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%I #38 Jun 20 2019 03:04:10
%S 123,234,345,456,8910,101112,230231232
%N Numbers that, for some x, are the concatenation of x, x+1 and x+2 and are divisible by at least two of x, x+1 and x+2.
%C 012 is not included because leading 0's are not allowed.
%C From _Charlie Neder_, Jun 05 2019: (Start)
%C If x = 10^k - 2, then x|x+1|x+2 (with | denoting concatenation) will be congruent to 22 modulo x, -9 modulo x+1, and 0 modulo x+2.
%C If x = 10^k - 1, then x|x+1|x+2 will be congruent to 12 modulo x and 1 modulo x+1.
%C Therefore, the only term such that x and x+2 have different lengths is 8910.
%C By reducing modulo x + {0,1,2} it can be shown that if at least two of x|10^k+2, x+1|10^2k-1, and x+2|2*10^2k-10^k are true - presuming x and x+2 are the same length - then x|x+1|x+2 is in this sequence. No further terms corresponding to x < 10^18. (End)
%C No further terms corresponding to x < 10^50. - _Chai Wah Wu_, Jun 19 2019
%e 230231232 is the concatenation of 230, 231 and 232 and is divisible by 231 and 232.
%p cat3:= proc(x)
%p local t;
%p t:= 10^length(x+2);
%p x*(1 + t*(1+10^length(x+1)))+t+2
%p end proc:
%p f:= proc(x) local q,a,b;
%p q:= cat3(x);
%p a:= (q/x)::integer;
%p b:= (q/(x+1))::integer;
%p if a and b then return q elif not(a) and not(b) then return NULL fi;
%p if (q/(x+2))::integer then q else NULL fi
%p end proc:
%p map(f, [$1..1000]);
%o (Python)
%o for k in range(1,8):
%o ..for x in range(10**(k-1),10**k-2): # will not find 8910
%o ....if sum([not (10**k+2)%x,not (10**(2*k)-1)%(x+1),\
%o ....not (2*10**(2*k)+10**k)%(x+2)]) >= 2:
%o ......print(str(x)+str(x+1)+str(x+2)) # _Charlie Neder_, Jun 05 2019
%Y Subsequence of A001703.
%Y Cf. A308527.
%K nonn,base,more
%O 1,1
%A _J. M. Bergot_ and _Robert Israel_, Jun 03 2019
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