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A080592
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My school teacher gave it to me as a challenge. But he died last week and I do not have the answer.
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1
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OFFSET
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1,1
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COMMENTS
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The next (and last) term may be 79. Add the distinct digits used in each prime factor of the prime factorization (ignoring exponents) to get these sums: 6, 10, 13, 15, 16 (i.e., A141346(a(n))).
For example, 66 = 2*3*11 --> {1,2,3} --> 1+2+3=6. Observe that (*) A141346(a(n+1)) = A141346(a(n)) - n + 5. There are an infinite number of k such that A141346(k) = a(n) above, so what pattern selects specifically these?
Take a(1) = 66 arbitrarily (or maybe because this is the smallest composite repdigit number ddd... (A010785) such that A141346(ddd...)=d [and d is, well, *the* "perfect number" (A000396(1)) with which to start]).
Then for n >= 2 a(n+1) is, whenever possible, the largest positive integer less than a(n) such that (*) and otherwise the smallest integer greater than a(n) such that (*). (This pattern could actually be continued through a(11) without modification but then A141346(a(12)) would have the impossible requirement to be -5.
Also problematic is that A141346(a(6)) would be 16 also so the question arises whether to use < or <= above.) On the other hand, from such few terms this pattern is quite possibly just a coincidence -- the intended pattern may be something entirely different (and may depend upon knowing information beyond the numbers themselves such as numerical scores on tests, atomic numbers, ASCII codes, keypad configurations, languages, etc.). (End)
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LINKS
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FORMULA
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One possible formula: For n = 1..4, a(n) = 60 + (((-n^2+n+4)/2) + 9*floor(n/4))*(3+4*floor(n/4)). - Wesley Ivan Hurt, Jan 03 2013
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CROSSREFS
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KEYWORD
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nonn,unkn
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AUTHOR
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DexXx (dexx(AT)pop.com.br), Feb 22 2003
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STATUS
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approved
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