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A322709 a(0)=0, a(1)=7 and a(n) = 30*a(n-1) - a(n-2) + 14 for n > 1.

%I A322709 #37 Mar 16 2023 08:18:54
%S A322709 0,7,224,6727,201600,6041287,181037024,5425069447,162571046400,
%T A322709 4871706322567,145988618630624,4374786852596167,131097616959254400,
%U A322709 3928553721925035847,117725514040791821024,3527836867501829594887,105717380511014096025600,3167993578462921051173127
%N A322709 a(0)=0, a(1)=7 and a(n) = 30*a(n-1) - a(n-2) + 14 for n > 1.
%C A322709 Also numbers k such that 7*A000217(k) is a square. - _Metin Sariyar_, Nov 16 2019
%H A322709 Colin Barker, <a href="/A322709/b322709.txt">Table of n, a(n) for n = 0..650</a>
%H A322709 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (31,-31,1).
%F A322709 sqrt(a(n)+1) + sqrt(a(n)) = (sqrt(8) + sqrt(7))^n.
%F A322709 sqrt(a(n)+1) - sqrt(a(n)) = (sqrt(8) - sqrt(7))^n.
%F A322709 a(n) = 31*a(n-1) - 31*a(n-2) + a(n-3) for n > 2.
%F A322709 From _Colin Barker_, Dec 25 2018: (Start)
%F A322709 G.f.: 7*x*(1 + x) / ((1 - x)*(1 - 30*x + x^2)).
%F A322709 a(n) = ((15+4*sqrt(14))^(-n) * (-1+(15+4*sqrt(14))^n)^2) / 4.
%F A322709 (End)
%F A322709 E.g.f.: (1/4)*(-2*exp(x) + exp((15-4*sqrt(14))*x) + exp((15+4*sqrt(14))*x)). - _Stefano Spezia_, Nov 16 2019
%F A322709 2*a(n) = A068203(n)-1. - _R. J. Mathar_, Mar 16 2023
%e A322709 (sqrt(8) + sqrt(7))^2 = 15 + 2*sqrt(56) = sqrt(225) + sqrt(224). So a(2) = 224.
%t A322709 LinearRecurrence[{31,-31,1}, {0, 7, 224}, 18] (* _Metin Sariyar_, Nov 23 2019 *)
%o A322709 (PARI) concat(0, Vec(7*x*(1 + x) / ((1 - x)*(1 - 30*x + x^2)) + O(x^20))) \\ _Colin Barker_, Dec 25 2018
%o A322709 (Magma) a:=[0,7]; [n le 2 select a[n] else 30*Self(n-1)-Self(n-2)+14: n in [1..18]]; // _Marius A. Burtea_, Nov 16 2019
%o A322709 (Magma) R<x>:=PowerSeriesRing(Integers(), 18); [0] cat Coefficients(R!(7*x*(1 + x) / ((1 - x)*(1-30*x + x^2))));  // _Marius A. Burtea_, Nov 16 2019
%Y A322709 Row 7 of A322699.
%Y A322709 Cf. A188932 (sqrt(7)+sqrt(8)).
%K A322709 nonn,easy
%O A322709 0,2
%A A322709 _Seiichi Manyama_, Dec 24 2018