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%I #41 Jul 06 2023 21:47:55
%S 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,
%T 1,73,501,833,489,121,13,1,1,99,985,2471,2241,891,169,15,1,1,129,1765,
%U 6321,8361,4961,1469,225,17,1
%N Square array, read by ascending antidiagonals, of the crystal ball sequences for the root lattices of type C_n.
%C 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.
%C 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).
%C 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.
%H Seiichi Manyama, <a href="/A142992/b142992.txt">Rows n = 0..139, flattened</a>
%H R. Bacher, P. de la Harpe and B. Venkov, <a href="https://doi.org/10.5802/aif.1689">Séries de croissance et séries d'Ehrhart associées aux réseaux de racines</a>, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 1137-1142.
%H D. Bump, K. Choi, P. Kurlberg and J. Vaaler, <a href="https://www.researchgate.net/publication/225542192_A_local_Riemann_hypothesis_I">A local Riemann hypothesis, I</a>, Math. Zeit. 233, (2000), 1-19.
%H J. H. Conway and N. J. A. Sloane, <a href="http://neilsloane.com/doc/Me220.pdf">Low dimensional lattices VII Coordination sequences</a>, Proc. R. Soc. Lond., Ser. A, 453 (1997), 2369-2389.
%F T(n,k) = Sum_{i = 0..n} C(2*n,2*i)*C(k+i,n).
%F 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.
%F 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 + ... .
%F 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.
%F 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).
%F 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.
%F 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.
%F 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.]).
%F 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) + ... .
%F 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)).
%F 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) - ... ).
%F 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)).
%F 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.
%e The square array begins
%e n\k|0...1....2.....3.....4......5
%e =================================
%e .0.|1...1....1.....1.....1......1
%e .1.|1...3....5.....7.....9.....11
%e .2.|1...9...25....49....81....121 A016754
%e .3.|1..19...85...231...489....891 A063496
%e .4.|1..33..225...833..2241...4961 A142993
%e .5.|1..51..501..2471..8361..22363 A142994
%e ...
%e Triangular array begins
%e n\k|0...1...2...3...4...5
%e =========================
%e .0.|1
%e .1.|1...1
%e .2.|1...3...1
%e .3.|1...9...5...1
%e .4.|1..19..25...7...1
%e .5.|1..33..85..49...9...1
%e 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, ... .
%e . . . . . . . . . . .
%e . . . . . 2 . . . . .
%e . . . . 2 . 2 . . . .
%e . . . 2 . 1 . 2 . . .
%e . . 2 . 1 . 1 . 2 . .
%e . 2 . 1 . 0 . 1 . 2 .
%e . . 2 . 1 . 1 . 2 . .
%e . . . 2 . 1 . 2 . . .
%e . . . . 2 . 2 . . . .
%e . . . . . 2 . . . . .
%e . . . . . . . . . . .
%p 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;
%t 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 *)
%Y Row 2-10 give A016754, A063496, A142993, A142994, A305549, A305721, A305722, A305723, A305724.
%Y Cf. A008288, A103884, A108625, A142979, A143007.
%K easy,nonn,tabl
%O 0,5
%A _Peter Bala_, Jul 18 2008
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