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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A142979 a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1) + (n+1)^2*a(n).

Original entry on oeis.org

1, 3, 13, 66, 406, 2868, 23220, 210192, 2116656, 23375520, 281792160, 3673814400, 51599514240, 775673176320, 12440524320000, 211848037632000, 3820318338816000, 72685037892096000, 1455838255452672000
Offset: 1

Views

Author

Peter Bala, Jul 17 2008

Keywords

Comments

This is the case m = 1 of the more general recurrence a(1) = 1, a(2) = 2*m + 1, a(n+2) = (2*m + 1)*a(n+1) + (n + 1)^2*a(n) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of Mercator's series for the constant log(2). For other cases see A024167 (m = 0), A142980 (m = 2), A142981 (m = 3) and A142982 (m = 4).
The solution to the general recurrence may be expressed as a sum: a(n) = n!*p_m(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p_m(k-1)*p_m(k)), where p_m(x) = Sum_{k = 0..m} 2^k*C(m,k)*C(x,k) = Sum_{k = 0..m} C(m,k)*C(x+k,m), is the Ehrhart polynomial of the m-dimensional cross polytope (the hyperoctahedron).
The first few values are p_0(x) = 1, p_1(x) = 2*x + 1, p_2(x) = 2*x^2 + 2*x + 1 and p_3(x) = (4*x^3 + 6*x^2 + 8*x + 3)/3.
The sequence {p_m(k)},k>=0 is the crystal ball sequence for the product lattice A_1 x... x A_1 (m copies). The table of values [p_m(k)]m,k>=0 is the array of Delannoy numbers A008288.
The polynomial p_m(x) is the unique polynomial solution of the difference equation (x + 1)*f(x+1) - x*f(x-1) = (2*m + 1)*f(x), normalized so that f(0) = 1. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane, that is, the polynomials p_m(x-1), m = 1,2,3,..., satisfy a Riemann hypothesis [BUMP et al., Theorems 4 and 6]. The o.g.f. for the p_m(x) is (1 + t)^x/(1 - t)^(x+1) = 1 + (2*x + 1)*t + (2*x^2 + 2*x + 1)*t^2 + ....
The general recurrence in the first paragraph above also has a second solution b(n) = n!*p_m(n) with initial conditions b(1) = 2*m + 1, b(2) = (2*m + 1)^2 + 1.
Hence the behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k*p_m(k-1)*p_m(k)) = 1/((2*m + 1) + 1^2/((2*m + 1) + 2^2/((2*m + 1) + 3^2/((2*m + 1) + ... + n^2/((2*m + 1) + ...))))) = (-1)^m * (log(2) - (1 - 1/2 + 1/3 - ... + (-1)^(m+1)/m)), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 29]).
For other sequences defined by similar recurrences and related to log(2) see A142983 and A142988. See also A142992 for the connection between log(2) and the C_n lattices. For corresponding results for the constants e, zeta(2) and zeta(3) see A000522, A142995 and A143003 respectively.

References

  • Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Crossrefs

Programs

  • Maple
    p := n -> 2*n+1: a := n -> n!*p(n)*sum ((-1)^(k+1)/(k*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 1..20)
  • Mathematica
    RecurrenceTable[{a[1]==1,a[2]==3,a[n+2]==3a[n+1]+(n+1)^2 a[n]},a,{n,20}] (* Harvey P. Dale, May 20 2012 *)

Formula

a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = 2*n + 1.
Recurrence: a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1) + (n + 1)^2*a(n).
The sequence b(n):= n!*p(n) satisfies the same recurrence with b(1) = 3, b(2) = 10.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(3 + 1^2/(3 + 2^2/(3 + 3^2/(3 + ... + (n-1)^2/3)))), for n >= 2.
Limit_{n -> oo} a(n)/b(n) = 1/(3 + 1^2/(3 + 2^2/(3 + 3^2/(3 + ... + (n-1)^2/(3 + ...))))) = Sum_{k >= 1} (-1)^(k+1)/(k*(4k^2 - 1)) = 1 - log(2).
Thus a(n) ~ c*n*n! as n -> oo, where c = 2*(1 - log(2)).
From Peter Bala, Dec 09 2024: (Start)
E.g.f.: A(x) = (2*x - (1 + x)*log(1 + x))/(1 - x)^2 satisfies the differential equation 1 + (x + 3)*A(x) + (x^2 - 1)*A'(x) = 0 with A(0) = 0.
Sum_{k = 1..n} Stirling_2(n, k) * a(k) = A317057(n+1). (End)

A142983 a(1) = 1, a(2) = 2, a(n+2) = 2*a(n+1) + (n + 1)*(n + 2)*a(n).

Original entry on oeis.org

1, 2, 10, 44, 288, 1896, 15888, 137952, 1419840, 15255360, 186693120, 2387093760, 33898314240, 502247692800, 8123141376000, 136785729024000, 2483065912320000, 46822564905984000, 942853671825408000, 19678282007924736000, 435355106182520832000
Offset: 1

Views

Author

Peter Bala, Jul 17 2008

Keywords

Comments

This is the case m = 1 of the general recurrence a(1) = 1, a(2) = 2*m, a(n+2) = 2*m*a(n+1) + (n + 1)*(n + 2)*a(n) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series 1/2 + 1/2*Sum_{k > 1} (-1)^(k+1)/(k*(k + 1)) = log(2). For other cases see A142984 (m = 2), A142985 (m = 3), A142986 (m = 4) and A142987 (m = 5).
The solution to the general recurrence may be expressed as a sum: a(n) = n!*p_m(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p_m(k)*p_m(k+1)), where p_m(x) = Sum_{k = 1..m} 2^(k-1)*C(m-1,k-1)*C(x,k) is the polynomial that gives the regular polytope numbers for the m-dimensional cross polytope as defined by [Kim] (see A142978). The first few values are p_1(x) = x, p_2(x) = x^2, p_3(x) = (2*x^3 + x)/3 and p_4(x) = (x^4 + 2*x^2)/3.
The polynomial p_m(x) is the unique polynomial solution of the difference equation x*(f(x+1) - f(x-1)) = 2*m*f(x), normalized so that f(1) = 1.
The o.g.f. for the p_m(x) is 1/2*((1 + t)/(1 - t))^x = 1/2 + x*t + x^2*t^2 + (2*x^3 + x)/3*t^3 + .... Thus p_m(x) is, apart from a constant factor, the Meixner polynomial of the first kind M_m(x;b,c) at b = 0, c = -1, also known as a Mittag-Leffler polynomial.
The general recurrence in the first paragraph above has a second solution b(n) = n!*p_m(n+1) with b(1) = 2*m, b(2) = m^2 + 2. Hence the behavior of a(n) for large n is given by Limit_{n-> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(p_m(k)*p_m(k+1)) = 1/((2*m) + 1*2/((2*m) + 2*3/((2*m) + 3*4/((2*m) + ... + n*(n + 1)/((2*m) + ...))))) = 1 + (-1)^(m+1) * (2*m)*(log(2) - (1 - 1/2 + 1/3 - ... + (-1)^(m+1)/m)), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).
See A142979, A142988 and A142992 for similar results. For corresponding results for Napier's constant e, the constant zeta(2) and Apery's constant zeta(3) refer to A000522, A142995 and A143003, respectively.

References

  • Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Crossrefs

Programs

  • Haskell
    a142983 n = a142983_list !! (n-1)
    a142983_list = 1 : 2 : zipWith (+)
                           (map (* 2) $ tail a142983_list)
                           (zipWith (*) (drop 2 a002378_list) a142983_list)
    -- Reinhard Zumkeller, Jul 17 2015
  • Maple
    a := n -> (n+1)!*sum ((-1)^(k+1)/(k*(k+1)), k = 1..n): seq(a(n), n = 1..20);
  • Mathematica
    Rest[CoefficientList[Series[(-x+2*Log[x+1])/(x-1)^2,{x,0,20}],x]*Range[0,20]!] (* Vaclav Kotesovec, Oct 21 2012 *)

Formula

a(n) = n!*p(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = n.
Recurrence: a(1) = 1, a(2) = 2, a(n+2) = 2*a(n+1) + (n + 1)*(n + 2)*a(n).
The sequence b(n) := n!*p(n+1) satisfies the same recurrence with b(1) = 2 and b(2) = 6.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(2 + 1*2/(2 + 2*3/(2 + 3*4/(2 + ... + (n - 1)*n/2)))), for n >= 2.
The behavior of a(n) for large n is given by Limit_{n -> oo} a(n)/b(n) = 1/(2 + 1*2/(2 + 2*3/(2 + 3*4/(2 + ... + n*(n+1)/(2 + ...))))) = Sum_{k >= 1} (-1)^(k+1)/(k*(k + 1)) = 2*log(2) - 1.
E.g.f.: (2*log(x+1)-x)/(x-1)^2. - Vaclav Kotesovec, Oct 21 2012

A145905 Square array read by antidiagonals: Hilbert transform of triangle A060187.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 9, 5, 1, 1, 27, 25, 7, 1, 1, 81, 125, 49, 9, 1, 1, 243, 625, 343, 81, 11, 1, 1, 729, 3125, 2401, 729, 121, 13, 1, 1, 2187, 15625, 16807, 6561, 1331, 169, 15, 1, 1, 6561, 78125, 117649, 59049, 14641, 2197, 225, 17, 1, 1, 19683, 390625, 823543
Offset: 0

Views

Author

Peter Bala, Oct 27 2008

Keywords

Comments

Definition of the Hilbert transform of a triangular array:
For many square arrays in the database the entries in a row are polynomial in the column index, of degree d say and hence the row generating function has the form P(x)/(1-x)^(d+1), where P is some polynomial function. Often the array whose rows are formed from the coefficients of these P polynomials is of independent interest. This suggests the following definition.
Let [L(n,k)]n,k>=0 be a lower triangular array and let R(n,x) := sum {k = 0 .. n} L(n,k)*x^k, denote the n-th row generating polynomial of L. Then we define the Hilbert transform of L, denoted Hilb(L), to be the square array whose n-th row, n >= 0, has the generating function R(n,x)/(1-x)^(n+1).
In this particular case, L is the array A060187, the array of Eulerian numbers of type B, whose row polynomials are the h-polynomials for permutohedra of type B. The Hilbert transform is an infinite Vandermonde matrix V(1,3,5,...).
We illustrate the Hilbert transform with a few examples:
(1) The Delannoy number array A008288 is the Hilbert transform of Pascal's triangle A007318 (view as the array of coefficients of h-polynomials of n-dimensional cross polytopes).
(2) The transpose of the array of nexus numbers A047969 is the Hilbert transform of the triangle of Eulerian numbers A008292 (best viewed in this context as the coefficients of h-polynomials of n-dimensional permutohedra of type A).
(3) The sequence of Eulerian polynomials begins [1, x, x + x^2, x + 4*x^2 + x^3, ...]. The coefficients of these polynomials are recorded in triangle A123125, whose Hilbert transform is A004248 read as square array.
(4) A108625, the array of crystal ball sequences for the A_n lattices, is the Hilbert transform of A008459 (viewed as the triangle of coefficients of h-polynomials of n-dimensional associahedra of type B).
(5) A142992, the array of crystal ball sequences for the C_n lattices, is the Hilbert transform of A086645, the array of h-vectors for type C root polytopes.
(6) A108553, the array of crystal ball sequences for the D_n lattices, is the Hilbert transform of A108558, the array of h-vectors for type D root polytopes.
(7) A086764, read as a square array, is the Hilbert transform of the rencontres numbers A008290.
(8) A143409 is the Hilbert transform of triangle A073107.

Examples

			Triangle A060187 (with an offset of 0) begins
1;
1, 1;
1, 6, 1;
so the entries in the first three rows of the Hilbert transform of
A060187 come from the expansions:
Row 0: 1/(1-x) = 1 + x + x^2 + x^3 + ...;
Row 1: (1+x)/(1-x)^2 = 1 + 3*x + 5*x^2 + 7*x^3 + ...;
Row 2: (1+6*x+x^2)/(1-x)^3 = 1 + 9*x + 25*x^2 + 49*x^3 + ...;
The array begins
n\k|..0....1.....2.....3......4
================================
0..|..1....1.....1.....1......1
1..|..1....3.....5.....7......9
2..|..1....9....25....49.....81
3..|..1...27...125...343....729
4..|..1...81...625..2401...6561
5..|..1..243..3125.16807..59049
...
		

Crossrefs

Cf. A008292, A039755, A052750 (first superdiagonal), A060187, A114172, A145901.

Programs

  • Maple
    T:=(n,k) -> (2*k + 1)^n: seq(seq(T(n-k,k),k = 0..n),n = 0..10);

Formula

T(n,k) = (2*k + 1)^n, (see equation 4.10 in [Franssens]). This array is the infinite Vandermonde matrix V(1,3,5,7, ....) having a LDU factorization equal to A039755 * diag(2^n*n!) * transpose(A007318).

A142993 Crystal ball sequence for the lattice C_4.

Original entry on oeis.org

1, 33, 225, 833, 2241, 4961, 9633, 17025, 28033, 43681, 65121, 93633, 130625, 177633, 236321, 308481, 396033, 501025, 625633, 772161, 943041, 1140833, 1368225, 1628033, 1923201
Offset: 0

Views

Author

Peter Bala, Jul 18 2008

Keywords

Comments

The lattice C_4 consists of all integer lattice points v = (a,b,c,d) in Z^4 such that a + b + c + d is even, equipped with the taxicab type norm ||v|| = (1/2) * (|a| + |b| + |c| + |d|). The crystal ball sequence of C_4 gives the number of lattice points v in C_4 with ||v|| <= n for n = 0,1,2,3,... [Bacher et al.].

Examples

			a(1) = 33. The origin has norm 0. The 32 lattice points in Z^4 of norm 1 (as defined above) are +-2*e_i, 1 <= i <= 4 and (+- e_i +- e_j), 1 <= i < j <= 4, where e_1, e_2, e_3 and e_4 denotes the standard basis of Z^4. These 32 vectors form a root system of type C_4. Hence sequence begins 1, 1 + 32 = 33, ... .
		

Crossrefs

Programs

  • Maple
    a := n -> (2*n+1)^2*(4*n^2+4*n+3)/3: seq(a(n), n = 0..24)

Formula

Partial sums of A019560. a(n) = (2*n+1)^2*(4*n^2+4*n+3)/3 = Sum_{k = 0..4} C(8,2k)*C(n+k,4) = Sum_{k = 0..4} C(8,2k+1)*C(n+k+1/2,4). O.g.f.: (1+28*x+70*x^2+28*x^3+x^4)/(1-x)^5 = (1/(1-x)) * T(4,(1+x)/(1-x)), where T(n,x) denotes the Chebyshev polynomial of the first kind. 2*log(2) = 17/12 - Sum_{n >= 1} 1/(n*a(n-1)*a(n)).
From Peter Bala, Mar 11 2024: (Start)
Sum_{k = 1..n+1} 1/(k*a(k)*a(k-1)) = 1/(33 - 3/(41 - 60/(57 - 315/(81 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*4^2))))).
E.g.f.: exp(x)*(1 + 32*x + 160*x^2/2! + 256*x^3/3! + 128*x^4/4!).
Note that T(8, i*sqrt(x)) = 1 + 32*x + 160*x^2 + 256*x^3 + 128*x^4. See A008310. (End)

A142994 Crystal ball sequence for the lattice C_5.

Original entry on oeis.org

1, 51, 501, 2471, 8361, 22363, 50973, 103503, 192593, 334723, 550725, 866295, 1312505, 1926315, 2751085, 3837087, 5242017, 7031507, 9279637, 12069447, 15493449, 19654139, 24664509, 30648559, 37741809, 46091811, 55858661, 67215511, 80349081
Offset: 0

Views

Author

Peter Bala, Jul 18 2008

Keywords

Comments

The lattice C_5 consists of all integer lattice points v = (x_1,...,x_5) in Z^5 such that (x_1 + ... + x_5) is even, equipped with the taxicab type norm ||v|| = (1/2) * (|x_1| + ... + |x_5|). The crystal ball sequence of C_5 gives the number of lattice points v in C_5 with ||v|| <= n for n = 0,1,2,3,... [Bacher et al.].
Partial sums of A019561.

Examples

			a(1) = 51. The origin has norm 0. The 50 lattice points in Z^5 of norm 1 (as defined above) are +-2*e_i, 1 <= i <= 5 and (+- e_i +- e_j), 1 <= i < j <= 5, where e_1, ... , e_5 denotes the standard basis of Z^5. These 50 vectors form a root system of type C_5. Hence the sequence begins 1, 1 + 50 = 51, ... .
		

Crossrefs

Row 5 of A142992. Cf. A019561, A063496, A142993.

Programs

  • Magma
    [(2*n+1)*(32*n^4+64*n^3+88*n^2+56*n+15)/15: n in [0..30]]; // Vincenzo Librandi, Dec 16 2015
  • Maple
    a := n -> (2*n+1)*(32*n^4+64*n^3+88*n^2+56*n+15)/15: seq(a(n), n = 0..20)
  • Mathematica
    CoefficientList[Series[(1 + 45 x + 210 x^2 + 210 x^3 + 45 x^4 + x^5)/(1 - x)^6, {x, 0, 33}], x] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1},{1, 51, 501, 2471, 8361, 22363}, 25] (* Vincenzo Librandi, Dec 16 2015 *)
  • Python
    A142994_list, m = [], [512, -768, 352, -48, 2, 1]
    for _ in range(10**2):
        A142994_list.append(m[-1])
        for i in range(5):
            m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015
    

Formula

a(n) = (2*n + 1)*(32*n^4 + 64*n^3 + 88*n^2 + 56*n + 15)/15.
a(n) = Sum_{k = 0..5} binomial(10, 2*k)*binomial(n+k, 5).
a(n) = Sum_{k = 0..5} binomial(10, 2*k+1)*binomial(n+k+1/2, 5).
O.g.f.: (1 + 45*x + 210*x^2 + 210*x^3 + 45*x^4 + x^5)/(1 - x)^6 = 1/(1 - x) * T(5, (1 + x)/(1 - x)), where T(n, x) denotes the Chebyshev polynomial of the first kind.
Sum_{n >= 1} 1/(n*a(n-1)*a(n)) = 2*log(2) - 41/30.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6), for n > 5. - Vincenzo Librandi, Dec 16 2015
From Peter Bala, Mar 11 2024: (Start)
Sum_{k = 1..n+1} 1/(k*a(k)*a(k-1)) = 1/(51 - 3/(59 - 60/(75 - 315/(99 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*5^2))))).
E.g.f.: exp(x)*(1 + 50*x + 400*x^2/2! + 1120*x^3/3! + 1280*x^4/4! + 512*x^5/5!).
Note that -T(10, i*sqrt(x)) = 1 + 50*x + 400*x^2 + 1120*x^3 + 1280*x^4 + 512*x^5. See A008310. (End)

A305549 Crystal ball sequence for the lattice C_6.

Original entry on oeis.org

1, 73, 985, 6321, 26577, 85305, 227305, 528865, 1110049, 2149033, 3898489, 6704017, 11024625, 17455257, 26751369, 39855553, 57926209, 82368265, 114865945, 157417585, 212372497, 282469881, 370879785, 481246113, 617731681, 785065321, 988591033, 1234319185
Offset: 0

Views

Author

Seiichi Manyama, Jun 09 2018

Keywords

Comments

Partial sums of A019562.

Crossrefs

Programs

  • PARI
    {a(n) = sum(k=0, 6, binomial(12,2*k)*binomial(n+k,6))}
    
  • PARI
    Vec((1 + 6*x + x^2)*(1 + 60*x + 134*x^2 + 60*x^3 + x^4) / (1 - x)^7 + O(x^40)) \\ Colin Barker, Jun 09 2018

Formula

a(n) = (128*n^6 + 384*n^5 + 800*n^4 + 960*n^3 + 692*n^2 + 276*n + 45)/45.
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7), for n > 6.
a(n) = Sum_{k = 0..6} binomial(12, 2*k)*binomial(n+k, 6).
G.f.: (1 + 6*x + x^2)*(1 + 60*x + 134*x^2 + 60*x^3 + x^4) / (1 - x)^7. - Colin Barker, Jun 09 2018
From Peter Bala, Mar 12 2024: (Start)
Sum_{k >= 1} 1/(k*a(k)*a(k-1)) = 7/5 - 2*log(2) = 1/(73 - 3/(81 - 60/(97 - 315/(121 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*6^2 - ...))))).
E.g.f.: exp(x)*(1 + 72*x + 840*x^2/2! + 3584*x^3/3! + 6912*x^4/4! + 6144*x^5/5! + 2048*x^6/6!).
Note that T(12, i*sqrt(x)) = 1 + 72*x + 840*x^2 + 3584*x^3 + 6912*x^4 + 6144*x^5 + 2048*x^6, where T(n, x) denotes the n-th Chebyshev polynomial of the first kind. See A008310.
Row 6 of A142992. (End)

A305721 Crystal ball sequence for the lattice C_7.

Original entry on oeis.org

1, 99, 1765, 14407, 74313, 284075, 880685, 2340495, 5529233, 11905267, 23784309, 44673751, 79684825, 136030779, 223619261, 355747103, 549905697, 828705155, 1220925445, 1762702695, 2498858857, 3484382923, 4786071885, 6484339631, 8675201969, 11472445971, 15009991829
Offset: 0

Views

Author

Seiichi Manyama, Jun 09 2018

Keywords

Comments

Partial sums of A019563.

Crossrefs

Programs

  • GAP
    b:=7;; List([0..30],n->Sum([0..b],k->Binomial(2*b,2*k)*Binomial(n+k,b))); # Muniru A Asiru, Jun 09 2018
  • Mathematica
    Array[Sum[Binomial[14, 2 k] Binomial[# + k, 7], {k, 0, 7}] &, 27, 0] (* Michael De Vlieger, Jun 11 2018 *)
    LinearRecurrence[{8,-28,56,-70,56,-28,8,-1},{1,99,1765,14407,74313,284075,880685,2340495},30] (* Harvey P. Dale, May 16 2023 *)
  • PARI
    {a(n) = sum(k=0, 7, binomial(14, 2*k)*binomial(n+k, 7))}
    
  • PARI
    Vec((1 + x)*(1 + 90*x + 911*x^2 + 2092*x^3 + 911*x^4 + 90*x^5 + x^6) / (1 - x)^8 + O(x^40)) \\ Colin Barker, Jun 09 2018
    

Formula

a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8), for n > 7.
a(n) = Sum_{k = 0..7} binomial(14, 2*k)*binomial(n+k, 7).
G.f.: (1 + x)*(1 + 90*x + 911*x^2 + 2092*x^3 + 911*x^4 + 90*x^5 + x^6) / (1 - x)^8. - Colin Barker, Jun 09 2018
From Peter Bala, Mar 12 2024: (Start)
Sum_{k >= 1} 1/(k*a(k)*a(k-1)) = 2*ln(2) - 289/210 = 1/(99 - 3/(107 - 60/(123 - 315/(147 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*7^2 - ...))))).
E.g.f.: exp(x)*(1 + 98*x + 1568*x^2/2! + 9408*x^3/3! + 26880*x^4/4! + 39424*x^5/5! + 28672*x^6/6! + 8192*x^7/7!).
Note that -T(14, i*sqrt(x)) = 1 + 98*x + 1568*x^2 + 9408*x^3 + 26880*x^4 + 39424*x^5 + 28672*x^6 + 8192*x^7, where T(n, x) denotes the n-th Chebyshev polynomial of the first kind. See A008310.
Row 7 of A142992. (End)

A305722 Crystal ball sequence for the lattice C_8.

Original entry on oeis.org

1, 129, 2945, 29953, 187137, 845185, 3032705, 9173505, 24331777, 58161793, 127791489, 261902081, 506298625, 931299201, 1641303169, 2786931713, 4580166657, 7312946305, 11379709825, 17304414465, 25772582657, 37668968833, 54121468545, 76551925249
Offset: 0

Views

Author

Seiichi Manyama, Jun 09 2018

Keywords

Comments

Partial sums of A019564.

Crossrefs

Programs

  • GAP
    b:=8;; List([0..25],n->Sum([0..b],k->Binomial(2*b,2*k)*Binomial(n+k,b))); # Muniru A Asiru, Jun 09 2018
  • PARI
    {a(n) = sum(k=0, 8, binomial(16, 2*k)*binomial(n+k, 8))}
    
  • PARI
    Vec((1 + 120*x + 1820*x^2 + 8008*x^3 + 12870*x^4 + 8008*x^5 + 1820*x^6 + 120*x^7 + x^8) / (1 - x)^9 + O(x^40)) \\ Colin Barker, Jun 09 2018
    

Formula

a(n) = 9*a(n-1) - 36*a(n-2) + 84*a(n-3) - 126*a(n-4) + 126*a(n-5) - 84*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9), for n>8.
a(n) = Sum_{k=0..8} binomial(16, 2k)*binomial(n+k, 8).
G.f.: (1 + 120*x + 1820*x^2 + 8008*x^3 + 12870*x^4 + 8008*x^5 + 1820*x^6 + 120*x^7 + x^8) / (1 - x)^9. - Colin Barker, Jun 09 2018

A305723 Crystal ball sequence for the lattice C_9.

Original entry on oeis.org

1, 163, 4645, 57799, 432073, 2286955, 9446125, 32398735, 96220561, 254831667, 614859189, 1373356887, 2874747225, 5693596923, 10751213181, 19475555103, 34015593249, 57523019715, 94516111685, 151342583015, 236760421097, 362658000011, 544937185805, 804585705647
Offset: 0

Views

Author

Seiichi Manyama, Jun 09 2018

Keywords

Comments

Partial sums of A035746.

Crossrefs

Programs

  • GAP
    b:=9;; List([0..25],n->Sum([0..b],k->Binomial(2*b,2*k)*Binomial(n+k,b))); # Muniru A Asiru, Jun 09 2018
  • Mathematica
    Table[Sum[Binomial[18,2k]Binomial[n+k,9],{k,0,9}],{n,0,40}] (* or *) LinearRecurrence[{10,-45,120,-210,252,-210,120,-45,10,-1},{1,163,4645,57799,432073,2286955,9446125,32398735,96220561,254831667},40] (* Harvey P. Dale, Jun 09 2023 *)
  • PARI
    {a(n) = sum(k=0, 9, binomial(18, 2*k)*binomial(n+k, 9))}
    
  • PARI
    Vec((1 + x)*(1 + 14*x + x^2)*(1 + 138*x + 975*x^2 + 1868*x^3 + 975*x^4 + 138*x^5 + x^6) / (1 - x)^10 + O(x^40)) \\ Colin Barker, Jun 09 2018
    

Formula

a(n) = Sum_{k=0..9} binomial(18, 2k)*binomial(n+k, 9).
From Colin Barker, Jun 09 2018: (Start)
G.f.: (1 + x)*(1 + 14*x + x^2)*(1 + 138*x + 975*x^2 + 1868*x^3 + 975*x^4 + 138*x^5 + x^6) / (1 - x)^10.
a(n) = 10*a(n-1) - 45*a(n-2) + 120*a(n-3) - 210*a(n-4) + 252*a(n-5) - 210*a(n-6) + 120*a(n-7) - 45*a(n-8) + 10*a(n-9) - a(n-10) for n>9.
(End)

A305724 Crystal ball sequence for the lattice C_10.

Original entry on oeis.org

1, 201, 7001, 104881, 927441, 5707449, 26986089, 104535009, 346615329, 1014889769, 2684641785, 6526963345, 14778775025, 31490462745, 63670078985, 122977987009, 228167048769, 408511495049, 708522994329, 1194315679089, 1962053519121, 3148993975161
Offset: 0

Views

Author

Seiichi Manyama, Jun 09 2018

Keywords

Comments

Partial sums of A035747.

Crossrefs

Programs

  • GAP
    b:=10;; List([0..25],n->Sum([0..b],k->Binomial(2*b,2*k)*Binomial(n+k,b))); # Muniru A Asiru, Jun 09 2018
  • PARI
    {a(n) = sum(k=0, 10, binomial(20, 2*k)*binomial(n+k, 10))}
    
  • PARI
    Vec((1 + 6*x + x^2)*(1 + 184*x + 3740*x^2 + 16136*x^3 + 25414*x^4 + 16136*x^5 + 3740*x^6 + 184*x^7 + x^8) / (1 - x)^11 + O(x^30)) \\ Colin Barker, Jun 09 2018
    

Formula

a(n) = 11*a(n-1) - 55*a(n-2) + 165*a(n-3) - 330*a(n-4) + 462*a(n-5) - 462*a(n-6) + 330*a(n-7) - 165*a(n-8) + 55*a(n-9) - 11*a(n-10) + a(n-11) for n>10.
a(n) = Sum_{k=0..10} binomial(20, 2k)*binomial(n+k, 10).
G.f.: (1 + 6*x + x^2)*(1 + 184*x + 3740*x^2 + 16136*x^3 + 25414*x^4 + 16136*x^5 + 3740*x^6 + 184*x^7 + x^8) / (1 - x)^11. - Colin Barker, Jun 09 2018
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