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%I #32 Sep 05 2019 08:45:28
%S 2,3,8,9,14,20,35,77,84,99,119,231,260,351,494,696,1665,1785,1845,
%T 3479,4059,6887,16302,17919,23183,23660,56979,58800,81968,83880,95930,
%U 137903,340955,358017,574925,803760,1336139,3375111,4684659,10316619,14935095,18610022
%N Numbers k with property that there is an m = m(k) such that m(m+1)/2 divides k(k+1)/2 and m(k) > m(i) for all i < k.
%C gcd(a(n),m(n)) < m(n), n > 1.
%C Let k = r * m, and r = (A000217(k) * (m+1))/((A000217(m) * (k+1)). For known terms 1 < r < 4.
%e 2 is a term because A000217(1) divides A000217(2).
%e 3 is a term because A000217(2) divides A000217(3).
%e 8 is a term because A000217(3) divides A000217(8).
%e 9 is a term because A000217(5) divides A000217(9).
%e 14 is a term because A000217(6) divides A000217(14).
%t m=1; L={2}; k=2; While[k < 10000, k++; tr = k (k + 1)/2; t = SelectFirst[ Reverse@ Divisors[2 tr], # != k && Mod[tr, # (# + 1)/2] == 0 &]; If[t > m, AppendTo[L, k]; m = t]]; L (* _Giovanni Resta_, Sep 05 2019 *)
%o (PARI) T(n) = {return((n * (n+1)) / 2)}
%o Tk(n, k) = {for (i = k, n - 1, if ((T(n)%T(i))==0, return(i+1)))}
%o Tn(n) = {phwm = 1;for (i = 2, n, nhwm = Tk(i, phwm); if(nhwm > phwm, phwm = nhwm; print1(i, ",")))}
%o Tn(5000000)
%Y Cf. A000217.
%K nonn
%O 1,1
%A _Torlach Rush_, Jul 30 2019
%E More terms from _Giovanni Resta_, Sep 05 2019
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