A375973 - OEIS

%I #18 Sep 14 2024 06:49:39

%S 1,2,5,70,99,195,240,323,2378,2716,15015,48505,80782,130662,510068,

%T 672210,1926615,2744210,4116315,10278759,31320850,87347695,93222358,

%U 155904960,177385520,189539896,250637778,272607725,486471832,647562465,1620820270

%N Record values in A375970.

%C a(n) is the largest number whose square divides A000330(A375971(n)).

%F a(n) = A375970(A375971(n)).

%e a(3) = 5 because A375971(3) = 650 and 5^2 is the largest square dividing 650.

%e From _David A. Corneth_, Sep 13 2024: (Start)

%e 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 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 g:= proc(n) local t, s, F; t:= n*(n+1)*(2*n+1)/6;

%p   F:= ifactors(t)[2];

%p   mul(s[1]^floor(s[2]/2), s=F)

%p end proc:

%p V:= NULL; m:= 0: count:= 0:

%p for k from 1 while count < 20 do

%p   v:= g(k);

%p   if v > m then m:= v; V:= V,v; count:= count+1; fi

%p od:

%p V;

%Y Cf. A000330, A375970, A375971.

%K nonn

%O 1,2

%A _Robert Israel_, Sep 04 2024

%E a(25) from _Michael S. Branicky_, Sep 06 2024

%E a(26)-a(31) from _David A. Corneth_, Sep 08 2024