A318236 a(n) = (3*2^(4*n+3) + 1)/5.

5, 77, 1229, 19661, 314573, 5033165, 80530637, 1288490189, 20615843021, 329853488333, 5277655813325, 84442493013197, 1351079888211149, 21617278211378381, 345876451382054093, 5534023222112865485, 88544371553805847757, 1416709944860893564109, 22667359117774297025741

Offset

0,1

Comments

a(n) is the smallest positive multiplicative inverse of 5 modulo 2^(4*n+3).

In binary, a(n) is written as 10011001...1001101 where "1001" appears n times. When n approaches infinity we get the 2-adic expansion of 1/5: ...10011001101. Similarly, the 2-adic expansion of 1/3 is ...101010101011.

Links

Jianing Song, Table of n, a(n) for n = 0..200

Index entries for linear recurrences with constant coefficients, signature (17,-16).

Formula

O.g.f.: (5 - 8*x)/((1 - x)*(1 - 16*x)).

E.g.f.: (24*exp(16*x) + exp(x))/5.

a(0) = 5, a(1) = 77; for n>1, a(n) = 17*a(n-1) - 16*a(n-2).

Example

The smallest solution to 5*x == 1 (mod 8) is x = (3*2^3 + 1)/5 = 5.
The smallest solution to 5*x == 1 (mod 128) is x = (3*2^7 + 1)/5 = 77.

Mathematica

Table[(3 2^(4 n + 3) + 1) / 5, {n, 0, 20}] (* Vincenzo Librandi, Aug 24 2018 *)
LinearRecurrence[{17, -16}, {5, 77}, 20] (* Harvey P. Dale, Sep 25 2020 *)

Prog

(PARI) a(n) = (3*2^(4*n + 3) + 1)/5
(Magma) [(3*2^(4*n + 3) + 1)/5: n in [0..20]]; // Vincenzo Librandi, Aug 24 2018
(Python)
def A318236(n): return (3*(1<<(n<<2)+3)+1)//5 # Chai Wah Wu, Jul 29 2022

Crossrefs

A007583 gives the smallest positive multiplicative inverse of 3 modulo 2^(2*n) and 2^(2*n+1), A299960 gives the smallest positive multiplicative inverse of 5 modulo 2^(4*n), 2^(4*n+1) and 2^(4*n+2).

Sequence in context: A186660 A287041 A360351 * A009485 A188455 A015056

Adjacent sequences: A318233 A318234 A318235 * A318237 A318238 A318239

Keyword

nonn,easy

Author

Jianing Song, Aug 21 2018

Status

approved