A389432 Smallest nonnegative integer with a constant congruence speed >= n (see A373387 for the definition of "constant congruence speed").

0, 2, 5, 15, 15, 65, 65, 255, 255, 1025, 1025, 4095, 4095, 16385, 16385, 65535, 65535, 262145, 262145, 1048575, 1048575, 4194305, 4194305, 16777215, 16777215, 67108865, 67108865, 268435455, 268435455, 1073741825, 1073741825, 4294967295, 4294967295, 17179869185

Offset

0,2

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, and (for all n >= 2) the present sequence can be easily derived from A337392 as a(n) = A337392(2*ceiling(n/2)) (for references, see Section 3 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

Andrew Howroyd, Table of n, a(n) for n = 0..1000

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

Index entries for linear recurrences with constant coefficients, signature (0,3,0,4).

Formula

For n >= 2, a(n) = 2^(n + (1 - (-1)^n)/2) - (-1)^(n*(n+1)/2).

G.f.: x*(12*x^4 + 9*x^2 + 5*x + 2)/((1 - 2*x)*(1 + 2*x)*(1 + x^2)). - Andrew Howroyd, Nov 06 2025

Example

a(1) = 2 since A373387(2) = 1 (and A373387(0) = A373387(1) = 0 < 1 = A373387(2)).

Mathematica

A389432[n_] := If[n < 2, 2*n, 2^(n + (1 - (-1)^n)/2) - I^(n*(n + 1))]; Array[A389432, 50, 0] (* or *)
LinearRecurrence[{0, 3, 0, 4}, {0, 2, 5, 15, 15, 65}, 50] (* Paolo Xausa, Jan 02 2026 *)

Prog

(Python)
def vp(n, p):
    c=0
    if n==0:
        return 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=[]
    m=0
    while True:
        t=V(m)
        while len(s)<=t:
            s.append(None)
        for k in range(t+1):
            if s[k] is None:
                s[k]=m
                print(m)
        m+=1
seq()

Crossrefs

Cf. A317905, A337392, A339313, A373387, A387664.

Sequence in context: A146110 A336385 A146109 * A146108 A032841 A058221

Adjacent sequences: A389429 A389430 A389431 * A389433 A389434 A389435

Keyword

nonn,base,easy

Author

Marco Ripà, Oct 03 2025

Status

approved