A387664 Smallest prime number with a constant convergence speed >= n.

2, 5, 193, 1249, 22943, 2218751, 4218751, 74218751, 574218751, 30000000001, 281907922943, 581907922943, 6581907922943, 123418092077057, 480163574218751, 19523418092077057, 40476581907922943, 2152996418333704193, 3640476581907922943, 3640476581907922943

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 > 1 and not a multiple of 10, this corresponds to A373387.

For any positive integer n, there are infinitely many prime numbers characterized by a constant congruence speed of exactly n.

Conjecture: the sequence contains infinitely many duplicates (e.g., a(19) = a(20), a(50) = a(51), and a(52) = a(53) = a(54): see Table 2 of "The congruence speed formula" in Links).

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..20.

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

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.

Formula

a(n) = min_{k >= n} A339313(k).

Example

a(19) = 3640476581907922943 since 3640476581907922943 is a prime number with a constant congruence speed of 20, and 3640476581907922943 < 23640476581907922943 (where 23640476581907922943 is the smallest prime with a constant congruence speed of exactly 19).

Prog

(Python)
from sympy import isprime
def vp(n, p):
    c=0
    while n%p==0:
        n//=p
        c+=1
    return c
def V(a):
    if a%10==0 or a==1:
        return 0
    r=a%20
    if r==1:
        return min(vp(a-1, 2), vp(a-1, 5))
    if r==11:
        return min(vp(a+1, 2), vp(a-1, 5))
    if a%10 in(2, 8):
        return vp(a*a+1, 5)
    if r in(3, 7):
        return min(vp(a+1, 2), vp(a*a+1, 5))
    if r in(13, 17):
        return min(vp(a-1, 2), vp(a*a+1, 5))
    if a%10==4:
        return vp(a+1, 5)
    if r==5:
        return vp(a-1, 2)
    if r==15:
        return vp(a+1, 2)
    if a%10==6:
        return vp(a-1, 5)
    if r==9:
        return min(vp(a-1, 2), vp(a+1, 5))
    if r==19:
        return min(vp(a+1, 2), vp(a+1, 5))
    return 0
def seq():
    s=[]
    k=2
    while True:
        if isprime(k):
            t=V(k)
            while len(s)<=t:
                s.append(None)
            for n in range(t+1):
                if s[n] is None:
                    s[n]=k
                    print(k)
        k+=1
seq()

Crossrefs

Cf. A000040, A317905, A337392, A337833, A339313, A373387, A389432.

Sequence in context: A226071 A319143 A100366 * A339313 A012975 A012954

Adjacent sequences: A387661 A387662 A387663 * A387665 A387666 A387667

Keyword

nonn,base,hard

Author

Marco Ripà, Oct 04 2025

Status

approved