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A056084 Numbers n such that n^8 = 1 (mod 9^3). 1

%I

%S 1,728,730,1457,1459,2186,2188,2915,2917,3644,3646,4373,4375,5102,

%T 5104,5831,5833,6560,6562,7289,7291,8018,8020,8747,8749,9476,9478,

%U 10205,10207,10934,10936,11663,11665,12392,12394,13121,13123,13850,13852

%N Numbers n such that n^8 = 1 (mod 9^3).

%C By definition, it is obvious that x is in the sequence iff x + 9^3 is. Since 1 and 728 (= -1 mod 9^3) are the only solutions in the interval [0,728], one has a(2n+1) = 729*n + 1 (n >= 0) and a(2n) = 729*n - 1, which can be combined in the given formulas for a(n). - _Robert Israel_, _Philipp Emanuel Weidmann_ and _M. F. Hasler_, Apr 13 2015

%H Colin Barker, <a href="/A056084/b056084.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).

%F a(2n+1) = 729*n + 1 (n >= 0), a(2n) = 729*n - 1 (n >= 1), a(n) = 729*floor(n/2)-(-1)^n. - _M. F. Hasler_, Apr 13 2015

%F a(n) = (1458*n-729+(-1)^n*725)/4. - Conjectured by _Philipp Emanuel Weidmann_, explained by _Robert Israel_, Apr 13 2015

%F a(n) = a(n-1) + a(n-2) - a(n-3) for n>2. - _Vincenzo Librandi_, Apr 14 2015

%F G.f.: x*(x^2+727*x+1) / ((x-1)^2*(x+1)). - _Colin Barker_, Apr 14 2015

%p map(n -> (n+1,n+728),729*[$0..100]); # _Robert Israel_, Apr 13 2015

%t x=9; Select[ Range[ 20000 ], PowerMod[ #, x-1, x^3 ]==1& ]

%o (PARI) a(n)=n\2*729-(-1)^bittest(n,0) \\ _M. F. Hasler_, Apr 13 2015

%o (MAGMA) I:=[1,728,730]; [n le 3 select I[n] else Self(n-1)+Self(n-2)-Self(n-3): n in [1..40]]; // _Vincenzo Librandi_, Apr 14 2015

%o (PARI) Vec(x*(x^2+727*x+1)/((x-1)^2*(x+1)) + O(x^100)) \\ _Colin Barker_, Apr 14 2015

%K nonn,easy

%O 1,2

%A _Robert G. Wilson v_, Jun 08 2000

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Last modified November 21 09:14 EST 2019. Contains 329362 sequences. (Running on oeis4.)