A365474 a(n) = A365339(10^n).

1, 7, 34, 193, 1276, 9656, 78562, 664643, 5761519, 50847598

Offset: 0

Author

Chai Wah Wu, Sep 04 2023

Keywords

• nonn
• hard
• more

Comments

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.

Links

• Paul Pollack, Carl Pomerance, and Enrique Treviño, Sets of monotonicity for Euler's totient function, preprint. See M(n).
• Paul Pollack, Carl Pomerance, and Enrique Treviño, Sets of monotonicity for Euler's totient function, Ramanujan J. 30 (2013), no. 3, pp. 379-398.
• Terence Tao, Monotone non-decreasing sequences of the Euler totient function, arXiv:2309.02325 [math.NT], 2023.

Crossrefs

Cf. A000010, A000720, A006880, A365339.

Programs

Python

from bisect import bisect
from sympy import totient
def A365474(n):
    m = 10**n
    plist, qlist, c = tuple(totient(i) for i in range(1,m+1)), [0]*(m+1), 0
    for i in range(m):
        qlist[a:=bisect(qlist,plist[i],lo=1,hi=c+1,key=lambda x:plist[x])]=i
        c = max(c,a)
    return c

Formula

a(n) = A006880(n)+64 for n >= 5 (conjectured).