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%I #28 Mar 16 2023 12:23:21
%S 0,0,3,3,14,7,17,20,78,26,55,87,252,63,129,264,620,124,251,635,1290,
%T 215,433,1308,2394,342,687,2415,4088,511,1025,4112,6552,728,1459,6579,
%U 9990,999,2001,10020,14630,1330,2663,14663,20724,1727,3457,20760,28548,2196,4395,28587
%N Quadruples (a,b,c,d) of the form ( n*(n^3-1), n^3-1, 2*n^3+1, n*(n^3+2) ).
%C Four consecutive (a,b,c,d) in the sequence are solutions to a^3+b^3+c^3 = d^3, that is a(4k+1)^3+a(4k+2)^3+a(4k+3)^3 = a(4k+4)^3.
%C Also, A058895(n)^3 + A068601(n)^3 + A033562(n)^3 = A185065(n)^3.
%C The sequence corresponds to the case m=1 in the identity (n*(n^3-m^3))^3+(m*(n^3-m^3))^3+(m*(2*n^3+m^3))^3 = (n*(n^3+2*m^3))^3.
%C G. H. Hardy and E. M. Wright gave this identity in their "An Introduction to the Theory of Numbers" together with (n*(n^3-2*m^3))^3+(m*(n^3+m^3))^3+(m*(2*n^3-m^3))^3 = (n*(n^3+m^3))^3 (see References). - _Bruno Berselli_, Mar 13 2012
%D G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Oxford University Press, 2008 (Sixth edition), Par. 13.7.
%H Vincenzo Librandi, <a href="/A204767/b204767.txt">Table of n, a(n) for n = 1..1000</a>
%t Flatten[Table[{n^4 - n, n^3 - 1, 2 n^3 + 1, n^4 + 2 n}, {n, 1, 40}]] (* _Vincenzo Librandi_, Jan 02 2014 *)
%o (Magma) &cat[[n*(n^3-1), n^3-1, 2*n^3+1, n*(n^3+2)]: n in [1..40]];
%Y Cf. A058895, A068601, A033562, A185065.
%K nonn,easy
%O 1,3
%A _Vincenzo Librandi_, Mar 04 2012
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