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%I #17 Jan 21 2015 13:03:26
%S 1,2,4,14,22,26,75,195,330,324,935,1598,3422,3663,10191,14066,12099,
%T 53661,121555,182169,235509,307615,633945,2097595,2072198,2643298,
%U 6544282,8675343,13670722,17573794,85112326,77295778,235873898,362150458,544042486,1457255474
%N Smallest k such that the simple continued fraction for Sum(d|k, 1/d) contains exactly n elements.
%e sum(d|195, 1/d) = 112/65 and 112/65 continued fraction is [1, 1, 2, 1, 1, 1, 1, 3] which contains 8 elements. There is no smaller number than 195 with this property hence a(8)=195.
%t a = Table[0, {50}]; Do[b = Length[ ContinuedFraction[ Apply[ Plus, 1/Divisors[n]]]]; If[ a[[b]] == 0, a[[b]] = n], {n, 1, 10^7}]
%o (PARI) for(n=1,21,s=1; while(length(contfrac(sumdiv(s,d,1/d)))<n,s++); print1(s,","))
%Y Cf. A071862.
%K easy,nonn
%O 1,2
%A _Benoit Cloitre_, Jun 09 2002
%E More terms from _Robert G. Wilson v_, Jun 11 2002
%E a(29)-a(36) from _Michel Marcus_, Sep 17 2012
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