%I #20 Nov 16 2023 12:24:31
%S 1,1,2,1,3,2,3,2,1,4,2,3,2,1,5,4,7,6,2,3,2,1,6,2,3,2,1,7,6,12,4,2,3,2,
%T 1,8,4,7,6,2,3,2,1,9,6,12,4,2,3,2,1,10,4,7,6,2,3,2,1,11,10,18,6,2,3,2,
%U 1,12,4,7,6,2,3,2,1,13,12,28,12,4,7,6,2,3,2,1,14,6,12,4,2,3,2,1
%N Irregular triangle read by rows: row n contains the sequence n, p(n), s(p(n)), p(s(p(n))), p(p(s(p(n)))), s(p(p(s(p(n))))), ..., repeatedy applying (p,s,p) to n, where p = phi (A000010), s = sigma = (A000203), stopping after the first 1 is reached. If 1 is never reached, row n contains n, -1.
%H Paolo Xausa, <a href="/A334523/b334523.txt">Table of n, a(n) for n = 1..9094</a> (rows 1..500 of the triangle, flattened)
%H L. Alaoglu and P. Erdős, <a href="http://dx.doi.org/10.1090/S0002-9904-1944-08257-8">A conjecture in elementary number theory</a>, Bull. Amer. Math. Soc. 50 (1944), 881-882.
%e Triangle begins:
%e 1, 1,
%e 2, 1,
%e 3, 2, 3, 2, 1,
%e 4, 2, 3, 2, 1,
%e 5, 4, 7, 6, 2, 3, 2, 1,
%e 6, 2, 3, 2, 1,
%e 7, 6, 12, 4, 2, 3, 2, 1,
%e 8, 4, 7, 6, 2, 3, 2, 1,
%e 9, 6, 12, 4, 2, 3, 2, 1,
%e 10, 4, 7, 6, 2, 3, 2, 1,
%e ...
%e For row n=5, for example, we get 5 -> phi(5) = 4 -> sigma(4) = 7 -> phi(7) = 6 -> phi(6) = 2 -> sigma(2) = 3 -> phi(3) = 2 -> phi(2) = 1 (stop).
%t A334523row[n_]:=Module[{i=0},NestWhileList[If[Mod[i++,3]==1,DivisorSigma[1,#],EulerPhi[#]]&,n,i==0||#>1&]];Array[A334523row,15] (* _Paolo Xausa_, Nov 16 2023 *)
%Y See A032452, A334725 for other versions.
%Y Cf. A000010, A000203, A334685, A334686.
%K nonn,tabf
%O 1,3
%A _N. J. A. Sloane_, May 09 2020, following a suggestion from _Amiram Eldar_.
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