OFFSET
0,3
COMMENTS
a(n)^(1/3) = A048365(n) is the index of the first occurrence of n in A269250. -- For n = 3k, obviously a(n) = 10^n. The first terms for indices n = 3k+1 and n = 3k+2 equals 4096*10^3k resp. 140608*10^3k. Is there an index from where on this is no longer true? - M. F. Hasler, Feb 20 2016
FORMULA
a(n) = A048365(n)^3; a(3n) = 10^(3n); a(3n+1) <= 4096*10^(3n) = (16*10^n)^3 for n>0; a(3n+2) <= 140608*10^(3n) = (52*10^n)^3, with equality for all known terms. - M. F. Hasler, Feb 20 2016
MATHEMATICA
nsmall = Table[Infinity, 20];
For[i = 0, i <= 10^6, i++, n0 = Count[IntegerDigits[i^3], 0];
If[nsmall[[n0 + 1]] > i^3, nsmall[[n0 + 1]] = i^3]];
Cases[nsmall, _?NumberQ] (* Robert Price, Mar 20 2020 *)
CROSSREFS
KEYWORD
nonn,base
AUTHOR
EXTENSIONS
Extended to a(0) = 1 and three lines of data completed by M. F. Hasler, Feb 20 2016
STATUS
approved