A336399 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).

1, 0, 2, 0, 0, 0, 360, 0, 1680, 0, 35280, 0, 332640, 0, 0, 0, 8648640, 0, 306306000, 0, 0, 0, 232792560, 0, 0, 0, 26771144400, 0, 481880599200, 0, 41923612130400, 0, 0, 0, 0, 0, 5487335009956800, 0, 0, 0, 245774847024907200, 0, 8105227020364874400, 0, 0, 0, 452140231622516236800, 0, 3984485791173424336800, 0

Offset

1,3

Links

Robert Israel, Table of n, a(n) for n = 1..2297

David A. Corneth, PARI program

Example

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

Maple

N:= 1: R:= 1: C:= 1:
for n from 2 to 60 do
  N:= ilcm(N, n);
  for d from 1 do
    x:= -C*10^d mod N;
    if x = 0 then lx:= 1 else lx:= 1+ilog10(x) fi;
    if lx = d then
       R:= R, x;
       C:= C*10^d+x;
       break
    elif lx < d then
       k:= ceil((10^(d-1)-x)/N);
       x:= x + k*N;
       if x < 10^d then
         R:= R, x;
         C:= C*10^d+x;
         break
    fi fi
od; od:
R; # Robert Israel, Sep 16 2020

Prog

(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])))))};
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)))), ", "))
(PARI) \\ See Corneth link. David A. Corneth, Jul 21 2020

Crossrefs

Cf. A336401 (corresponding numbers), A003418 (LCM's).

Cf. A045874, A051883, A078282, A078283, A214437.

Sequence in context: A169772 A193542 A193545 * A370701 A086260 A124505

Adjacent sequences: A336396 A336397 A336398 * A336400 A336401 A336402

Keyword

base,nonn,changed

Author

Eder Vanzei, Jul 20 2020

Extensions

a(27)-a(50) from David A. Corneth, Jul 20 2020

Status

approved