A059999 a(n) = (1/6)*n^5 - (19/8)*n^4 + (51/4)*n^3 - (253/8)*n^2 + (445/12)*n - 14.

2, 3, 5, 7, 11, 42, 168, 520, 1312, 2861, 5607, 10133, 17185, 27692, 42786, 63822, 92398, 130375, 179897, 243411, 323687, 423838, 547340, 698052, 880236, 1098577, 1358203, 1664705, 2024157, 2443136, 2928742, 3488618, 4130970, 4864587, 5698861, 6643807, 7710083

Offset

1,1

Comments

Deliberately contrived to begin with first five primes; illustrates absurdity of many "guess the next term" puzzles.

The 4th-order formula a(n) = (1/8)*n^4 - (17/12)*n^3 + (47/8)*n^2 - (103/12)*n + 6 is sufficient to yield a(n) = prime(n) for n = 1..5. - Jon E. Schoenfield, Sep 15 2018

Links

Harry J. Smith, Table of n, a(n) for n = 1..1000

de.rec.denksport, FAQ.

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Formula

From Elmo R. Oliveira, Sep 02 2025: (Start)

G.f.: x*(14*x^5 + 14*x^4 - 18*x^3 + 17*x^2 - 9*x + 2)/(x-1)^6.

E.g.f.: 14 + exp(x)*(4*x^5 - 17*x^4 + 64*x^3 - 180*x^2 + 384*x - 336)/24.

a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n > 6. (End)

Mathematica

Table[(n^5)/6 - (19/8)n^4 + (51/4)n^3 - (253/8)n^2 + (445/12)n - 14, {n, 40}] (* Alonso del Arte, Jan 25 2017 *)

Prog

(PARI) a(n) = (4*n^5 - 57*n^4 + 306*n^3 - 759*n^2 + 890*n)/24 - 14 \\ Harry J. Smith, Jul 01 2009

Crossrefs

Sequence in context: A195337 A141500 A215162 * A237438 A040130 A127727

Adjacent sequences: A059996 A059997 A059998 * A060000 A060001 A060002

Keyword

dumb,easy,nonn

Author

Rainer Rosenthal, Mar 10 2001

Status

approved