A367367 Largest k such that k*(n-1)^k >= n^k-1.

1, 2, 6, 10, 14, 19, 23, 28, 33, 38, 43, 48, 53, 59, 64, 70, 75, 81, 87, 92, 98, 104, 110, 116, 122, 128, 134, 141, 147, 153, 159, 166, 172, 178, 185, 191, 198, 204, 211, 218, 224, 231, 238, 244, 251, 258, 265, 271, 278, 285, 292, 299, 306, 313, 320, 327, 334

Offset

2,2

Comments

Inspired by the book "Getallentheorie - Een inleiding" from Frits Beukers, pp. 103, 104 (in Dutch). Given some base b, a number can be equal to the sum of powers of its digits, where the power is equal to the number of digits of the number represented in base b.

For any base b, the number of numbers satisfying this condition is finite, and the upper bound of any number m satisfying this condition is given by m < n^a(n); i.e., a(n) is the maximum number of digits k in base n such that (d_1)^k + (d_2)^k + ... + (d_k)^k = d_1*n^(k-1) + d_2*n^(k-2) + ... + d_k.

References

Frits Beukers, "Getallen - Een inleiding" (In Dutch), Epsilon Uitgaven, Amsterdam (2015).

Links

Table of n, a(n) for n=2..58.

Example

a(2) = 1 since 0 = 0_2, 1 = 1_2, and for any number n larger than 1, the number of 1's in the binary representation of n is less than n.
a(3) = 2 since the largest number satisfying the condition is 8, i.e., 2^2 + 2^2 = 22_3 = 8, and thus two digits long in base 3.
Examples of numbers satisfying the condition in base 10: 1^4 + 6^4 + 3^4 + 4^4 = 1634, 9^4 + 4^4 + 7^4 + 4^4 = 9474 and 8^11 + 2^11 + 6^11 + 9^11 + 3^11 + 9^11 + 1^11 + 6^11 + 5^11 + 7^11 + 8^11 = 82693916578. However, such examples in base 10 can have up to 33 digits as an upper bound.

Mathematica

result = Reap[  b = 1;  While[b <= 57, {b, k, lhs, rhs} = {b + 1, 1, b - 1, b - 1}; While[lhs >= rhs, k += 1; {lhs, rhs} = {k (b - 1)^k, b^k - 1}; ]; Sow[k - 1]; ]][[2, 1]] (* James C. McMahon, Dec 19 2023 *)

Prog

(Python)
b = 1
while b <= 57:
    b, k, lhs, rhs = b+1, 1, b-1, b-1
    while lhs >= rhs:
        k += 1
        lhs, rhs = k*(b-1)**k, b**k-1
    print(k-1, end = ", ")
(Python)
def A367367(n):
    kmin, kmax = 1, 1
    while kmax*(n-1)**kmax >= n**kmax-1:
        kmax <<= 1
    while True:
        kmid = kmax+kmin>>1
        if kmid*(n-1)**kmid < n**kmid-1:
            kmax = kmid
        else:
            kmin = kmid
        if kmax-kmin <= 1:
            break
    return kmin # Chai Wah Wu, Dec 21 2023

Crossrefs

Sequence in context: A290727 A190883 A284678 * A185548 A351553 A354715

Adjacent sequences: A367364 A367365 A367366 * A367368 A367369 A367370

Keyword

nonn

Author

A.H.M. Smeets, Dec 18 2023

Status

approved