# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/ Search: id:a246921 Showing 1-1 of 1 %I A246921 #8 Sep 13 2014 14:55:02 %S A246921 5,13,5,13,21,5,13,17,65,5,13,21,165,725,2805,5,13,445,1525,1853,5933, %T A246921 7925,30629,5,13,17,21,65,136,288,960,1260,4224,16128,21608,83520,5, %U A246921 13,3029,10205,11029,12773,28157,34973,42853,47965,53365,136165,184045 %N A246921 Irregular triangular array: row n gives numbers D, each being the discriminant of the minimal polynomial of a quadratic irrational represented by a continued fraction with period an n-tuple of 1s and 3s. %e A246921 First 5 rows: %e A246921 5 ... 13 %e A246921 5 ... 13 ... 21 %e A246921 5 ... 13 ... 17 .. 65 %e A246921 5 ... 13 ... 21 .. 165 .. 725 .. 2805 %e A246921 5 ... 13 ... 445 . 1525 . 1853 . 5933 . 7925 . 30629 %e A246921 The following list shows for n = 3 the purely periodic continued fractions (with period an n-tuple of 1s and 2s), each followed by the number r it represents, the minimal polynomial a*x^2 + b*x + c of r, and the discriminant, D = b^2 - 4*a*c. %e A246921 [(1,1,1)] = (1+sqrt(5))/2, -1 - x + x^2, D = 5 %e A246921 [(1,1,3)] = (-1 + sqrt(17)/2, -4 + x + x^2, D = 17 %e A246921 [(1,3,1)] = (3 + sqrt(17)/4, -1 - 3 x + 2 x^2, D = 17 %e A246921 [(3,1,1)] = (-1 + Sqrt[65])/4, -8 + x + 2 x^2, D = 65 %e A246921 [(1,3,3)] = (1 + sqrt(17)/4, -2 - x + 2 x^2, D = 17 %e A246921 [(3,1,3)] = (-3 + sqrt(65))/4, -7 + 3 x + 2 x^2, D = 65 %e A246921 [(3,3,3)] = (5 + sqrt(65))/10, -2 - 5 x + 5 x^2, D = 65 %e A246921 [(3,3,3)] = (-1 + sqrt(13))/2, -3 + x + x^2, D = 13 %e A246921 The distinct values of D are 5, 13, 17, 65, as in row 3. %t A246921 z = 7; u[n_] := u[n] = Table[MinimalPolynomial[Map[FromContinuedFraction[{1, #}] &, Tuples[{1, 3}, k]], x], {k, 1, n}]; d = Discriminant[u[z], x]; %t A246921 t = Table[Union[d[[n]]], {n, 1, z}]; TableForm[t] (* A246921 array *) %t A246921 Flatten[t] (* A246921 sequence *) %Y A246921 Cf. A246904, A246922. %K A246921 nonn,tabf,easy %O A246921 1,1 %A A246921 _Clark Kimberling_, Sep 07 2014 # Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE