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Array T(n,k) read by antidiagonals, where row n is the increasing sequence of numbers m for which the simple continued fraction of sqrt(m) has period n, n >= 0, k >= 1.
3

%I #11 Nov 24 2024 09:25:41

%S 1,2,4,3,5,9,41,6,10,16,7,130,8,17,25,13,14,269,11,26,36,19,29,23,370,

%T 12,37,49,58,21,53,28,458,15,50,64,31,73,22,74,32,697,18,65,81,106,44,

%U 202,45,85,33,986,20,82,100,43,113,69,250,52,89,34,1313,24,101,121

%N Array T(n,k) read by antidiagonals, where row n is the increasing sequence of numbers m for which the simple continued fraction of sqrt(m) has period n, n >= 0, k >= 1.

%C A permutation of the positive integers.

%e Array begins:

%e n\k| 1 2 3 4 5 6 7 8 9 10 11

%e ---+------------------------------------------------

%e 0 | 1 4 9 16 25 36 49 64 81 100 121

%e 1 | 2 5 10 17 26 37 50 65 82 101 122

%e 2 | 3 6 8 11 12 15 18 20 24 27 30

%e 3 | 41 130 269 370 458 697 986 1313 1325 1613 1714

%e 4 | 7 14 23 28 32 33 34 47 55 60 62

%e 5 | 13 29 53 74 85 89 125 173 185 218 229

%e 6 | 19 21 22 45 52 54 57 59 70 77 88

%e 7 | 58 73 202 250 274 314 349 425 538 761 1010

%e 8 | 31 44 69 71 91 92 108 135 153 158 160

%e 9 | 106 113 137 149 265 389 493 610 698 754 970

%e 10 | 43 67 86 93 115 116 118 129 154 159 161

%e The least n for which CF(sqrt(n)) has period of length 4 is n=7, with CF=[2;1,1,1,4,1,1,1,4,1,1,1,4,...]; thus T(4,1)=7.

%e [The array T(n,k) is indexed by n=0,1,2,3,..., k=1,2,3... .]

%e Row 0 consists of squares: 1,4,9,...

%Y Cf. A002522, A003285, A013642, A091450, A091451, A091453.

%Y Rows 0-100 are: A000290 (except the initial 0), A002522 (except the initial 1), A013642, A013643, A013644, A010337, A020347, A010338, A020348, A010339, A020349-A020439.

%K nonn,tabl

%O 0,2

%A _Clark Kimberling_, Feb 03 2004

%E a(17) = T(3,3) corrected by _Pontus von Brömssen_, Nov 23 2024