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

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, 14579, 14581, 15308

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

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Colin Barker)

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

Cf. A056082, A056083, A056085, A056086, A056087, A056088, A056089.

Sequence in context: A178654 A094733 A321822 * A191345 A345744 A368590

Adjacent sequences: A056081 A056082 A056083 * A056085 A056086 A056087

Keyword

nonn,easy

Author

Robert G. Wilson v, Jun 08 2000

Status

approved