P(n,0,5) for n in range(41)] # G. C. Greubel, May 21 2023

Formula

Legendre polynomial evaluated at 5. - Michael Somos, Dec 04 2001
G.f.: 1/sqrt(1 - 10*x + x^2).
a(n) equals the central coefficient of (1 + 5*x + 6*x^2)^n. - Paul D. Hanna, Jun 03 2003
a(n) equals the (n+1)-th term of the binomial transform of 1/(1-2x)^(n+1). - Paul D. Hanna, Sep 29 2003
a(n) = Sum_{k=0..n} 2^k*binomial(n, k)*binomial(n+k, k). - Benoit Cloitre, Apr 13 2004
a(n) = Sum_{k=0..n} binomial(n,k)^2 * 2^k * 3^(n-k). - Paul D. Hanna, Feb 04 2012
E.g.f.: exp(5*x) * Bessel_I(0, 2*sqrt(6)*x). - Paul Barry, May 25 2005
D-finite with recurrence: n*a(n) - 5*(2n-1)*a(n-1) + (n-1)*a(n-2) = 0 [Eq (4) in the T. D. Noe article]. R. J. Mathar, Jun 26 2012
a(n) ~ (5 + 2*sqrt(6))^n/(2*sqrt(Pi*n)*sqrt(5*sqrt(6) - 12)). - Vaclav Kotesovec, Oct 05 2012
a(n) = hypergeom([-n, n+1], [1], -2). - Peter Luschny, May 23 2014
a(n) = Sum_{k=0..n} 2^k * C(2*k, k) * C(n+k, n-k). - Paul D. Hanna, Aug 17 2014
a(n) = Sum_{k=0..n} (k+1) * 3^k * (-1)^(n-k) * binomial(n,k) * binomial(n+k+1,n) / (n+k+1). - Vladimir Kruchinin, Nov 23 2014
From Peter Bala, Nov 28 2021: (Start)
a(n) = (1/3)*(1/2)^n*Sum_{k >= n} binomial(k,n)^2*(2/3)^k.
a(n) = (1/3)^(n+1)*hypergeom([n+1, n+1], [1], 2/3).
a(n) = (2^n)*hypergeom([-n, -n], [1], 3/2).
a(n) = [x^n] ((x - 1)*(3 - 2*x))^n
a(n) = (1/2)^n*A098270(n). (End)
a(n) = (-1)^n * Sum_{k=0..n} (1/10)^(n-2*k) * binomial(-1/2,k) * binomial(k,n-k). - Seiichi Manyama, Aug 28 2025
a(n) = Sum_{k=0..floor(n/2)} 6^k * 5^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). - Seiichi Manyama, Aug 30 2025

A243949 Squares of the central Delannoy numbers: a(n) = A001850(n)^2.

Original entry on oeis.org

1, 9, 169, 3969, 103041, 2832489, 80802121, 2365752321, 70611901441, 2139090528969, 65568745087209, 2029206892664961, 63300531617048961, 1987912809986437161, 62787371136571152009, 1992942254830520803329, 63531842302018973818881, 2033004661359005674887561
Offset: 0

Views

Author

Paul D. Hanna, Aug 17 2014

Keywords

Comments

In general, we have the binomial identity:
if b(n) = Sum_{k=0..n} t^k * C(2*k, k) * C(n+k, n-k), then b(n)^2 = Sum_{k=0..n} (t^2+t)^k * C(2*k, k)^2 * C(n+k, n-k), where the g.f. of b(n) is 1/sqrt(1 - (4*t+2)*x + x^2), and the g.f. of b(n)^2 is 1 / AGM(1-x, sqrt((1+x)^2 - (4*t+2)^2*x)), where AGM(x,y) = AGM((x+y)/2,sqrt(x*y)) is the arithmetic-geometric mean.
Note that the g.f. of A001850 is 1/sqrt(1 - 6*x + x^2).
Limit_{n -> oo} a(n+1)/a(n) = (3 + 2*sqrt(2))^2 = 17 + 12*sqrt(2).
From Gheorghe Coserea, Jul 05 2016: (Start)
Diagonal of the rational function 1/(1 - x - y - z - x*y + x*z - y*z - x*y*z).
Annihilating differential operator: x*(x-1)*(x+1)*(x^2-34*x+1)*Dx^2 + (3*x^4-66*x^3-70*x^2+70*x-1)*Dx + x^3-7*x^2-35*x+9.
(End).
The sequence b(n) mentioned above is the sequence of shifted Legendre polynomials P(n,2*t + 1) (see A063007). See Zudilin for a g.f. for the sequence b(n)^2. - Peter Bala, Mar 02 2017

Examples

			G.f.: A(x) = 1 + 9*x + 169*x^2 + 3969*x^3 + 103041*x^4 + 2832489*x^5 +...
		

Crossrefs

Sequences of the form LegendreP(n, 2*m+1)^2: A000012 (m=0), this sequence (m=1), A243943 (m=2), A243944 (m=3), A243007 (m=4).
Related to diagonal of rational functions: A268545 - A268555.

Programs

  • Magma
    [Evaluate(LegendrePolynomial(n), 3)^2 : n in [0..40]]; // G. C. Greubel, May 17 2023
    
  • Mathematica
    Table[Sum[2^k *Binomial[2*k, k]^2 *Binomial[n+k, n-k], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Aug 18 2014 *)
    a[n_]:= HypergeometricPFQ[{1/2, -n, n+1}, {1, 1}, -8];
    Table[a[n], {n, 0, 17}] (* Peter Luschny, Mar 14 2018 *)
    LegendreP[Range[0, 30], 3]^2 (* G. C. Greubel, May 17 2023 *)
  • PARI
    {a(n) = sum(k=0, n, 2^k * binomial(2*k, k)^2 * binomial(n+k, n-k) )}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    {a(n)=polcoeff( 1 / agm(1-x, sqrt((1+x)^2 - 36*x +x*O(x^n))), n)}
    for(n=0, 20, print1(a(n), ", "))
    
  • Python
    from math import comb
    def A243949(n): return sum(comb(n,k)*comb(n+k,k) for k in range(n+1))**2 # Chai Wah Wu, Mar 23 2023
    
  • SageMath
    [gen_legendre_P(n,0,3)^2 for n in range(41)] # G. C. Greubel, May 17 2023

Formula

G.f.: 1 / AGM(1-x, sqrt(1-34*x+x^2)). - Paul D. Hanna, Aug 30 2014
a(n) = Sum_{k=0..n} 2^k * C(2*k, k)^2 * C(n+k, n-k).
a(n)^(1/2) = Sum_{k=0..n} C(2*k, k) * C(n+k, n-k).
Recurrence: n^2*(2*n-3)*a(n) = (2*n-1)*(35*n^2 - 70*n + 26)*a(n-1) - (2*n-3)*(35*n^2 - 70*n + 26)*a(n-2) + (n-2)^2*(2*n-1)*a(n-3). - Vaclav Kotesovec, Aug 18 2014
a(n) ~ (4 + 3*sqrt(2)) * (3 + 2*sqrt(2))^(2*n) / (8*Pi*n). - Vaclav Kotesovec, Aug 18 2014
From Gheorghe Coserea, Jul 05 2016: (Start)
G.f.: hypergeom([1/12, 5/12],[1],27648*x^4*(x^2-34*x+1)*(x-1)^2/(1-36*x+134*x^2-36*x^3+x^4)^3)/(1-36*x+134*x^2-36*x^3+x^4)^(1/4).
0 = x*(x-1)*(x+1)*(x^2-34*x+1)*y'' + (3*x^4-66*x^3-70*x^2+70*x-1)*y' + (x^3-7*x^2-35*x+9)*y, where y is g.f.
(End)
a(n) = Sum_{k = 0..n} 4^k*binomial(n+k,2*k)^2*binomial(2*k,k). - Peter Bala, Mar 02 2017
a(n) = hypergeom([1/2, -n, n + 1], [1, 1], -8). - Peter Luschny, Mar 14 2018
G.f.: Sum_{n >= 0} (2^n)*binomial(2*n,n)^2 *x^n/(1-x)^(2*n+1). - Peter Bala, Feb 07 2022

A243007 a(n) = A084769(n)^2.

Original entry on oeis.org

1, 81, 14641, 3272481, 806616801, 210358905201, 56912554609681, 15800522430616641, 4471485120646226881, 1284238494711502355601, 373195323236525968732401, 109489964937514282794301281, 32378265673661271315300820641, 9639042117142706280223219663281
Offset: 0

Views

Author

Paul D. Hanna, Aug 18 2014

Keywords

Comments

In general, we have the binomial identity:
if b(n) = Sum_{k=0..n} t^k * C(2*k, k) * C(n+k, n-k),
then b(n)^2 = Sum_{k=0..n} (t^2+t)^k * C(2*k, k)^2 * C(n+k, n-k),
where the g.f. of b(n) is 1/sqrt(1 - (4*t+2)*x + x^2),
and the g.f. of b(n)^2 is 1 / AGM(1-x, sqrt((1+x)^2 - (4*t+2)^2*x)), where AGM(x,y) = AGM((x+y)/2,sqrt(x*y)) is the arithmetic-geometric mean.
Note that the g.f. of A084769 is 1/sqrt(1 - 18*x + x^2).
Limit_{n -> oo} a(n+1)/a(n) = (9 + 4*sqrt(5))^2 = 161 + 72*sqrt(5).

Examples

			G.f.: A(x) = 1 + 81*x + 14641*x^2 + 3272481*x^3 + 806616801*x^4 +...
		

Crossrefs

Sequences of the form LegendreP(n, 2*m+1)^2: A000012 (m=0), A243949 (m=1), A243943 (m=2), A243944 (m=3), this sequence (m=4).
Cf. A084769.

Programs

  • Magma
    [Evaluate(LegendrePolynomial(n),9)^2 : n in [0..30]]; // G. C. Greubel, May 17 2023
    
  • Mathematica
    Table[SeriesCoefficient[1/Sqrt[1 -18x +x^2], {x,0,n}], {n,0,20}]^2 (* Vincenzo Librandi, Feb 14 2018 *)
    LegendreP[Range[0,30], 9]^2 (* G. C. Greubel, May 17 2023 *)
  • PARI
    {a(n) = sum(k=0, n, 20^k * binomial(2*k, k)^2 * binomial(n+k, n-k) )}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    {a(n) = sum(k=0, n, 4^k * binomial(2*k, k) * binomial(n+k, n-k) )^2}
    for(n=0, 20, print1(a(n), ", "))
    {a(n)=polcoeff( 1 / agm(1-x, sqrt((1+x)^2 - 18^2*x +x*O(x^n))), n)}
    for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Aug 30 2014
    
  • SageMath
    [gen_legendre_P(n,0,9)^2 for n in range(41)] # G. C. Greubel, May 17 2023

Formula

G.f.: 1 / AGM(1-x, sqrt(1- 322*x + x^2)). - Paul D. Hanna, Aug 30 2014
a(n) = Sum_{k=0..n} 20^k * C(2*k, k)^2 * C(n+k, n-k).
a(n)^(1/2) = Sum_{k=0..n} 4^k * C(2*k, k) * C(n+k, n-k).
a(n) ~ (2 + sqrt(5))^(4*n+2) / (8*sqrt(5)*Pi*n). - Vaclav Kotesovec, Sep 28 2019

A243944 a(n) = A084768(n)^2.

Original entry on oeis.org

1, 49, 5329, 717409, 106523041, 16735820689, 2727812288881, 456250924320961, 77788137919752001, 13459803510972477169, 2356471368269511061009, 416518496068852312607521, 74207592486779379593752801, 13309569813247406938272432721, 2400816685486139045360488325809
Offset: 0

Views

Author

Paul D. Hanna, Aug 18 2014

Keywords

Comments

In general, we have the binomial identity:
if b(n) = Sum_{k=0..n} t^k * C(2*k, k) * C(n+k, n-k),
then b(n)^2 = Sum_{k=0..n} (t*(t+1))^k * C(2*k, k)^2 * C(n+k, n-k),
where the g.f. of b(n) is 1/sqrt(1 - (4*t+2)*x + x^2),
and the g.f. of b(n)^2 is 1 / AGM(1-x, sqrt((1+x)^2 - (4*t+2)^2*x)), where AGM(x,y) = AGM((x+y)/2,sqrt(x*y)) is the arithmetic-geometric mean.
Note that the g.f. of A084768 is 1/sqrt(1 - 14*x + x^2).
Limit a(n+1)/a(n) = (7 + 4*sqrt(3))^2 = 97 + 56*sqrt(3).

Examples

			G.f.: A(x) = 1 + 49*x + 5329*x^2 + 717409*x^3 + 106523041*x^4 +...
		

Crossrefs

Sequences of the form LegendreP(n, 2*m+1)^2: A000012 (m=0), A243949 (m=1), A243943 (m=2), this sequence (m=3), A243007 (m=4).
Cf. A084768.

Programs

  • Magma
    [Evaluate(LegendrePolynomial(n),7)^2 : n in [0..40]]; // G. C. Greubel, May 17 2023
    
  • Mathematica
    Table[Sum[12^k * Binomial[2*k, k]^2 * Binomial[n+k, n-k], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Sep 28 2019 *)
    CoefficientList[Series[2*EllipticK[1 - (1-x)^2/(1 - 194*x + x^2)] / (Pi*Sqrt[1 - 194*x + x^2]), {x, 0, 20}], x] (* Vaclav Kotesovec, Sep 28 2019 *)
    LegendreP[Range[0, 40], 7]^2 (* G. C. Greubel, May 17 2023 *)
  • PARI
    {a(n) = sum(k=0, n, 12^k * binomial(2*k, k)^2 * binomial(n+k, n-k) )}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    {a(n) = sum(k=0, n, 3^k * binomial(2*k, k) * binomial(n+k, n-k) )^2}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    /* Using AGM: */
    {a(n)=polcoeff( 1 / agm(1-x, sqrt((1+x)^2 - 14^2*x +x*O(x^n))), n)}
    for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Aug 30 2014
    
  • SageMath
    [gen_legendre_P(n,0,7)^2 for n in range(41)] # G. C. Greubel, May 17 2023

Formula

G.f.: 1 / AGM(1-x, sqrt(1-194*x+x^2)). - Paul D. Hanna, Aug 30 2014
a(n) = Sum_{k=0..n} 12^k * C(2*k, k)^2 * C(n+k, n-k).
a(n)^(1/2) = Sum_{k=0..n} 3^k * C(2*k, k) * C(n+k, n-k).
a(n) ~ (1+sqrt(3))^(8*n+4) / (sqrt(3) * Pi * n * 2^(4*n+5)). - Vaclav Kotesovec, Sep 28 2019
a(n) = (LegendreP(n, 7))^2. - G. C. Greubel, May 17 2023
Showing 1-4 of 4 results.