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A348126 a(n) is the least k for which x(k) > Sum_{j=1..k} 1/j, where x(k) is the Newton iterate (starting from x(1)=1) for exp(-P(n,x)) and P(n,x) is the n-th Maclaurin polynomial for exp(x). 0

%I #33 Dec 13 2021 17:23:38

%S 2,3,11,36,104,287,776,2074,5519,14672,38999,103709,275970,734862,

%T 1958108,5220797,13927895,37176046,99277815,265238573,708928640,

%U 1895558566,5070252360

%N a(n) is the least k for which x(k) > Sum_{j=1..k} 1/j, where x(k) is the Newton iterate (starting from x(1)=1) for exp(-P(n,x)) and P(n,x) is the n-th Maclaurin polynomial for exp(x).

%C Conjecture: a(n) is finite for every positive integer n, and a(n) approaches infinity with n.

%e Let H(k) = Sum_{j=1..k} 1/j, i.e., the k-th harmonic number.

%e a(1)=2: Newton's method applied to f(x)=exp(-1-x) starting at x(1)=1 gives x(1)=1, x(2)=2, whereas H(1)=1 and H(2)=1.5.

%e a(3)=11: Newton's method applied to f(x) = exp(-1 - x - x^2/2 - x^3/6) starting at x(1)=1 gives x(10)=2.91631 and x(11)=3.03873, whereas H(10)=2.92897 and H(11)=3.01988.

%o (PARI)

%o a(n)=

%o {default(seriesprecision,n);

%o default(realprecision,max(38,2*#digits(n!,10)));

%o Tp = truncate(taylor(exp(x),x,n));

%o t=1.0;

%o hsum=1.0;

%o j=1;

%o while(t<=hsum, j=j+1; t=t+1.0/subst(Tp,x,t); hsum = hsum + 1.0/j);

%o j;}

%Y Cf. A001008, A002805.

%K nonn,more

%O 1,1

%A _Jennifer Sirkin_, Oct 01 2021

%E a(15) inserted by _Jinyuan Wang_, Dec 10 2021

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