A272298 -id:A272298 - cp's OEIS frontend

A057781 a(n) = n^4+4 = (n^2-2n+2)(n^2+2n+2) = ((n-1)^2+1)((n+1)^2+1).

4, 5, 20, 85, 260, 629, 1300, 2405, 4100, 6565, 10004, 14645, 20740, 28565, 38420, 50629, 65540, 83525, 104980, 130325, 160004, 194485, 234260, 279845, 331780, 390629, 456980, 531445, 614660, 707285, 810004, 923525, 1048580, 1185925

Offset: 0

Henry Bottomley, Nov 04 2000

Donald E. Knuth, The Art of Computer Programming, Addison-Wesley, Reading, MA, 1997, Vol. 1, exercise 1.2.1, Nr. 11, p. 19. [From Reinhard Zumkeller, Apr 11 2010]

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A016755, A033999, A053755.

Cf. A000583, A002522, A002523, A272298.

[n^4+4: n in [0..40]]; // Vincenzo Librandi, Sep 07 2011

Table[n^4+4,{n,0,60}] (* Vladimir Joseph Stephan Orlovsky, Apr 15 2011 *)
LinearRecurrence[{5,-10,10,-5,1},{4,5,20,85,260},40] (* Harvey P. Dale, Aug 20 2020 *)

first(m)=vector(m,i,i--;i^4+4) \\ Anders Hellström, Sep 15 2015

G.f.: -(5*x^4-5*x^3+35*x^2-15*x+4) / (x-1)^5. - Colin Barker, Mar 29 2013
a(n) = A002523(n) + 3.
a(n) = A002522(n-1) * A002522(n+1).
Sum_{k=0..n} A033999(k)*A016755(k)/a(k) = A033999(n)*(n+1)/A053755(n+1), see Knuth reference. - Reinhard Zumkeller, Apr 11 2010
a(n) = (n^2)^2 + 2^2 = (n^2-2)^2 + (2*n)^2. - Thomas Ordowski, Sep 15 2015
a(n) = A272298(3*n)/3^4. - Bruno Berselli, Apr 29 2016
Sum_{n>=0} 1/a(n) = (Pi*coth(Pi) + 1)/8. - Amiram Eldar, Oct 04 2021

A272297 a(n) = n^4 + 64.

Original entry on oeis.org

64, 65, 80, 145, 320, 689, 1360, 2465, 4160, 6625, 10064, 14705, 20800, 28625, 38480, 50689, 65600, 83585, 105040, 130385, 160064, 194545, 234320, 279905, 331840, 390689, 457040, 531505, 614720, 707345, 810064, 923585, 1048640, 1185985, 1336400, 1500689, 1679680, 1874225, 2085200

Offset: 0

Bruno Berselli, Apr 25 2016

This is the case k=2 of Sophie Germain's Identity n^4+(2*k^2)^2 = ((n-k)^2+k^2)*((n+k)^2+k^2).

Bruno Berselli, Table of n, a(n) for n = 0..1000
Wikipedia, Sophie Germain's Identity.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A005917.
Subsequence of A227855.
Cf. A000583 (k=0), A057781 (k=1), A272298 (k=3).

Magma
```
[n^4+64: n in [0..40]];
```
Mathematica
```
Table[n^4 + 64, {n, 0, 40}]
```
Maxima
```
makelist(n^4+64, n, 0, 40);
```
PARI
```
vector(40, n, n--; n^4+64)
```
Python
```
[n**4+64 for n in range(40)]
```

for n in range(0,10**5):print(n**4+64) # Soumil Mandal, Apr 30 2016

Sage
```
[n^4+64 for n in (0..40)]
```

O.g.f.: (64 - 255*x + 395*x^2 - 245*x^3 + 65*x^4)/(1 - x)^5.
E.g.f.: (64 + x + 7*x^2 + 6*x^3 + x^4)*exp(x).
a(n) = (n^2 - 8)^2 + (4*n)^2.

cp's OEIS Frontend

A057781 a(n) = n^4+4 = (n^2-2n+2)(n^2+2n+2) = ((n-1)^2+1)((n+1)^2+1).

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A272297 a(n) = n^4 + 64.

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cp's OEIS Frontend

A057781 a(n) = n^4+4 = (n^2-2*n+2)*(n^2+2*n+2) = ((n-1)^2+1)*((n+1)^2+1).

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Author

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References

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Programs

Magma

Mathematica

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Formula

A272297 a(n) = n^4 + 64.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Maxima

PARI

Python

Python

Sage

Formula

A057781 a(n) = n^4+4 = (n^2-2n+2)(n^2+2n+2) = ((n-1)^2+1)((n+1)^2+1).