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%I #11 Nov 11 2017 17:27:25
%S 1,2,1,3,3,1,4,4,3,1,5,6,5,3,1,6,7,7,5,3,1,7,9,8,8,5,3,1,8,10,11,9,8,
%T 5,3,1,9,12,12,12,10,8,5,3,1,10,13,14,13,13,10,8,5,3,1,11,15,16,16,14,
%U 14,10,8,5,3,1,12,16,18,18,17,15,14,10,8,5,3,1,13,18,19,20,19
%N Rectangular array by antidiagonals, n >= 1, k >= 1: R(n,k) = n + [n/2] + ... + [n/k], where [ ]=floor.
%C R(n,k) is the number of ordered pairs (x,y) of integers x,y satisfying 1 <= x <= k, 1 <= y <= k, and x*y <= n.
%C Limiting row: A000618=(1,3,5,8,10,14,16,20,...).
%C Row 1: A000027
%C Row 2: A032766
%C Row 3: A106252
%C Row 4: A211703
%C Row 5: A211704
%C R(n,n) = A000618(n)
%C ...
%C For n > =1, row n is a homogeneous linear recurrence sequence of order A005728(n), and it exemplifies a certain class, C, of recurrences which are palindromic (in the sense given below). The class depends on sequences s having n-th term [(n^k)/j], where k and j are arbitrary fixed positive integers and [ ] = floor. The characteristic polynomial of s is (x^j-1)(x-1)^k, which is a palindromic polynomial (sometimes called a reciprocal polynomial). The class C consists of sequences u given by the form
%C ...
%C u(n) = c(1)*[r(1)*n^k(1)] + ... + c(m)*[r(m)*n^k(m)],
%C ...
%C where c(i) are integers and r(i) are rational numbers. Assume that r(i) is in lowest terms, and let j(i) be its denominator. Then the characteristic polynomial of u is the least common multiple of all the irreducible (over the integers) factors of all the polynomials (x^j(i)-1)(x-1)^k(i). As all such factors are palindromic (indeed, they are all cyclotomic polynomials), the characteristic polynomial of u is also palindromic. In other words, if the generating function of u is written as p(x)/q(x), then q(x) is a palindromic polynomial.
%C Thus, if q(x) = q(h)x^h + ... + q(1)x + q(0),
%C then (q(h), q(h-1), ..., q(1), q(0)) is palindromic, and consequently, the recurrence coefficients for u, after excluding q(0); i.e., (- q(h-1), ... - q(1)), are palindromic. For example, row 3 of A211701 has the following recurrence: u(n) = u(n-2) + u(n-3) - u(n-5), for which q(x) = x^5 - x^3 - x^2 + 1, with recurrence coefficients (0,1,1,0,-1).
%C Recurrence coefficients (palindromic after excluding the last term) are shown here:
%C for row 1: (2, -1)
%C for row 2: (1 ,1, -1)
%C for row 3: (0, 1, 1, 0, -1)
%C for row 4: (0, 0, 1, 1, 0, 0, -1)
%C for row 5: (-1, -1, 0, 1, 2, 2, 1, 0, -1, -1, -1)
%C for row 6: (0, -1, 0, 0, 1, 1, 1, 1, 0, 0, -1, 0, -1)
%C for row 7: (-1, -2, -2, -2, -1, 0, 2, 3, 4, 4, 3, 2,
%C 0, -1, -2, -2, -2, -1, -1)
%C for row 13: (-2,-4,-7,-12,-18,-27,-37,-50,-64,-80,-95,
%C -111,-123,-133,-137,-136,-126,-110,-84,-52,
%C -12,32,80,127,173,213,246,269,281,281,269,
%C 246,213,173,127,80,32,-12,-52,-84,-110,
%C -126,-136,-137,-133,-123,-111,-95,-80,-64,
%C -50,-37,-27,-18,-12,-7,-4,-2,-1)
%e Northwest corner:
%e 1 2 3 4 5 6 7 8 9 10
%e 1 3 4 6 7 9 10 12 13 15
%e 1 3 5 7 8 11 12 14 16 18
%e 1 3 5 8 9 12 13 16 18 19
%t f[n_, m_] := Sum[Floor[n/k], {k, 1, m}]
%t TableForm[Table[f[n, m], {m, 1, 20}, {n, 1, 16}]]
%t Flatten[Table[f[n + 1 - m, m], {n, 1, 14}, {m, 1, n}]]
%Y Cf. A211702, A211703, A211704, A211705.
%K nonn,tabl
%O 1,2
%A _Clark Kimberling_, Apr 19 2012
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