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A060273
Hard numbers: a(n) = smallest positive number m with f(m) = n, where f(m) is the smallest number of digits that are needed to construct m using only 1's, 2's and any number of +, -, *, ^ signs, allowing concatenation of the digits.
3
1, 3, 5, 7, 29, 51, 151, 601, 1631, 7159, 19145, 71515, 378701
OFFSET
1,2
COMMENTS
It seems that to obtain this sequence we need to impose two additional rules: 1. x-y is allowed only when x-y > 0 (which also applies to A060274). 2. "Allowing concatenation of the digits" *only* applies to the base digits, 1 and 2, not to the decimal representations of subexpression values. So for example, 13 cannot be obtained via: 13 = 1 concat 3 = 1 concat (1+2) because "3" is not a string consisting solely of 1's or 2's, but can be obtained via: 13 = 11 + 2 = (1 concat 1) + 2. Then the example 151 really does have complexity 7 under this measure.
REFERENCES
C. Pickover, Mazes for the Mind, St. Martin's Press, NY, 1992, p. 376.
C. Pickover, "Wonders of Numbers", Chapter 78, 'Creator Numbers', Oxford University Press, NY, 2001. pp. 187-189, 343-345.
Ken Shirriff, University of California, personal communication.
LINKS
Sean A. Irvine, Java program (github)
C. A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Zentralblatt review
EXAMPLE
a(7) = 151 because 151 is the smallest number that requires 7 digits for its expression.
CROSSREFS
Sequence in context: A058047 A098860 A106920 * A124077 A288891 A358699
KEYWORD
nonn,base
AUTHOR
Jason Earls, Mar 23 2001
EXTENSIONS
Entry improved by comments from Tim Peters (tim.one(AT)comcast.net), Nov 14 2004
Missing a(1)=1 inserted and existing terms verified by Sean A. Irvine, Nov 06 2022
STATUS
approved