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Least prime p such that the continued fraction expansion of its square root contains the first n natural numbers, but does not contain n+1.
1

%I #24 Oct 19 2024 08:33:44

%S 13,2,127,19,211,463,919,1741,1951,2539,4861,8521,8719,9811,10651,

%T 21319,25309,19609,29527,42379,61879,58171,89959,97579,144271,135319,

%U 164431,217519,201919,230101,216451,289111,307759,323359,558979,468199,488791

%N Least prime p such that the continued fraction expansion of its square root contains the first n natural numbers, but does not contain n+1.

%C Version of A187261 for prime numbers.

%C a(n) >= A187261(n) and a(n) = A187261(n) if A187261(n) is prime.

%C a(n) = A187261(n) for n's: 2,4,5,6,9,14,15,16,18,19,20,22,23,24,25,26,27,28,29,30,31,33,..

%C Among first 100 terms, the largest is a(96)=48169339, less than this there are also a(102)=44302171 and a(105)=47106151.

%H Zak Seidov, <a href="/A185808/b185808.txt">Table of n, a(n) for n = 1..100</a>

%e a(1) = 13 because the c.f. (c.f.=continued fraction) of sqrt(13) = 3,{1,1,1,1,6}, and c.f. contains 1.

%e a(2) = 2 because the c.f. of sqrt(2) = 1,{2}, and c.f. contains 1..2.

%e a(3) = 127 because the c.f. of sqrt(127) = 11,{3,1,2,2,7,11,7,2,2,1,3,22}, and c.f. contains 1..3.

%e a(4) = 19 because the c.f. of sqrt(19) = 4, {2, 1, 3, 1, 2, 8}, and c.f. contains 1..4.

%Y Cf. A187261.

%K nonn

%O 1,1

%A _Zak Seidov_, Mar 08 2011