A208125 Number of distinct n-th rows in arrays whose columns are running modulus recurrence sequences.

1, 1, 3, 10, 41, 46, 277, 566, 1601, 1834, 18341, 17802, 213625, 230870, 225365, 465790, 7452641, 8129874, 146337733

Offset: 1

Author

Joel Iiams, Mar 28 2012

Keywords

• nonn
• more

Comments

For each positive integer b, construct an array of integers with columns indexed by {0,1,2,3,...} and rows indexed by {1,2,3,4,...}. Identify the entry of the n-th row and k-th column by z_{n,k}(b). Set z_{1,k}(b) = 0 for all k. Compute z_{n,k}(b) = (b*z_{n-1,k}(b) + k) mod n, always using least nonnegative residues. The columns of these arrays are RuMoR (Running Modulus Recurrence) sequences. The sequence given is the number of distinct n-th rows among all arrays.

In the Mathematica dynamic program that accompanies the paper by Dearden et al. (2013), the output table is the transpose of the array alluded in the name of the sequence, the comments above, and the example below. The same holds for the table on pp. 3-4 in the paper (where b = 2 for the table on p. 3 and b = 6 for the table on p. 4). - Petros Hadjicostas, Dec 13 2019

Examples

			a(3) = 3. Every distinct row is periodic with a period dividing 6. b=1 generates 0, 2, 2, 1, 1, 0 repeating, b=2 generates 0, 0, 2, 2, 1, 1 repeating, and b=3 generates 0, 1, 2 repeating. All other values of b give one of these.
From _Petros Hadjicostas_, Dec 13 2019: (Start)
Using the dynamic Mathematica program provided with the paper by Dearden et al. (2013) (but taking the transpose of the output table), we see that for all b >= 1 the first row is always 0, 0, 0, 0, ..., so a(1) = 1.
By looking at the second rows, we see that for all b >= 1 the 2nd row is always 0, 1, 0, 1, ..., so a(2) = 1.
By looking at the 3rd rows, we see that for all b with 1 = b mod 3, we get 0, 2, 2, 1, 1, 0 repeating (with period 6); for all b with 2 = b mod 3, we get 0, 0, 2, 2, 1, 1 repeating (with period 6); and with 0 = b mod 3, we get 0, 1, 2 repeating (with period 3). (See also the example above.) Thus, a(3) = 3.
By looking at the 4th rows, we see that for all b with 1 = b mod 12, we get 0, 3, 0, 0, 1, 1, 2, 1, 2, 2, 3, 3 repeating (with period 12); for 2 = b mod 12, we get 0, 1, 2, 3, 2, 3, 2, 3, 0, 1, 0, 1 repeating (with period 12); for 3 = b mod 12, we get 0, 0, 0, 3, 3, 3, 2, 2, 2, 1, 1, 1 repeating (with period 12); for (b mod 12) = 0, 4, or 8, we get 0, 1, 2, 3 repeating (with period 4); for 5 = b mod 12, we get 0, 1, 0, 1, 1, 2, 2, 3, 2, 3, 3, 0 repeating (with period 12); for 6 = b mod 12, we get 0, 3, 2, 3, 2, 1, 2, 1, 0, 1, 0, 3 repeating (with period 12); for 7 = b mod 12, we get 0, 3, 0, 2, 3, 1, 2, 1, 2, 0, 1, 3 repeating (with period 12); for 9 = b mod 12, we get 0, 2, 0, 3, 1, 3, 2, 0, 2, 1, 3, 1 repeating (with period 12); with 10 = b mod 12, we get 0, 1, 2, 1, 2, 1, 2, 3, 0, 3, 0, 3 repeating (with period 12); and for 11 = b mod 12, we get 0, 1, 0, 3, 0, 2, 3, 2, 3, 1, 2 repeating (with period 12). Thus, a(4) = 10. (End)
		

Links

• B. Dearden, J. Iiams, and J. Metzger, A Function Related to the Rumor Sequence Conjecture, J. Integer Seq. 14(1) (2011), Article 11.2.3.
• B. Dearden, J. Iiams, and J. Metzger, Rumor Arrays, J. Integer Seq. 16 (2013), Article 13.9.3.
• B. Dearden and J. Metzger, Running Modulus Recursions, J. Integer Seq. 13(1) (2010), Article 10.1.6.

Crossrefs

Cf. A177356, A330411.

Programs

Mathematica

(* the following code computes a(n) for parameter n - it is slow for n >= 8 *)
x[, , 0] = 0; x[b_, c_, n_] := x[b, c, n] = Mod[b*x[b, c, n - 1] + c, n]; Table[mtab = Table[Table[x[b, c, n], {c, 0, Apply[LCM, Range[n]]}], {b, 1, Apply[LCM, Range[n]]}]; Length[Union[mtab]], {n, 6}]