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 1-3 of 3 results.

A063496 a(n) = (2*n - 1)*(8*n^2 - 8*n + 3)/3.

Original entry on oeis.org

1, 19, 85, 231, 489, 891, 1469, 2255, 3281, 4579, 6181, 8119, 10425, 13131, 16269, 19871, 23969, 28595, 33781, 39559, 45961, 53019, 60765, 69231, 78449, 88451, 99269, 110935, 123481, 136939, 151341, 166719, 183105, 200531, 219029, 238631, 259369, 281275, 304381
Offset: 1

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Author

N. J. A. Sloane, Aug 01 2001

Keywords

Comments

Number of potential flows in a 2 X 2 matrix with integer velocities in -n..n, i.e., number of 2 X 2 matrices with adjacent elements differing by no more than n, counting matrices differing by a constant only once. - R. H. Hardin, Feb 27 2002
Number of ordered quadruples (a,b,c,d), -(n-1) <= a,b,c,d <= n-1, such that a+b+c+d = 0. - Benoit Cloitre, Jun 14 2003
If Y and Z are 2-blocks of a (2n+1)-set X then a(n-1) is the number of 5-subsets of X intersecting both Y and Z. - Milan Janjic, Oct 28 2007
Equals binomial transform of [1, 18, 48, 32, 0, 0, 0, ...]. - Gary W. Adamson, Jul 19 2008

Crossrefs

(1/12)*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.

Programs

  • Magma
    [(2*n-1)*(8*n^2-8*n+3)/3: n in [1..40]]; // Wesley Ivan Hurt, May 09 2014
  • Maple
    A063496:=n->(2*n-1)*(8*n^2-8*n+3)/3; seq(A063496(n), n=1..40); # Wesley Ivan Hurt, May 09 2014
  • Mathematica
    Table[(2*n - 1)*(8*n^2 - 8*n + 3)/3, {n, 40}] (* Wesley Ivan Hurt, May 09 2014 *)
    LinearRecurrence[{4,-6,4,-1}, {1,19,85,231}, 30] (* G. C. Greubel, Dec 01 2017 *)
  • PARI
    a(n) = { (2*n - 1)*(8*n^2 - 8*n + 3)/3 } \\ Harry J. Smith, Aug 23 2009
    
  • PARI
    my(x='x+O('x^30)); Vec(serlaplace((-3+6*x+24*x^2+16*x^3)*exp(x)/3 + 1)) \\ G. C. Greubel, Dec 01 2017
    

Formula

From Peter Bala, Jul 18 2008: (Start)
The following remarks about the C_3 lattice assume the sequence offset is 0.
Partial sums of A010006. So this sequence is the crystal ball sequence for the C_3 lattice - row 3 of A142992. The lattice C_3 consists of all integer lattice points v = (a,b,c) in Z^3 such that a + b + c is even, equipped with the taxicab type norm ||v|| = (1/2) * (|a| + |b| + |c|).
The crystal ball sequence of C_3 gives the number of lattice points v in C_3 with ||v|| <= n for n = 0,1,2,3,... [Bacher et al.].
For example, a(1) = 19 because the origin has norm 0 and the 18 lattice points in Z^3 of norm 1 (as defined above) are +-(2,0,0), +-(0,2,0), +-(0,0,2), +-(1,1,0), +-(1,0,1), +-(0,1,1), +-(1,-1,0), +-(1,0,-1) and +-(0,1,-1). These 18 vectors form a root system of type C_3.
O.g.f.: x*(1 + 15*x + 15*x^2 + x^3)/(1 - x)^4 = x/(1 - x) * T(3, (1 + x)/(1 - x)), where T(n, x) denotes the Chebyshev polynomial of the first kind.
2*log(2) = 4/3 + Sum_{n >= 1} 1/(n*a(n)*a(n+1)). (End)
a(n+1) = (1/Pi) * Integral_{x=0..Pi} (sin((n+1/2)*x)/sin(x/2))^4. - Yalcin Aktar, Nov 02 2011, corrected by R. J. Mathar, Dec 01 2011
From G. C. Greubel, Dec 01 2017: (Start)
G.f.: x*(1 + 15*x + 15*x^2 + x^3)/(1 - x)^4.
E.g.f.: (-3 + 6*x + 24*x^2 + 16*x^3)*exp(x)/3 + 1. (End)
a(n) = A005900(2n-1). - Ivan N. Ianakiev, Mar 27 2022
From Peter Bala, Mar 11 2024: (Start)
Sum_{k = 1..n+1} 1/(k*a(k)*a(k+1)) = 1/(19 - 3/(27 - 60/(43 - 315/(67 - ... -n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*3^2))))).
E.g.f.: exp(x)*(1 + 18*x + 48*x^2/2! + 32*x^3/3!). Note that -T(6, i*sqrt(x)) = 1 + 18*x + 48*x^2 + 32*x^3, where T(n, x) denotes the n-th Chebyshev polynomial of the first kind. See A008310. (End)

A142992 Square array, read by ascending antidiagonals, of the crystal ball sequences for the root lattices of type C_n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 9, 5, 1, 1, 19, 25, 7, 1, 1, 33, 85, 49, 9, 1, 1, 51, 225, 231, 81, 11, 1, 1, 73, 501, 833, 489, 121, 13, 1, 1, 99, 985, 2471, 2241, 891, 169, 15, 1, 1, 129, 1765, 6321, 8361, 4961, 1469, 225, 17, 1
Offset: 0

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Author

Peter Bala, Jul 18 2008

Keywords

Comments

The lattice C_n consists of all integer lattice points v = (x_1,...,x_n) in Z^n such that the sum x_1 + ... + x_n is even. Let ||v|| = 1/2 * Sum_{i = 1..n} |x_i|; this defines a norm on C_n. The k-th term of the crystal ball sequence of C_n gives the number of lattice points v in C_n with ||v|| <= k [Bacher et al.]. The case n = 2 is illustrated in the Example section below.
This array has a remarkable relationship with the constant log(2). The row, column and (conjecturally) the diagonal entries of the array occur in series acceleration formulas for log(2) (see the Formula section below for some examples).
See A103884 for the table of coordination sequences of the C_n lattices. For the crystal ball sequences for the A_n and D_n lattices see A108625 and A108553 respectively. For the crystal ball sequences for the product lattices A_1 x ... x A_1(n copies) and A_n x A_n see A008288 and A143007 respectively.

Examples

			The square array begins
n\k|0...1....2.....3.....4......5
=================================
.0.|1...1....1.....1.....1......1
.1.|1...3....5.....7.....9.....11
.2.|1...9...25....49....81....121 A016754
.3.|1..19...85...231...489....891 A063496
.4.|1..33..225...833..2241...4961 A142993
.5.|1..51..501..2471..8361..22363 A142994
...
Triangular array begins
n\k|0...1...2...3...4...5
=========================
.0.|1
.1.|1...1
.2.|1...3...1
.3.|1...9...5...1
.4.|1..19..25...7...1
.5.|1..33..85..49...9...1
Case n = 2: The C_2 lattice consists of all integer lattice points v = (x,y) in Z x Z such that x + y is even, equipped with the taxicab type norm ||v|| = 1/2 * (|x| + |y|). There are 8 lattice points (marked with a 1 on the figure below) satisfying ||v|| = 1 and 16 lattice points (marked with a 2 on the figure) satisfying ||v|| = 2. Hence the crystal ball sequence for the C_2 lattice (row 2 of the table) begins 1, 1+8 = 9, 1+8+16 = 25, ... .
. . . . . . . . . . .
. . . . . 2 . . . . .
. . . . 2 . 2 . . . .
. . . 2 . 1 . 2 . . .
. . 2 . 1 . 1 . 2 . .
. 2 . 1 . 0 . 1 . 2 .
. . 2 . 1 . 1 . 2 . .
. . . 2 . 1 . 2 . . .
. . . . 2 . 2 . . . .
. . . . . 2 . . . . .
. . . . . . . . . . .
		

Crossrefs

Programs

  • Maple
    with combinat: T := (n,k) -> add(binomial(2n,2i)*binomial(k+i,n),i = 0..n): for n from 0 to 9 do seq(T(n,k), k = 0..9) end do;
  • Mathematica
    t[n_, k_] := Sum[ Binomial[2*n, 2*i]*Binomial[k+i, n], {i, 0, n}]; Table[t[n-k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 06 2013 *)

Formula

T(n,k) = Sum_{i = 0..n} C(2*n,2*i)*C(k+i,n).
O.g.f. for row n: 1/(1-x)^(n+1) * Sum_{k = 0..n} C(2*n,2*k)*x^k = 1/(1-x) * T(n,(1+x)/(1-x)), where T(n,x) denotes the Chebyshev polynomial of the first kind.
O.g.f. for the array: 1/(1-x) * {(1-t) - x*(1+t)}/{(1-t)^2 - x*(1+t)^2} = (1+x+x^2+x^3+...) + (1+3*x+5*x^2+7*x^3+...)*t + (1+9*x+25*x^2+49*x^3+...)*t^2 + ... .
Row n of the array has the form [p_n(0),p_n(1),p_n(2),...], where the polynomial function p_n(x) = Sum_{k = 0..n} C(2*n,2*k)*C(x+k,n). The first few are p_0(x) = 1, p_1(x) = 2*x+1, p_2(x) = (2*x+1)^2, p_3(x) = (2*x+1)*(8*x^2+8*x+3)/3 and p_4(x) = (2*x+1)^2*(4*x^2+4*x+3)/3.
Alternative expressions for p_n(x) include p_n(x) = Sum_{k = 0..n} 2^(2*k)*n/(n+k)*C(n+k,2*k)*C(x,k) and p_n(x) = Sum_{k = 1..n} 2^(k-1)*C(n-1,k-1)*C(2*x+1,k).
The polynomials p_n(x) satisfy the 3-term recurrence relation n*p_n(x) = 2*(2*x+1)*p_(n-1)(x)+(n-2)*p(n-2)(x) for n >= 2; their generating function is 1/2*((1+t)/(1-t))^(2*x+1) = 1/2 + (2*x+1)*t + (2*x+1)^2*t^2 + (2*x+1)*(8*x^2+8*x+3)/3*t^3 + ... . Thus p_n(x) is, apart from a constant factor, the Meixner polynomial of the first kind M_n(2*x+1;b,c) at b = 0, c = -1. Compare with A142979.
The polynomial p_n(x) is the unique polynomial solution to the difference equation (2*x+1)*{f(x+1/2) - f(x-1/2)} = 2*n*f(x), normalized so that f(0) = 1. The function p_n(x) is also the unique polynomial solution to the difference equation (2*x+1)*{(x+1)*f(x+1) + x*f(x-1)} = ((2*x+1)^2 + 2*n^2)*f(x), normalized so that f(0) = 1.
The zeros of p_n(x) lie on the vertical line Re x = -1/2 in the complex plane, that is, the polynomials p_n(x-1), n = 1,2,3,..., satisfy a Riemann hypothesis (adapt the proof of the lemma on p.4 of [BUMP et al.]).
For n > 0, the entries in row n of the array occur in series acceleration formulas for log(2): 2*log(2) = 1 + (1/2 - 1/6 +...+(-1)^n/(n*(n-1))) + (-1)^(n+1)*Sum_{k >= 1} 1/(k*T(n,k-1)*T(n,k)). For example, the fourth row of the table (n = 3) gives 2*log(2) = 4/3 + 1/(1*1*19) + 1/(2*19*85) + 1/(3*85*231) + ... .
The corresponding result for column k is 2*log(2) = 1 + (1/(1*3) + 1/(2*3*5) +...+ 1/(k*(2*k-1)*(2k+1)) + (2*k+1)*Sum_{n >= 1} (-1)^(n+1)/(n*(n+1)*T(n,k)* T(n+1,k)).
For example, the third column of the table (k = 2) gives 2*log(2) = 41/30 + 5*(1/(1*2*5*25) - 1/(2*3*25*85) + 1/(3*4*85*225) - ... ).
For the main diagonal calculation suggests the result: 2*log(2) = 4/3 + Sum_{n >= 1} (-1)^(n+1)*(5*n+3)/(n*(n+1)*T(n,n)*T(n+1,n+1)).
Similar series acceleration formulas for log(2) come from the row, column and diagonal entries of the square array of Delannoy numbers, A008288 (which may viewed as the array of crystal ball sequences for the product lattices A_1 x...x A_1). For corresponding results for the constants zeta(2) and zeta(3) see A108625 and A143007 respectively.

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)
Showing 1-3 of 3 results.