| Displaying 1-3 of 3 results found.
page
1
Numbers k such that the continued fraction for sqrt(k) has period 2.
+10
9
3, 6, 8, 11, 12, 15, 18, 20, 24, 27, 30, 35, 38, 39, 40, 42, 48, 51, 56, 63, 66, 68, 72, 80, 83, 84, 87, 90, 99, 102, 104, 105, 110, 120, 123, 132, 143, 146, 147, 148, 150, 152, 156, 168, 171, 182, 195, 198, 200, 203, 210, 224, 227, 228, 230, 231, 235, 240, 255, 258, 260, 264
COMMENTS
This sequence is identical to the sequence of numbers of the form k = a^2 + b, where a and b are positive integers and b is a factor of 2a greater than 1, in which case the continued fraction expansion of sqrt(k) is [a; [2a/b, 2a]]. - David Terr, Jun 11 2004
REFERENCES
Kenneth H. Rosen, Elementary Number Theory and Its Applications, Addison-Wesley, 1984, page 426 (but beware of errors!)
MATHEMATICA
cf2Q[n_]:=Module[{s=Sqrt[n]}, If[IntegerQ[s], 1, Length[ ContinuedFraction[ s][[2]]]]==2]; Select[Range[300], cf2Q] (* Harvey P. Dale, Jun 21 2017 *)
Dimensions of split simple Lie algebras over any field of characteristic zero.
(Formerly M2712)
+10
3
3, 8, 10, 14, 15, 21, 24, 28, 35, 36, 45, 48, 52, 55, 63, 66, 78, 80, 91, 99, 105, 120, 133, 136, 143, 153, 168, 171, 190, 195, 210, 224, 231, 248, 253, 255, 276, 288, 300, 323, 325, 351, 360, 378, 399, 406, 435, 440, 465, 483, 496, 528, 561, 575, 595, 624, 630
REFERENCES
Freeman J. Dyson, Missed opportunities, Bull. Amer. Math. Soc. 78 (1972), 635-652.
N. Jacobson, Lie Algebras. Wiley, NY, 1962; pp. 141-146.
I. G. Macdonald, Some conjectures for root systems, SIAM J. Math. Anal., 13 (1982), 988-1007.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
EXAMPLE
The Lie algebras in question and their dimensions are the following:
A_l: l(l+2), l >= 1,
B_l: l(2l+1), l >= 2,
C_l: l(2l+1), l >= 3,
D_l: l(2l-1), l >= 4,
G_2: 14, F_4: 52, E_6: 78, E_7: 133, E_8: 248.
MAPLE
M:=4200; M2:=M^2; sa:=[seq(l*(l+2), l=1..M)]; sb:=[seq(l*(2*l+1), l=2..M)]; sd:=[seq(l*(2*l-1), l=4..M)]; se:=[14, 52, 78, 133, 248]; s:=convert(sa, set) union convert(sb, set) union convert(sd, set) union convert(se, set); t:=convert(s, list); for i from 1 to nops(t) do if t[i] <= M2 then lprint(i, t[i]); fi; od:
MATHEMATICA
max = 26; sa = Table[ k*(k+2), {k, 1, max}]; sb = Table[ k*(2k+1), {k, 2, max}]; sd:= Table[ k*(2k-1), {k, 4, max}]; se = {14, 52, 78, 133, 248}; Select[ Union[sa, sb, sd, se], # <= max^2 &](* Jean-François Alcover, Nov 18 2011, after Maple *)
PROG
(Haskell)
import Data.Set (deleteFindMin, fromList, insert)
a003038 n = a003038_list !! (n-1)
a003038_list = f (fromList (3 : [14, 52, 78, 133, 248]))
(drop 2 a005563_list) (drop 4 a000217_list) where
f s (x:xs) (y:ys) = m : f (x `insert` (y `insert` s')) xs ys where
(m, s') = deleteFindMin s
EXTENSIONS
More terms from Pab Ter (pabrlos(AT)yahoo.com), May 09 2004
1, 1, 1, 2, 1, 2, 4, 2, 2, 5, 9, 4, 4, 5, 13, 21, 9, 8, 10, 13, 35, 51, 21, 18, 20, 26, 35, 96, 127, 51, 42, 45, 52, 70, 96, 267, 323, 127, 102, 105, 117, 140, 192, 267, 750, 835, 323, 254, 255, 273, 315, 384, 534, 720, 2123, 2188, 835, 646, 635, 663, 735, 864, 1068
COMMENTS
Left border = Motzkin numbers, A001006. Row sums = A005773 shifted: (1, 2, 5, 13, 35, 96, 267,...). Sum of n-th row terms = rightmost term of next row.
EXAMPLE
First few rows of the triangle =
1;
1, 1;
2, 1, 2;
4, 2, 2, 5;
9, 4, 4, 5, 13;
21, 9, 8, 10, 13, 35;
51, 21, 18, 20, 26, 35, 96;
127, 51, 42, 45, 52, 70, 96, 267;
323, 127, 102, 105, 117, 140, 192, 267, 750;
835, 323, 254, 255, 273, 315, 384, 534, 720, 2123;
...
Row 3 = (4, 2, 2, 5) = termwise product of (4, 2, 1, 1) and the first 4 terms of A005773: (1, 1, 2, 5) = (4*1, 2*1, 1*2, 1*5). (4, 2, 1, 1) = the first 4 terms of A001066, reversed.
Search completed in 0.010 seconds
| 