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Numerators of b(n) where b(1) = 1, b(i) = b(i-1) + 1/b(i-1).
+10
8
1, 2, 5, 29, 941, 969581, 1014556267661, 1099331737522548368039021, 1280590510388959061548230114212510564911731118541, 1726999038066943724857508638586386504281539279376091034086485112150121338989977841573308941492781
OFFSET
1,2
COMMENTS
a(n) is also the numerator of the fractional chromatic number of the Mycielski graph M_n. - Eric W. Weisstein, Mar 05 2011
It appears that lim_{n->infinity} (1/n)*exp(2*(b(n)^2-2n)) = c1 = 0.57...... - Benoit Cloitre, Oct 16 2002
c1 = 0.574810274671785...; see A232975. - Jon E. Schoenfield, Nov 30 2013
b(n)^2 = t/2 + u + (u - 1/2)/t + (-u^2 + 2*u - 11/12)/t^2 + (4*u^3/3 - 5*u^2 + 17*u/3 - 65/36)/t^3 + ... where t = 4*n, u = (log n)/2 + c, and c = -0.2768576248625765389364372...; see A233770. - Jon E. Schoenfield, Dec 15 2013
a(n) is also the numerator of b(n) where b(0) = b(1) = 1 and b(n) = (b(n-1)^2 + b(n-2)^2) / b(n-2) for n > 1 where the denominator of b(n) is partial products of A073834. - Michael Somos, Aug 16 2014
a(n) is also the numerator of b(n) where b(1) = 1 and b(2) = 2 and b(n) = b(n-2) + b(n-1) - (b(n-2)^2/b(n-1)) for n > 2. This has a geometric interpretation: One can prove, given two half lines starting at the center of a series of concentric circles, and a set of triangles each defined by the intersections of the two half lines with any given circle and one of the intersections of the rays with the next circle, that if the circles have radii specified by b(n), the triangle areas are all equal. - Sjoerd C. de Vries, Aug 13 2015
REFERENCES
H. L. Montgomery, Ten Lectures on the Interface Between Analytic Number Theory and Harmonic Analysis, Amer. Math. Soc., 1996, p. 187.
D. J. Newman, A Problem Seminar, Springer; see Problem #60.
J. H. Silverman, The arithmetic of dynamical systems, Springer, 2007, see p. 113 Table 3.1.
LINKS
Clark Kimberling, Polynomials associated with reciprocation, JIS 12 (2009) 09.3.4.
Eric Weisstein's World of Mathematics, Fractional Chromatic Number
Eric Weisstein's World of Mathematics, Mycielski Graph
FORMULA
a(n) = a(n-1)^2 + A073834(n-1)^2; A073834(n) = a(n-1) * A073834(n-1). - Franklin T. Adams-Watters, Aug 04 2008
0 = a(n)^2*(a(n+1) - a(n)^2) - (a(n+2) - a(n+1)^2) for all n > 0. - Michael Somos, Aug 16 2014
EXAMPLE
1, 2, 5/2, 29/10, 941/290, 969581/272890, 1014556267661/264588959090, 1099331737522548368039021/268440386798659418988490, ...
MATHEMATICA
f[n_]:=n+1/n; Prepend[Numerator[NestList[f, 2, 9]], 1] (* Vladimir Joseph Stephan Orlovsky, Nov 19 2010 *)
Numerator[NestList[# + 1/# &, 1, 10]] (* Eric W. Weisstein, Mar 05 2001 *)
a[ n_] := If[ n<1, 0, If[ n<3, n, With[{x = a[n-2]^2, y = a[n-1]}, y y + x y - x x]]]; (* Michael Somos, Aug 16 2014 *)
Numerator@RecurrenceTable[{b[n] == b[-2 + n] - b[-2 + n]^2/b[-1 + n] + b[-1 + n], b[1] == 1,
b[2] == 2}, b, {n, 1, 10}] (* Sjoerd C. de Vries, Aug 13 2015 *)
PROG
(PARI) {a(n) = if( n<1, 0, if( n<3, n, my(x = a(n-2)^2, y = a(n-1)); y^2 + x*y -x^2))}; /* Michael Somos, Mar 05 2012 */
CROSSREFS
See A073834 for denominators. See A232975 for c1; see A233770 for c.
KEYWORD
frac,nonn
AUTHOR
Alex Fink, Aug 12 2002
STATUS
approved
Coefficients of denominator polynomials Q(n,x) associated with reciprocation.
+10
5
1, 1, 0, 1, 0, 1, 0, 1, 0, 4, 0, 4, 0, 1, 0, 1, 0, 11, 0, 45, 0, 88, 0, 88, 0, 45, 0, 11, 0, 1, 0, 1, 0, 26, 0, 293, 0, 1896, 0, 7866, 0, 22122, 0, 43488, 0, 60753, 0, 60753, 0, 43488, 0, 22122, 0, 7866, 0, 1896, 0, 293, 0, 26, 0, 1, 0, 1, 0, 57, 0, 1512, 0, 24858, 0, 284578, 0
OFFSET
1,10
COMMENTS
1. Q(n,1)=A073834(n) for n>=1.
2. For n>=3, Q(n)=Q(n,x)=i*T(n,i*x), where T(n) is the polynomial at A147986.
Thus all the zeros of Q(n,x), for n>=2, are nonreal.
LINKS
Clark Kimberling, Polynomials associated with reciprocation, Journal of Integer Sequences 12 (2009, Article 09.3.4) 1-11.
FORMULA
The basic idea is to iterate the reciprocation-sum mapping x/y -> x/y+y/x.
Let x be an indeterminate, P(1)=x, Q(1)=1 and for n>1, define P(n)=P(n-1)^2+Q(n-1)^2 and Q(n)=P(n-1)*Q(n-1), so that P(n)/Q(n)=P(n-1)/Q(n-1)-Q(n-1)/P(n-1).
EXAMPLE
Q(1) = 1
Q(2) = x
Q(3) = x^3+x
Q(4) = x^7+4*x^5+4*x^3+1
so that, as an array, the sequence begins with:
1
1 0
1 0 1 0
1 0 4 0 4 0 1
KEYWORD
nonn,tabf
AUTHOR
Clark Kimberling, Nov 24 2008
STATUS
approved
Let b(0) = 1, b(n) = Sum_{k=0..n-1} b(k) / b(n-k-1), then a(n) is the denominator of b(n).
+10
1
1, 1, 1, 2, 7, 308, 1065372, 5699432573835, 742435596532024691458409520, 1770094160863794205114840009375146894748207874734794924
OFFSET
0,4
COMMENTS
The next term is too large to include.
FORMULA
G.f. for fractions satisfies: 1 / Sum_{n>=0} b(n) * x^n = 1 - x * Sum_{n>=0} x^n / b(n).
EXAMPLE
1, 1, 2, 7/2, 44/7, 3459/308, 21398845/1065372, 204701870532176/5699432573835, ...
MATHEMATICA
b[0] = 1; b[n_] := b[n] = Sum[b[k]/b[n - k - 1], {k, 0, n - 1}]; a[n_] := Denominator[b[n]]; Table[a[n], {n, 0, 9}]
CROSSREFS
Cf. A000108, A022857, A022858, A073834, A367787 (numerators).
KEYWORD
nonn,frac
AUTHOR
Ilya Gutkovskiy, Nov 30 2023
STATUS
approved
Primes dividing some member of A073833.
+10
0
2, 5, 29, 41, 89, 101, 109, 269, 421, 509, 521, 709, 929, 941, 1549, 1861, 2281, 2521, 2749, 2801, 2909, 3121, 3169, 3469, 5821, 5881, 7109, 8069, 8969, 9041, 9181, 10061, 10601, 11549, 15121, 16061, 16889, 16981, 21929, 30089, 30169, 32561, 41149
OFFSET
1,1
COMMENTS
Primes that divide A073833(n) will divide A073834(m) for any m > n, and this is all the prime divisors of A073834(m).
Iterating f(x) = x + 1/x modulo p will eventually either produce a zero (in which case p is in this sequence), or it will loop to an earlier term (in which case it is not). Since f(-x) = -f(x), encountering the negation of an earlier term means that the iteration is looping.
Note that A073833(6) = 969581 = 521 * 1861 is the first composite member of that sequence.
PROG
(PARI) ina(p)=local(m, k, v); m=Mod(1, p); v=vector(p\2); while(m!=0, k=lift(m); if(2*k>p, k=p-k); if(v[k], return(0)); v[k]=1; m+=1/m); 1
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved

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