OFFSET
1,3
COMMENTS
An obvious upper bound for this sequence is a(n) <= n-1 because 0^0 + (n-1) = n.
Another upper bound can be defined recursively: a(n) <= a(n-1) + 1 because if n-1 = a^b + c, then n = a^b + c + 1, thus one possible sum is a+b+c+1 or a(n-1) + 1.
EXAMPLE
a(1) = 0 because 0^0 + 0 = 1 and 0 + 0 + 0 = 0.
a(9) = 5 because 3^2 + 0 = 9 and 3 + 2 + 0 = 5 and there is no ordered triple (a,b,c) such that a^b + c = 9 with a+b+c < 5.
PROG
(Python)
def a(n):
minSum = n-1
for a in range(n-1):
for b in range(n-a-1):
if a**b>n:
break
c = n-a**b
if a+b+c<minSum:
minSum = a+b+c
return minSum
CROSSREFS
KEYWORD
nonn
AUTHOR
Joshua R. Tint, May 27 2022
STATUS
approved