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
Keywords
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
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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
Comments