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A096654
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Denominators of self-convergents to 1/(e-2).
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9
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1, 2, 8, 38, 222, 1522, 11986, 106542, 1054766, 11506538, 137119578, 1772006854, 24681524038, 368577425634, 5874202721042, 99515904921182, 1785757627196766, 33835407673201882, 675016383080377546, 14143200407398386678, 310507536216973671158, 7128173005328786885714
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OFFSET
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0,2
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COMMENTS
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The self-continued fraction of r>0 is here introduced as the sequence (b(0), b(1), b(2), ...) defined as follows: put r(0)=r, b(0)=[r(0)] and for n>=1, put r(n)=b(n-1)/(r(n-1)-b(n-1)) and b(n)=[r(n)]. This differs from simple continued fraction, for which r(n)=1/(r(n-1)-b(n-1)). Now r=lim(p(n)/q(n)), where p(0)=b(1), q(0)=1, p(1)=b(0)(b(1)+1), q(1)=b(1) and for n>=2, p(n)=b(n)*p(n-1)+b(n-1)*p(n-2), q(n)=b(n)*q(n-1)+b(n-1)*q(n-2); p(0),p(1),... are the numerators of the self-convergents to r; q(0),q(1),... are the denominators of the self-convergents to r. Thus A096654 is given by a(n)=(n+1)*a(n-1)+n*a(n-2), a(0)=1, a(1)=2.
Number of increasing runs of odd length in all permutations of [n+1]. Example: a(2) = 8 because we have (123), 13(2), (3)12, (2)13, 23(1), (3)(2)(1) (the runs of odd length are shown between parentheses). - Emeric Deutsch, Aug 29 2004
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LINKS
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FORMULA
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a(n) = (n+1)*a(n-1) + n*a(n-2), with a(0)=1, a(1)=2. - Alex Ratushnyak, Aug 05 2012
a(n) = (n+1)! - 2*floor(((n+1)!+1)/e) + (n+2)!-2*floor(((n+2)!+1)/e). (End)
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EXAMPLE
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a(2)=q(2)=3*2+2*1=8, a(3)=q(3)=4*8+3*2=38. The convergents p(0)/q(0) to p(4)/q(4) are 1/1, 3/2, 11/8, 53/38, 309/222.
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MAPLE
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G:=(3-x-2*(1+x)*exp(-x))/(1-x)^3: Gser:=series(G, x=0, 22): 1, seq(n!*coeff(Gser, x^n), n=1..21);
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PROG
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(Python)
prpr = 1
prev = 2
for n in range(2, 77):
print(prpr, end=', ')
curr = (n+1)*prev + n*prpr
prpr = prev
prev = curr
(PARI) x='x+O('x^66); Vec(serlaplace((3-x-2*(1+x)*exp(-x))/(1-x)^3)) /* Joerg Arndt, Aug 06 2012 */
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CROSSREFS
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KEYWORD
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nonn,frac,easy,changed
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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