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%I #39 May 19 2021 01:34:31
%S 1,0,2,0,0,0,360,0,1680,0,35280,0,332640,0,0,0,8648640,0,306306000,0,
%T 0,0,232792560,0,0,0,26771144400,0,481880599200,0,41923612130400,0,0,
%U 0,0,0,5487335009956800,0,0,0,245774847024907200,0,8105227020364874400,0,0,0,452140231622516236800,0,3984485791173424336800,0
%N a(1) = 1, a(n) is the smallest number such that the concatenation a(1)a(2)...a(n) is divisible by lcm(1..n).
%H Robert Israel, <a href="/A336399/b336399.txt">Table of n, a(n) for n = 1..2297</a>
%H David A. Corneth, <a href="/A336399/a336399.gp.txt">PARI program</a>
%e a(7) = 360 as the smallest positive integer k such that the concatenation a(1)a(2)..a(6)k is divisible by lcm(1..7) = 420. - _David A. Corneth_, Jul 21 2020
%p N:= 1: R:= 1: C:= 1:
%p for n from 2 to 60 do
%p N:= ilcm(N,n);
%p for d from 1 do
%p x:= -C*10^d mod N;
%p if x = 0 then lx:= 1 else lx:= 1+ilog10(x) fi;
%p if lx = d then
%p R:= R,x;
%p C:= C*10^d+x;
%p break
%p elif lx < d then
%p k:= ceil((10^(d-1)-x)/N);
%p x:= x + k*N;
%p if x < 10^d then
%p R:= R,x;
%p C:= C*10^d+x;
%p break
%p fi fi
%p od; od:
%p R; # _Robert Israel_, Sep 16 2020
%o (PARI) a(n) = {if(n==1,return(1));for(n1 = 0, oo, ; k[n]=eval(concat(Str(k[n-1]), n1)); n2=0; for(n3 = 1, n, if(k[n] % n3 == 0, n2+=1; if(n2==n, return(k[n])))))};
%o k = vector(10000);print1(k[1]=1,", ");for(j=1, 20, print1(a(j+1) - a(j)*10^(length(Str(a(j+1))) - length(Str(a(j)))), ", "))
%o (PARI) See Corneth link \\ _David A. Corneth_, Jul 21 2020
%Y Cf. A336401 (corresponding numbers), A003418 (LCM's).
%Y Cf. A045874, A051883, A078282, A078283, A214437.
%K base,nonn
%O 1,3
%A _Eder Vanzei_, Jul 20 2020
%E a(27)-a(50) from _David A. Corneth_, Jul 20 2020
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