A365397 a(n) = 64 + A000720(n) - A365398(n).
63, 63, 63, 62, 63, 62, 63, 62, 62, 61, 62, 61, 62, 62, 61, 60, 61, 60, 61, 60, 60, 60, 61, 60, 60, 60, 60, 59, 60, 59, 60, 60, 60, 60, 60, 59, 60, 60, 60, 59, 60, 59, 60, 60, 60, 60, 61, 60, 60, 60, 60, 60, 61, 60, 60, 59, 59, 59, 60, 59, 60, 60, 59, 58, 58
Offset: 1
Keywords
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.
Programs
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Python
from bisect import bisect from sympy import divisor_sigma, primepi def A365397(n): plist, qlist, c = tuple(divisor_sigma(i) for i in range(1,n+1)), [0]*(n+1), 0 for i in range(n): qlist[a:=bisect(qlist,plist[i],lo=1,hi=c+1,key=lambda x:plist[x])]=i c = max(c,a) return 64+primepi(n)-c # Chai Wah Wu, Sep 08 2023
Formula
a(n)<=63. This is due to the fact that A000203(p) = p+1 for p prime, and therefore A365398(n) >= A000720(n)+1. - Chai Wah Wu, Sep 08 2023
Comments