A272870 Real part of (n + i)^4.

1, -4, -7, 28, 161, 476, 1081, 2108, 3713, 6076, 9401, 13916, 19873, 27548, 37241, 49276, 64001, 81788, 103033, 128156, 157601, 191836, 231353, 276668, 328321, 386876, 452921, 527068, 609953, 702236, 804601, 917756, 1042433, 1179388, 1329401, 1493276

Offset

0,2

Comments

a(1) and a(2) are the only two negative terms in the sequence. Since (n + i)^2 = (n^2 - 1) + 2ni, it follows that (n + i)^4 = (n^2 - 1 + 2ni)^2 = (n^4 - 6n^2 + 1) + (4n^3 - 4n)i, so the real part of (n + i)^4 is n^4 - 6n^2 + 1. Then n^4 + 1 > 6n^2 for all n > 2, ensuring a(n) is positive for all n > 2. - Alonso del Arte, Jun 04 2016

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

a(n) = n^4 - 6*n^2 + 1.

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 4.

G.f.: (1-9*x+23*x^2+13*x^3-4*x^4) / (1-x)^5.

E.g.f.: (1 - 5*x + x^2 + 6*x^3 + x^4)*exp(x). - Ilya Gutkovskiy, May 08 2016

Example

a(5) = 476 because (5 + i)^4 = 476 + 480*i.

Mathematica

Table[Re[(n + I)^4], {n, 0, 35}] (* or *)
Table[n^4 - 6 n^2 + 1, {n, 0, 35}] (* or *)
CoefficientList[Series[(1 - 9 x + 23 x^2 + 13 x^3 - 4 x^4)/(1 - x)^5, {x, 0, 35}], x] (* Michael De Vlieger, May 08 2016 *)

Prog

(PARI) a(n) = n^4-6*n^2+1
(PARI) vector(50, n, n--; real((n+I)^4)).
(PARI) Vec((1-9*x+23*x^2+13*x^3-4*x^4)/(1-x)^5 + O(x^50))

Crossrefs

Cf. A272871 (imaginary part).

Cf. A121670 ((n+i)^3), A121672 ((n+i)^5).

Sequence in context: A366314 A176966 A117977 * A362650 A176750 A030687

Adjacent sequences: A272867 A272868 A272869 * A272871 A272872 A272873

Keyword

sign,easy

Author

Colin Barker, May 08 2016

Status

approved