A288572 a(n) = smallest positive integer k such that (n^3)^k == 1 mod (n+1)^3.

1, 6, 16, 50, 6, 98, 64, 54, 50, 242, 48, 338, 98, 150, 256, 578, 54, 722, 400, 294, 242, 1058, 192, 1250, 338, 486, 784, 1682, 150, 1922, 1024, 726, 578, 2450, 432, 2738, 722, 1014, 1600, 3362, 294, 3698, 1936, 1350, 1058, 4418, 768, 4802, 1250, 1734, 2704, 5618, 486, 6050, 3136, 2166, 1682, 6962

Offset: 1

Author

N. J. A. Sloane, Jun 28 2017

Keywords

• nonn

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)

Examples

			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.
		

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.

Crossrefs

Cf. A053191.

Programs

Maple

seq(numtheory:-order(n^3, (n+1)^3), n=1..300); # Robert Israel, Jun 28 2017

PARI

a(n) = my(k=1); while(Mod(n^3, (n+1)^3)^k!=1, k++); k \\ Felix Fröhlich, Jun 28 2017

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

Extensions

Corrected and more terms from Robert Israel, Jun 28 2017