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a(1)=1, a(n) is the smallest integer > a(n-1) such that the sum of elements of the simple continued fraction for S(n)=1/a(1)+1/a(2)+...+1/a(n) equals n^3.
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%I #4 Mar 30 2012 17:30:44

%S 1,7,45,401,719,1136,5613,6358,12448,24739,28082,42850,59604,78928,

%T 81119,169213,214725,309015,432821,496399,706170,725188,1163780,

%U 2284457,2941839,3857806,4133465,5890433,6190258,6286719,6888119

%N a(1)=1, a(n) is the smallest integer > a(n-1) such that the sum of elements of the simple continued fraction for S(n)=1/a(1)+1/a(2)+...+1/a(n) equals n^3.

%e 1/a(1)+1/a(2)+1/a(3)+1/a(4) = (1+1/7+1/45+1/401) which continued fraction is {1, 5, 1, 29, 1, 4, 3, 1, 1, 18} and 1+5+1+29+1+4+3+1+1+18 = 64 = 4^3.

%t a[1] = 1; a[n_] := a[n] = (s = Sum[1/a[i], {i, 1, n - 1}]; While[Plus @@ ContinuedFraction[s + 1/k] != n^3, k++ ]; k); k = 1; Do[ Print[ a[n]], {n, 1, 31}]

%o (PARI) s=1; t=1; for(n=2,31,s=s+1/t; while(abs(n^3+1-sum(i=1,length(contfrac(s+1/t)), component(contfrac(s+1/t), i)))>0,t++); print1(t,","))

%Y Cf. A071183.

%K nonn

%O 1,2

%A _Robert G. Wilson v_, Jun 17 2002