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A208060 a(n) = 1 + 2*n + 2^2*n*[n/2] + 2^3*n*[n/2]*[n/3] + 2^4*n*[n/2]*[n/3]*[n/4] + ... where [x]=floor(x).

%I A208060 #13 Jan 12 2020 12:46:27
%S A208060 1,3,13,43,233,611,4405,10515,64145,218755,1215821,2689083,28162105,
%T A208060 61179795,307475813,1236997051,8042542625,17101581699,146671231501,
%U A208060 309740445795,2415132010441,8877053064643,40919003272005,85564885298027,1068638260341937,2783025471994851
%N A208060 a(n) = 1 + 2*n + 2^2*n*[n/2] + 2^3*n*[n/2]*[n/3] + 2^4*n*[n/2]*[n/3]*[n/4] + ... where [x]=floor(x).
%C A208060 Compare the definition of a(n) to the exponential series:
%C A208060 exp(2*n) = 1 + 2*n + 2^2*n*(n/2) + 2^3*n*(n/2)*(n/3) + 2^4*n*(n/2)*(n/3)*(n/4) + ...
%C A208060 Conjecture: limit a(n)^(1/n) = 2*L where L = 2.200161058099... is the geometric mean of Luroth expansions, where log(L) = Sum_{n>=1} log(n)/(n*(n+1)) = 0.7885305659115... (cf. A085361).
%H A208060 Seiichi Manyama, <a href="/A208060/b208060.txt">Table of n, a(n) for n = 0..1000</a>
%F A208060 a(n) = 1 + Sum_{m=1..n} Product_{k=1..m} 2^k*floor(n/k).
%e A208060 a(5) = 1 + 2*5+ 4*5[5/2] + 8*5[5/2][5/3] + 16*5[5/2][5/3][5/4] + 32*5[5/2][5/3][5/4][5/5] = 1 + 2*5 + 4*5*2 + 8*5*2*1 + 16*5*2*1*1 + 32*5*2*1*1*1 = 611.
%o A208060 (PARI) {a(n)=1+sum(m=1, n, prod(k=1, m, 2*floor(n/k)))}
%o A208060 (PARI) /* More efficient: variant of a program by Charles R Greathouse IV */
%o A208060 {a(n)=my(k=1); 1+sum(m=1, n, k*=2*(n\m))}
%o A208060 for(n=0, 60, print1(a(n), ", "))
%Y A208060 Cf. A075885.
%K A208060 nonn
%O A208060 0,2
%A A208060 _Paul D. Hanna_, Feb 23 2012