OFFSET
1,2
COMMENTS
The reference gives the sequence without the initial 1 and then it is "alternate 3n+2 and 2n+3".
An alternative solution for this sequence is to consider the first second differences repeat giving 2, 3, 5, 5, 8, 7, 11, 12, 14, 14, 17, 16, 20, 21. - Lorraine Gregory, Ed.D. (lgregory(AT)lssu.edu), Jan 19 2009
REFERENCES
D. Dolan, J. Williamson and M. Muri, Mathematics Activities for Elementary School Teachers: A Problem Solving Approach, Addison Wesley Longman, Inc., NY, Fourth Edition (ISBN 0-201-61321-2), Chapter 1, Activity 2, #9.
LINKS
Harry J. Smith, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).
FORMULA
From Colin Barker, Mar 28 2012: (Start)
a(n) = 2*a(n-2)-a(n-4).
G.f.: x*(1+2*x+x^2+x^3)/((1-x)^2*(1+x)^2). (End)
a(n) = (5*n - 2 + (-1)^n*(n - 2))/4. - Andrew Howroyd, Dec 26 2024
MAPLE
for n from 0 to 80 do printf(`%d, %d, `, 2*n+1, 3*n+2) od:
MATHEMATICA
nn=50; Riffle[Range[1, 2*nn, 2], 3*Range[0, nn]+2] (* Harvey P. Dale, Jan 16 2013 *)
PROG
(PARI) a(n) = { if (n%2, n, 3*n/2 - 1) } \\ Harry J. Smith, Sep 02 2009
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
Nancy Shaffer (nancys(AT)rose.net), Aug 30 2001
EXTENSIONS
Better description from Brian Galebach, Sep 05, 2001
More terms from James A. Sellers, Sep 25 2001
STATUS
approved