A249916 a(n) = 4*(n - 1) - a(n-3), n >= 3, a(0) = a(1) = 1, a(2) = 5.

1, 1, 5, 7, 11, 11, 13, 13, 17, 19, 23, 23, 25, 25, 29, 31, 35, 35, 37, 37, 41, 43, 47, 47, 49, 49, 53, 55, 59, 59, 61, 61, 65, 67, 71, 71, 73, 73, 77, 79, 83, 83, 85, 85, 89, 91, 95, 95, 97, 97, 101, 103, 107, 107, 109, 109, 113, 115, 119, 119, 121, 121, 125, 127

Offset

0,3

Comments

Conjecture: These are the natural numbers of the form 6*j +- 1 in which those of the form 12*k +- 1 are repeated.

From Jianing Song, Jan 28 2019: (Start)

The second conjecture in Formula section is correct.

We can see this from the recurrence: a(n) = 4*(n - 1) - a(n-3) (1)

Replace n with n+1: a(n+1) = 4*n - a(n-2) (2)

Subtract (1) from (2): a(n+1) = 4 + a(n) - a(n-2) + a(n-3) (3)

Replace n with n+1: a(n+2) = 4 + a(n+1) - a(n-1) + a(n-2) (4)

Subtract (3) from (4): a(n+2) = 2*a(n+1)-a(n)-a(n-1)+2*a(n-2)-a(n-3) (5)

This also confirms the conjecture in Comment section and the conjecture on the g.f. in Formula section. (End)

Links

Table of n, a(n) for n=0..63.

Index entries for linear recurrences with constant coefficients, signature (2,-1,-1,2,-1).

Formula

G.f.: (1 - x + 4*x^2 - x^3 + x^4)/((1 - x)^2*(1 + x^3)). [Confirmed, see Jianing Song in Comment section.]

Recurrence: a(n) = 2*a(n-1) - a(n-2) - a(n-3) + 2*a(n-4) - a(n-5) for n > 4, a(0)=a(1)=1, a(2)=5, a(3)=7, a(4)=11. [Confirmed, see Jianing Song in Comment section.]

Mathematica

a[0] = a[1] = 1; a[2] = 5; a[n_] := 4*(n - 1) - a[n - 3]; Table[a[n], {n, 0, 63}]
RecurrenceTable[{a[0]==a[1]==1, a[2]==5, a[n]==4(n-1)-a[n-3]}, a, {n, 70}] (* Harvey P. Dale, Jan 26 2019 *)

Crossrefs

Cf. A007310, A091998.

Sequence in context: A254930 A317769 A104200 * A115044 A206770 A107257

Adjacent sequences: A249913 A249914 A249915 * A249917 A249918 A249919

Keyword

nonn,easy

Author

L. Edson Jeffery, Jan 14 2015

Status

approved