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%I #22 Mar 05 2019 21:40:52
%S 4,14,33,64,110,174,260,369,506,672,872,1108,1382,1699,2061,2472,2933,
%T 3448,4021,4653,5349,6110,6941,7844,8822,9878,11015,12237,13545,14943,
%U 16435,18023,19709,21498,23392,25394,27507,29734,32079,34543,37131,39844,42687,45662,48772
%N 1/(Integral_{x=0..1} x^(x^(x^n)) dx - 1/2), rounded to the nearest integer.
%C Linked to the problem of sorting parenthesized expressions (x^x....^x) (cf. A000081 and A222379, A222380) according to the value of their integral from 0 to 1: This value is minimal, for a given number n of x's, for G[n](x) := x^((...(x^x)^x....)^x) = x^(x^(x^(n-2))), which converges pointwise to x^(x^0) = x^1 = x for all x in [0,1], as n -> oo. The corresponding integrals therefore tend to 1/2 as n -> oo. This sequence is a convenient measure of the distance of these integrals from 1/2.
%C See A322008 for the maximal values of such integrals.
%H Vladimir Reshetnikov, <a href="https://mathoverflow.net/questions/324203/integrals-of-power-towers/324382#324382">Integrals of power towers</a>, on MathOverflow.net, Feb. 26, 2019.
%e For n=0, Integral_{x=0..1} x^(x^(x^0)) dx = Integral_{x=0..1} x^x dx = A083648 = 0.7834..., and 1/(0.7834... - 0.5) = 1 / 0.2834... = 3.528..., so a(0) = round(3.528...) = 4.
%e For n=1, Integral_{x=0..1} x^(x^(x^1)) dx = Integral_{x=0..1} x^(x^x) dx = 0.5731..., and 1/(0.5731... - 0.5) = 1 / 0.0731... = 13.67..., so a(1) = round(13.67...) = 14.
%p Digits:= 20:
%p a:= n-> round(evalf(1/(int(x^(x^(x^n)), x=0..1)-1/2))):
%p seq(a(n), n=0..44); # _Alois P. Heinz_, Mar 01 2019
%t f[n_] := Round[1/(NIntegrate[x^(x^(x^n)), {x, 0, 1}, WorkingPrecision -> 24] - 1/2)]; Array[f, 45, 0] (* _Robert G. Wilson v_, Mar 01 2019 *)
%o (PARI) A322009(n)=1\/intnum(x=0, 1, x^x^x^n-x)
%Y Cf. A322008; A000081, A222379, A222380; A083648.
%K nonn
%O 0,1
%A _M. F. Hasler_, Mar 01 2019
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