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The list of the B values in the common solutions to the 2 equations 15*k + 1 = A^2, 19*k + 1 = B^2.
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%I #18 Jan 30 2024 13:44:30

%S 1,18,305,5167,87534,1482911,25121953,425590290,7209912977,

%T 122142930319,2069219902446,35054595411263,593858902089025,

%U 10060546740102162,170435435679647729,2887341859813909231,48914376181156809198,828657053219851847135,14038255528556324592097

%N The list of the B values in the common solutions to the 2 equations 15*k + 1 = A^2, 19*k + 1 = B^2.

%C The case C=15 of finding k such that C*k+1 and (C+4)*k+2 are both perfect squares (A160682).

%C The 2 equations are equivalent to the Pell equation x^2 - 285*y^2 = 1, with x = (285*k+17)/2 and y = A*B/2.

%H Andersen, K., Carbone, L. and Penta, D., <a href="https://pdfs.semanticscholar.org/8f0c/c3e68d388185129a56ed73b5d21224659300.pdf">Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields</a>, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (17, -1).

%F B(t+2) = 17*B(t+1) - B(t).

%F B(t) = ((285+19*w)*((17+w)/2)^(t-1)+(285-19*w)*((17-w)/2)^(t-1))/570 where w=sqrt(285).

%F G.f.: (1+x)*x/(1-17*x+x^2).

%p t:=0: for b from 1 to 1000000 do a:=sqrt((15*b^2+4)/19):

%p if (trunc(a)=a) then t:=t+1: n:=(b^2-1)/19: print(t,a,b,n): end if: end do:

%t LinearRecurrence[{17,-1},{1,18},30] (* _Harvey P. Dale_, Jan 30 2024 *)

%o (Sage) [(lucas_number2(n,17,1)-lucas_number2(n-1,17,1))/15 for n in range(1, 20)] # _Zerinvary Lajos_, Nov 10 2009

%Y Cf. A160682, A161595 (sequence of A), A161583 (sequence of k).

%K nonn

%O 1,2

%A _Paul Weisenhorn_, Jun 14 2009

%E Edited, extended by _R. J. Mathar_, Sep 02 2009