OFFSET
1,1
COMMENTS
Like Keith numbers but starting from n/2 digits to reach n.
Consider the digits of n/2. Take their sum and repeat the process deleting the first addend and adding the previous sum. The sequence lists the numbers that after some iterations reach a sum equal to themselves.
If it exists, a(20) > 10^12. - Lars Blomberg Mar 13 2017
EXAMPLE
642/2 = 321:
3 + 2 + 1 = 6;
2 + 1 + 6 = 9;
1 + 6 + 9 = 16;
6 + 9 + 16 = 31;
9 + 16 + 31 = 56;
16 + 31 + 56 = 103;
31 + 56 + 103 = 190;
56 + 103 + 190 = 349;
103 + 190 + 349 = 642.
MAPLE
with(numtheory): P:=proc(q, h, w) local a, b, k, n, t, v; v:=array(1..h);
for n from 1/w by 1/w to q do a:=w*n; b:=ilog10(a)+1; if b>1 then
for k from 1 to b do v[b-k+1]:=(a mod 10); a:=trunc(a/10); od; t:=b+1; v[t]:=add(v[k], k=1..b);
while v[t]<n do t:=t+1; v[t]:=add(v[k], k=t-b..t-1); od;
if v[t]=n then print(n); fi; fi; od; end: P(10^6, 1000, 1/2);
MATHEMATICA
With[{n = 2}, Select[Range[10 n, 10^6, n], Function[k, Last@ NestWhile[Append[Rest@ #, Total@ #] &, IntegerDigits[k/n], Total@ # <= k &] == k]]] (* Michael De Vlieger, Feb 27 2017 *)
CROSSREFS
KEYWORD
nonn,base,more
AUTHOR
Paolo P. Lava, Feb 27 2017
EXTENSIONS
a(15)-a(19) from Lars Blomberg, Mar 13 2017
STATUS
approved