OFFSET
1,2
COMMENTS
From Robert Israel, Jun 28 2017: (Start)
Multiplicative order of n^3 mod (n+1)^3.
a(n) divides A053191(n+1).
a(n) is even for n >= 2.
If n == 3 (mod 4) then a(n) == 0 (mod 16), otherwise it appears that a(n) == 2 or 6 (mod 16).
If n == 0 or 4 (mod 6), then a(n) = 2*n^2.
(End)
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
K. Kashihara, Comments and Topics on Smarandache Notions and Problems, Erhus University Press, 1996, 50 pages. See page 45.
K. Kashihara, Comments and Topics on Smarandache Notions and Problems, Erhus University Press, 1996, 50 pages. [Cached copy] See page 45.
FORMULA
Empirical: a(n+36) = 3*a(n+24) - 3*a(n+12) + a(n) for n >= 2. - Robert Israel, Jun 28 2017
Empirical g.f.: x*(x^36 + 2*x^35 + 2*x^33 + 2*x^32 + 6*x^31 + 16*x^30 + 50*x^29 + 6*x^28 + 98*x^27 + 64*x^26 + 54*x^25 + 47*x^24 + 236*x^23 + 48*x^22 + 332*x^21 + 92*x^20 + 132*x^19 + 208*x^18 + 428*x^17 + 36*x^16 + 428*x^15 + 208*x^14 + 132*x^13 + 95*x^12 + 338*x^11 + 48*x^10 + 242*x^9 + 50*x^8 + 54*x^7 + 64*x^6 + 98*x^5 + 6*x^4 + 50*x^3 + 16*x^2 + 6*x + 1)/(-x^36 + 3*x^24 - 3*x^12 + 1). - Colin Barker, Jun 30 2017
EXAMPLE
5^3 = 125, 6^3 = 216, 125^6 = 14551915228366851806640625 == 1 mod 216, so a(5) = 6.
(6^3)^98 == 1 mod 7^3, so a(6) = 98.
MAPLE
seq(numtheory:-order(n^3, (n+1)^3), n=1..300); # Robert Israel, Jun 28 2017
PROG
(PARI) a(n) = my(k=1); while(Mod(n^3, (n+1)^3)^k!=1, k++); k \\ Felix Fröhlich, Jun 28 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Jun 28 2017
EXTENSIONS
Corrected and more terms from Robert Israel, Jun 28 2017
STATUS
approved