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

Author

Mattias Dooreman, May 23 2016

Keywords

• nonn

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.

Examples

			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
		

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).

Crossrefs

Cf. A143689.

Programs

Mathematica

Accumulate@ Table[If[Mod[n, 3] == 1 && n > 1, -n, n], {n, 50}] (* Michael De Vlieger, May 23 2016 *)
LinearRecurrence[{1,0,2,-2,0,-1,1},{1,3,6,2,7,13,6},80] (* Harvey P. Dale, Oct 13 2024 *)

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)

Extensions

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