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Denominators of the continued fraction convergents to sqrt(7)/2.
4

%I #23 Nov 22 2017 08:57:47

%S 1,3,31,96,223,765,7873,24384,56641,194307,1999711,6193440,14386591,

%T 49353213,507918721,1573109376,3654137473,12535521795,129009355423,

%U 399563588064,928136531551,3183973182717,32767868358721,101487578258880,235743024876481,808716652888323,8322909553759711,25777445314167456,59877800182094623,205410845860451325

%N Denominators of the continued fraction convergents to sqrt(7)/2.

%C The numerators are given in A294972.

%C The continued fraction of sqrt(7)/2 is [1, repeat(3,10,3,2)].

%H Colin Barker, <a href="/A294973/b294973.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,254,0,0,0,-1).

%F From _Colin Barker_, Nov 19 2017: (Start)

%F G.f.: (1 + 3*x - x^2)*(1 + 32*x^2 + x^4) / ((1 - 16*x^2 + x^4)*(1 + 16*x^2 + x^4)).

%F a(n) = 254*a(n-4) - a(n-8) for n > 7.

%F (End)

%F The g.f. is correct: denominator recurrence a(n) = b(n)*a(n-1) + a(n-2), a(-1) = 0, a(0) = 1, (a(-2) = a(0) = 1) with b(n) modulo 4 from the continued fraction given above: b(0) = 1, b(4*(k+1)) = 2, b(4*k+1) = 3 = b(4*k+3) and b(4*k+2) = 10, for k >= 0. The 4-section is G(x) = Sum_{k>=0} a(k)*x^k = G_0(x^4) + x*G_1(x^4) + x^2*G_2(x^4) + x^3*G_3(x^4) with G_j(x) = Sum_{k>=0} a(4*k+j)*x^k, for j=0..3. The recurrence leads to four equations (omit the x here): G_1 = 3*G_0 + x*G_3, G_2 = 10*G_1 + G_4, G_3 = 3*G_2 + G_1, G_0 = 2*x*G_3 +1 + x*G_2 (using a(-2) = 1). This can be solved to obtain for G(x) = (1 + 3*x + 31*x^2 + 96*x^3 - 31*x^4 + 3*x^5 - x^6)/(1 - 254*x^4 + x^8), and the numerator and denominator factorize like given in the above conjecture. - _Wolfdieter Lang_, Nov 19 2017

%t Denominator[Convergents[Sqrt[7]/2, 30]] (* _Vaclav Kotesovec_, Nov 19 2017 *)

%o (PARI) Vec((1 + 3*x - x^2)*(1 + 32*x^2 + x^4) / ((1 - 16*x^2 + x^4)*(1 + 16*x^2 + x^4)) + O(x^40)) \\ _Colin Barker_, Nov 21 2017

%Y Cf. A242703, A294972.

%K nonn,cofr,frac,easy

%O 0,2

%A _Wolfdieter Lang_, Nov 18 2017