OFFSET
0,1
COMMENTS
Comment (entirely taken from Cugiani's text - see References) from _Vincenco Librandi_, Aug 23 2011: (Start)
This deals with primitive polynomials in GF_k(p). There are p^k monic k-th order polynomials J(p) = x^k + a(k-1)*x^(k-1) + ... + a(0), because there are k independent coefficients a(.), each restricted modulo the prime p. phi(p^k-1)/k of these polynomials are primitive, where phi=A000010. [Example for p=7 and k=2: phi(7^2-1)/2 = phi(48)/2 = 16/2=8. See A011260 for p=2, A027385 for p=3, A027741 for p=5 etc.] Of these sets of primitive polynomials we select with p=1031 the polynomial x^3+73*x^2+x+67 for k=3 in A163303 and x^4+984*x^3+90*x^2+394-x+858 for k=4 in A163304 by the following criteria (This could be extended to k=5, 6,...): Let r = (p^k -1)/(p-1). We demand (see Theorem 1 in Hansen-Mullen)
i) (-1)^k a(0) is a primitive element of J(p).
ii) The remainder of the division of x^r through the polynomial equals (-1)^k a(0).
iii) The remainder of the division of x^(r/q) through the polynomial must have positive degree for each prime divisor q|r.
(End)
REFERENCES
Marco Cugiani, Metodi numerico statistici (Collezione di Matematica applicata n.7), UTET Torino, 1980, pp.78-84
LINKS
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).
FORMULA
G.f.: (858-1151*x+6027*x^2-6093*x^3+383*x^4)/(1-x)^5. - Bruno Berselli, Aug 24 2011
From G. C. Greubel, Dec 18 2016: (Start)
a(n) = 5*a(n-) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
E.g.f.: (858 + 2281*x + 3861*x^2 + 990*x^3 + x^4)*exp(x). (End)
MATHEMATICA
Table[n^4+984n^3+902n^2+394n+858, {n, 0, 30}] (* Harvey P. Dale, Aug 16 2011 *)
PROG
(Magma) [n^4+984*n^3+902*n^2+394*n+858: n in [0..30]]; // Vincenzo Librandi, Aug 17 2011
(PARI) a(n) = n^4+984*n^3+902*n^2+394*n+858 \\ Charles R Greathouse IV, Aug 17 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Jul 24 2009
EXTENSIONS
Corrected and extended by Harvey P. Dale, Aug 16 2011
Offset changed from 1 to 0 by Vincenzo Librandi, Aug 17 2011
STATUS
approved