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 1-4 of 4 results.

A106253 First difference of A106252.

Original entry on oeis.org

2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2
Offset: 1

Views

Author

John W. Layman, Apr 27 2005

Keywords

Comments

It appears that this sequence is periodic with period 6.

Crossrefs

Cf. A106252.
Cf. A083039. - R. J. Mathar, Aug 24 2008

Programs

  • Mathematica
    LinearRecurrence[{-1, 0, 1, 1},{2, 2, 2, 1},105] (* Ray Chandler, Aug 26 2015 *)

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

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Apr 19 2012

Keywords

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)

Examples

			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
		

Crossrefs

Programs

  • 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}]]

A212536 T(n,k)=Number of nondecreasing sequences of n 1..k integers with every element dividing the sequence sum.

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 4, 3, 3, 1, 5, 4, 5, 3, 1, 6, 5, 7, 5, 4, 1, 7, 6, 8, 10, 8, 4, 1, 8, 7, 11, 12, 15, 8, 5, 1, 9, 8, 12, 17, 21, 15, 12, 5, 1, 10, 9, 14, 18, 30, 21, 24, 12, 6, 1, 11, 10, 16, 23, 33, 40, 33, 29, 16, 6, 1, 12, 11, 18, 26, 46, 44, 69, 40, 39, 16, 7, 1, 13, 12, 19, 30, 53, 64, 83, 91
Offset: 1

Views

Author

R. H. Hardin May 20 2012

Keywords

Comments

Table starts
.1.2..3..4..5...6...7...8...9..10..11...12...13...14...15...16...17...18...19
.1.2..3..4..5...6...7...8...9..10..11...12...13...14...15...16...17...18...19
.1.3..5..7..8..11..12..14..16..18..19...22...23...25...27...29...30...33...34
.1.3..5.10.12..17..18..23..26..30..31...40...41...43...47...52...53...59...60
.1.4..8.15.21..30..33..46..53..66..67...87...88...95..111..125..126..143..144
.1.4..8.15.21..40..44..64..76.103.104..148..149..165..197..229..230..271..272
.1.5.12.24.33..69..83.116.145.188.193..290..293..332..428..496..497..606..607
.1.5.12.29.40..91.106.161.202.266.272..474..478..561..747..874..876.1141.1142
.1.6.16.39.57.130.157.245.331.439.455..867..878.1034.1417.1646.1651.2236.2240
.1.6.16.45.70.166.200.334.451.644.665.1424.1440.1713.2384.2785.2793.3927.3932

Examples

			Some solutions for n=8 k=4
..1....1....2....1....1....1....1....2....1....1....1....1....2....2....1....1
..1....2....2....2....1....1....1....3....1....1....1....1....2....2....2....1
..2....3....3....2....1....1....1....3....1....2....1....2....2....4....2....2
..2....3....3....2....1....1....1....3....1....2....1....2....2....4....3....2
..2....3....3....2....1....2....2....3....1....2....1....2....4....4....4....3
..4....4....3....3....2....2....2....3....1....2....1....2....4....4....4....3
..4....4....4....3....2....4....2....3....2....2....2....2....4....4....4....3
..4....4....4....3....3....4....2....4....2....2....4....4....4....4....4....3
		

Crossrefs

Column 3 is A000212(floor((n+5)/2))
Row 3 is A106252

A299647 Positive solutions to x^2 == -2 (mod 11).

Original entry on oeis.org

3, 8, 14, 19, 25, 30, 36, 41, 47, 52, 58, 63, 69, 74, 80, 85, 91, 96, 102, 107, 113, 118, 124, 129, 135, 140, 146, 151, 157, 162, 168, 173, 179, 184, 190, 195, 201, 206, 212, 217, 223, 228, 234, 239, 245, 250, 256, 261, 267, 272, 278, 283, 289, 294, 300, 305, 311, 316
Offset: 1

Views

Author

Bruno Berselli, Mar 06 2018

Keywords

Comments

Positive numbers congruent to {3, 8} mod 11.
Equivalently, interleaving of A017425 and A017485.

Crossrefs

Subsequence of A106252, A279000.
Cf. A017497: positive solutions to x == -2 (mod 11).
Cf. A017437: positive solutions to x^3 == -2 (mod 11).
Nonnegative solutions to x^2 == -2 (mod j): A005843 (j=2), A001651 (j=3), A047235 (j=6), A156638 (j=9), this sequence (j=11).

Programs

  • GAP
    List([1..60], n -> 5*n-2+(2*n-(-1)^n-3)/4);
    
  • Julia
    [(11(2n-1)-(-1)^n)>>2 for n in 1:60] # Peter Luschny, Mar 07 2018
  • Magma
    [5*n-2+(2*n-(-1)^n-3)/4: n in [1..60]];
    
  • Mathematica
    Table[5 n - 2 + (2 n - (-1)^n - 3)/4, {n, 1, 60}]
    CoefficientList[ Series[(3 + 5x + 3x^2)/((x - 1)^2 (x + 1)), {x, 0, 57}], x] (* or *)
    LinearRecurrence[{1, 1, -1}, {3, 8, 14}, 58] (* Robert G. Wilson v, Mar 08 2018 *)
  • Maxima
    makelist(5*n-2+(2*n-(-1)^n-3)/4, n, 1, 60);
    
  • PARI
    vector(60, n, nn; 5*n-2+(2*n-(-1)^n-3)/4)
    
  • Python
    [5*n-2+(2*n-(-1)**n-3)/4 for n in range(1, 60)]
    
  • Sage
    [5*n-2+(2*n-(-1)^n-3)/4 for n in (1..60)]
    

Formula

O.g.f.: x*(3 + 5*x + 3*x^2)/((1 + x)*(1 - x)^2).
E.g.f.: (-1 + 12*exp(x) - 11*exp(2*x) + 22*x*exp(2*x))*exp(-x)/4.
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = 5*n - 2 + (2*n - (-1)^n - 3)/4.
a(n) = 4*n - 1 + floor((n - 1)/2) + floor((3*n - 1)/3).
a(n+k) - a(n) = 11*k/2 + (1 - (-1)^k)*(-1)^n/4.
a(n+k) + a(n) = 11*(2*n + k - 1)/2 - (1 + (-1)^k)*(-1)^n/4.
E.g.f.: 3 + ((22*x - 11)*exp(x) - exp(-x))/4. - David Lovler, Aug 08 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(5*Pi/22)*Pi/11. - Amiram Eldar, Feb 27 2023
From Amiram Eldar, Nov 23 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = cosec(3*Pi/22)/2.
Product_{n>=1} (1 + (-1)^n/a(n)) = sec(5*Pi/22)*sin(2*Pi/11). (End)
Showing 1-4 of 4 results.