OFFSET
1,1
COMMENTS
The initial term a(1)=3 seems to be the least one that leads to a sequence that does not have a polynomial closed form.
The first cycles mod p of this sequence are:
p Cycle of a mod p
- ----------------
1 0
2 1, 0
3 0, 0, 1
4 3, 0, 1, 0
5 3, 0, 1, 0, 3
6 3, 0, 1, 0, 3, 4
7 3, 0, 1, 0, 3, 0, 3
8 3, 0, 1, 0, 3, 4, 5, 4
9 3, 0, 1, 0, 3, 1, 0, 6, 1
For k>=0, let c_k denote the variant with initial term k.
Naturaly, we have a=c_3.
For some values of k, c_k has a polynomial closed form.
The first such values to be known are:
- k=0: c_0(n) = 0 = A000004(n),
- k=1: c_1(n) = (n-2)^2 = A000290(n-2),
- k=2: c_2(n) = (n-2)*(n-3) = A002378(n-3),
- k=19: c_19(n) = (n-2)*(n^3 - 14*n^2 + 63*n - 88)/2,
- k=20: c_20(n) = (n-2)*(n-3)*(n-5)*(n-6)/2,
- k=22: c_22(n) = (n-2)*(n-3)*(n^2 - 11*n + 32)/2,
- k=40: c_40(n) = (n-2)*(n-3)*(n-5)*(n-6),
- k=172: c_172(n) = (n-2)*(n-3)*(n-5)*(n^3 - 23*n^2 + 172*n - 408)/12.
We notice that c_40 = 2*c_20.
As for A281409, this sequence is the first of a family (of sequences parametrized by their initial term) showing some kind of irregularity.
For k>=0 and n>0, let d_n(k)=c_k(n):
- In particular: d_1(k)=k, and a(n)=d_n(3),
- For any n>1, d_n is periodic.
The cycles for the first d_n (with n>1) are:
n Cycle of d_n
- ------------
2 0
3 0, 1
4 0, 4, 2
5 0, 9, 6, 3
6 0, 16, 12, 28, 24, 40, 36, 52, 48, 4
7 0, 25, 20, 45, 40, 5
LINKS
Rémy Sigrist, Table of n, a(n) for n = 1..2000
Rémy Sigrist, PARI program for A284148
EXAMPLE
By definition, a(1)=3.
a(2) must equal 3 mod 1; a(2)=0 is suitable.
a(3) must equal 3 mod 2 and 0 mod 1; a(3)=1 is suitable.
a(4) must equal 3 mod 3 and 0 mod 2 and 1 mod 1; a(4)=0 is suitable.
a(5) must equal 3 mod 4 and 0 mod 3 and 1 mod 2 and 0 mod 1; a(5)=3 is suitable.
CROSSREFS
KEYWORD
nonn
AUTHOR
Rémy Sigrist, Mar 21 2017
STATUS
approved