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%I #18 Jan 07 2014 04:04:48
%S 1,8,15,127,128,144,255,403,448,512,1023,29127,47360
%N Terms of the cycles reached after iterations of numerator(sigma(n)/n) = A017665(n).
%C If all integers were in A014567, then this sequence would not exist and we would be looking at A216200; but some are in A069059, allowing the trajectories of A017665 to go down.
%C The term of the sequence correspond to the 5 cycles: [1], [15, 8], [448, 127, 128, 255, 144, 403], [1023, 512], [47360, 29127].
%C Are there some starting x's whose fate will remain unknown, like 276 for A098007?
%C Are there other cycles to be found?
%C No other cycles found with largest member less than 10^9.
%C There are no other cycles with the smallest member < 10^11. All numbers < 10^11 reach one of the five known cycles. - _Donovan Johnson_, Jan 07 2014
%e Obviously 1 is a fixed point for A017665, so 1 is in the sequence.
%e A017665(8) = 15 and A017665(15) = 8, so both 8 and 15 are in the sequence.
%o (PARI) iscycle(v, nextn) = {for (i=1, #v, if (v[i] == nextn, return (1););); return (0);}
%o fcycle(n, known) = {v = vector(1); v[1] = n; first = n; while ((nextn = numerator(sigma(n)/n)) <= first, if (vecsearch(known, nextn), return([])); if (iscycle(v, nextn), return (v)); v = concat(v, nextn); n = nextn;); return ([]);}
%o fcycles(na, nb) = {known = []; known = [1, 8, 127, 512, 29127]; for (n = na, nb, v = fcycle(n, known); if (#v, print(v, ", "); return();););} \\ use empty vector for known to search for cycles from start; when a new cycle is found, insert its smallest term to vector known.
%Y Cf. A000203, A017665.
%K nonn,more
%O 1,2
%A _Michel Marcus_, Dec 27 2013
%E Missing terms 512 and 1023 noticed by _Donovan Johnson_ added by _Michel Marcus_, Jan 02 2014
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