OFFSET
1,3
COMMENTS
It is known (Murthy 2001) that the sequence is infinite. (1) The number {3(10^(k+2)+1)}^3 for all k produces such numbers. (2) Less trivially, {10^(n+2) - 4}^3 is a member of this sequence for n = 4*{(10^(3k)-1)/27}-1, for all k, for which the sum of the cubes of the digits is {6*10^k}^3.
REFERENCES
Amarnath Murthy, Smarandache Fermat Additive Cubic Sequence, 2011. (To be published in the Smarandache Notions Journal.)
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
EXAMPLE
474552 = 78^3 is a term since 4^3+7^3+4^3+5^3+5^3+2^3 = 729 = 9^3.
MAPLE
R:= NULL: count:= 0:
for x from 0 while count < 100 do
v:= x^3;
t:= add(s^3, s=convert(v, base, 10));
if surd(t, 3)::integer then
R:= R, v; count:= count+1;
fi;
od:
R; # Robert Israel, Apr 15 2025
MATHEMATICA
Select[Range[0, 2500]^3, IntegerQ[Total[IntegerDigits[#]^3]^(1/3)]&] (* Harvey P. Dale, Jun 03 2012 *)
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Amarnath Murthy, Apr 21 2001
EXTENSIONS
Corrected and extended by Gaurav Kumar, Aug 29 2009
STATUS
approved