A084377 a(n) = n^3 + 7.

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.

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

Cf. A000578, A084378, A084381.

Sequence in context: A020690 A191413 A066971 * A041675 A041098 A041427

Adjacent sequences: A084374 A084375 A084376 * A084378 A084379 A084380

Keyword

easy,nonn

Author

Cino Hilliard, Jun 23 2003

Extensions

More terms from Franklin T. Adams-Watters, Aug 29 2006

Status

approved