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%I #20 Apr 29 2020 20:59:18
%S 5,53,1060,697,14027,7830448093388,278269,7794416,1784625167675,
%T 217659538,7299226328,58429863516468189,2265720635440119410,
%U 11301046374119,5483279396166772909558757483,9796440127236265192879361141313874782657110677228434,24212615127434834,1506888944866952574
%N a(n) is the least k >= n that written in base n and then interpreted in base n+1 is a multiple of k.
%C 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
%e a(5) = 697 = (10242)_5 and (10242)_6 = 1394 = 2 * 697.
%e a(7) = 7830448093388 = (1435505542406624)_7, which when interpreted in base 8 is equal to 7 * 7830448093388.
%t a[n_] := Block[{k = n}, While[ Mod[ FromDigits[ IntegerDigits[k, n], n + 1], k] > 0, k++]; k]; a /@ Range[2, 6]
%o (Python)
%o def BaseUp(n,b):
%o up, b1 = 0, 1
%o while n > 0:
%o up, b1, n = up+(n%b)*b1, b1*(b+1), n//b
%o return up
%o n = 2
%o while n < 20:
%o k = n
%o while BaseUp(k,n)%k != 0:
%o k = k+1
%o print(n,k)
%o n = n+1 # _A.H.M. Smeets_, Mar 31 2020
%Y Cf. A062845, A062853.
%K nonn,base
%O 2,1
%A _Giovanni Resta_, Dec 14 2019
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