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A056084 Numbers n such that n^8 = 1 (mod 9^3). 1
1, 728, 730, 1457, 1459, 2186, 2188, 2915, 2917, 3644, 3646, 4373, 4375, 5102, 5104, 5831, 5833, 6560, 6562, 7289, 7291, 8018, 8020, 8747, 8749, 9476, 9478, 10205, 10207, 10934, 10936, 11663, 11665, 12392, 12394, 13121, 13123, 13850, 13852 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

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

LINKS

Colin Barker, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

FORMULA

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

a(n) = (1458*n-729+(-1)^n*725)/4. - Conjectured by Philipp Emanuel Weidmann, explained by Robert Israel, Apr 13 2015

a(n) = a(n-1) + a(n-2) - a(n-3) for n>2. - Vincenzo Librandi, Apr 14 2015

G.f.: x*(x^2+727*x+1) / ((x-1)^2*(x+1)). - Colin Barker, Apr 14 2015

MAPLE

map(n -> (n+1, n+728), 729*[$0..100]); # Robert Israel, Apr 13 2015

MATHEMATICA

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

PROG

(PARI) a(n)=n\2*729-(-1)^bittest(n, 0) \\ M. F. Hasler, Apr 13 2015

(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

(PARI) Vec(x*(x^2+727*x+1)/((x-1)^2*(x+1)) + O(x^100)) \\ Colin Barker, Apr 14 2015

CROSSREFS

Sequence in context: A178654 A094733 A068162 * A191345 A023704 A043487

Adjacent sequences:  A056081 A056082 A056083 * A056085 A056086 A056087

KEYWORD

nonn,easy

AUTHOR

Robert G. Wilson v, Jun 08 2000

STATUS

approved

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Last modified June 25 00:41 EDT 2017. Contains 288708 sequences.