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a(n) = smallest positive integer k such that (n^3)^k == 1 mod (n+1)^3.
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%I #26 Jun 30 2017 05:27:47

%S 1,6,16,50,6,98,64,54,50,242,48,338,98,150,256,578,54,722,400,294,242,

%T 1058,192,1250,338,486,784,1682,150,1922,1024,726,578,2450,432,2738,

%U 722,1014,1600,3362,294,3698,1936,1350,1058,4418,768,4802,1250,1734,2704,5618,486,6050,3136,2166,1682,6962

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

%C From _Robert Israel_, Jun 28 2017: (Start)

%C Multiplicative order of n^3 mod (n+1)^3.

%C a(n) divides A053191(n+1).

%C a(n) is even for n >= 2.

%C If n == 3 (mod 4) then a(n) == 0 (mod 16), otherwise it appears that a(n) == 2 or 6 (mod 16).

%C If n == 0 or 4 (mod 6), then a(n) = 2*n^2.

%C (End)

%H Robert Israel, <a href="/A288572/b288572.txt">Table of n, a(n) for n = 1..10000</a>

%H K. Kashihara, <a href="http://www.gallup.unm.edu/~smarandache/Kashihara.pdf">Comments and Topics on Smarandache Notions and Problems</a>, Erhus University Press, 1996, 50 pages. See page 45.

%H K. Kashihara, <a href="/A011772/a011772.pdf">Comments and Topics on Smarandache Notions and Problems</a>, Erhus University Press, 1996, 50 pages. [Cached copy] See page 45.

%F Empirical: a(n+36) = 3*a(n+24) - 3*a(n+12) + a(n) for n >= 2. - _Robert Israel_, Jun 28 2017

%F 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

%e 5^3 = 125, 6^3 = 216, 125^6 = 14551915228366851806640625 == 1 mod 216, so a(5) = 6.

%e (6^3)^98 == 1 mod 7^3, so a(6) = 98.

%p seq(numtheory:-order(n^3, (n+1)^3), n=1..300); # _Robert Israel_, Jun 28 2017

%o (PARI) a(n) = my(k=1); while(Mod(n^3, (n+1)^3)^k!=1, k++); k \\ _Felix Fröhlich_, Jun 28 2017

%Y Cf. A053191.

%K nonn

%O 1,2

%A _N. J. A. Sloane_, Jun 28 2017

%E Corrected and more terms from _Robert Israel_, Jun 28 2017