A273465 Numbers generated by starting at 1 and adding twice and subtracting once following the sequence of positive integers.

1, 3, 6, 2, 7, 13, 6, 14, 23, 13, 24, 36, 23, 37, 52, 36, 53, 71, 52, 72, 93, 71, 94, 118, 93, 119, 146, 118, 147, 177, 146, 178, 211, 177, 212, 248, 211, 249, 288, 248, 289, 331, 288, 332, 377, 331, 378, 426, 377, 427

Offset

1,2

Comments

This sequence is generated by listing the successive results of 1 + 2 + 3 - 4 + 5 + 6 - 7 + 8 + 9 - 10 + ... etc.

By adding twice and subtracting once, the results at positions i, i+2 and i+4 are linked by the relationship: n, n+1, n.

Links

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (1,0,2,-2,0,-1,1).

Formula

Let the sequence be written 1, 3, a, 2, a+1, b, a, b+1, c, b, c+1, d, c, ... The (a, a+1, a) triple at positions i, i+2, i+4 can be determined from the previous two sequences: (c = 2*b - a + 3). This new sequence starts at position i+3.

G.f.: (-2*z^6 - z^4 + 6*z^3 - 3*z^2 - 2*z - 1)/((z - 1)^3*(z^2 + z + 1)^2).

a(3*n-2) = A143689(n-1).

From Franck Maminirina Ramaharo, Dec 27 2018: (Start)

a(n) = a(n-1) + 2*a(n-3) - 2*a(n-4) - a(n-6) + a(n-7), n >= 8.

E.g.f.: (1/18)*(exp(3*x/2)*(32 + 6*x + 3*x^2) + (4 - 12*x)*cos(sqrt(3)*x/2) - 4*sqrt(3)*(1 + x)*sin(sqrt(3)*x/2))/exp(x/2). (End)

Example

n1 = 0 + 1 = 1
n2 = 1 + 2 = 3
n3 = 1 + 2 + 3 = 6
n4 = 1 + 2 + 3 - 4 = 2
n5 = 1 + 2 + 3 - 4 + 5 = 7
n6 = 1 + 2 + 3 - 4 + 5 + 6 = 13
n7 = 1 + 2 + 3 - 4 + 5 + 6 - 7 = 6
...
->
(n3, n5, n7) = (6, 7, 6)
(n6, n8, n10) = (13, 14, 13)
...
->
n9 = 2.n6 - n7 + 3 = 23

Mathematica

Accumulate@ Table[If[Mod[n, 3] == 1 && n > 1, -n, n], {n, 50}] (* Michael De Vlieger, May 23 2016 *)

Crossrefs

Cf. A143689.

Sequence in context: A249558 A072007 A078783 * A328503 A333826 A125717

Adjacent sequences: A273462 A273463 A273464 * A273466 A273467 A273468

Keyword

nonn

Author

Mattias Dooreman, May 23 2016

Extensions

a(25) corrected by Seiichi Manyama, May 26 2016

Status

approved