A379953 - OEIS

%I #24 Jan 19 2025 17:50:43

%S 0,2,6,4,0,12,0,8,18,10,0,24,0,42,30,16,0,36,0,20,42,22,0,48,0,26,54,

%T 84,0,60,0,32,66,34,0,72,0,38,78,40,0,210,0,44,90,46,0,96,0,50,102,52,

%U 0,108,0,168,456,58,0,120,0,62,126,64,260,132,0,68,138,1330,0,144,0,74,150,76,0,1794,0,80,162,82,0,420

%N Largest k >= 0 such that (n*k)^3/(n^3+k^3) is an integer.

%C For all n, a(n) < n^2, as for k^3 + n^3 to divide k^3 * n^3, it must also divide n^3 * (k^3 + n^3) - k^3*n^3 = n^6, so k^3 <= n^6 - n^3, and in particular k < n^2. - _Robert Israel_, Jan 16 2025

%C Not every a(n) is a multiple of n: a(231) = 616 is the first case where n does not divide a(n).

%C First odd terms are a(1474) = 6633, a(1628) = 2849 and a(1860) = 5115.

%C For all odd n, a(n) is even.

%C If n = p^j where p is a prime >= 5, then a(n) = 0 (see link for proof). - _Robert Israel_, Jan 16 2025

%H Antti Karttunen, <a href="/A379953/b379953.txt">Table of n, a(n) for n = 1..10000</a>

%H Robert Israel, <a href="/A379953/a379953.pdf">Proof of comment</a> (which applies to A119612, A379953 and A379954)

%p f:= proc(n) local k;

%p   for k from n^2 by -1 do

%p     if (n*k)^3 mod (n^3 + k^3) = 0 then return k fi

%p   od

%p end proc:

%p map(f, [$1..100]); # _Robert Israel_, Jan 16 2025

%t A379953[n_] := Module[{k = n^2}, While[PowerMod[--k*n, 3, # + k^3] > 0] & [n^3]; k];

%t Array[A379953, 100] (* _Paolo Xausa_, Jan 19 2025 *)

%o (PARI) A379953(n) = forstep(k=n^2, 0, -1, if(!(((n*k)^3)%(k^3+n^3)), return(k)));

%Y Cf. A119612, A379954.

%K nonn

%O 1,2

%A _Antti Karttunen_, Jan 16 2025