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%I #14 Dec 16 2019 10:25:07
%S 2,1,3,5,4,24,2,22,21,20,19,6,17,16,15,14,4,6,20,22,99,8,97,3,95,94,
%T 93,92,13,393,89,391,9,389,85,84,83,82,384,80,382,60,59,58,75,377,376,
%U 375,53,373,372,68,370,66,368,796,24,365,793,363,362,361,789,788,787,786,14,784
%N a(n) is the index of the first 0 term in the rumor sequence with initial 0th term 1 and parameters b = 3 and n.
%C "A rumor sequence (running modulus recurrence sequence) is defined as follows: fix integer parameters b > 1 and n > 0. Set z[0] = any integer, and, for k > 0, define z[k] to be the least nonnegative residue of b*z[k-1] modulo (k+n). The rumor sequence conjecture states that all such rumor sequences are eventually 0." [Copied from the comments for A177356]
%H B. Dearden, J. Iiams, and J. Metzger, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL16/Dearden/dearden4.html">Rumor Arrays</a>, J. Integer Seq. 16 (2013), #13.9.3.
%H B. Dearden and J. Metzger, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL13/Dearden/dearden3.html">Running Modulus Recursions</a>, J. Integer Seq. 13(1) (2010), #10.1.6.
%F a(n) = inf{m > 0 | z[0] = 1, z[m] = 0, and z[k] = (3*z[k-1] mod (k + n)) for k = 1..m}.
%e For n = 1, we have z[0] = 1, z[1] = 1, and z[2] = 0, so a(1) = 2.
%e For n = 2, we have z[0] = 1 and z[1] = 0, so a(2) = 1.
%e For n = 3, we have z[0] = 1, z[1] = 3, z[2] = 4, and z[3] = 0, so a(3) = 3.
%e For n = 4, we have z[0] = 1, z[1] = 3, z[2] = 3, z[3] = 2, z[4] = 6, and z[5] = 0, so a(4) = 5.
%e For n = 5, we have z[0] = 1, z[1] = 3, z[2] = 2, z[3] = 6, and z[4] = 0, so a(5) = 4.
%t For[n=1, n<50, n++, k=0; Clear[z]; z[0]=1; z[k_]:=z[k]=Mod[3z[k-1], k+n];
%t While[z[k]>0, k++]; Print[k]; ] (* _Petros Hadjicostas_, Dec 14 2019, modification of a program in A177356 *)
%Y Cf. A177356, A208125.
%K nonn
%O 1,1
%A _Petros Hadjicostas_, Dec 13 2019
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