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%I #17 May 06 2022 21:08:56
%S 2,1,4,1,3,1,8,2,1,2,1,12,2,1,1,3,1,1,5,2,1,1,9,2,1,20,6,2,2,1,3,2,1,
%T 1,4,3,1,1,1,6,2,2,1,1,28,9,1,1,15,7,3,2,1,1,32,4,2,2,1,1,1,8,1,1,1,1,
%U 5,2,1,1,1,1,6,3,2,1,1,1,40,4,2,1,1,1,1,21,5,2,2,1,1,1,14,6,2,2,1,1,1
%N a(n) is the second partial quotient in the simple continued fraction for sqrt(prime(n)).
%F a(n) = floor(1/(sqrt(prime(n))-floor(sqrt(prime(n))))), where prime(n) is the n-th prime.
%F a(n) = floor(2*s/r) where s = floor(sqrt(p)) = A000006(n), r = p - s^2 = A056892(n), and p = prime(n). - _Kevin Ryde_, May 06 2022
%e For n=8, prime(n)=19, floor(sqrt(19))=4 and 1/(sqrt(19)-4) = 2.786..., so a(8)=2.
%t a[n_] := Floor[1/(Sqrt[Prime[n]]-Floor[Sqrt[Prime[n]]])]
%o (PARI) a(n) = my(r); sqrtint(prime(n),&r)<<1 \ r; \\ _Kevin Ryde_, May 06 2022
%Y Cf. A000006, A000040, A056892.
%K nonn,easy
%O 1,1
%A _Roger L. Bagula_, Feb 01 2002
%E Edited by _Dean Hickerson_, Feb 14 2002
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