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%I #21 Sep 06 2023 16:10:43
%S 1,7,34,193,1276,9656,78562,664643,5761519,50847598
%N a(n) = A365339(10^n).
%C The Pollack et al. reference lists a(4)-a(7) and conjectures that A365339(n) = A000720(n)+64 for n >= 31957 which in turns implies the conjecture that a(n) = A006880(n)+64 for n >= 5.
%H Paul Pollack, Carl Pomerance, and Enrique Treviño, <a href="https://math.dartmouth.edu/~carlp/MonotonePhi.pdf">Sets of monotonicity for Euler's totient function</a>, preprint. See M(n).
%H Paul Pollack, Carl Pomerance, and Enrique Treviño, <a href="https://doi.org/10.1007/s11139-012-9386-6">Sets of monotonicity for Euler's totient function</a>, Ramanujan J. 30 (2013), no. 3, pp. 379-398.
%H Terence Tao, <a href="https://arxiv.org/abs/2309.02325">Monotone non-decreasing sequences of the Euler totient function</a>, arXiv:2309.02325 [math.NT], 2023.
%F a(n) = A006880(n)+64 for n >= 5 (conjectured).
%o (Python)
%o from bisect import bisect
%o from sympy import totient
%o def A365474(n):
%o m = 10**n
%o plist, qlist, c = tuple(totient(i) for i in range(1,m+1)), [0]*(m+1), 0
%o for i in range(m):
%o qlist[a:=bisect(qlist,plist[i],lo=1,hi=c+1,key=lambda x:plist[x])]=i
%o c = max(c,a)
%o return c
%Y Cf. A000010, A000720, A006880, A365339.
%K nonn,hard,more
%O 0,2
%A _Chai Wah Wu_, Sep 04 2023
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