A178617 a(n) = n^4 - (n+1)^3.

-1, -7, -11, 17, 131, 409, 953, 1889, 3367, 5561, 8669, 12913, 18539, 25817, 35041, 46529, 60623, 77689, 98117, 122321, 150739, 183833, 222089, 266017, 316151, 373049, 437293, 509489, 590267, 680281, 780209, 890753, 1012639, 1146617

Offset

0,2

Links

G. C. Greubel, Table of n, a(n) for n = 0..5000

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

Formula

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), where a(0)=-1, a(1)=-7, a(2)=-11, a(3)=17, a(4)=131. - Harvey P. Dale, Jan 30 2013

From G. C. Greubel, Jan 29 2019: (Start)

G.f.: (-1 - 2*x + 14*x^2 + 12*x^3 + x^4)/(1-x)^5.

E.g.f.: (-1 - 6*x + x^2 + 5*x^3 + x^4)*exp(x). (End)

Example

-7 is an element since 1^4 - 2^3 = 1 - 8 = -7,
-11 is an element since 2^4 - 3^3 = 16 - 27 = -11.

Mathematica

Table[n^4-(n+1)^3, {n, 0, 40}]
LinearRecurrence[{5, -10, 10, -5, 1}, {-1, -7, -11, 17, 131}, 40] (* Harvey P. Dale, Jan 30 2013 *)

Prog

(PARI) vector(40, n, n--; n^4-(n+1)^3) \\ G. C. Greubel, Jan 29 2019
(Magma) [n^4-(n+1)^3: n in [0..40]]; // G. C. Greubel, Jan 29 2019
(SageMath) [n^4-(n+1)^3 for n in (0..40)] # G. C. Greubel, Jan 29 2019
(GAP) List([0..40], n -> n^4-(n+1)^3); # G. C. Greubel, Jan 29 2019

Crossrefs

Cf. A028387, A153257.

Sequence in context: A199327 A260892 A020455 * A188074 A190689 A048215

Adjacent sequences: A178614 A178615 A178616 * A178618 A178619 A178620

Keyword

sign,easy

Author

Vladimir Joseph Stephan Orlovsky, May 30 2010

Status

approved