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a(0)=1 and, for n >= 1, (BM)a(n) = 2a(n-1), where BM is the BinomialMean transform. BM is defined by (BM)a(n) = (M^n)a(0) where (M)a(n) is the mean (a(n) + a(n+1))/2, or, alternatively, by (BM)a(n) = (Sum_{k=0..n} C(n,k)a(k))/(2^n).
a(0) = 1 and, for n >= 1, (BM)a(n) = 2*a(n-1), where BM is the BinomialMean transform.
The BinomialMean transform BM is defined by (BM)a(n) = (M^n)a(0) where (M)a(n) is the mean (a(n) + a(n+1))/2, or, alternatively, by (BM)a(n) = (Sum_{k=0..n} binomial(n,k)*a(k))/(2^n).
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a(0)=1 and, for n >= 1, (BM)a(n) = 2a(n-1), where BM is the BinomialMean transform. BM is defined by (BM)a(n) = (M^n)a(0) where (M)a(n) is the mean (a(n) + a(n+1))/2, or, alternatively, by (BM)a(n) = (Sum[_{k=0..n} C(n,k)a(k),k=0..n])/(2^n).
Given that a(0)=1 and a(1)=3. Then (BM)a(2) = (1 + 2*3 + a(2))/4 = 2a(1) = 6, hence a(2)=17.
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O.g.f. as a continued fraction: A(x) = 1/(1 + x - 2^2*x/(1 - 2*(2 - 1)^2*x/(1 + x - 2^4*x/(1 - 2*(2^2 - 1)^2*x/(1 + x - 2^6*x/(1 - 2*(2^3 - 1)^2*x/(1 + x - 2^8*x/(1 - 2*(2^4 - 1)^2*x/(1 + x - ... ))))))))). Cf. A075272. - Peter Bala, Nov 10 2017
Cf. A075272.
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