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COMMENTS
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Start with n=1, sigma(1)/1 is 1/1; the set of distinct values is just 1, so the first term of sequence is 1.
Start with n=2, sigma(2)/2 is 3/2; the set of distinct values is {2, 3}. A new value has appeared: 3. Repeat the process. sigma(3)/3 is 4/3; the set of distinct values is now {2, 3, 4}. A new value has appeared: 4. Repeat the process until no new value appears. The resulting final array is {1, 2, 3, 4, 5, 6, 7, 8, 15}. The maximum value of this array is 15, our second value in sequence.
Then from n=3 to 8, the resulting array is the same; and from n=9 to 15 the resulting arrays albeit different still have a maximum value 15.
The next array for which a maximum occurs is obtained when starting with n=16: 1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 13, 14, 15, 16, 21, 24, 26, 27, 31, 32, 35, 40, 48, 49, 52, 56, 57, 63, 64, 65, 70, 72, 80, 84, 85, 93, 104, 105, 108, 127, 128, 144, 255, 403, 448.
The indices at which these maxima occur are: 1, 2, 16, 81, 343, 490, 935, 1029, 5061, 8661, 18049, 39981, ...
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PROG
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(PARI) rab(ns) = {vnew = []; vold = [ns]; vtot = [ns]; until (! #vnew, vnew = []; for (i=1, #vold, vi = vold[i]; ab = sigma(vi)/vi; if (! vecsearch(vtot, num=numerator(ab)), vtot = vecsort(concat(vtot, num)); vnew = vecsort(concat(vnew, num))); if (! vecsearch(vtot, den=denominator(ab)), vtot = vecsort(concat(vtot, den)); vnew = vecsort(concat(vnew, den))); ); vold = vnew; ); return(vtot); }
lista(nn) = {m = 0; for (n=1, nn, if ((mm = vecmax(rab(n))) > m, m = mm; print1(m, ", ")); ); }
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