A084378 -id:A084378 - cp's OEIS frontend

A084381 a(n) = n^3 + 5.

5, 6, 13, 32, 69, 130, 221, 348, 517, 734, 1005, 1336, 1733, 2202, 2749, 3380, 4101, 4918, 5837, 6864, 8005, 9266, 10653, 12172, 13829, 15630, 17581, 19688, 21957, 24394, 27005, 29796, 32773, 35942, 39309, 42880, 46661, 50658, 54877, 59324, 64005

Offset: 0

Cino Hilliard, Jun 23 2003

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).

Cf. A000578, A084378.

[n^3+5: n in [0..50]]; // Vincenzo Librandi, Jun 09 2016

Table[n^3 + 5, {n, 0, 60}] (* Vladimir Joseph Stephan Orlovsky, Apr 15 2011 *)

PARI
```
a(n) = n^3 + 5;
```

G.f.: (5 - 14*x + 19*x^2 - 4*x^3)/(1 - x)^4. - Vincenzo Librandi, Jun 09 2016

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. - Vincenzo Librandi, Jun 09 2016

Offset 0 and a(0) = 5 by Vincenzo Librandi, Jun 09 2016

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

Original entry on oeis.org

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

Cino Hilliard, Jun 23 2003

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.

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).

Cf. A000578, A084378, A084381.

[n^3+7: n in [0..50]]; // Vincenzo Librandi, Jun 10 2016

Table[n^3 + 7, {n, 0, 60}] (* Vladimir Joseph Stephan Orlovsky, Apr 15 2011 *)

PARI
```
a(n) = n^3 + 7;
```

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

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

A084382 a(n) = n^3 + 6.

Original entry on oeis.org

6, 7, 14, 33, 70, 131, 222, 349, 518, 735, 1006, 1337, 1734, 2203, 2750, 3381, 4102, 4919, 5838, 6865, 8006, 9267, 10654, 12173, 13830, 15631, 17582, 19689, 21958, 24395, 27006, 29797, 32774, 35943, 39310, 42881, 46662, 50659, 54878, 59325, 64006, 68927, 74094

Offset: 0

Cino Hilliard, Jun 23 2003

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).

Cf. A000578, A084378, A084381.

[n^3+6: n in [0..50]]; // Vincenzo Librandi, Jun 10 2016

Table[n^3 + 6, {n, 0, 60}] (* Vladimir Joseph Stephan Orlovsky, Apr 15 2011 *)
LinearRecurrence[{4, -6, 4, -1}, {6, 7, 14, 33}, 60] (* Harvey P. Dale, Aug 08 2013 *)

PARI
```
a(n) = n^3 + 6;
```

a(0)=6, a(1)=7, a(2)=14, a(3)=33; for n >= 4, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Aug 08 2013
G.f.: (6 - 17*x + 22*x^2 - 5*x^3)/(1 - x)^4. - Vincenzo Librandi, Jun 10 2016
E.g.f.: (6 + x + 3*x^2 + x^3)*exp(x). - Elmo R. Oliveira, Apr 20 2025

Offset 0 and a(0) = 6 from Vincenzo Librandi, Jun 10 2016

A034324 a(n) = (n-1)(n-2)(n-3) + n.

Original entry on oeis.org

1, 2, 3, 10, 29, 66, 127, 218, 345, 514, 731, 1002, 1333, 1730, 2199, 2746, 3377, 4098, 4915, 5834, 6861, 8002, 9263, 10650, 12169, 13826, 15627, 17578, 19685, 21954, 24391, 27002, 29793, 32770, 35939, 39306, 42877, 46658, 50655, 54874, 59321

Offset: 1

Laurence Michaels (guardian(AT)ntplx.net)

(n*a(n+1)^3+1)/(n^3+1) is the smallest integer of the form (n*k^3+1)/(n^3+1). - Benoit Cloitre, May 02 2002

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Dorin Andrica and Ovidiu Bagdasar, On k-partitions of multisets with equal sums, The Ramanujan J. (2021) Vol. 55, 421-435.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000578, A084378, A084381.

List([1..50], n-> (n-2)^3 +2); # G. C. Greubel, Aug 23 2019

[(n-2)^3 +2: n in [1..50]]; // G. C. Greubel, Aug 23 2019

seq( (n-2)^3 +2, n=1..50); # G. C. Greubel, Aug 23 2019

Table[(n-3)(n-2)(n-1)+n, {n,50}] (* or *) Table[n^3+2, {n,-1,50}]  (* Vladimir Joseph Stephan Orlovsky, Apr 15 2011 *)
CoefficientList[Series[(1 -2x +x^2 +6x^3)/(1-x)^4, {x, 0, 50}], x] (* Vincenzo Librandi, Feb 24 2014 *)
LinearRecurrence[{4,-6,4,-1},{1,2,3,10},50] (* Harvey P. Dale, Aug 06 2018 *)

a(n)=(n-3)*(n-2)*(n-1)+n \\ Charles R Greathouse IV, Jul 02 2016

[(n-2)^3 +2 for n in (1..50)] # G. C. Greubel, Aug 23 2019

a(n) = (n-2)^3 + 2 = A084380(n-2). - Philippe Deléham, Feb 23 2014
a(n+1) = A002061(n)*(n-2) + 3. - Philippe Deléham, Feb 23 2014
G.f.: x*(1-2*x+x^2+6*x^3)/(1-x)^4. - Philippe Deléham, Feb 23 2014
E.g.f.: 6 + (x^3-3*x^2+7*x-6)*exp(x). - Nikolaos Pantelidis, Feb 06 2023

Extended and corrected by Erich Friedman

A274077 a(n) = n^3 + 4.

Original entry on oeis.org

4, 5, 12, 31, 68, 129, 220, 347, 516, 733, 1004, 1335, 1732, 2201, 2748, 3379, 4100, 4917, 5836, 6863, 8004, 9265, 10652, 12171, 13828, 15629, 17580, 19687, 21956, 24393, 27004, 29795, 32772, 35941, 39308, 42879, 46660, 50657, 54876, 59323, 64004, 68925

Offset: 0

Vincenzo Librandi, Jun 09 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Sequences of the type n^3+k: A000578 (k=0), A001093 (k=1), A084380 (k=2), A084378 (k=3), this sequence (k=4), A084381 (k=5), A084382 (k=6), A084377 (k=7).

Magma
```
[n^3+4: n in [0..50]];
```

seq(n^3+4, n=0..100); # Robert Israel, Jun 09 2016

Table[n^3 + 4, {n, 0, 60}]
Range[0,50]^3+4 (* or *) LinearRecurrence[{4,-6,4,-1},{4,5,12,31},50] (* Harvey P. Dale, Jul 01 2017 *)

a(n) = n^3 + 4 \\ Felix Fröhlich, Jun 09 2016

O.g.f.: (4 - 11*x + 16*x^2 - 3*x^3)/(1 - x)^4.
E.g.f.: (x^3 + 3*x^2 + x + 4)*exp(x). - Robert Israel, Jun 09 2016
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).

cp's OEIS Frontend

A084381 a(n) = n^3 + 5.

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

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A084382 a(n) = n^3 + 6.

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A034324 a(n) = (n-1)*(n-2)*(n-3) + n.

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A274077 a(n) = n^3 + 4.

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Magma

Maple

Mathematica

PARI

Formula

A034324 a(n) = (n-1)(n-2)(n-3) + n.