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Values of y in x^2 - 49 = 2*y^2.
2

%I #20 Aug 29 2022 10:34:21

%S 4,6,14,30,40,84,176,234,490,1026,1364,2856,5980,7950,16646,34854,

%T 46336,97020,203144,270066,565474,1184010,1574060,3295824,6900916,

%U 9174294,19209470,40221486,53471704,111960996,234428000,311655930,652556506

%N Values of y in x^2 - 49 = 2*y^2.

%C The expression 2*n^2 + c with c = 49 yields more squares than any other value of c in the range 1 < c < 100 and n < 5*10^4. - _K. D. Bajpai_, Nov 04 2013

%H G. C. Greubel, <a href="/A106526/b106526.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,6,0,0,-1).

%F a(n) = 6*a(n-3) - a(n-6), with initial terms 4, 6, 14, 30, 40, 84. - _T. D. Noe_, Nov 04 2013

%F From _G. C. Greubel_, Aug 12 2021: (Start)

%F a(n) = 2*A276600(n+1).

%F G.f.: (2*x)*(2 + 3*x + 7*x^2 + 3*x^3 + 2*x^4)/(1 - 6*x^3 + x^6). (End)

%e a(12) = 2856; as 12 mod 3 = 0, a(12) = 14*A001109(12/3) = 204*14 = 2856; also 2*2856^2 = 4039^2 - 49, i.e., A106525(12)^2 - 49;

%e a(13) = 5980; as 13 mod 3 = 1, a(13) = A001109(4+2) - A001109(4+1) + A001109(4) + A001109(4-1) = 6930 - 1189 + 204 + 35 = 5980; also 2*5980^2 = 8457^2 - 49, i.e., A106525(13)^2 - 49;

%e a(14) = 7950; as 14 mod 3 = 2, a(14) = A001109(4+2) + A001109(4+1) - A001109(4) + A001109(4-1) = 6930 + 1189 - 204 + 35 = 7950; also 2*7950^2 = 11243^2 - 49, i.e., A106525(14)^2 - 49.

%t LinearRecurrence[{0,0,6,0,0,-1}, {4,6,14,30,40,84}, 40] (* _T. D. Noe_, Nov 04 2013 *)

%o (Magma) I:=[4,6,14,30,40,84]; [n le 6 select I[n] else 6*Self(n-3) - Self(n-6): n in [1..41]]; // _G. C. Greubel_, Aug 12 2021

%o (Sage)

%o def A106526_list(prec):

%o P.<x> = PowerSeriesRing(ZZ, prec)

%o return P( (2*x)*(2 +3*x +7*x^2 +3*x^3 +2*x^4)/(1 -6*x^3 +x^6) ).list()

%o a=A106526_list(41); a[1:] # _G. C. Greubel_, Aug 12 2021

%Y Cf. A001109, A106525, A276600.

%K nonn

%O 1,1

%A Andras Erszegi (erszegi.andras(AT)chello.hu), May 07 2005