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A294973
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Denominators of the continued fraction convergents to sqrt(7)/2.
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4
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1, 3, 31, 96, 223, 765, 7873, 24384, 56641, 194307, 1999711, 6193440, 14386591, 49353213, 507918721, 1573109376, 3654137473, 12535521795, 129009355423, 399563588064, 928136531551, 3183973182717, 32767868358721, 101487578258880, 235743024876481, 808716652888323, 8322909553759711, 25777445314167456, 59877800182094623, 205410845860451325
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OFFSET
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0,2
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COMMENTS
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The numerators are given in A294972.
The continued fraction of sqrt(7)/2 is [1, repeat(3,10,3,2)].
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LINKS
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FORMULA
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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)).
a(n) = 254*a(n-4) - a(n-8) for n > 7.
(End)
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
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MATHEMATICA
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,cofr,frac,easy
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AUTHOR
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STATUS
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approved
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