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.
102 has three digits, 102 is divisible by 3, and 10 is also magic, so 102 is a member.
P1:={seq(i,i=1..9)}:
for i from 2 to 25 do
P||i:={}:
for n from 1 to nops(P||(i-1)) do
for j from 0 to 9 do
if P||(i-1)[n]*10+j mod i = 0 then P||i:={op(P||i),P||(i-1)[n]*10+j}: fi:
od:
od:
od:
`union`({0},seq(P||i,i=1..25)); # Martin Renner, Mar 05 2016
divQ[n_]:=Divisible[n,IntegerLength[n]]; lessQ[n_]:=FromDigits[Most[IntegerDigits[n]]]; pdQ[n_]:=If[Or[n<10,And[divQ[n],divQ[lessQ[n]]]],True]; Select[Range[0,180],pdQ[#]&] (* Ivan N. Ianakiev, Aug 23 2016 *)
def agen(): # generator of terms
yield 0
magic, biggermagic, digits = list(range(1, 10)), [], 2
while len(magic) > 0:
yield from magic
for i in magic:
for d in range(10):
t = 10*i + d
if t%digits == 0:
biggermagic.append(t)
magic, biggermagic, digits = biggermagic, [], digits+1
print([an for an in agen()][:70]) # Michael S. Branicky, Feb 07 2022
spi[{ctn_,n_,a_}]:=Module[{k=1},While[Mod[ctn*10^IntegerLength[k]+k,n+1] != 0,k++];{ctn*10^IntegerLength[k]+k,n+1,k}]; NestList[spi,{1,1,1},80][[All,3]] (* Harvey P. Dale, Nov 30 2022 *)
num = 1; print(1); for (n = 2, 9, c = 10*num%n; d = n - c; print(d); num = 10*num + d); for (n = 10, 90, c = 10*num%n; d = n - c; if (d < 10, print(d); num = 10*num + d, c = 100*num%n; d = n - c; if (d < 10, d = d + n); print(d); num = 100*num + d));
Concatenation of a(0),...,a(6) is 1024569, not used so far are 3, 7, 8, 10, 11, 12, ..., the smallest of these that appended to 1024569 gives a multiple of 8 is 12: 102456912 = 8*12807114, hence a(7) = 12.
After we have the first 11 terms, 1,0,2,0,0,0,5,6,4,0,5, the next number x must be chosen so that 102000564050 + x is divisible by 12; this implies that x = 10.
M:=80; a[1]:=1; N:=1; for n from 2 to M do N:=10*N; t2:=N mod n; if t2 = 0 then a[n]:=0; else a[n]:=n-t2; fi; N:=N+a[n]; od: [seq(a[n],n=1..M)];
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
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
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)))), ", "))
\\ See Corneth link. David A. Corneth, Jul 21 2020
a(3) = 19838179232721317143537 because 19 == 3 mod 2, 198 == 3 mod 3, 1983 == 3 mod 4,..., 19838179232721317143537 == 3 mod 23; but no additional digit makes a 3 mod 24 number.
a(1) = 1121114531 because 11 == 1 mod 2, 112 == 1 mod 3, 1211 == 1 mod 4, ..., 1121114531 == 1 mod 10 but there is no digit such that 1121114531d == 1 mod 11. (10 is not a digit.)
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