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A167775 Subsequence of A167708 whose indices are congruent to 1 mod 5, i.e., a(n) = A167708(5n+1).

%I A167775 #23 Sep 05 2025 00:50:10
%S A167775 10,2441,829930,282173759,95938248130,32618722190441,
%T A167775 11090269606501810,3770659047488424959,1282012985876457984250,
%U A167775 435880644538948226220041,148198137130256520456829690,50386930743642678007095874559,17131408254701380265892140520370
%N A167775 Subsequence of A167708 whose indices are congruent to 1 mod 5, i.e., a(n) = A167708(5n+1).
%C A167775 Solutions to the Pell equation a(n)^2 - 19*b(n)^2 = 81. The corresponding b(n) are in A167822. - _Klaus Purath_, Aug 31 2025
%H A167775 G. C. Greubel, <a href="/A167775/b167775.txt">Table of n, a(n) for n = 0..250</a>
%H A167775 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (340,-1).
%F A167775 a(n+2) = 340*a(n+1) - a(n).
%F A167775 a(n+1) = 170*a(n) + 39*sqrt(19*a(n)^2 - 1539).
%F A167775 G.f.: (10 + 2441*z - 10*z*340)/(1 - 340*z + z^2).
%F A167775 a(n) = ((10 + sqrt(19))/2)*(170 + 39*sqrt(19))^(n) + ((10 - sqrt(19))/2)* (170 - 39*sqrt(19))^(n).
%e A167775 a(0) = A167708(1) = 10, a(1) = A167708(6) = 2441, ...
%p A167775 u(0):=10:u(1):=2441:for n from 0 to 20 do u(n+2):=340*u(n+1)-u(n):od:seq(u(n),n=0..20); taylor(((10+2441*z-10*z*340)/(1-340*z+z^2)),z=0,20);
%t A167775 LinearRecurrence[{340, -1}, {10, 2441}, 50] (* _G. C. Greubel_, Jun 23 2016 *)
%o A167775 (Magma) I:=[10,2441]; [n le 2 select I[n] else 340*Self(n-1)-Self(n-2): n in [1..20]]; // _Vincenzo Librandi_, Jun 24 2016
%Y A167775 Cf. A167708, A167709, A167774, A167778, A167779, A167780, A167822.
%K A167775 easy,nonn,changed
%O A167775 0,1
%A A167775 _Richard Choulet_, Nov 11 2009