

A047838


a(n) = floor(n^2/2)  1.


11



1, 3, 7, 11, 17, 23, 31, 39, 49, 59, 71, 83, 97, 111, 127, 143, 161, 179, 199, 219, 241, 263, 287, 311, 337, 363, 391, 419, 449, 479, 511, 543, 577, 611, 647, 683, 721, 759, 799, 839, 881, 923, 967, 1011, 1057, 1103, 1151, 1199, 1249, 1299, 1351, 1403
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OFFSET

2,2


COMMENTS

Define the organization number of a permutation pi_1, pi_2, ..., pi_n to be the following. Start at 1, count the steps to reach 2, then the steps to reach 3, etc. Add them up. Then the maximal value of the organization number of any permutation of [1..n] for n = 0, 1, 2, 3, ... is given by 0, 1, 3, 7, 11, 17, 23, ... (this sequence). This was established by Graham Cormode (graham(AT)research.att.com), Aug 17 2006, see link below, answering a question raised by Tom Young (mcgreg265(AT)msn.com) and Barry Cipra, Aug 15 2006
From Dmitry Kamenetsky, Nov 29 2006: (Start)
This is the length of the longest nonselfintersecting spiral drawn on an n X n grid. E.g., for n=5 the spiral has length 17:
10111
10101
10101
10001
11111 (End)
It appears that a(n+1) is the maximum number of consecutive integers (beginning with 1) that can be placed, one after another, on an npeg Towers of Hanoi, such that the sum of any two consecutive integers on any peg is a square. See the problem: http://onlinejudge.uva.es/p/v102/10276.html.  Ashutosh Mehra, Dec 06 2008
a(n) = number of (w,x,y) with all terms in {0,...,n} and w=x+yw.  Clark Kimberling, Jun 11 2012
The same sequence represents also the solution to the "pigeons problem": maximal value of the sum of the lengths of n1 line segments (connected at their endpoints) required to pass through n trail dots, with unit distance between adjacent points, visiting all of them without overlap two or more segments. In this case, a(0)=0, a(1)=1, a(2)=3, and so on.  Marco Ripà, Jan 28 2014


LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 2..10000
Graham Cormode, Notes on the organization number of a permutation.
Marco Ripà, The rectangular spiral solution for the n_1xn_2x...xn_k points problem, viXra.org.
Index entries for linear recurrences with constant coefficients, signature (2,0,2,1).


FORMULA

a(2)=1; for n>2, a(n) = a(n1) + n  1 + (n1 mod 2).  Benoit Cloitre, Jan 12 2003
T(n1) + floor(n/2)  1 = T(n)  floor((n+3)/2), where T(n) is the nth triangular number (A000217).  Robert G. Wilson v, Aug 31 2006
Equals (n1)th row sums of triangles A134151 and A135152. Also, = binomial transform of [1, 2, 2, 2, 4, 8, 16, 32, ...].  Gary W. Adamson, Nov 21 2007
G.f.: x^2(1+x+x^2x^3)/((1x)^3*(1+x)).  R. J. Mathar, Sep 09 2008
a(n) = floor((n^2+4*n+2)/2).  Gary Detlefs, Feb 10 2010
a(n) = abs(A188653(n)).  Reinhard Zumkeller, Apr 13 2011
a(n) = (2*n^2+(1)^n5)/4.  Bruno Berselli, Sep 14 2011
a(n) = a(n) = A007590(n)  1.
a(n) = A080827(n)  2.  Kevin Ryde, Aug 24 2013
a(n) = 2*a(n1)2*a(n3)+a(n4), n>4.  Wesley Ivan Hurt, Aug 06 2015


EXAMPLE

x^2 + 3*x^3 + 7*x^4 + 11*x^5 + 17*x^6 + 23*x^7 + 31*x^8 + 39*x^9 + 49*x^10 + ...


MAPLE

seq(floor((n^2+4*n+2)/2), n=0..20) # Gary Detlefs, Feb 10 2010


MATHEMATICA

Table[ Floor[n^2/2]  1, {n, 2, 60}] (* Robert G. Wilson v, Aug 31 2006 *)
LinearRecurrence[{2, 0, 2, 1}, {1, 3, 7, 11}, 60] (* Harvey P. Dale, Jan 16 2015 *)


PROG

(PARI) a(n) = n^2\2  1
(MAGMA) [Floor(n^2/2)1 : n in [2..100]]; // Wesley Ivan Hurt, Aug 06 2015


CROSSREFS

Complement of A047839. First difference is A052928.
Cf. A000217, A007590, A052928, A080827, A134151, A135151, A135152, A188653.
Sequence in context: A023234 A237662 A134707 * A188653 A262502 A211076
Adjacent sequences: A047835 A047836 A047837 * A047839 A047840 A047841


KEYWORD

nonn,easy


AUTHOR

Michael Somos, May 07 1999


EXTENSIONS

Edited by Charles R Greathouse IV, Apr 23 2010


STATUS

approved



