A389979 Positive integers (not multiples of 10) whose constant congruence speed is greater than the sum of the constant congruence speeds of all their prime factors counted with multiplicity.

15, 57, 65, 95, 145, 249, 255, 305, 335, 385, 415, 545, 575, 626, 655, 705, 807, 815, 818, 865, 895, 943, 1025, 1055, 1068, 1126, 1135, 1185, 1318, 1345, 1535, 1585, 1665, 1695, 1751, 1807, 1855, 1874, 1943, 1985, 2015, 2095, 2126, 2175, 2182, 2249, 2305, 2318, 2335

Offset

1,1

Comments

For the definition of "constant congruence speed", see 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)).

Here, prime factors are considered with their multiplicity (e.g., the prime factors of 75 are 3, 5, and 5 and thus the sum of their constant congruence speeds equals 1+2+2=5).

Furthermore, 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..49.

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, April 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(1) = 15 since the constant congruence speeds of 3 and 5 equal 1, while the constant congruence speed of 3*5 is 4 (and 4 > 2).

Prog

(Python)
from sympy import factorint
def upto(n):
    s=[]
    for k in range(2, n+1):
        if k%10==0:
            continue
        Vk=A373387(k)
        if Vk is None:
            continue
        sumVp=0
        for p, e in factorint(k).items():
            Vp=A373387(p)
            if Vp is not None:
                sumVp+=e*Vp
        if Vk>sumVp:
            s.append(k)
    return s
print(upto(10000))

Crossrefs

Cf. A317905, A373387, A389432, A389980, A389981.

Sequence in context: A157856 A082820 A184231 * A039341 A043164 A043944

Adjacent sequences: A389976 A389977 A389978 * A389980 A389981 A389982

Keyword

nonn,base

Author

Marco Ripà, Oct 20 2025

Status

approved