A204767 Quadruples (a,b,c,d) of the form ( n*(n^3-1), n^3-1, 2*n^3+1, n*(n^3+2) ).

0, 0, 3, 3, 14, 7, 17, 20, 78, 26, 55, 87, 252, 63, 129, 264, 620, 124, 251, 635, 1290, 215, 433, 1308, 2394, 342, 687, 2415, 4088, 511, 1025, 4112, 6552, 728, 1459, 6579, 9990, 999, 2001, 10020, 14630, 1330, 2663, 14663, 20724, 1727, 3457, 20760, 28548, 2196, 4395, 28587

Offset: 1

Author

Vincenzo Librandi, Mar 04 2012

Keywords

• nonn
• easy

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.

Also, A058895(n)^3 + A068601(n)^3 + A033562(n)^3 = A185065(n)^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

Crossrefs

Cf. A058895, A068601, A033562, A185065.

Programs

Magma

&cat[[n*(n^3-1), n^3-1, 2*n^3+1, n*(n^3+2)]: n in [1..40]];

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 *)