A013642
Numbers k such that the continued fraction for sqrt(k) has period 2.
Original entry on oeis.org
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
Offset: 1
- Kenneth H. Rosen, Elementary Number Theory and Its Applications, Addison-Wesley, 1984, page 426 (but beware of errors!).
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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 *)
A003038
Dimensions of split simple Lie algebras over any field of characteristic zero.
Original entry on oeis.org
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
Offset: 1
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.
- 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).
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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
-- Reinhard Zumkeller, Dec 16 2012
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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:
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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 *)
More terms from Pab Ter (pabrlos(AT)yahoo.com), May 09 2004
A144025
Eigentriangle by rows, A001006(n-k)*A005773(k); 0<=k<=n.
Original entry on oeis.org
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, 750, 2123, 2188, 835, 646, 635, 663, 735, 864, 1068, 1500, 2123, 6046
Offset: 0
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, 750, 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.
Showing 1-3 of 3 results.
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