%I #26 Mar 07 2019 14:57:14
%S 2,5,10,17,26,37,50,65,82,101,123,146,171,198,227,258,291,326,364,403,
%T 444,487,532,579,628,679,733,788,845,904,965,1028,1093,1160,1230,1301,
%U 1374,1449,1526,1605,1686,1769,1855,1942,2031,2122,2215,2310,2407,2506,2608
%N 1/(1 - Integral_{x=0..1} x^(x^n) dx), 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 maximal, for a given number n of x's, for F[n](x) := (...(x^x)^x....)^x = x^(x^(n-1)), which converges pointwise to x^0 = x for all x < 1, as n -> oo. The corresponding integrals therefore tend to 1 as n -> oo. This sequence is a convenient measure of the distance of these integrals from 1.
%C See A322009 for the minimal 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.
%F Conjectures from _Colin Barker_, Mar 07 2019: (Start)
%F G.f.: (2 + x + 2*x^2 + 2*x^3 + 2*x^4 + 2*x^5 + 2*x^6 + 2*x^7 + x^9 + x^10 - x^11) / ((1 - x)^3*(1 + x)*(1 + x^2)*(1 + x^4)).
%F a(n) = 2*a(n-1) - a(n-2) + a(n-8) - 2*a(n-9) + a(n-10) for n>11.
%F (End)
%e For n=0, Integral_{x=0..1} x^(x^0) dx = Integral_{x=0..1} x^1 dx = 1/2, so a(0) = 1/(1 - 1/2) = 1 / 0.5 = 2.
%e For n=1, Integral_{x=0..1} x^(x^1) dx = Integral_{x=0..1} x^x dx = A083648 = 0.78343..., so a(1) = round( 1 / (1 - 0.78343...)) = round( 1 / 0.21656...) = 5.
%p a:= n-> round(evalf(1/(1-(int(x^(x^n), x=0..1))))):
%p seq(a(n), n=0..50); # _Alois P. Heinz_, Mar 01 2019
%t f[n_] := Round[1/(1 - NIntegrate[x^(x^n), {x, 0, 1}])]; Array[f, 51, 0] (* _Robert G. Wilson v_, Mar 01 2019 *)
%o (PARI) apply( A322008(n)=1\/intnum(x=0,1,1-x^x^n), [0..50])
%Y Cf. A322009; A000081, A222379, A222380, A306679; A083648.
%K nonn
%O 0,1
%A _M. F. Hasler_, Mar 01 2019