OFFSET
1,1
COMMENTS
a(1)=3, a(2)=2, a(a(1)+a(2)+...+a(n)) = a(n) and a(a(1)+a(2)+...+a(n)+1) = 5-a(n).
More generally, sequence a(n) = floor(r*(n+2))-floor(r*(n+1)), r = (1/2) *(z+sqrt(z^2+4)), z integer >=1, is defined with a(1), a(2) and a(a(1)+a(2)+...+a(n)+f(z)) = a(n); a(a(1)+a(2)+...+a(n)+f(z)+1) = (2z+1)-a(n) where f(1)=0, f(z)=z-2 for z>=2.
Conjecture: a(n) = A097509(n+1). - Benedict W. J. Irwin, Mar 13 2016. [See the discussion in A097509. - N. J. A. Sloane, Mar 09 2021]
Theorem: Referring to the solution to Problem B6 in the 81st William Lowell Putnam Mathematical Competition (see link), in the notation of the first solution, the sequence a(n) = c_{n+1} indexed from 1 equals the present sequence, A082844. - Manjul Bhargava, Kiran Kedlaya, and Lenny Ng, Sep 09 2021.
LINKS
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng, Solutions to the 81st William Lowell Putnam Mathematical Competition
Luke Schaeffer, Jeffrey Shallit, and Stefan Zorcic, Beatty Sequences for a Quadratic Irrational: Decidability and Applications, arXiv:2402.08331 [math.NT], 2024. See pp. 17-19.
N. J. A. Sloane, Families of Essentially Identical Sequences, Mar 24 2021 (Includes this sequence)
FORMULA
a(n) = floor(r*(n+2))-floor(r*(n+1)) where r=1+sqrt(2).
MAPLE
A082844:=n->floor((1+sqrt(2))*(n+2))-floor((1+sqrt(2))*(n+1)): seq(A082844(n), n=1..100); # Wesley Ivan Hurt, Mar 13 2016
MATHEMATICA
With[{r=1+Sqrt[2]}, Table[Floor[r*(n+2)]-Floor[r*(n+1)], {n, 110}]] (* Harvey P. Dale, Oct 10 2012 *)
PROG
(Magma) [Floor((1+Sqrt(2))*(n+2))-Floor((1+Sqrt(2))*(n+1)) : n in [1..100]]; // Wesley Ivan Hurt, Mar 13 2016
CROSSREFS
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A003151 as the parent: A003151, A001951, A001952, A003152, A006337, A080763, A082844 (conjectured), A097509, A159684, A188037, A245219 (conjectured), A276862. - N. J. A. Sloane, Mar 09 2021
KEYWORD
nonn,easy
AUTHOR
Benoit Cloitre, Apr 15 2003; revised Jun 07 2003
STATUS
approved