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A205497
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Rectangular array M read by antidiagonals in which entry M_{n-k,k} in row n-k and column k, 0 <= k <= n, gives the coefficient of x^k in the numerator of the conjectured generating function for row n+3 of the tabular form of A050446.
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5
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1, 1, 1, 1, 3, 1, 1, 7, 7, 1, 1, 14, 31, 14, 1, 1, 26, 109, 109, 26, 1, 1, 46, 334, 623, 334, 46, 1, 1, 79, 937, 2951, 2951, 937, 79, 1, 1, 133, 2475, 12331, 20641, 12331, 2475, 133, 1, 1, 221, 6267, 47191, 123216, 123216, 47191, 6267, 221, 1
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OFFSET
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0,5
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COMMENTS
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Christopher H. Gribble kindly calculated the first 100 antidiagonals of the array M which starts as
1, 1, 1, 1, 1, 1, ...
1, 3, 7, 14, 26, 46, ...
1, 7, 31, 109, 334, 937, ...
1, 14, 109, 623, 2951, 12331, ...
1, 26, 334, 2951, 20641, 123216, ...
1, 46, 937, 12331, 123216, 1019051, ...
...
In the following, let M_{n,k} denote the entry in row n and column k of M, n,k in {0,1,...}.
Conjecture: 1. M_{n,k)=M_{k,n), for all n and k; that is, M is symmetric about the central terms {1,3,31,623,...}. (This has been verified for the first 100 antidiagonals of M.)
Conjecture: 2. For m in {3,4,...}, row m of array A050446 has generating function of the form H_m(x)/(1-x)^m, in which the numerator H_m(x) is a polynomial of degree m-3 in x with coefficients given by the entries of the (m-3)-th antidiagonal of M containing the sequence of entries {M_{m-3-j,j}}, j=0..m-3 (see the example below). It is known that H_1(x) = H_2(x) = 1.
Conjecture: 3. Define the Chebyshev polynomials of the second kind by U_0(t)=1, U_1(t)=2*t and U_r(t)=2*t*U_(r-1)(t)-U_(r-2)(t) (r>1). Assuming Conjecture 1, lim_{n -> infinity} M_{n+1,k}/M_{n,k} = U_k(cos(Pi/(2*k+3))) = spectral radius of the (k+1) X (k+1) unit-primitive matrix (see [Jeffery]) A_{2*k+3,k} = [0,...,0,1; 0,...,0,1,1; ...; 0,1,...,1; 1,...,1], with identical limits for the columns of the transpose M^T of M.
Conjecture: 4. Let S(u,v) denote the entry in row u and column v of triangle S=A187660, 0<=v<=u. Define the polynomials P_u(x) = Sum[S(u,v)*x^v]. Assuming Conjecture 1, then (i) the generating function for row (or column) n of M is of the form
G_n(x)/((P_1(x))^(n+1) * (P_2(x))^n * ... * (P_n(x))^2 * P_(n+1)(x)),
in which (ii) the numerator G_n(x) is a polynomial of degree A005586(n), and (iii) the denominator is a polynomial of degree A000292(n+1).
Remarks: The coefficients in the numerators G_n(x) appear to have no pattern. The polynomial P_j(x), j in {1,...,n+1}, of Conjecture 4 is also obtained from the characteristic polynomial of the unit-primitive matrix A_{2*j+3,j} of Conjecture 3 by taking the exponents of the latter in reverse order; and P_j(x) is otherwise identical to the characteristic polynomial of the unit-primitive matrix A_{2*j+3,1}.
It appears that the sums of antidiagonals are the Euler numbers: 1, 2, 5, 16, 61, ... given in entry A000111, i.e., the number of up-down permutations of [n]. - Kyle Petersen, May 10 2012
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LINKS
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FORMULA
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Conjecture: 5.1. G.f. for column 0 of M is 1/(1-x) (A000012).
Conjecture: 5.2. G.f. for column 1 of M is 1/((1-x)^2*(1-x-x^2)) (A001924).
Conjecture: 5.3. G.f. for column 2 of M is (1 - x^2 - x^3 - x^4 + x^5)/((1-x)^3*(1-x-x^2)^2*(1 - 2*x - x^2 + x^3)) (A205492).
Conjecture: 5.4. G.f. for column 3 of M is (1 + x - 6*x^2 - 15*x^3 + 21*x^4 + 35*x^5 - 13*x^6 - 51*x^7 + 3*x^8 + 21*x^9 + 5*x^10 + x^11 - 5*x^12 - x^13 - x^14)/((1-x)^4*(1-x-x^2)^3*(1 - 2*x - x^2 + x^3)^2*(1 - 2*x - 3*x^2 + x^3 + x^4)) (A205493).
Conjecture: 5.5. G.f. for column 4 of M is (1 + 4*x - 31*x^2 - 67*x^3 + 348*x^4 + 418*x^5 - 1893*x^6 - 1084*x^7 + 4326*x^8 + 4295*x^9 - 7680*x^10 - 9172*x^11 + 9104*x^12 + 11627*x^13 - 5483*x^14 - 10773*x^15 + 1108*x^16 + 7255*x^17 + 315*x^18 - 3085*x^19 - 228*x^20 + 669*x^21 + 102*x^22 - 23*x^23 - 45*x^24 - 16*x^25 + 11*x^26 + 2*x^27 - x^28)/((1-x)^5*(1-x-x^2)^4*(1 - 2*x - x^2 + x^3)^3*(1 - 2*x - 3*x^2 + x^3 + x^4)^2*(1 - 3*x - 3*x^2 + 4*x^3 + x^4 - x^5)) (A205494).
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EXAMPLE
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For clarity, the antidiagonals of M may also be written as the rows of a triangle, yielding
1;
1, 1;
1, 3, 1;
1, 7, 7, 1;
1, 14, 31, 14, 1;
1, 26, 109, 109, 26, 1;
1, 46, 334, 623, 334, 46, 1;
...
then, by the conjectures and the definition of H_m(x), row m=7 of table A050446 has generating function H_7(x)/(1-x)^7 = (Sum_{j=0..4} M_{4-j,j}*x^j)/(1-x)^7 = (1 + 14*x + 31*x^2 + 14*x^3 + x^4)/(1-x)^7.
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MATHEMATICA
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nmax = 9; t[n_, 0] = 1;
t[n_, m_?Positive] := t[n, m] = t[n, m-1] + Sum[t[2k, m-1] t[n-1-2k, m], {k, 0, (n-1)/2}];
f[n_] := f[n] = Table[t[n, k], {k, 0, 2nmax}] // FindGeneratingFunction[#, x]& // Numerator;
Do[M[n-k, k] = Coefficient[f[n+2], x, k]//Abs, {n, 0, nmax}, {k, 0, nmax}];
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PROG
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(Python) Replace prepended dots with spaces.
print("\n")
t_rows = 100
t_cols = 101
c_rows = 102
c_cols = 102
r_rows = 101
r_cols = 101
# Calculate t(n, m)=t(n, m-1)+Sum(t(2k, m-1)*t(n-1-2k, m), {k, 0, (n-1)/2}):
.t = [] for a in range(0, t_rows*t_cols):
..t.append(0)
for n in range(0, t_rows):
..t[n*t_cols]=1
for m in range(0, t_cols):
..t[m]=1
for n in range(1, t_rows):
..for m in range(1, t_cols):
....s=0
....for k in range(0, int((n-1)/2)+1):
......s=s+t[2*k*t_cols+m-1]*t[(n-1-2*k)*t_cols+m]
....t[n*t_cols+m]=t[n*t_cols+m-1]+s
# Calculate c(n, k)=c(n-1, k)+c(n-1, k-1): Pascal's triangle.
c = [] for a in range(0, c_rows*c_cols):
..c.append(0) c[0] = 1
for n in range(1, c_rows):
..for k in range(1, n + 1):
....c[n*c_cols+k]=c[(n-1)*c_cols+k]+c[(n-1)*c_cols+k-1]
# Negate even terms to form coefficients of expansion of 1/(1-x)^n.
for n in range(1, c_rows):
..for k in range(1, n + 1):
....if k%2 == 0:
......c[n*c_cols+k]=-c[n*c_cols+k]
# Form Cauchy product r of t and c for each n.
r = [] for a in range(0, r_rows*r_cols):
..r.append(0)
for n in range(1, r_rows):
..for k in range(0, n+1):
....for j in range(0, k+1):
......r[n * r_cols + k] = r[n * r_cols + k] + c[(n + 1) * c_cols + j + 1] * t[(n - 1) * t_cols + k - j]
for n in range(1, _rows):
..a = ""
..for k in range(0, n-2):
....a=a+str(r[n*r_cols+k])+" "
..print(a)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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