A168198 a(n) = 3*n - a(n-1) + 1 with n > 1, a(1)=1.

1, 6, 4, 9, 7, 12, 10, 15, 13, 18, 16, 21, 19, 24, 22, 27, 25, 30, 28, 33, 31, 36, 34, 39, 37, 42, 40, 45, 43, 48, 46, 51, 49, 54, 52, 57, 55, 60, 58, 63, 61, 66, 64, 69, 67, 72, 70, 75, 73, 78, 76, 81, 79, 84, 82, 87, 85, 90, 88, 93, 91, 96, 94, 99, 97, 102, 100, 105, 103, 108

Offset: 1

Author

Vincenzo Librandi, Nov 20 2009

Keywords

• nonn
• easy

Comments

Alternately add 5 and subtract 2, starting with 1. Apparently this was a test question: Find the next two numbers after 1,6,4,9,7,12,10. - N. J. A. Sloane, Dec 18 2010

Examples

			From _Muniru A Asiru_, Mar 20 2018: (Start)
For n = 2, a(2) = 3*2 - a[2-1] + 1 = 6 - a[1] + 1 = 6 - 1 + 1 = 6.
For n = 3, a(3) = 3*3 - a[3-1] + 1 = 9 - a[2] + 1 = 9 - 6 + 1 = 4.
For n = 4, a(4) = 3*4 - a[4-1] + 1 = 12 - a[3] + 1 = 12 - 4 + 1 = 9.
... (End)
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• David Klein, What do the NAEP math tests really measure?, Notices Amer. Math. Soc., 58 (No. 1, 2011), 53-55.
• Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Crossrefs

Cf. A084964, A103889, A168199, A168200.

Programs

GAP

a:=[1];; for n in [2..80] do a[n]:=3*n-a[n-1]+1; od; a; # Muniru A Asiru, Mar 20 2018

Magma

I:=[1,6,4]; [n le 3 select I[n] else Self(n-1)+Self(n-2)-Self(n-3): n in [1..60]]; // Vincenzo Librandi, Feb 28 2012

Maple

a:= proc(n) option remember: if n = 1 then 1 elif n >= 2 then 3*n - procname(n-1) + 1 fi; end:
seq(a(n), n = 1..70); # Muniru A Asiru, Mar 20 2018

Mathematica

LinearRecurrence[{1,1,-1},{1,6,4},100] (* Vincenzo Librandi, Feb 28 2012 *)

PARI

a(n)=(6*n+5+7*(-1)^n)/4 \\ Charles R Greathouse IV, Jan 11 2012

Formula

a(n) = (6*n + 5 + 7*(-1)^n)/4. - Jon E. Schoenfield, Jun 24 2010

G.f.: x*(1+5*x-3*x^2)/((1+x)(1-x)^2). - Bruno Berselli, Feb 28 2012

Sum_{n>=1} (-1)^(n+1)/a(n) = 1/3 + Pi/(6*sqrt(3)) + log(3)/2. - Amiram Eldar, Feb 23 2023