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A084377
a(n) = n^3 + 7.
3
7, 8, 15, 34, 71, 132, 223, 350, 519, 736, 1007, 1338, 1735, 2204, 2751, 3382, 4103, 4920, 5839, 6866, 8007, 9268, 10655, 12174, 13831, 15632, 17583, 19690, 21959, 24396, 27007, 29798, 32775, 35944, 39311, 42882, 46663, 50660, 54879, 59326
OFFSET
0,1
COMMENTS
These numbers cannot be perfect squares. - Cino Hilliard, Sep 03 2006
[The following short proof was supplied by Don Reble. - N. J. A. Sloane, Apr 10 2023]
Proof that n^3+7 <> k^2 for all integers n,k.
Assume y^2 - x^3 = 7 has an integer solution.
Modulo 4, we have {0,1,0,1} - {0,1,0,3} == 3; y is even and x is odd.
y^2+1 = x^3+8 = (x+2) [(x-1)^2+3]. Let z = (x-1)^2+3 == 3 mod 4.
The 1-mod-4 numbers are closed under multiplication, so z has a prime factor p == 3 mod 4.
That p divides y^2+1; y^2 == -1 mod p.
But (quadratic reciprocity) there is no square root of -1 modulo p.
That refutes the assumption.
FORMULA
G.f.: (7 - 20*x + 25*x^2 - 6*x^3)/(1 - x)^4. - Vincenzo Librandi, Jun 10 2016
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. - Vincenzo Librandi, Jun 10 2016
MATHEMATICA
Table[n^3 + 7, {n, 0, 60}] (* Vladimir Joseph Stephan Orlovsky, Apr 15 2011 *)
PROG
(PARI) a(n) = n^3 + 7;
(Magma) [n^3+7: n in [0..50]]; // Vincenzo Librandi, Jun 10 2016
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Cino Hilliard, Jun 23 2003
EXTENSIONS
More terms from Franklin T. Adams-Watters, Aug 29 2006
STATUS
approved