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A146337
Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 14.
3
118, 154, 179, 201, 212, 244, 251, 262, 286, 292, 307, 340, 347, 388, 403, 418, 422, 430, 467, 471, 474, 494, 497, 500, 519, 543, 548, 566, 587, 594, 598, 670, 683, 687, 692, 698, 699, 703, 713, 722, 742, 745, 754, 831, 833, 847, 873, 879, 932, 939, 945
OFFSET
1,1
COMMENTS
For primes in this sequence see A146359.
LINKS
EXAMPLE
a(1) = 421 because continued fraction of (1+sqrt(421))/2 = 17, 5, 3, 1, 1, 1, 2, 26, 2, 1, 1, 1, 3, 5, 13, 5, 3, 1, 1, 1, 2, 26, 2, 1, 1, 1, 3, 5, 13, 5, 3, 1, 1, 1, 2, 26... has period (5, 3, 1, 1, 1, 2, 26, 2, 1, 1, 1, 3, 5, 13) length 14.
MAPLE
A := proc(n) option remember ; local c; try c := numtheory[cfrac](1/2+sqrt(n)/2, 'periodic', 'quotients') ; RETURN(nops(c[2]) ); catch: RETURN(-1) end try ; end: isA146337 := proc(n) if A(n) = 14 then RETURN(true); else RETURN(false); fi; end: for k from 1 do if isA146337(k) then printf("%d, ", k) ; fi; od: # R. J. Mathar, Nov 08 2008
MATHEMATICA
cf14Q[n_]:=Module[{s=(1+Sqrt[n])/2}, !IntegerQ[s]&&Length[ ContinuedFraction[ s][[2]]] == 14]; Select[Range[1000], cf14Q] (* Harvey P. Dale, Oct 15 2015 *)
KEYWORD
nonn
AUTHOR
Artur Jasinski, Oct 30 2008
EXTENSIONS
More terms from R. J. Mathar, Nov 08 2008
STATUS
approved