A330435 a(n) is the least k >= n that written in base n and then interpreted in base n+1 is a multiple of k.

5, 53, 1060, 697, 14027, 7830448093388, 278269, 7794416, 1784625167675, 217659538, 7299226328, 58429863516468189, 2265720635440119410, 11301046374119, 5483279396166772909558757483, 9796440127236265192879361141313874782657110677228434, 24212615127434834, 1506888944866952574

Offset

2,1

Comments

Theorem: a(n) > n^(n*log(2)). Proof: if k=(d_m...d_2d_1d_0)_n, and K=(d_m...d_2d_1d_0)_{n+1} then K <= k*(1+1/n)^m. Suppose k=a(n). Then K >= 2*k, hence (1+1/n)^m >= 2, and therefore m >= log(2)/log(1+1/n) > n*log(2). Since d_m can be assumed to be >= 1, we may assume that k >= n^m, and this yields k > n^(n*log(2)). - Dimiter Skordev, Mar 14 2020

Links

Table of n, a(n) for n=2..19.

Example

a(5) = 697 = (10242)_5 and (10242)_6 = 1394 = 2 * 697.
a(7) = 7830448093388 = (1435505542406624)_7, which when interpreted in base 8 is equal to 7 * 7830448093388.

Mathematica

a[n_] := Block[{k = n}, While[ Mod[ FromDigits[ IntegerDigits[k, n], n + 1], k] > 0, k++]; k]; a /@ Range[2, 6]

Prog

(Python)
def BaseUp(n, b):
    up, b1 = 0, 1
    while n > 0:
        up, b1, n = up+(n%b)*b1, b1*(b+1), n//b
    return up
n = 2
while n < 20:
    k = n
    while BaseUp(k, n)%k != 0:
        k = k+1
    print(n, k)
    n = n+1 # A.H.M. Smeets, Mar 31 2020

Crossrefs

Cf. A062845, A062853.

Sequence in context: A180351 A337151 A171192 * A113068 A099086 A304999

Adjacent sequences: A330432 A330433 A330434 * A330436 A330437 A330438

Keyword

nonn,base

Author

Giovanni Resta, Dec 14 2019

Status

approved