A056530 Sequence remaining after third round of Flavius Josephus sieve; remove every fourth term of A047241.

1, 3, 7, 13, 15, 19, 25, 27, 31, 37, 39, 43, 49, 51, 55, 61, 63, 67, 73, 75, 79, 85, 87, 91, 97, 99, 103, 109, 111, 115, 121, 123, 127, 133, 135, 139, 145, 147, 151, 157, 159, 163, 169, 171, 175, 181, 183, 187, 193, 195, 199, 205, 207, 211, 217, 219, 223, 229, 231

Offset: 1

Author

Henry Bottomley, Jun 19 2000

Keywords

• easy
• nonn

Comments

Numbers {1, 3, 7} mod 12: A017533, A017557, A017605 interleaved.

Links

• Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).
• Index entries for sequences related to the Josephus Problem

Crossrefs

We have A000027 after 0 rounds of sieving, A005408 after 1 round of sieving, A047241 after 2 rounds, A056530 after 3 rounds, A056531 after 4 rounds, A000960 after all rounds. After n rounds the remaining sequence comprises A002944(n) numbers mod A003418(n+1), i.e. 1/(n+1) of them.

Programs

Mathematica

LinearRecurrence[{1,0,1,-1},{1,3,7,13},60] (* Harvey P. Dale, Oct 19 2022 *)

Formula

From Chai Wah Wu, Jul 24 2016: (Start)

a(n) = a(n-1) + a(n-3) - a(n-4) for n > 4.

G.f.: x*(5*x^3 + 4*x^2 + 2*x + 1)/(x^4 - x^3 - x + 1). (End)

a(n) = 4*n - (13 + 2*A131713(n))/3. - R. J. Mathar, Jun 22 2020