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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). 3
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 (list; graph; refs; listen; history; text; internal format)
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 * A086260 A124505 A326855

Adjacent sequences:  A336396 A336397 A336398 * A336400 A336401 A336402

KEYWORD

base,nonn

AUTHOR

Eder Vanzei, Jul 20 2020

EXTENSIONS

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

STATUS

approved

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Last modified October 6 16:26 EDT 2022. Contains 357270 sequences. (Running on oeis4.)