cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A078283 a(1) = 1, a(n) is the smallest multiple of n which is obtained by inserting/prefixing or suffixing at least one digit in a(n-1).

Original entry on oeis.org

1, 10, 102, 1012, 10120, 101202, 1012025, 10120256, 101120256, 1011202560, 10112002560, 100112002560, 1001120025360, 10010120025360, 100010120025360, 1000010120025360, 10000101020025360, 100001010200253606
Offset: 1

Views

Author

Amarnath Murthy, Nov 25 2002

Keywords

Links

Crossrefs

Cf. A078282, A078284.

Programs

  • Maple
    f:= proc(m,n)
 local L,nL,d,r,K,Kp,tmin,t,V,Vt,x;
 L:= convert(m,base,10);
 nL:= nops(L);
 L:= Vector(L);
 for d from 1 do
   r:= infinity;
   V:= Vector(nL+d);
   for K in combinat:-choose(nL+d,nL) do
     V[K]:= L;
     Kp:= sort(convert({$1..nL+d} minus convert(K,set), list));
     if Kp[-1] = nL+d then tmin:= 10^(d-1) else tmin:= 0 fi;
     for t from tmin to 10^d-1 do
       Vt:= convert(10^d+t,base,10);
       V[Kp]:= Vector(Vt[1..-2]);
       x:= add(10^(i-1)*V[i],i=1..nL+d);
       if x > r then break fi;
       if x mod n = 0 then r:= x; break fi;
     od;
   od;
   if r < infinity then return r fi
 od;
end proc:
A[1]:= 1:
for n from 2 to 30 do A[n]:= f(A[n-1],n) od:
seq(A[i],i=1..30); # Robert Israel, Jul 21 2020

Extensions

More terms from Sascha Kurz, Jan 04 2003

A214437 Least numbers whose groups of 2,3,...,n digits taken from the left are divisible by 2,3,...,n.

Original entry on oeis.org

1, 10, 102, 1020, 10200, 102000, 1020005, 10200056, 102000564, 1020005640, 10200056405, 102006162060, 1020061620604, 10200616206046, 102006162060465, 1020061620604656, 10200616206046568, 108054801036000018, 1080548010360000180, 10805480103600001800
Offset: 1

Views

Author

Robin Garcia, Jul 17 2012

Keywords

Comments

The first 11 terms of the sequence are coincident with A078282.
a(6) is formed with 66,7 % zeros; A(5) with 60 %; a(7) with 57,1 %; a(4), a(8), a(10) and a(20) with 50 %.
a(n) is the first term of A144688 with n digits, except that A144688 includes zero as first term. - Franklin T. Adams-Watters, Jul 18 2012
There are 25 terms in the sequence; the 25-digit number 3608528850368400786036725 is the last number to satisfy the requirements. - Shyam Sunder Gupta, Aug 04 2013

Examples

			a(6) = 102000 because 10, 102, 1020, 10200 and 102000 are divisible by 2, 3, 4, 5 and 6.
There are nine one-digit numbers that are divisible by 1; the smallest is 1, so a(1)=1.
For two-digit numbers, the second digit must be even, i.e., 0,2,4,6,8 to make it divisible by 2, which gives 10 as the smallest number to satisfy the requirement, so a(2)=10. - _Shyam Sunder Gupta_, Aug 04 2013

Links

Crossrefs

Cf. A078282, A158242, A144688.

Programs

  • Mathematica
    a=Table[j, {j, 9}]; r=2; t={};
While[!a == {}, n=Length[a]; nmin=Last[a]; k=1; b={};
While[!k>n, z0=a[[k]]; Do[z=10*z0+j; If[Mod[z, r]==0, b=Append[b, z]], {j, 0, 9}]; k++]; AppendTo[t, nmin]; a=b; r++]; t (* Shyam Sunder Gupta, Aug 04 2013 *)

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).

Original entry on oeis.org

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

Views

Author

Eder Vanzei, Jul 20 2020

Keywords

Examples

			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

Links

Crossrefs

Cf. A336401 (corresponding numbers), A003418 (LCM's).
Cf. A045874, A051883, A078282, A078283, A214437.

Programs

  • 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
  • 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

Extensions

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

A078284 a(n) = A078283(n)/n.

Original entry on oeis.org

1, 5, 34, 253, 2024, 16867, 144575, 1265032, 11235584, 101120256, 919272960, 8342666880, 77009232720, 715008573240, 6667341335024, 62500632501585, 588241236472080, 5555611677791867, 52632110631712424, 500005051001268028
Offset: 1

Views

Author

Amarnath Murthy, Nov 25 2002

Keywords

Crossrefs

Cf. A078282, A078283.

Extensions

More terms from Sascha Kurz, Jan 04 2003

A336401 a(n) = a(n-1) concatenated with the smallest number k, such that a(n) is divisible by lcm(1..n).

Original entry on oeis.org

1, 10, 102, 1020, 10200, 102000, 102000360, 1020003600, 10200036001680, 102000360016800, 10200036001680035280, 102000360016800352800, 102000360016800352800332640, 1020003600168003528003326400
Offset: 1

Views

Author

Eder Vanzei, Jul 20 2020

Keywords

Crossrefs

Cf. A045874, A078282, A078283, A214437, A336399.

Programs

  • PARI
    a(n)={if(n==1,return(1));for(n1=0,oo,k=eval(concat(Str(a(n-1)),n1));n2=0;for(n3=1,n,if(k%n3==0,n2+=1;if(n2==n,return(k)))))};

A090490 For n > 1, a(n) is the least multiple of n that can be obtained by adding one digit to each end of a(n-1).

Original entry on oeis.org

1, 110, 11100, 1111000, 111110000, 11111100000, 1111111000006, 111111110000064, 11111111100000648, 1111111111000006480, 111111111110000064800, 11111111111100000648000, 1111111111111000006480005, 111111111111110000064800052, 11111111111111100000648000525
Offset: 1

Views

Author

Amarnath Murthy, Dec 03 2003

Keywords

Comments

Sequence has 15 terms. No multiple of 16 can be obtained in this way from a(15). - David Wasserman, Dec 06 2005

Examples

			a(8) = 111111110000064 is a multiple of 8 and is the concatenation of 1, a(7) and 4.

Crossrefs

Cf. A078282.

Extensions

More terms from David Wasserman, Dec 06 2005
Showing 1-6 of 6 results.

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