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 1-7 of 7 results.

A103881 Square array T(n,k) (n >= 1, k >= 0) read by antidiagonals: coordination sequence for root lattice A_n.

Original entry on oeis.org

1, 1, 2, 1, 6, 2, 1, 12, 12, 2, 1, 20, 42, 18, 2, 1, 30, 110, 92, 24, 2, 1, 42, 240, 340, 162, 30, 2, 1, 56, 462, 1010, 780, 252, 36, 2, 1, 72, 812, 2562, 2970, 1500, 362, 42, 2, 1, 90, 1332, 5768, 9492, 7002, 2570, 492, 48, 2, 1, 110, 2070, 11832, 26474, 27174, 14240, 4060, 642, 54, 2, 1, 132, 3080, 22530, 66222, 91112, 65226, 26070, 6040, 812, 60, 2
Offset: 1

Views

Author

Ralf Stephan, Feb 20 2005

Keywords

Comments

T(n,k) is the number of integer sequences of length n+1 with sum zero and sum of absolute values 2k. - R. H. Hardin, Feb 23 2009

Examples

			Array begins:
  1,   2,     2,      2,       2,        2,         2,          2, ... A040000;
  1,   6,    12,     18,      24,       30,        36,         42, ... A008458;
  1,  12,    42,     92,     162,      252,       362,        492, ... A005901;
  1,  20,   110,    340,     780,     1500,      2570,       4060, ... A008383;
  1,  30,   240,   1010,    2970,     7002,     14240,      26070, ... A008385;
  1,  42,   462,   2562,    9492,    27174,     65226,     137886, ... A008387;
  1,  56,   812,   5768,   26474,    91112,    256508,     623576, ... A008389;
  1,  72,  1332,  11832,   66222,   271224,    889716,    2476296, ... A008391;
  1,  90,  2070,  22530,  151560,   731502,   2777370,    8809110, ... A008393;
  1, 110,  3080,  40370,  322190,  1815506,   7925720,   28512110, ... A008395;
  1, 132,  4422,  68772,  643632,  4197468,  20934474,   85014204, ... A035837;
  1, 156,  6162, 112268, 1219374,  9129276,  51697802,  235895244, ... A035838;
  1, 182,  8372, 176722, 2206932, 18827718, 120353324,  614266354, ... A035839;
  1, 210, 11130, 269570, 3838590, 37060506, 265953170, 1511679210, ... A035840;
  ...
Antidiagonals:
  1;
  1,  2;
  1,  6,    2;
  1, 12,   12,    2;
  1, 20,   42,   18,    2;
  1, 30,  110,   92,   24,    2;
  1, 42,  240,  340,  162,   30,    2;
  1, 56,  462, 1010,  780,  252,   36,   2;
  1, 72,  812, 2562, 2970, 1500,  362,  42,  2;
  1, 90, 1332, 5768, 9492, 7002, 2570, 492, 48,  2;
		

Crossrefs

Programs

  • GAP
    T:=Flat(List([1..12],n->Concatenation([1],List([1..n-1],k->Sum([1..n],i->Binomial(n-k+1,i)*Binomial(k-1,i-1)*Binomial(n-i,k)))))); # Muniru A Asiru, Oct 14 2018
    
  • Magma
    A103881:= func< n,k | k le 0 select 1 else (&+[Binomial(n-k+1, j)*Binomial(k-1, j-1)*Binomial(n-j, k): j in [1..n-k]]) >;
    [A103881(n,k): k in [0..n-1], n in [1..15]]; // G. C. Greubel, Oct 16 2018; May 24 2023
    
  • Maple
    T:=proc(n,k) option remember; local i;
    if k=0 then 1 else
    add( binomial(n+1,i)*binomial(k-1,i-1)*binomial(n-i+k,k),i=1..n); fi;
    end:
    g:=n->[seq(T(n-i,i),i=0..n-1)]:
    for n from 1 to 14 do lprint(op(g(n))); od:
  • Mathematica
    T[n_, k_]:= (n+1)*(n+k-1)!*HypergeometricPFQ[{1-k,1-n,-n}, {2,-n-k+1}, 1]/(k!*(n-1)!); T[, 0]=1; Flatten[Table[T[n-k, k], {n,12}, {k,0,n-1}]] (* _Jean-François Alcover, Dec 27 2012 *)
  • PARI
    A103881(n,k) = if(k==0, 1, sum(j=1, n-k, binomial(n-k+1, j)*binomial(k-1, j-1)*binomial(n-j, k)));
    for(n=1, 15, for(k=0, n-1, print1(A103881(n,k), ", "))) \\ G. C. Greubel, Oct 16 2018; May 24 2023
    
  • SageMath
    def A103881(n,k): return 1 if k==0 else (n-k+1)*binomial(n-1,k)*hypergeometric([k-n,1+k-n,1-k], [2,1-n], 1).simplify()
    flatten([[A103881(n,k) for k in range(n)] for n in range(1,16)]) # G. C. Greubel, May 24 2023

Formula

T(n,k) = Sum_{i=1..n} C(n+1, i)*C(k-1, i-1)*C(n-i+k, k), T(n,0)=1.
G.f. of n-th row: (Sum_{i=0..n} C(n, i)^2*x^i)/(1-x)^n.
From G. C. Greubel, May 24 2023: (Start)
T(n, k) = Sum_{j=0..n} binomial(n,j)^2 * binomial(n+k-j-1, n-1) (array).
T(n, k) = (n+1)*binomial(n+k-1,k)*hypergeometric([-n,1-n,1-k], [2,1-n-k], 1), with T(n, k) = 1 (array).
t(n, k) = (n-k+1)*binomial(n-1,k)*hypergeometric([k-n,1+k-n,1-k], [2,1-n], 1), with t(n, 0) = 1 (antidiagonals).
Sum_{k=0..n-1} t(n, k) = A047085(n). (End)
From Peter Bala, Jul 09 2023: (Start)
T(n,k) = [x^k] Legendre_P(n, (1 + x)/(1 - x)).
(n+1)*T(n+1,k) = (n+1)*T(n+1,k-1) + (2*n+1)*(T(n,k) + T(n,k-1)) - n*(T(n-1,k) - T(n-1,k-1)). (End)

Extensions

Corrected by N. J. A. Sloane, Dec 15 2012, at the suggestion of Manuel Blum

A103885 a(n) = [x^(2*n)] ((1 + x)/(1 - x))^n.

Original entry on oeis.org

1, 2, 16, 146, 1408, 14002, 142000, 1459810, 15158272, 158611106, 1669752016, 17664712562, 187641279616, 2000029880786, 21380213588848, 229129634462146, 2460955893981184, 26482855453375042, 285475524009208720, 3082024598888203090, 33319523640218177408
Offset: 0

Views

Author

Ralf Stephan, Feb 20 2005

Keywords

Comments

From Peter Bala, Mar 01 2020: (Start)
The recurrence given below can be rewritten in the form
(2*n+1)*(2*n+2)*P(2,n)*a(n+1) - (2*n-1)*(2*n-2)*P(2,-n)*a(n-1) = Q(2,n^2)*a(n), where the polynomial Q(2,n) = 4*(55*n^2 - 34*n + 3) and the polynomial P(2,n) = 5*n^2 - 5*n + 1 satisfies the symmetry condition P(2,n) = P(2,1-n) and has real zeros.
More generally, for fixed m = 1,2,3,..., we conjecture that the sequence b(n) := a(m*n) satisfies a recurrence of the form ( Product_{k = 1..2*m} (2*m*n + k) ) * P(2*m,n)*b(n+1) + (-1)^m*( Product_{k = 1..2*m} (2*m*n - k) ) * P(2*m,-n)*b(n-1) = Q(2*m,n^2)*b(n), where the polynomials P(2*m,n) and Q(2*m,n) have degree 2*m. Conjecturally, the polynomial P(2*m,n) = P(2*m,1-n) and has real zeros in the interval [0, 1]. The 4*m zeros of the polynomial Q(2*m,n^2) seem to belong to the interval [-1, 1] and 4*m - 2 of these zeros appear to be approximated by the rational numbers +- k/(3*m), where 1 <= k <= 3*m - 2, k not a multiple of 3. (End)

Crossrefs

Programs

  • Magma
    A103885:= func< n | n eq 0 select 1 else (&+[ Binomial(n, k)*Binomial(2*n+k-1, n-1): k in [0..n]]) >;
    [A103885(n): n in [0..40]]; // G. C. Greubel, Oct 27 2024
    
  • Maple
    a := n -> `if`(n=0, 1, 2*n*hypergeom([1 - 2*n, 1 - n], [2], 2)):
    seq(simplify(a(n)), n=0..17); # Peter Luschny, Dec 30 2019
    # Alternative (after Peter Bala ):
    gf := n -> ( (1 + x)/(1 - x) )^n: ser := n -> series(gf(n), x, 40):
    seq(coeff(ser(n), x, 2*n), n=0..17); # Peter Luschny, Mar 20 2020
  • Mathematica
    Prepend[Table[Sum[2^i Binomial[n, i] Binomial[2n-1, i-1], {i, 1, 2n}], {n,1,20}], 1] (* Vaclav Kotesovec, Jul 01 2015 *)
  • PARI
    a(n) = if (n==0, 1, sum(i=0, n, 2^i * binomial(n, i) * binomial(2*n-1, i-1))); \\ Michel Marcus, Mar 21 2020
    
  • SageMath
    def A103885(n): return 1 if n==0 else sum(binomial(n, k)*binomial(2*n+k-1, n-1) for k in range(n+1))
    [A103885(n) for n in range(41)] # G. C. Greubel, Oct 27 2024

Formula

a(n) = Sum_{i=0..n} 2^i * binomial(n,i) * binomial(2*n-1,i-1). [Original definition, with summation range {i=1..n}.]
a(n) = A103884(n, n).
G.f.: A(x) = x*B(x)'/B(x), where B(x) is g.f. of A027307. - Vladimir Kruchinin, Jun 30 2015
From Vaclav Kotesovec, Jul 01 2015: (Start)
Recurrence: n*(2*n-1)*(5*n^2 - 15*n + 11)*a(n) = 2*(55*n^4 - 220*n^3 + 296*n^2 - 152*n + 24)*a(n-1) + (n-2)*(2*n-3)*(5*n^2 - 5*n + 1)*a(n-2).
a(n) ~ ((11 + 5*sqrt(5))/2)^n / (2 * 5^(1/4) * sqrt(Pi*n)). (End)
a(n) = [x^n] (1/(1 - x - x/(1 - x - x/(1 - x - x/(1 - x - x/(1 - ...))))))^n, a continued fraction. - Ilya Gutkovskiy, Sep 29 2017
a(n) = 2*n*hypergeom([1 - 2*n, 1 - n], [2], 2) for n >= 1. - Peter Luschny, Dec 30 2019
From Peter Bala, Mar 01 2020: (Start)
a(n) = Sum_{k = 0..n} C(n, k)*C(2*n+k-1, n-1), with a(0) = 1.
a(n) = Sum_{k = 0..n} C(2*n, 2*k)*C(2*n-k-1, n-1), with a(0) = 1.
a(n) = (1/2)*Sum_{k = 0..n} C(2*n, n-k)*C(2*n+k-1, k). Cf. A156894.
a(n) = [x^n] S(x)^n, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the sequence of large Schröder numbers A006318.
a(n) = (1/2) * [x^(n)] ( (1 + x)/(1 - x) )^(2*n). Cf. A002003(n) = [x^n] ( (1 + x)/(1 - x) )^n.
Conjecture: a(n) = - [x^n] G(x)^(-n), where G(x) = 1 + 2*x + 14*x^2 + 134*x^3 + 1482*x^4 + ... is the o.g.f. of A144097.
a(p) == 2 ( mod p^3 ) for prime p >= 5. (End)
From Peter Bala, Sep 22 2021: (Start)
a(n) = Sum_{k = 0..n} 4^k*binomial(n+k-1,n)*binomial(n,k)^2 / binomial(2*k,k).
Equivalently, a(n) = [x^n] T(n,(1+x)/(1-x)), where T(n,x) is the n-th Chebyshev polynomial of the first kind. Cf. A103882. (End)
For n>0, a(n) = (1/3) * [x^n] (1/S(-x))^(3*n), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the sequence of large Schröder numbers A006318. Cf. A370102. - Peter Bala, Jul 29 2024

Extensions

a(0) = 1 added and new name by Peter Bala, Mar 01 2020

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

Views

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.

A103903 Square array T(n,k) read by antidiagonals: coordination sequence for lattice D_n.

Original entry on oeis.org

1, 1, 24, 1, 40, 144, 1, 60, 370, 456, 1, 84, 792, 1640, 1056, 1, 112, 1498, 4724, 4930, 2040, 1, 144, 2592, 11620, 18096, 11752, 3504, 1, 180, 4194, 25424, 55650, 52716, 24050, 5544, 1, 220, 6440, 50832, 149568, 195972, 127816, 44200, 8256, 1, 264
Offset: 4

Views

Author

Ralf Stephan, Feb 21 2005

Keywords

Examples

			1,24,144,456,1056,2040,3504,5544,8256,11736,
1,40,370,1640,4930,11752,24050,44200,75010,119720,
1,60,792,4724,18096,52716,127816,271908,524640,938652,
1,84,1498,11620,55650,195972,559258,1371316,2999682,6003956,
1,112,2592,25424,149568,629808,2100832,5910288,14610560,32641008,
1,144,4194,50832,361602,1801872,6976866,22413456,62407170,155242640,
		

Crossrefs

Programs

  • Mathematica
    nmin = 4; nmax = 13; f[x_, n_] := ((1/2)*((-1+Sqrt[x])^(2n)+(1+Sqrt[x])^(2n))*(1-x)^n) / (-1+x)^(2n)-(2n*x*(1+x)^(n-2)) / (1-x)^n; t = Table[ CoefficientList[ Series[ f[x, n], {x, 0, nmax-nmin} ], x], {n, nmin, nmax} ]; Flatten[ Table[ t[[n-k+1, k]], {n, 1, nmax-nmin+1}, {k, 1, n} ] ] (* Jean-François Alcover, Jan 24 2012, after g.f. *)
  • PARI
    T(n,k)={polcoeff(sum(i=0, n, (binomial(2*n, 2*i) - 2*n*binomial(n-2, i-1))*x^i)/(1-x)^n + O(x*x^k), k)} \\ Andrew Howroyd, Jul 03 2018

Formula

G.f. of n-th row: (Sum_{i=0..n} (binomial(2*n, 2*i) - 2*n*binomial(n-2, i-1))*x^i)/(1-x)^n.

A019560 Coordination sequence for C_4 lattice.

Original entry on oeis.org

1, 32, 192, 608, 1408, 2720, 4672, 7392, 11008, 15648, 21440, 28512, 36992, 47008, 58688, 72160, 87552, 104992, 124608, 146528, 170880, 197792, 227392, 259808, 295168, 333600, 375232, 420192, 468608
Offset: 0

Views

Author

mbaake(AT)sunelc3.tphys.physik.uni-tuebingen.de (Michael Baake)

Keywords

Crossrefs

Cf. A103884 (row 4). For coordination sequences of other C_n lattices see A022144 (C_2), A010006 (C3), A019560 - A019564 (C_4 through C_8), A035746 - A035787 (C_9 through C_50).

Programs

  • Magma
    [1] cat [(32/3)*n*(1 + 2*n^2): n in [1..40]]; // Vincenzo Librandi, Apr 10 2017
  • Mathematica
    Join[{1}, Table[(32/3) n (1 + 2 n^2), {n, 30}]] (* Vincenzo Librandi, Apr 10 2017 *)

Formula

a(n) = (32/3)*n*(1 + 2*n^2) for n>0.
G.f.: (1 + 28*x + 70*x^2 + 28*x^3 + x^4)/(1 - x)^4.
G.f. for sequence with interpolated zeros: cosh(8*arctanh(x)) = 1/2*(((1 + x)/(1 - x))^4 + ((1 - x)/(1 + x))^4) = 1 + 32*x^2 + 192*x^4 + 608*x^6 + .... Cf. A057813. - Peter Bala, Apr 09 2017
a(n) = A008412(2*n). - Seiichi Manyama, Jun 08 2018

A019561 Coordination sequence for C_5 lattice.

Original entry on oeis.org

1, 50, 450, 1970, 5890, 14002, 28610, 52530, 89090, 142130, 216002, 315570, 446210, 613810, 824770, 1086002, 1404930, 1789490, 2248130, 2789810, 3424002, 4160690, 5010370, 5984050, 7093250, 8350002
Offset: 0

Views

Author

mbaake(AT)sunelc3.tphys.physik.uni-tuebingen.de (Michael Baake)

Keywords

Crossrefs

Cf. A103884 (row 5). For coordination sequences of other C_n lattices see A022144 (C_2), A010006 (C3), A019560 - A019564 (C_4 through C_8), A035746 - A035787 (C_9 through C_50).

Programs

  • Mathematica
    LinearRecurrence[{5,-10,10,-5,1},{1,50,450,1970,5890,14002},30] (* Harvey P. Dale, Nov 21 2021 *)

Formula

G.f.: (1+45*x+210*x^2+210*x^3+45*x^4+x^5)/(1-x)^5 = 1+2*x*(5+10*x+x^2)^2/(1-x)^5.
G.f. for sequence with interpolated zeros: cosh(10*arctanh(x)) = 1/2*( ((1 + x)/(1 - x))^5 + ((1 - x)/(1 + x))^5 ) = 1 + 50*x^2 + 450*x^4 + 1970*x^6 + .... - Peter Bala, Apr 09 2017
a(n) = A008413(2*n). - Seiichi Manyama, Jun 08 2018

A103883 Square array A(n,k) read by antidiagonals: coordination sequence for lattice B_n.

Original entry on oeis.org

1, 1, 8, 1, 18, 16, 1, 32, 74, 24, 1, 50, 224, 170, 32, 1, 72, 530, 768, 306, 40, 1, 98, 1072, 2562, 1856, 482, 48, 1, 128, 1946, 6968, 8130, 3680, 698, 56, 1, 162, 3264, 16394, 28320, 20082, 6432, 954, 64, 1, 200, 5154, 34624, 83442, 85992, 42130, 10304, 1250, 72
Offset: 2

Views

Author

Ralf Stephan, Feb 20 2005

Keywords

Examples

			Array, A(n, k), begins:
  1,   8,    16,     24,      32,       40,        48, ... A022144;
  1,  18,    74,    170,     306,      482,       698, ... A022145;
  1,  32,   224,    768,    1856,     3680,      6432, ... A022146;
  1,  50,   530,   2562,    8130,    20082,     42130, ... A022147;
  1,  72,  1072,   6968,   28320,    85992,    214864, ... A022148;
  1,  98,  1946,  16394,   83442,   307314,    907018, ... A022149;
  1, 128,  3264,  34624,  216448,   954880,   3301952, ... A022150;
  1, 162,  5154,  67266,  507906,  2653346,  10666146, ... A022151;
  1, 200,  7760, 122264, 1099040,  6728168,  31208560, ... A022152;
  1, 242, 11242, 210474, 2224178, 15804866,  83999962, ... A022153;
  1, 288, 15776, 346304, 4254912, 34792672, 210482016, ... A022154;
  ...
Antidiagonals, T(n, k), begin as:
  1;
  1,   8;
  1,  18,   16;
  1,  32,   74,    24;
  1,  50,  224,   170,    32;
  1,  72,  530,   768,   306,    40;
  1,  98, 1072,  2562,  1856,   482,   48;
  1, 128, 1946,  6968,  8130,  3680,  698,  56;
  1, 162, 3264, 16394, 28320, 20082, 6432, 954, 64;
		

Crossrefs

Programs

  • Magma
    A103883:= func< n,k | (&+[Binomial(n-j-1,n-k-1)*(Binomial(2*n-2*k+1,2*j) - 2*j*Binomial(n-k,j)) : j in [0..k]]) >;
    [A103883(n,k): k in [0..n-2], n in [2..14]]; // G. C. Greubel, May 24 2023
    
  • Mathematica
    offset = 2;
    T[n_, k_] := SeriesCoefficient[Sum[(Binomial[2n + 1, 2i] - 2i Binomial[n, i]) x^i, {i, 0, n}]/(1 - x)^n, {x, 0, k}];
    Table[T[n - k, k], {n, offset, 11}, {k, 0, n - offset}] // Flatten (* Jean-François Alcover, Feb 13 2019 *)
  • SageMath
    def A103883(n,k): return sum(binomial(n-j-1,n-k-1)*(binomial(2*n-2*k+1,2*j) - 2*j*binomial(n-k,j)) for j in range(k+1))
    flatten([[A103883(n,k) for k in range(n-1)] for n in range(2,15)]) # G. C. Greubel, May 24 2023

Formula

G.f. of n-th row: (Sum_{i=0..n} (C(2n+1, 2*i) - 2*i*C(n, i))*x^i)/(1-x)^n.
From G. C. Greubel, May 24 2023: (Start)
A(n, k) = Sum_{j=0..k} binomial(n+k-j-1, n-1)*(binomial(2*n+1, 2*j) - 2*j*binomial(n, j)) (array).
T(n, k) = Sum_{j=0..k} binomial(n-j-1, n-k-1)*(binomial(2*n-2*k+1, 2*j) - 2*j*binomial(n-k, j)) (antidiagonals). (End)
Showing 1-7 of 7 results.