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.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Cino Hilliard, Proof that a cube plus 7 cannot be a square
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
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