A396781 - OEIS

%I #27 Jun 11 2026 18:24:12

%S 1,4,81,16384,1953125,2176782336,4747561509943,18014398509481984,

%T 109418989131512359209,10000000000000000000000000,

%U 144209936106499234037676064081,34182189187166852111368841966125056,972786042517719014174576083150881262357,500026926136478545004035990584179037499817984

%N a(n) is the largest integer of the form r^k (with 0 < r < n and k > 0) whose base-n digital root is r, and whose number of digits in base-n is k.

%e a(2) = 1: the only solution is 1 (base 2: "1", r=1, k=1).

%e a(3) = 4: the solutions are 1, 2, and 4. The maximum is 4 (base 3: "11", r=2, k=2).

%e a(4) = 81: the solutions are 1, 2, 3, 9, 27, and 81. The maximum is 81 (base 4: "1101", r=3, k=4).

%e a(10) = 109418989131512359209: this is the maximum value found in base 10, corresponding to 9^21.

%o (Python)

%o import numpy as np

%o def f(n, b):

%o     if n == 0: return 1

%o     d = []

%o     t = abs(n)

%o     while t > 0:

%o         d.append(t % b)

%o         t //= b

%o     return len(d)

%o def r(n, b):

%o     if n == 0: return 0

%o     return 1 + ((n - 1) % (b - 1))

%o def s(m=19):

%o     o = []

%o     for b in range(2, m + 1):

%o         v = [1]

%o         for i in range(2, b):

%o             k_max = int(np.floor(1 / (1 - (np.log(i) / np.log(b)))))

%o             for k in range(1, k_max + 1):

%o                 n = int(i**k)

%o                 if f(n, b) == k and r(n, b) == i:

%o                     v.append(n)

%o         o.append(str(max(v)))

%o     print(", ".join(o))

%o s(19)

%Y Cf. A394570.

%K nonn,base

%O 2,2

%A _Ali Sidheek_, Jun 05 2026