OFFSET
1,1
COMMENTS
Like A332580 a heuristic argument, based on the divergent sum of reciprocals which approximates the probability that the concatenation of n and n+1 will divide the concatenation of n+2, n+3, ..., suggests that k should always exist.
LINKS
Scott R. Shannon, Table of n, a(n) for n = 1..10000
EXAMPLE
a(1) = 3 as '1'||'2' = 12 and '3'||'4'||'5'||'6' = 3456, which is divisible by 12 (where '||' denotes decimal concatenation).
a(4) = 24 as '4'||'5' = 45 and '6'||'7'||....||'29'||'30' = 6789101112131415161718192021222324252627282930, which is divisible by 45.
MAPLE
a:= proc(n) local i, t, m; t, m:= parse(cat(n, n+1)), 0;
for i from n+2 do m:= parse(cat(m, i)) mod t;
if m=0 then break fi od; i-n-2
end:
seq(a(n), n=1..50); # Alois P. Heinz, Feb 29 2020
PROG
(PARI) a(n) = {my(k=1, small=eval(concat(Str(n), Str(n+1))), big=n+2); while( big % small, big = eval(concat(Str(big), Str(n+2+k))); k++); k--; } \\ Michel Marcus, Feb 29 2020
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Scott R. Shannon, Feb 25 2020
STATUS
approved