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%I #13 Jan 25 2022 20:09:31
%S 1,2,3,4,5,8,9,10,15,16,20,22,25,38,40,44,50,55,66,68,74,75,80,95,100,
%T 110,111,125,144,175,200,230,240,250,260,265,267,268,269,300,301,316,
%U 358,375,400,440,453,476,500,514,542,620,688,875,1000,1096,1120,1250
%N Concatenation of n in base 2 up to base 10 is divisible by at least one of these base b numbers, all numbers interpreted as decimals.
%e n=9 -> 1001{2}=100{3}=21{4}=14{5}=13{6}=12{7}=11{8}=10{9}=9{10} -> 10011002114131211109 which is divisible by 21 (from 9 in base 4).
%o (Python)
%o from sympy.ntheory.digits import digits
%o def ok(n):
%o sb = ["".join(str(d) for d in digits(n, b)[1:]) for b in range(2, 11)]
%o cn = int("".join(sb))
%o return any(cn%int(sbi) == 0 for sbi in sb)
%o print([k for k in range(1, 1251) if ok(k)]) # _Michael S. Branicky_, Jan 25 2022
%Y Cf. A054219.
%K nonn,base
%O 1,2
%A _Patrick De Geest_, Feb 15 2000
%E Data corrected by _Sean A. Irvine_, Jan 25 2022