|
| |
|
|
A054925
|
|
Ceiling(n*(n-1)/4).
|
|
5
| |
|
|
0, 0, 1, 2, 3, 5, 8, 11, 14, 18, 23, 28, 33, 39, 46, 53, 60, 68, 77, 86, 95, 105, 116, 127, 138, 150, 163, 176, 189, 203, 218, 233, 248, 264, 281, 298, 315, 333, 352, 371, 390, 410, 431, 452, 473, 495, 518, 541, 564, 588, 613, 638, 663, 689, 716, 743, 770, 798
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,4
|
|
|
COMMENTS
| Number of edges in "median" graph - gives positions of largest entries in rows of table in A054924.
Form the clockwise spiral starting 0,1,2,....; then A054925(n+1) interleaves 2 horizontal (A033951, A033991) and 2 vertical (A007742, A054552) branches. A bisection is A014848. - Paul Barry (pbarry(AT)wit.ie), Oct 08 2007
|
|
|
FORMULA
| Euler transform of length 6 sequence [ 2, 0, 1, 1, 0, -1]. - Michael Somos Sep 02 2006
G.f.: x^2(x^2-x+1)/((1-x)^3(1+x^2))=x^2(1-x^6)/((1-x)^2(1-x^3)(1-x^4)). a(1-n)=a(n). - Michael Somos Feb 11 2004
|
|
|
MAPLE
| seq(ceil(binomial(n, 2)/2), n=0..57); # [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jan 12 2009]
|
|
|
PROG
| (PARI) a(n)=ceil(n*(n-1)/4)
(Other) sage: [ceil(binomial(n, 2)/2) for n in xrange(0, 58)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Dec 01 2009]
|
|
|
CROSSREFS
| Cf. A054924, A011848(n)=a(-n). A054925 + A011848 = C(n, 2).
Sequence in context: A071894 A078444 A194221 * A194248 A126097 A024611
Adjacent sequences: A054922 A054923 A054924 * A054926 A054927 A054928
|
|
|
KEYWORD
| nonn
|
|
|
AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com), May 24 2000
|
| |
|
|
