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%I #12 Mar 12 2024 20:14:17
%S 7,8,213,214,2325,2532,4625,30282,32358,32384,60098,570816,572884,
%T 575190,9732128,243513275
%N Starting from k=7, each subsequent term is the next larger k such that the ratio A276086(k)/A003415(k) is nearer to 1 than for the previous k in the sequence.
%C Note that A276086(6) / A003415(6) = 5/5 = 1. If there are any x > 6, for which the ratio is 1, then the least one of them will terminate this sequence. Question: Could this sequence actually be infinite?
%C If it exists, a(17) > 1207959552.
%H <a href="/index/Pri#primorialbase">Index entries for sequences related to primorial base</a>
%e k A049345(k) A276086(k)/A003415(k) A276086(k)-A003415(k)
%e ----------------------------------------------------------------------
%e 7, 101, 10/1 = 10, 9
%e 8, 110, 15/12 = 1.25, 3
%e 213, 10011, 66/74 = 0.89189189, -8
%e 214, 10020, 99/109 = 0.90825688, -10
%e 2325, 100211, 1950/1780 = 1.0955056, 170
%e 2532, 110200, 3575/3388 = 1.0551948, 187
%e 4625, 200021, 3042/2900 = 1.0489655, 142
%e 30282, 1011200, 32725/34181 = 0.95740324, -1456
%e 32358, 1100300, 27625/26971 = 1.0242483, 654
%e 32384, 1101210, 116025/117696 = 0.98580241, -1671
%e 60098, 2001110, 30345/30749 = 0.98686136, -404
%e 570816, 12011100, 2114035/2093568 = 1.0097761, 20467
%e 572884, 12100020, 642447/643056 = 0.99905296, -609
%e 575190, 12200000, 927979/927483 = 1.0005348, 496
%e 9732128, 101103110, 26152035/26148912 = 1.0001194, 3123
%e 243513275, 1220000021, 99685818/99683810 = 1.0000201, 2008.
%o (PARI)
%o A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
%o A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
%o print1(7, ", "); r = A276086(7)/A003415(7); for(n=7, oo, t=A276086(n)/A003415(n); if(abs(1-t) < abs(1-r), r=t; print1(n, ", ")))
%Y Cf. A003415, A049345, A276086, A351228.
%K nonn,hard,more
%O 1,1
%A _Antti Karttunen_, Mar 12 2024