A083658 a(n) = a(n-1) + a(n-2) + gcd(a(n-1), a(n-2)) for n > 1; a(0)=1, a(1)=1.

1, 1, 3, 5, 9, 15, 27, 45, 81, 135, 243, 405, 729, 1215, 2187, 3645, 6561, 10935, 19683, 32805, 59049, 98415, 177147, 295245, 531441, 885735, 1594323, 2657205, 4782969, 7971615, 14348907, 23914845, 43046721, 71744535, 129140163, 215233605, 387420489

Offset

0,3

Comments

Record high values in A003961 (except for the duplicated 1). - Nicolas Bělohoubek, Jun 18 2022

Apart from a(0), this sequence is the answer to Question 21 in the 2022 Shanghai College Entrance Mathematics Examination: a(1) = 1, a(2*m) = 3^m for all m; for any n >= 2, there exists 1 <= i <= n-1 such that a(n+1) = 2*a(n)-a(i). Find a(n). - Yifan Xie, Jul 20 2022

a(n) n>1 are a subset of the record values formed by the odd composite numbers (A071904) divided by their largest prime factor. For example, A071904[2434] = 6561 with largest pf = 3. 6561/3 = 2187 and appears in A083658. - Bill McEachen, Jul 06 2024

Links

Michael De Vlieger, Table of n, a(n) for n = 0..4191

Yulu Education, 2022 Shanghai College Entrance Examination Mathematics Paper and Answer Analysis (Examinee Recall Version) (In Chinese)

Index entries for linear recurrences with constant coefficients, signature (0,3).

Formula

a(2n) = 3^n, a(2n+1) = 5*3^(n-1) for n>0; a(0)=1, a(1)=1.

G.f.: (2*x^3+1+x)/(1-3*x^2). - R. J. Mathar, Feb 27 2010

a(n) = 3 * a(n-2), n>3, a(2)=3, a(3)=5. - Bill McEachen, Jul 06 2024

Mathematica

CoefficientList[Series[(-2*x^3 - x - 1)/(3*x^2 - 1), {x, 0, 200}], x] (* Vladimir Joseph Stephan Orlovsky, Jun 10 2011 *)

Crossrefs

Cf. A003961.

Sequence in context: A147087 A140190 A298352 * A018436 A018298 A017913

Adjacent sequences: A083655 A083656 A083657 * A083659 A083660 A083661

Keyword

nonn,easy

Author

Paul D. Hanna, Jun 13 2003

Status

approved