OFFSET
1,3
COMMENTS
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.
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.
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
REFERENCES
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Oxford University Press, 2008 (Sixth edition), Par. 13.7.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
MATHEMATICA
Flatten[Table[{n^4 - n, n^3 - 1, 2 n^3 + 1, n^4 + 2 n}, {n, 1, 40}]] (* Vincenzo Librandi, Jan 02 2014 *)
PROG
(Magma) &cat[[n*(n^3-1), n^3-1, 2*n^3+1, n*(n^3+2)]: n in [1..40]];
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 04 2012
STATUS
approved