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A380645 The expansion of the Stieltjes continued fraction 1/(1 - x/(1 - a(A053645(0))*x/(1 - a(A053645(1))*x/(1 - a(A053645(2))*x/...)))) gives the sequence itself.

%I A380645 #26 Feb 07 2025 15:58:12
%S A380645 1,1,2,5,15,52,201,857,4370,34365,478287,9095996,189526537,4036216585,
%T A380645 87129122290,2478683501397,2450240534552191,12482183328151728692,
%U A380645 65634092872761268943625,345370818796643845031835465,1817414952852912380501431441282
%N A380645 The expansion of the Stieltjes continued fraction 1/(1 - x/(1 - a(A053645(0))*x/(1 - a(A053645(1))*x/(1 - a(A053645(2))*x/...)))) gives the sequence itself.
%H A380645 Wikipedia, <a href="https://en.wikipedia.org/wiki/Moment_problem">Moment problem</a>.
%F A380645 Let s(n) be the Hankel transform of this sequence. Starting: 1, 1, 1, 2, 8, 2400, 1440000,... .
%F A380645 s(n) = Product_{m=0..n-2} Product_{k=1..m} a(A053645(2*k))*a(A053645(2*k+1)). This formula is particular interesting here, because if we expand the Hankel determinants based on A380645: s(0) = 1, s(1) = a(0), s(2) = a(0)*a(2) - a(1)^2, ..., then we will be able to obtain A380645 directly from a system of equations. This means the product formula for the Hankel determinant is also definition of this sequence.
%F A380645 The Hankel transforms of this sequence starting with offset 0 and starting with offset 1 are both strictly positive, this means this sequence can be considered as a sequence of moments in the Stieltjes moment problem: a(n) = Integral_{x>=0} x^n*V(n, x) dx. It is conjectured that the measure V(n, x) is related to some kind of interesting distribution which shows self similarity over intervals of powers of two.
%e A380645 The sequence 1, a(A053645(0)), a(A053645(1)), ... begins:
%e A380645 1, 1, 1, 2, 1, 2, 5, 15, 1, 2, 5, 15, 52, 201, 857, 4370, 1, 2, 5, 15, 52, 201, 857, 4370, 34365, 478287, 9095996, 189526537, 4036216585, 87129122290, 2478683501397, 2450240534552191, 1, 2, 5, 15, ...
%e A380645 This also forms a fractal sequence as it is invariant under the removal of each term's first occurrence:
%e A380645  , 1, 1,  , 1, 2,  ,  , 1, 2, 5, 15,  ,  ,  ,  , 1, 2, 5, 15, 52, 52, 201, 857, ...
%e A380645 We insert this fractal sequence into the Stilties continued fractions and expand:
%e A380645   1/(1 - 1*x/(1 - 1*x/(1 - 1*x/(1 - 2*x/(1 - 1*x/(1 - 2*x/(1 - 5*x/(...)))))))) =
%e A380645   1 + 1*x + 2*x^2 + 5*x^3 + 15*x^4 + 52*x^5 + 201*x^6 + 857*x^7 + 4070*x^8 + 21765*x^9 + ... .
%o A380645 (PARI)
%o A380645 cf(v) = {my(f = O(x)); for(k=1, #v, f=1/(1-v[#v+1-k]*x*f)); Vec(f+O(x^(#v)))}
%o A380645 vectorA(numiter) = {my(v = [1]); for(k=0, numiter, v=concat(v, cf(v))); v[1+#v/2..#v]}
%Y A380645 Cf. A053645.
%K A380645 nonn
%O A380645 0,3
%A A380645 _Thomas Scheuerle_, Feb 06 2025