OFFSET
1,1
COMMENTS
a(7) >= 192. - Rick L. Shepherd, May 08 2003
From Lars Blomberg, Apr 08 2018: (Start)
The terms a(6)-a(9) have been computed with these rules:
Division by 0 or exponentiation 0^0 is not allowed.
Concatenation where at least one operand is a fraction or where the second operand is negative are skipped.
Exponentiations yielding > 100000 digits or with exponent > 32-bit signed integer are skipped. (End)
EXAMPLE
With three 3's one can form 1=(3/3)^3, 2=3-3/3, 3=3+3-3, 4=3+3/3, but not 5, so a(3)=5.
With four 4's one can get 1=44/44, 2=4/4+4/4, 3=4-(4/4)^4, 4=4+(4-4)^4, 5=4+(4/4)^4, 6=(4+4)/4+4, 7=44/4-4, 8=4+4+4-4, 9=4+4+4/4, 10=(44-4)/4, 11=(4/4) | (4/4), 12=(44+4)/4, but not 13, so a(4)=13 (| denotes concatenation).
With five 5's one can get 1 = (((5 | 5) | 5)^(5 - 5)), 2 = (((5 | 5) - 5) / (5 * 5)), 3 = (((5 * 5) + 5) / (5 + 5)), 4 = (((5 / 5) * 5) - (5 / 5)), 5 = (((5 | 5) * 5) / (5 | 5)), 6 = (((5 | 5) + 5) / (5 + 5)),
7 = (((5 / 5) + 5) + (5 / 5)), 8 = (( 5 + 5) - ((5 + 5) / 5)), 9 = (( 5 + 5) - ((5 / 5)^5)), 10 = (((5 | 5) / 5) - (5 / 5)), 11 = (((5 | 5) / 5)^(5 / 5)), 12 = (((5 | 5) / 5) + (5 / 5)),
13 = (((5 | 5) + (5 + 5)) / 5), 14 = (((5 / 5) | 5) - (5 / 5)), 15 = (((5 / 5) | 5)^(5 / 5)), 16 = (((5 / 5) | 5) + (5 / 5)), 17 = (( 5 | (5^5)) / (5^5)), but not 18, so a(5) = 18.
CROSSREFS
KEYWORD
nonn,base,more
AUTHOR
N. J. A. Sloane, Dec 28 2002
EXTENSIONS
a(5) from Frank Ellermann, Dec 30 2002, who finds that a(6) >= 89
a(6)-a(9) from Lars Blomberg, Apr 08 2018
STATUS
approved