A391434 Composite numbers (not multiples of 10) whose (radix-10) constant congruence speed equals the maximum of the constant congruence speeds of all their prime factors.

4, 6, 8, 9, 12, 16, 22, 27, 33, 34, 35, 36, 38, 39, 44, 45, 46, 48, 49, 52, 54, 58, 62, 64, 66, 69, 72, 75, 78, 81, 85, 87, 88, 92, 94, 96, 102, 104, 106, 108, 111, 114, 115, 116, 117, 121, 122, 123, 125, 128, 134, 136, 138, 141, 142, 144, 146, 148, 152, 153

Offset

1,1

Comments

For the definition of "constant congruence speed", see A373387 and Def. 1.2 (and also Def. 1.1) of "Number of stable digits of any integer tetration" in Links (for all positive integers m > 1 and not a multiple of 10, this corresponds to A373387(m)).

If the product of two positive integers m' and m'' is not divisible by 10, then the constant congruence speed of m'*m'' is necessarily greater than or equal to the minimum of the constant congruence speeds of m' and m'' (see Equation 2.4 of "A Compact Notation for Peculiar Properties Characterizing Integer Tetration" in Links).

By definition, this sequence contains no prime number.

References

Marco Ripà, La strana coda della serie n^n^...^n, Trento, UNI Service, Nov 2011. ISBN 978-88-6178-789-6

Links

Table of n, a(n) for n=1..60.

Marco Ripà, The congruence speed formula, Notes on Number Theory and Discrete Mathematics, 2021, 27(4), 43-61.

Marco Ripà and Gabriele Di Pietro, A Compact Notation for Peculiar Properties Characterizing Integer Tetration, Zenodo, 2025.

Marco Ripà and Luca Onnis, Number of stable digits of any integer tetration, Notes on Number Theory and Discrete Mathematics, 2022, 28(3), 441-457.

Example

a(2) = 6 since the constant congruence speed of 6 is 1, the constant congruence speed of 2 is 1, the constant congruence speed of 3 is 1, and 1 equals max{1,1}.

Prog

(Python)
def upto(n):
    s=[]
    for k in range(4, n+1):
        if k%10==0:
            continue
        Vk=A373387(k)
        if Vk is None:
            continue
        pf=factorint(k)
        if sum(pf.values())==1:
            continue
        maxVp=0
        for p, e in pf.items():
            Vp=A373387(p)
            if Vp is not None and Vp>maxVp:
                maxVp=Vp
        if Vk==maxVp:
            s.append(k)
    return s
print(upto(10000))

Crossrefs

Cf. A317905, A373387, A389432, A389979, A389980, A389981, A390320, A392231, A392233.

Subsequence of A067251.

Sequence in context: A344030 A350706 A036310 * A392233 A046760 A115684

Adjacent sequences: A391431 A391432 A391433 * A391435 A391436 A391437

Keyword

nonn,base

Author

Marco Ripà and Gabriele Di Pietro, Jan 12 2026

Status

approved