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%I #19 Mar 22 2017 21:55:08
%S 3,0,1,0,3,28,45,276,595,1128,1953,3160,4851,264540,190333,254268,
%T 18915,3366496,32385,125391168,199588483,64620,174673821,5039370820,
%U 1784859363,16908230328,165025,34420237176,58409997075,1367074573228,2294838551853,15289788305820
%N Lexicographically earliest sequence of nonnegative integers such that a(1)=3 and the sequence is p-periodic mod p for any p > 0.
%C 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.
%C The first cycles mod p of this sequence are:
%C p Cycle of a mod p
%C - ----------------
%C 1 0
%C 2 1, 0
%C 3 0, 0, 1
%C 4 3, 0, 1, 0
%C 5 3, 0, 1, 0, 3
%C 6 3, 0, 1, 0, 3, 4
%C 7 3, 0, 1, 0, 3, 0, 3
%C 8 3, 0, 1, 0, 3, 4, 5, 4
%C 9 3, 0, 1, 0, 3, 1, 0, 6, 1
%C For k>=0, let c_k denote the variant with initial term k.
%C Naturaly, we have a=c_3.
%C For some values of k, c_k has a polynomial closed form.
%C The first such values to be known are:
%C - k=0: c_0(n) = 0 = A000004(n),
%C - k=1: c_1(n) = (n-2)^2 = A000290(n-2),
%C - k=2: c_2(n) = (n-2)*(n-3) = A002378(n-3),
%C - k=19: c_19(n) = (n-2)*(n^3 - 14*n^2 + 63*n - 88)/2,
%C - k=20: c_20(n) = (n-2)*(n-3)*(n-5)*(n-6)/2,
%C - k=22: c_22(n) = (n-2)*(n-3)*(n^2 - 11*n + 32)/2,
%C - k=40: c_40(n) = (n-2)*(n-3)*(n-5)*(n-6),
%C - k=172: c_172(n) = (n-2)*(n-3)*(n-5)*(n^3 - 23*n^2 + 172*n - 408)/12.
%C We notice that c_40 = 2*c_20.
%C As for A281409, this sequence is the first of a family (of sequences parametrized by their initial term) showing some kind of irregularity.
%C For k>=0 and n>0, let d_n(k)=c_k(n):
%C - In particular: d_1(k)=k, and a(n)=d_n(3),
%C - For any n>1, d_n is periodic.
%C The cycles for the first d_n (with n>1) are:
%C n Cycle of d_n
%C - ------------
%C 2 0
%C 3 0, 1
%C 4 0, 4, 2
%C 5 0, 9, 6, 3
%C 6 0, 16, 12, 28, 24, 40, 36, 52, 48, 4
%C 7 0, 25, 20, 45, 40, 5
%H Rémy Sigrist, <a href="/A284148/b284148.txt">Table of n, a(n) for n = 1..2000</a>
%H Rémy Sigrist, <a href="/A284148/a284148.gp.txt">PARI program for A284148</a>
%e By definition, a(1)=3.
%e a(2) must equal 3 mod 1; a(2)=0 is suitable.
%e a(3) must equal 3 mod 2 and 0 mod 1; a(3)=1 is suitable.
%e a(4) must equal 3 mod 3 and 0 mod 2 and 1 mod 1; a(4)=0 is suitable.
%e a(5) must equal 3 mod 4 and 0 mod 3 and 1 mod 2 and 0 mod 1; a(5)=3 is suitable.
%Y Cf. A000004, A000290, A002378, A281409.
%K nonn
%O 1,1
%A _Rémy Sigrist_, Mar 21 2017
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