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A375973 Record values in A375970.

%I A375973 #18 Sep 14 2024 06:49:39
%S A375973 1,2,5,70,99,195,240,323,2378,2716,15015,48505,80782,130662,510068,
%T A375973 672210,1926615,2744210,4116315,10278759,31320850,87347695,93222358,
%U A375973 155904960,177385520,189539896,250637778,272607725,486471832,647562465,1620820270
%N A375973 Record values in A375970.
%C A375973 a(n) is the largest number whose square divides A000330(A375971(n)).
%F A375973 a(n) = A375970(A375971(n)).
%e A375973 a(3) = 5 because A375971(3) = 650 and 5^2 is the largest square dividing 650.
%e A375973 From _David A. Corneth_, Sep 13 2024: (Start)
%e A375973 70 is in the sequence as A000330(24) = 24 * 25 * 49 / 6 = 4 * 25 * 49. The largest square dividing 4 is 4, the largest square dividing 25 is 25 and the largest square dividing 49 is 49.
%e A375973 So the largest k such that k^2 divides 4 * 25 * 49 is sqrt(4)*sqrt(25)*sqrt(49) = 2*5*7 = 70, a record. (End)
%p A375973 g:= proc(n) local t, s, F; t:= n*(n+1)*(2*n+1)/6;
%p A375973   F:= ifactors(t)[2];
%p A375973   mul(s[1]^floor(s[2]/2), s=F)
%p A375973 end proc:
%p A375973 V:= NULL; m:= 0: count:= 0:
%p A375973 for k from 1 while count < 20 do
%p A375973   v:= g(k);
%p A375973   if v > m then m:= v; V:= V,v; count:= count+1; fi
%p A375973 od:
%p A375973 V;
%Y A375973 Cf. A000330, A375970, A375971.
%K A375973 nonn
%O A375973 1,2
%A A375973 _Robert Israel_, Sep 04 2024
%E A375973 a(25) from _Michael S. Branicky_, Sep 06 2024
%E A375973 a(26)-a(31) from _David A. Corneth_, Sep 08 2024