

A008590


Multiples of 8.


41



0, 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184, 192, 200, 208, 216, 224, 232, 240, 248, 256, 264, 272, 280, 288, 296, 304, 312, 320, 328, 336, 344, 352, 360, 368, 376, 384, 392, 400, 408, 416, 424, 432
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OFFSET

0,2


COMMENTS

For n > 3, the number of squares on the infinite 4column halfstrip chessboard at <= n knight moves from any fixed point on the short edge.
First differences of odd squares: a(n) = A016754(n)  A016754(n1) for n > 0.  Reinhard Zumkeller, Nov 08 2009
Complement of A047592; A168181(a(n)) = 0.  Reinhard Zumkeller, Nov 30 2009
For n >= 1, number of pairs (x, y) of Z^2, such that max(abs(x), abs(y)) = n.  Michel Marcus, Nov 28 2014
These terms are the area of square frames (using integer lengths), with specific instances where the area equals the sum of inner and outer perimeters (see example and formula below). The thickness of the frames are always 2, which is of further significance when considering that all regular polygons have an area that is equal to perimeter when apothem is 2.  Peter M. Chema, Apr 03 2016
From Lechoslaw Ratajczak, Sep 03 2017: (Start)
Conjecture: let gcd_2(b,c) be the second greatest common divisor and lcd_2(b,c) be the second least common divisor of not coprime integers b and c. Consecutive elements of this sequence (for a(n) > 0) are consecutive integers m for which both Sum_{k=1..m, gcd(k,m)<>1} gcd_2(k,m) and Sum_{k=1..m, gcd(k,m) <>1} lcd_2(k,m) are even numbers.
a(1) = 8 because 1+2+1+4 = 8 (8 is even) and 2+2+2+2 = 8 (8 is even).
a(2) = 16 because 1+2+1+4+1+2+1+8 = 20 (20 is even) and 2+2+2+2+2+2+2+2 = 16 (16 is even).
a(3) = 24 because 1+1+2+3+4+1+1+6+1+1+4+3+2+1+1+12 = 44 (44 is even) and 2+3+2+2+2+3+2+2+2+3+2+2+2+3+2+2 = 36 (36 is even).
The conjecture was checked for 5*10^4 consecutive integers. (End)


LINKS

Ivan Panchenko, Table of n, a(n) for n = 0..200
Tanya Khovanova, Recursive Sequences
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 320
Luis Manuel Rivera, Integer sequences and kcommuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014.
Index entries for linear recurrences with constant coefficients, signature (2,1)


FORMULA

a(n) = (2*n+1)^2  (2*n1)^2.  Xavier Acloque, Oct 22 2003
From Vincenzo Librandi, Dec 24 2010: (Start)
a(n) = 8*n = 2*a(n1)  a(n2).
G.f.: 8*x/(x1)^2. (End)
a(n) = Sum_{k=1..4n} (i^k + 1)*(i^(4nk) + 1), where i=sqrt(1).  Bruno Berselli, Mar 19 2012
a(n) = (n+2)^2  (n2)^2 = 4*(n+2) + 4*(n2), as exemplified below.  Peter M. Chema, Apr 03 2016


EXAMPLE

Beginning with n = 2, illustration of the terms as the area of square frames, where area equals the sum of inner and outer perimeters:
_ _ _ _ _ _ _ _
_ _ _ _ _ _ _  
_ _ _ _ _ _    _ _ _ _ 
_ _ _ _ _    _ _ _     
_ _ _ _    _ _         
   _             
   _   _ _   _ _ _   _ _ _ _ 
         
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
a(2) = 16 a(3) = 24 a(4) = 32 a(5) = 40 a(6) = 48
The inner square has side n2 and outer square side n+2, pursuant to the above and related formula. Note that a(2) is simply the square 4*4, with the inner square having side 0; considering the inner square as a center point, this frame also has thickness of 2.
E.g., for a(4), the square frame is formed by a 6 X 6 outer square and a 2 X 2 inner square, with the area (6 X 6 minus 2 X 2) equal to the perimeter (4*6 + 4*2) at 32.  Peter M. Chema, Apr 03 2016


MATHEMATICA

Table[8*n, {n, 0, 5!}] (* Vladimir Joseph Stephan Orlovsky, Mar 03 2010 *)


PROG

(Haskell)
a008590 = (* 8)
a008590_list = [0, 8..]  Reinhard Zumkeller, Apr 02 2012
(PARI) a(n) = 8*n; \\ Altug Alkan, Apr 08 2016


CROSSREFS

Cf. A010014.
Essentially the same as A022144.
Subsequence of A185359, apart initial 0.
Sequence in context: A185359 A022144 A181390 * A186544 A061824 A085131
Adjacent sequences: A008587 A008588 A008589 * A008591 A008592 A008593


KEYWORD

nonn,easy,changed


AUTHOR

N. J. A. Sloane


STATUS

approved



