|
%I #11 Jan 25 2022 20:09:20
%S 1,2,3,4,6,8,9,10,12,14,15,16,20,24,32,36,40,64,70,72,80,96,100,104,
%T 116,128,131,144,160,168,192,200,216,256,288,292,293,300,320,336,344,
%U 360,362,370,380,400,432,512,536,544,560,576,640,648,727,768,800,864
%N Concatenation of n in base 10 down to base 2 is divisible by at least one of these base b numbers, all numbers interpreted as decimals.
%e n=9 -> 9{10}=10{9}=11{8}=12{7}=13{6}=14{5}=21{4}=100{3}=1001{2} -> 91011121314211001001 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[::-1]))
%o return any(cn%int(sbi) == 0 for sbi in sb)
%o print([k for k in range(1, 865) if ok(k)]) # _Michael S. Branicky_, Jan 25 2022
%Y Cf. A054220.
%K nonn,base
%O 1,2
%A _Patrick De Geest_, Feb 15 2000
%E Data corrected by _Sean A. Irvine_, Jan 25 2022
|
