A211701 Rectangular array by antidiagonals, n >= 1, k >= 1: R(n,k) = n + [n/2] + ... + [n/k], where [ ]=floor.

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, 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, 14, 10, 8, 5, 3, 1, 12, 16, 18, 18, 17, 15, 14, 10, 8, 5, 3, 1, 13, 18, 19, 20, 19

Offset

1,2

Comments

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.

Limiting row: A000618=(1,3,5,8,10,14,16,20,...).

Row 1: A000027

Row 2: A032766

Row 3: A106252

Row 4: A211703

Row 5: A211704

R(n,n) = A000618(n)

...

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

...

u(n) = c(1)*[r(1)*n^k(1)] + ... + c(m)*[r(m)*n^k(m)],

...

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.

Thus, if q(x) = q(h)x^h + ... + q(1)x + q(0),

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).

Recurrence coefficients (palindromic after excluding the last term) are shown here:

for row 1: (2, -1)

for row 2: (1 ,1, -1)

for row 3: (0, 1, 1, 0, -1)

for row 4: (0, 0, 1, 1, 0, 0, -1)

for row 5: (-1, -1, 0, 1, 2, 2, 1, 0, -1, -1, -1)

for row 6: (0, -1, 0, 0, 1, 1, 1, 1, 0, 0, -1, 0, -1)

for row 7: (-1, -2, -2, -2, -1, 0, 2, 3, 4, 4, 3, 2,

0, -1, -2, -2, -2, -1, -1)

for row 13: (-2,-4,-7,-12,-18,-27,-37,-50,-64,-80,-95,

-111,-123,-133,-137,-136,-126,-110,-84,-52,

-12,32,80,127,173,213,246,269,281,281,269,

246,213,173,127,80,32,-12,-52,-84,-110,

-126,-136,-137,-133,-123,-111,-95,-80,-64,

-50,-37,-27,-18,-12,-7,-4,-2,-1)

Links

Table of n, a(n) for n=1..83.

Example

Northwest corner:
  1   2   3   4   5   6   7   8   9  10
  1   3   4   6   7   9  10  12  13  15
  1   3   5   7   8  11  12  14  16  18
  1   3   5   8   9  12  13  16  18  19

Mathematica

f[n_, m_] := Sum[Floor[n/k], {k, 1, m}]
TableForm[Table[f[n, m], {m, 1, 20}, {n, 1, 16}]]
Flatten[Table[f[n + 1 - m, m], {n, 1, 14}, {m, 1, n}]]

Crossrefs

Cf. A211702, A211703, A211704, A211705.

Sequence in context: A210258 A181108 A211782 * A183110 A117895 A188002

Adjacent sequences: A211698 A211699 A211700 * A211702 A211703 A211704

Keyword

nonn,tabl

Author

Clark Kimberling, Apr 19 2012

Status

approved