A259911 - OEIS

%I #11 Nov 28 2024 11:14:18

%S 5,12,12,21,8,21,8,60,60,8,5,24,13,24,5,60,140,12,12,140,60,77,12,285,

%T 5,285,12,77,24,28,44,120,120,44,28,24,13,5,21,168,29,168,21,5,13,140,

%U 44,168,56,1020,1020,56,168,44,140,165,120,93,8,1365,40,1365,8,93,120,165

%N Triangular array; row k shows the discriminant of the field of the number having purely periodic continued fraction with period (j,k+1-j), for j=1..k.

%H Clark Kimberling, <a href="/A259911/b259911.txt">Table of n, a(n) for n = 1..1000</a>

%e First eight rows:

%e   5

%e   12    12

%e   21     8    21

%e    8    60    60     8

%e    5    24    13    24     5

%e   60   140    12    12   140    60

%e   77    12   285     5   285    12    77

%e   24    28    44   120   120    44    28    24

%e The number whose continued fraction is periodic with period (1,1) is the golden ratio, (1+sqrt(5))/2, so that the number in row 1 is 5.

%e As a square array A(n,k) read by antidiagonals, where A(n,k) corresponds to the continued fraction with pure period (n,k):

%e    5,  12,  21,   8,    5,   60,   77,  24, ...

%e   12,   8,  60,  24,  140,   12,   28,   5, ...

%e   21,  60,  13,  12,  285,   44,   21, 168, ...

%e    8,  24,  12,   5,  120,  168,   56,   8, ...

%e    5, 140, 285, 120,   29, 1020, 1365, 440, ...

%e   60,  12,  44, 168, 1020,   40, 1932, 156, ...

%e   77,  28,  21,  56, 1365, 1932,   53, 840, ...

%e   24,   5, 168,   8,  440,  156,  840,  17, ...

%e   ...

%t v = Table[FromContinuedFraction[{j, {k + 1 - j, j}}], {k, 1, 20}, {j, 1, k}];

%t TableForm[NumberFieldDiscriminant[v]]

%Y Cf. A259912 (main diagonal of square array), A259913 (first column).

%K nonn,tabl,easy

%O 1,1

%A _Clark Kimberling_, Jul 20 2015