A370688 Numbers k such that A052410(k) = A010888(k).

0, 1, 2, 3, 5, 6, 7, 128, 2401, 8192, 78125, 524288, 823543, 33554432, 282475249, 1220703125, 2147483648, 96889010407, 137438953472, 8796093022208, 19073486328125, 33232930569601, 562949953421312, 11398895185373143, 36028797018963968, 298023223876953125, 2305843009213693952

Offset: 1

Author

Stefano Spezia, Feb 27 2024

Keywords

• nonn
• base

Comments

From Chai Wah Wu, Mar 02 2024: (Start)

Theorem: k is a term if and only if k is 0, 1, 3, 6 or of the form 2^(6*m+1), 5^(6*m+1), or 7^(3*m+1), m >= 0.

Proof: 0, 1, 3, 6 can be shown to be terms by direct computation. Since 1 <= A010888(k) <= 9 for k > 0, all terms > 0 are of the form q^m, where 2 <= q <= 9. This implies that all terms > 1 are either 6^m or of the form p^m, where p is a prime <= 7. If k = 6^m for m > 1, then k is divisible by 9 and A010888(k) = 9, whereas A052410(k) = 6 and thus k is not a term. Similarly, if k = 3^m for m > 1, then k is divisible by 9 and A010888(k) = 9 whereas A052410(k) = 3 and thus k is not a term. Lastly, for p = 2, 5 or 7, A010888(p^m) = 1 + ((p^m-1) mod 9), and since p^m and 9 are coprime, this implies that A010888(p^m) = p^m mod 9 whereas A052410(p^m) = p, thus m must satisfy p^m mod 9 = p mod 9. Since 6, 6, 3 are the multiplicative orders of 2, 5, 7 modulo 9 respectively, implying 2^6 == 5^6 == 7^3 == 1 (mod 9), the result follows.

(End)

Crossrefs

Cf. A010888, A052409, A052410.

Programs

Mathematica

A052409[n_]:=GCD@@Last/@FactorInteger[n]; A010888[n_]:=If[n==0, 0, n-9 Floor[(n-1)/9]]; a={}; kmax = 10^9; For[k=0, k<=kmax, k++, If[k^(1/A052409[k])==A010888[k], AppendTo[a,k]]]; a

Python

from itertools import count, islice
from math import gcd
from sympy import factorint, integer_nthroot
def A370688_gen(startvalue=0): # generator of terms >= startvalue
    if startvalue <=0: yield 0
    if startvalue <=1: yield 1
    for k in count(max(startvalue,2)):
        r = 1 + (k - 1) % 9
        if r>1:
            kmin, kmax = 0, 1
            while r**kmax <= k:
                kmax <<= 1
            while True:
                kmid = kmax+kmin>>1
                if r**kmid > k:
                    kmax = kmid
                else:
                    kmin = kmid
                if kmax-kmin <= 1:
                    break
            if r**kmin==k:
                m = integer_nthroot(k,gcd(*factorint(k).values()))[0]
                if m == r:
                    yield k
A370688_list = list(islice(A370688_gen(),10)) # Chai Wah Wu, Mar 02 2024

Python

# faster program based on theorem
from itertools import islice
def A370688_gen(): # generator of terms
    kmax, mlist, dlist = 10, [7,7,4], [6,6,3]
    yield from (0,1,2,3,5,6,7)
    while True:
        klist = []
        for i, p in enumerate((2,5,7)):
            while (k:=p**mlist[i]) <= kmax:
                klist.append(k)
                mlist[i] += dlist[i]
        yield from sorted(klist)
        kmax *= 10
A370688_list = list(islice(A370688_gen(),10)) # Chai Wah Wu, Mar 02 2024

Extensions

a(18)-a(27) from Chai Wah Wu, Mar 02 2024