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.

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A245925 G.f.: Sum_{n>=0} x^n*Sum_{k=0..n} (-1)^k * C(n,k)^2 * Sum_{j=0..k} C(k,j)^2 * x^j.

Original entry on oeis.org

1, -3, 25, -243, 2601, -29403, 344569, -4141875, 50737129, -630663003, 7930793025, -100681224075, 1288236350025, -16592960274075, 214939203248025, -2797935722568243, 36578032462268649, -480000660000226875, 6320012816203363489, -83462977778600141643, 1105193229806740453201
Offset: 0

Views

Author

Paul D. Hanna, Aug 15 2014

Keywords

Comments

The g.f.s formed from a(2*n)^(1/2) and (-a(2*n+1)/3)^(1/2) are:
A245926: sqrt( (1-x + sqrt(1-14*x+x^2)) / (2*(1-14*x+x^2)) );
A245927: sqrt( (1-x - sqrt(1-14*x+x^2)) / (6*x*(1-14*x+x^2)) ).
Lim_{n->infinity} a(n+1)/a(n) = -(7 + 4*sqrt(3)).

Examples

			G.f.: A(x) = 1 - 3*x^2 + 25*x^4 - 243*x^6 + 2601*x^8 - 29403*x^10 + ...
where the g.f. is given by the binomial series:
A(x) = 1 + x*(1 - (1+x)) + x^2*(1 - 2^2*(1+x) + (1+2^2*x+x^2))
+ x^3*(1 - 3^2*(1+x) + 3^2*(1+2^2*x+x^2) - (1+3^2*x+3^2*x^2+x^3))
+ x^4*(1 - 4^2*(1+x) + 6^2*(1+2^2*x+x^2) - 4^2*(1+3^2*x+3^2*x^2+x^3) + (1+4^2*x+6^2*x^2+4^2*x^3+x^4))
+ x^5*(1 - 5^2*(1+x) + 10^2*(1+2^2*x+x^2) - 10^2*(1+3^2*x+3^2*x^2+x^3) + 5^2*(1+4^2*x+6^2*x^2+4^2*x^3+x^4) - (1+5^2*x+10^2*x^2+10^2*x^3+5^2*x^4+x^5))
+ x^6*(1 - 6^2*(1+x) + 15^2*(1+2^2*x+x^2) - 20^2*(1+3^2*x+3^2*x^2+x^3) + 15^2*(1+4^2*x+6^2*x^2+4^2*x^3+x^4) - 6^2*(1+5^2*x+10^2*x^2+10^2*x^3+5^2*x^4+x^5) + (1+6^2*x+15^2*x^2+20^2*x^3+15^2*x^4+6^2*x^5+x^6)) + ...
in which the coefficients of odd powers of x vanish.
We can also express the g.f. by the binomial series identity:
A(x) = 1/(1+x) + x/(1+x)^3*(1-x)^2 + x^2/(1+x)^5*(1 - 2^2*x + x^2)^2
+ x^3/(1+x)^7*(1 - 3^2*x + 3^2*x^2 - x^3)^2
+ x^4/(1+x)^9*(1 - 4^2*x + 6^2*x^2 - 4^2*x^3 + x^4)^2
+ x^5/(1+x)^11*(1 - 5^2*x + 10^2*x^2 - 10^2*x^3 + 5^2*x^4 - x^5)^2
+ x^6/(1+x)^13*(1 - 6^2*x + 15^2*x^2 - 20^2*x^3 + 15^2*x^4 - 6^2*x^5 + x^6)^2 + ...
		

Crossrefs

Programs

  • Maple
    A245925 := n -> (-1)^n*add(binomial(2*(n-k), n-k)*binomial(2*n-k, k)^2, k=0..n); seq(A245925(n), n=0..20); # Peter Luschny, Aug 17 2014
  • Mathematica
    Table[Sum[Sum[(-1)^(j+k) * Binomial[2*n - k, j + k]^2 * Binomial[j + k, k]^2, {j, 0, 2*n - 2*k}], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Aug 16 2014 after Paul D. Hanna *)
    a[n_] := (-1)^n*HypergeometricPFQ[{-n, -n, n + 1, n + 1}, {1/2, 1, 1}, 1/4];
    Table[a[n], {n, 0, 20}] (* Peter Luschny, Mar 14 2018 *)
  • PARI
    /* By definition: */
    {a(n)=polcoeff(sum(m=0, n, x^m*sum(k=0, m, (-1)^k*binomial(m, k)^2*sum(j=0, k, binomial(k, j)^2*x^j)+x*O(x^n))), n)}
    for(n=0, 20, print1(a(2*n), ", "))
    
  • PARI
    /* From alternate g.f.: */
    {a(n)=local(A=1);A=sum(m=0,n,x^m/(1+x)^(2*m+1)*sum(k=0,m,binomial(m,k)^2*(-x)^k)^2+x*O(x^n));polcoeff(A,n)}
    for(n=0,20,print1(a(2*n),", "))
    
  • PARI
    /* From formula for a(n); printing only nonzero terms: */
    {a(n)=sum(k=0, n\2, sum(j=0, n-2*k, (-1)^(j+k)*binomial(n-k, j+k)^2*binomial(j+k, k)^2))}
    for(n=0, 20, print1(a(2*n), ", "))
    
  • PARI
    /* From formula for a(n) (nonzero terms): */
    {a(n)=sum(k=0, n, sum(j=0, 2*n-2*k, (-1)^(j+k)*binomial(2*n-k,j+k)^2*binomial(j+k, k)^2))}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    /* Formula for a(n), after Peter Luschny and Robert Israel: */
    {a(n) = (-1)^n * sum(k=0,n, binomial(2*k, k) * binomial(n+k, n-k)^2)}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    /* Simpler formula for a(n): */
    {a(n) = sum(k=0, n, (-1)^k * binomial(2*k, k)^2 * binomial(n+k, n-k) )}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    /* Using AGM: */
    {a(n)=polcoeff( 1 / agm(1-x, sqrt(1+14*x+x^2 +x*O(x^n))), n)}
    for(n=0,20,print1(a(n),", "))
    
  • Sage
    A245925 = lambda n: (-1)^n*sum(binomial(2*(n-k), n-k)*binomial(2*n-k, k)^2 for k in (0..n))
    [A245925(n) for n in range(21)] # Peter Luschny, Aug 17 2014

Formula

G.f.: Sum_{n>=0} x^n / (1+x)^(2*n+1) * ( Sum_{k=0..n} C(n,k)^2*(-x)^k )^2.
G.f.: 1 / AGM(1-x^2, sqrt(1+14*x^2+x^4)), where AGM(x,y) = AGM((x+y)/2, sqrt(x*y)) is the arithmetic-geometric mean.
a(2*n) = A245926(n)^2.
a(2*n+1) = (-3)*A245927(n)^2.
a(n) = Sum_{k=0..n} Sum_{j=0..2*n-2*k} (-1)^(j+k) * C(2*n-k,j+k)^2 * C(j+k,k)^2.
D-finite with recurrence: n^2*(2*n-3)*a(n) = -(2*n-1)*(13*n^2 - 26*n + 10)*a(n-1) + (2*n-3)*(13*n^2 - 26*n + 10)*a(n-2) + (n-2)^2*(2*n-1)*a(n-3). - Vaclav Kotesovec, Aug 16 2014
a(n) ~ (-1)^n * (2+sqrt(3)) * (7+4*sqrt(3))^n / (4*Pi*n). - Vaclav Kotesovec, Aug 16 2014
a(n) = (-1)^n*Sum_{k=0..n} binomial(2*(n-k), n-k)*binomial(2*n-k, k)^2. - Peter Luschny, Aug 17 2014
a(n) = (-1)^n*binomial(2*n,n)*hyper4F3([-n,-n,-n,-n+1/2],[1,-2*n,-2*n], 4). - Peter Luschny, Aug 17 2014
a(n) = Sum_{k=0..n} (-1)^k * C(2*k, k)^2 * C(n+k, n-k). - Paul D. Hanna, Aug 17 2014
a(n) = (-1)^n*hypergeom([-n, -n, n + 1, n + 1], [1/2, 1, 1], 1/4). - Peter Luschny, Mar 14 2018
a(n) = Legendre_P(n, sqrt(-3))^2. - Peter Bala, Dec 22 2020
G.f.: Sum_{n >= 0} (-1)^n*binomial(2*n,n)^2*x^n/(1-x)^(2*n+1). - Peter Bala, Feb 07 2022
From Peter Bala, Apr 05 2022: (Start)
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n-k,k)*binomial(2*n-2*k,n-k)^2.
a(n) = (-1)^n * Sum_{k = 0..n} binomial(2*k,k)*binomial(n+k,n-k)^2.
a(n) = (-1)^n*binomial(2*n,n)^2*hypergeom([-n,-n,-n,], [-2*n,-n+1/2], 1/4). (End)

A245926 Expansion of g.f. sqrt( (1-x + sqrt(1-14*x+x^2)) / (2*(1-14*x+x^2)) ).

Original entry on oeis.org

1, 5, 51, 587, 7123, 89055, 1135005, 14660805, 191253843, 2513963567, 33244446601, 441772827105, 5894323986301, 78912561223553, 1059543126891027, 14261959492731387, 192392702881384275, 2600355510685245087, 35206018016510388345, 477377227987055971905
Offset: 0

Views

Author

Paul D. Hanna, Aug 15 2014

Keywords

Comments

Square each term to form a bisection of A245925.
Limit a(n+1)/a(n) = 7 + 4*sqrt(3).

Examples

			G.f.: A(x) = 1 + 5*x + 51*x^2 + 587*x^3 + 7123*x^4 + 89055*x^5 +...
where
A(x)^2 = (1-x + sqrt(1-14*x+x^2)) / (2*(1-14*x+x^2)).
Explicitly,
A(x)^2 = 1 + 10*x + 127*x^2 + 1684*x^3 + 22717*x^4 + 309214*x^5 + 4231675*x^6 + 58117672*x^7 + 800173945*x^8 +...+ A245923(n)*x^n +...
		

Crossrefs

Column k=3 of A337389.

Programs

  • Maple
    A245926 := n -> sqrt(add(binomial(4*n-2*k, 2*n-k)*binomial(4*n-k, k)^2, k=0..2*n)); seq(A245926(n), n=0..20); # Peter Luschny, Aug 17 2014
  • Mathematica
    CoefficientList[Series[Sqrt[(1 - x + Sqrt[1 - 14*x + x^2])/(2*(1 - 14*x + x^2))], {x,0,50}], x] (* G. C. Greubel, Jan 29 2017 *)
    a[n_] := (-1)^n Hypergeometric2F1[-n, n + 1/2, 1, 4];
    Table[a[n], {n, 0, 19}] (* Peter Luschny, Mar 16 2018 *)
  • PARI
    /* From definition: */
    {a(n)=polcoeff( sqrt( (1-x + sqrt(1-14*x+x^2 +x*O(x^n))) / (2*(1-14*x+x^2 +x*O(x^n))) ), n)}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    /* From formula for a(n)^2: */
    {a(n)=sqrtint(sum(k=0, 2*n, sum(j=0, 4*n-2*k, (-1)^(j+k)*binomial(4*n-k,j+k)^2*binomial(j+k, k)^2)))}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    /* From formula for a(n)^2: */
    {a(n) = sqrtint( sum(k=0, 2*n, binomial(2*k, k)^2*binomial(2*n+k, 2*n-k)*(-1)^k) )}
    for(n=0, 20, print1(a(n), ", "))

Formula

a(n)^2 = Sum_{k=0..2*n} Sum_{j=0..4*n-2*k} (-1)^(j+k) * C(4*n-k,j+k)^2 * C(j+k,k)^2.
a(n) ~ (3*sqrt(3)-5) * (7+4*sqrt(3))^(n+1) / (4*sqrt(Pi*n)). - Vaclav Kotesovec, Aug 16 2014
a(n)^2 = C(4*n,2*n)*hyper4F3([-2*n,-2*n,-2*n,-2*n+1/2],[1,-4*n,-4*n],4). - Peter Luschny, Aug 17 2014
a(n)^2 = Sum_{k=0..2*n} binomial(4*n-2*k, 2*n-k)*binomial(4*n-k, k)^2. - Peter Luschny, Aug 17 2014
a(n)^2 = Sum_{k=0..2*n} (-1)^k * C(2*k, k)^2 * C(2*n+k, 2*n-k). - Paul D. Hanna, Aug 17 2014
From Peter Bala, Mar 14 2018: (Start)
a(n) = (-1)^n*P(2*n,sqrt(-3)), where P(n,x) denotes the n-th Legendre polynomial. See A008316.
a(n) = (1/C(2*n,n))*Sum_{k = 0..n} (-1)^(n+k)*C(n,k)*C(n+k,k)* C(2*n+2*k,n+k) = Sum_{k = 0..n} (-1)^(n+k)*C(2*k,k)*C(n,k) *C(2*n+2*k,2*n)/C(n+k,n). In general, P(2*n,sqrt(1+4*x)) = (1/C(2*n,n))*Sum_{k = 0..n} C(n,k)*C(n+k,k)*C(2*n+2*k,n+k) *x^k.
a(n) = Sum_{k = 0..2*n} C(2*n,k)^2 * u^(n-k), where u = (1 - sqrt(-3))/2 is a primitive sixth root of unity.
a(n) = (-1)^n*Sum_{k = 0..2*n} C(2*n,k)*C(2*n+k,k)*u^(2*k).
(End)
a(n) = (-1)^n*hypergeom([-n, n + 1/2], [1], 4). - Peter Luschny, Mar 16 2018
a(0) = 1, a(1) = 5 and n * (2*n-1) * (4*n-5) * a(n) = (4*n-3) * (28*n^2-42*n+9) * a(n-1) - (n-1) * (2*n-3) * (4*n-1) * a(n-2) for n > 1. - Seiichi Manyama, Aug 28 2020
From Peter Bala, May 03 2022: (Start)
Conjecture: a(n) = [x^n] ( (1 + x + x^2)*(1 + x)^2/(1 - x)^2 )^n. Equivalently, a(n) = Sum_{k = 0..n} Sum_{i = 0..k} Sum_{j = 0..n-k-i} C(n, k)*C(k, i)*C(2*n, j)*C(3*n-k-i-j-1, n-k-i-j).
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. Calculation suggests that the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for primes p >= 5 and positive integers n and k. (End)

A243947 Expansion of g.f. sqrt( (1+x - sqrt(1-18*x+x^2)) / (10*x*(1-18*x+x^2)) ).

Original entry on oeis.org

1, 11, 155, 2365, 37555, 610897, 10098997, 168894355, 2849270515, 48395044705, 826479148001, 14177519463191, 244109912494525, 4216385987238575, 73024851218517275, 1267712063327871245, 22052786911315216595, 384321597582115655825, 6708530714274563938225
Offset: 0

Views

Author

Paul D. Hanna, Aug 17 2014

Keywords

Comments

Multiply the square of each term by 5 to form a bisection of A243945.
Limit_{n->oo} a(n+1)/a(n) = 9 + 4*sqrt(5).

Examples

			G.f.: A(x) = 1 + 11*x + 155*x^2 + 2365*x^3 + 37555*x^4 + 610897*x^5 + ...
where
A(x)^2 = (1+x - sqrt(1-18*x+x^2)) / (10*x*(1-18*x+x^2)).
		

Crossrefs

Programs

  • Maple
    seq(add(1/2*binomial(2*k+1,k)*binomial(n,k)*binomial(2*n+2*k+2,2*n+1)/binomial(n+k+1,n), k = 0..n), n = 0..20); # Peter Bala, Mar 15 2018
  • Mathematica
    CoefficientList[Series[Sqrt[((1+x-Sqrt[1-18x+x^2]))/(10x(1-18x+x^2))],{x,0,20}],x] (* Harvey P. Dale, Jul 31 2016 *)
    a[n_] := Hypergeometric2F1[-n, n + 3/2, 1, -4];
    Table[a[n], {n, 0, 18}] (* Peter Luschny, Mar 16 2018 *)
  • PARI
    /* From definition: */
    {a(n)=polcoeff( sqrt( (1+x - sqrt(1-18*x+x^2 +x^2*O(x^n))) / (10*x*(1-18*x+x^2 +x*O(x^n))) ), n)}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    /* From a(n) = sqrt( A243945(2*n+1)/5 ): */
    {a(n)=sqrtint( (1/5)*sum(k=0, 2*n+1, binomial(2*k, k)^2*binomial(2*n+k+1, 2*n-k+1)) )}
    for(n=0, 20, print1(a(n), ", "))
    
  • Python
    from math import comb
    def A243947(n): return sum(5**(n-k)*comb(m:=k<<1,k)*comb((n<<1)+1,m) for k in range(n+1)) # Chai Wah Wu, Mar 23 2023

Formula

a(n)^2 = (1/5) * Sum_{k=0..2*n+1} C(2*k, k)^2 * C(2*n+k+1, 2*n-k+1).
a(n) ~ (9+4*sqrt(5))^(n+1) / (2*5^(1/4)*sqrt(2*Pi*n) * sqrt(5+2*sqrt(5))). - Vaclav Kotesovec, Aug 18 2014. Equivalently, a(n) ~ phi^(6*n + 9/2) / (2^(3/2) * sqrt(5*Pi*n)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 08 2021
From Peter Bala, Mar 15 2018: (Start)
a(n) = (1/sqrt(5))*P(2*n+1,sqrt(5)), where P(n,x) denotes the n-th Legendre polynomial. See A008316.
a(n) = Sum_{k = 0..n} (1/2)*C(2*k+1,k)*C(n,k)*C(2*n+2*k+2,2*n+1)/C(n+k+1,n). In general, (1/sqrt(1 + 4*x))*P(2*n+1,sqrt(1+4*x)) = (1/(2*C(2*n+1,n))) * Sum_{k = 0..n} C(n,k)*C(n+k+1,k)*C(2*n+2*k+2,n+k+1)*x^k.
a(n) = (1/sqrt(5))*Sum_{k = 0..2*n+1} C(2*n+1,k)^2 * phi^(2*n-2*k+1), where phi = (sqrt(5) + 1)/2.
a(n) = (1/sqrt(5))*Sum_{k = 0..2*n+1} C(2*n+1,k)*C(2*n+1+k,k) * Phi^k, where Phi = (sqrt(5) - 1)/2. (End)
a(n) = hypergeom([-n, n + 3/2], [1], -4). - Peter Luschny, Mar 16 2018
From Peter Bala, Mar 17 2018: (Start)
a(n) = Sum_{k = 0..n} C(2*n+1,2*k)*C(2*k,k)*5^(n-k).
D-finite with recurrence: n*(4*n-3)*(2*n+1)*a(n) = (4*n-1)*(36*n^2-18*n-7)*a(n-1) - (n-1)*(2*n-1)*(4*n+1)*a(n-2). (End)

A300946 Rectangular array A(n, k) = (-1)^k*hypergeom([-k, k + n/2 - 1/2], [1], 4) with row n >= 0 and k >= 0, read by ascending antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 19, 1, 5, 33, 239, 1, 7, 51, 387, 3011, 1, 9, 73, 587, 4737, 38435, 1, 11, 99, 847, 7123, 59523, 496365, 1, 13, 129, 1175, 10321, 89055, 761121, 6470385, 1, 15, 163, 1579, 14499, 129367, 1135005, 9854211, 84975315
Offset: 0

Views

Author

Peter Luschny, Mar 16 2018

Keywords

Examples

			Array starts:
[0] 1,  1,  19,  239,  3011,  38435,  496365,  6470385, ... [A299864]
[1] 1,  3,  33,  387,  4737,  59523,  761121,  9854211, ... [A299507]
[2] 1,  5,  51,  587,  7123,  89055, 1135005, 14660805, ... [A245926]
[3] 1,  7,  73,  847, 10321, 129367, 1651609, 21360031, ... [A084768]
[4] 1,  9,  99, 1175, 14499, 183195, 2351805, 30539241, ... [A245927]
[5] 1, 11, 129, 1579, 19841, 253707, 3284737, 42924203, ...
[6] 1, 13, 163, 2067, 26547, 344535, 4508877, 59402397, ...
		

Crossrefs

Programs

  • Mathematica
    Arow[n_, len_] := Table[(-1)^k Hypergeometric2F1[-k, k + n/2 - 1/2, 1, 4], {k, 0, len}]; Table[Print[Arow[n, 7]], {n, 0, 6}];

A337369 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt(2 / ( (1-2*(k+4)*x+((k-4)*x)^2) * (1+(k-4)*x+sqrt(1-2*(k+4)*x+((k-4)*x)^2)) )).

Original entry on oeis.org

1, 1, 6, 1, 7, 30, 1, 8, 51, 140, 1, 9, 74, 393, 630, 1, 10, 99, 736, 3139, 2772, 1, 11, 126, 1175, 7606, 25653, 12012, 1, 12, 155, 1716, 14499, 80464, 212941, 51480, 1, 13, 186, 2365, 24310, 183195, 864772, 1787607, 218790, 1, 14, 219, 3128, 37555, 352716, 2351805, 9400192, 15134931, 923780
Offset: 0

Views

Author

Seiichi Manyama, Aug 25 2020

Keywords

Examples

			Square array begins:
     1,     1,     1,      1,      1,      1, ...
     6,     7,     8,      9,     10,     11, ...
    30,    51,    74,     99,    126,    155, ...
   140,   393,   736,   1175,   1716,   2365, ...
   630,  3139,  7606,  14499,  24310,  37555, ...
  2772, 25653, 80464, 183195, 352716, 610897, ...
		

Crossrefs

Columns k=0..5 give A002457, A273055, A337370, A245927, A002458, A243947.
Main diagonal gives A337387.

Programs

  • Mathematica
    T[n_, k_] := Sum[If[k == 0, Boole[n == j], k^(n - j)] * Binomial[2*j, j] * Binomial[2*n + 1, 2*j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, Aug 25 2020 *)
  • PARI
    {T(n, k) = sum(j=0, n, k^(n-j)*binomial(2*j, j)*binomial(2*n+1, 2*j))}

Formula

T(n,k) = Sum_{j=0..n} k^(n-j) * binomial(2*j,j) * binomial(2*n+1,2*j).
T(0,k) = 1, T(1,k) = k+6 and n * (2*n+1) * (4*n-3) * T(n,k) = (4*n-1) * (4*(k+4)*n^2-2*(k+4)*n-k-2) * T(n-1,k) - (k-4)^2 * (n-1) * (2*n-1) * (4*n+1) * T(n-2,k) for n > 1. - Seiichi Manyama, Aug 29 2020
For fixed k > 0, T(n,k) ~ (2 + sqrt(k))^(2*n + 3/2) / sqrt(8*k*Pi*n). - Vaclav Kotesovec, Aug 31 2020

A245924 Expansion of (1-x - sqrt(1 - 14*x + x^2)) / (6*x*(1 - 14*x + x^2)).

Original entry on oeis.org

1, 18, 279, 4132, 59949, 860022, 12252547, 173756232, 2456093529, 34634926810, 487525847535, 6852798238572, 96216461002117, 1349689029354558, 18918661407653979, 265016591806251664, 3710426585319049905, 51924984423522889122, 726369947645489367751, 10157588028419864394420
Offset: 0

Views

Author

Paul D. Hanna, Aug 16 2014

Keywords

Comments

Self-convolution of A245927.
Limit a(n+1)/a(n) = 7 + 4*sqrt(3).

Examples

			G.f.: A(x) = 1 + 18*x + 279*x^2 + 4132*x^3 + 59949*x^4 + 860022*x^5 +...
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1 - x - Sqrt[1 - 14*x + x^2])/(6*x*(1 - 14*x + x^2)), {x,0,50}], x] (* G. C. Greubel, Feb 14 2017 *)
  • PARI
    {a(n)=polcoeff( (1-x - sqrt(1-14*x+x^2 +x^2*O(x^n))) / (6*x*(1-14*x+x^2 +x*O(x^n))), n)}
    for(n=0, 20, print1(a(n), ", "))

Formula

a(n) ~ (26 + 15*sqrt(3)) * (7 + 4*sqrt(3))^n / 24 * (1 - 1/(3^(1/4)*sqrt(Pi*n/2))). - Vaclav Kotesovec, Aug 17 2014
D-finite with recurrence: (n+1)*a(n) +7*(-4*n-1)*a(n-1) +99*(2*n-1)*a(n-2) +7*(-4*n+5)*a(n-3) +(n-2)*a(n-4)=0. - R. J. Mathar, Jan 23 2020

A299507 a(n) = (-1)^n*hypergeom([-n, n], [1], 4).

Original entry on oeis.org

1, 3, 33, 387, 4737, 59523, 761121, 9854211, 128772609, 1694927619, 22437369633, 298419470979, 3984500221569, 53376363001731, 717044895641121, 9656091923587587, 130310873022310401, 1761872309456567811, 23861153881099854369, 323634591584064809859
Offset: 0

Views

Author

Peter Luschny, Mar 16 2018

Keywords

Crossrefs

Programs

  • Maple
    seq(simplify( (-1)^n*hypergeom([-n, n], [1], 4)), n = 0..20); # Peter Bala, Apr 18 2024
  • Mathematica
    a[n_] := (-1)^n Hypergeometric2F1[-n, n, 1, 4]; Table[a[n], {n, 0, 19}]

Formula

From Vaclav Kotesovec, Jul 05 2018: (Start)
Recurrence: n*(2*n-3)*a(n) = 2*(14*n^2 - 28*n + 11)*a(n-1) - (n-2)*(2*n-1)*a(n-2).
a(n) ~ 2^(-3/2) * 3^(1/4) * (7 + 4*sqrt(3))^n / sqrt(Pi*n). (End)
From Peter Bala, Apr 18 2024: (Start)
a(n) = Sum_{k = 0..n} binomial(n, k)*binomial(n+k-1, k-1)*3^k = R(n, 3) for n >= 1, where R(n, x) denotes the n-th row polynomial of A253283.
a(n) = 3*n* hypergeom([1 - n, n + 1], [2], -3) for n >= 1.
a(n) = (1/2)*(LegendreP(n, 7) - LegendreP(n-1, 7)) for n >= 1.
a(n) = [x^n] ( (1 - x)/(1 - 4*x) )^n.
It follows that the Gauss congruences hold: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
G.f.: (sqrt(x^2 - 14*x + 1) - x + 1)/(2*sqrt(x^2 - 14*x + 1)) = 1 + 3*x + 33*x^2 + 387*x^3 + .... (End)

A299864 a(n) = (-1)^n*hypergeom([-n, n - 1/2], [1], 4).

Original entry on oeis.org

1, 1, 19, 239, 3011, 38435, 496365, 6470385, 84975315, 1122708899, 14906800361, 198740733581, 2658870294349, 35677678567549, 479965685669059, 6471364940381007, 87425255326277907, 1183139999323074963, 16036589185819644633, 217668383345249016045
Offset: 0

Views

Author

Peter Luschny, Mar 16 2018

Keywords

Crossrefs

Programs

  • Maple
    seq((-1)^n*orthopoly[P](n,0,-3/2,-7),n=0..100); # Robert Israel, Mar 21 2018
  • Mathematica
    a[n_] := (-1)^n Hypergeometric2F1[-n, n - 1/2, 1, 4]; Table[a[n], {n, 0, 19}]

Formula

From Robert Israel, Mar 21 2018: (Start)
a(n) = JacobiP(n,0,-3/2,-7).
n*(2*n-3)*(4*n-7)*a(n)+(2*n-5)*(n-1)*(4*n-3)*a(n-2)-(4*n-5)*(28*n^2-70*n+39)*a(n-1) = 0. (End)
a(n) ~ sqrt(3) * (1 + sqrt(3))^(4*n - 1) / (2^(2*n + 1) * sqrt(Pi*n)). - Vaclav Kotesovec, Jul 05 2018

A337467 Expansion of sqrt(2 / ( (1-2*x+49*x^2) * (1-7*x+sqrt(1-2*x+49*x^2)) )).

Original entry on oeis.org

1, 3, -21, -139, 531, 6489, -9723, -292293, -135117, 12514313, 29905809, -501239553, -2310673379, 18245192679, 140574917259, -562805403867, -7557237645741, 11275709877369, 371974318253601, 201852054629631, -16932135947326551, -42530838930147813, 709138646702505999
Offset: 0

Views

Author

Seiichi Manyama, Aug 28 2020

Keywords

Crossrefs

Column k=3 of A337464.

Programs

  • Mathematica
    a[n_] := Sum[(-3)^(n-k) * Binomial[2*k, k] * Binomial[2*n+1, 2*k], {k, 0, n}]; Array[a, 23, 0] (* Amiram Eldar, Apr 29 2021 *)
  • PARI
    N=40; x='x+O('x^N); Vec(sqrt(2/((1-2*x+49*x^2)*(1-7*x+sqrt(1-2*x+49*x^2)))))
    
  • PARI
    {a(n) = sum(k=0, n, (-3)^(n-k)*binomial(2*k, k)*binomial(2*n+1, 2*k))}

Formula

a(n) = Sum_{k=0..n} (-3)^(n-k) * binomial(2*k,k) * binomial(2*n+1,2*k).
a(0) = 1, a(1) = 3 and n * (2*n+1) * (4*n-3) * a(n) = (4*n-1) * (4*n^2-2*n+1) * a(n-1) - 49 * (n-1) * (2*n-1) * (4*n+1) * a(n-2) for n > 1. - Seiichi Manyama, Aug 29 2020
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