A080592 My school teacher gave it to me as a challenge. But he died last week and I do not have the answer.

66, 63, 57, 95

Offset

1,1

Comments

From Rick L. Shepherd, Jun 28 2008: (Start)

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)

Links

Table of n, a(n) for n=1..4.

Formula

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

Crossrefs

Cf. A141346.

Sequence in context: A033386 A008896 A071712 * A059175 A031960 A376289

Adjacent sequences: A080589 A080590 A080591 * A080593 A080594 A080595

Keyword

nonn,unkn

Author

DexXx (dexx(AT)pop.com.br), Feb 22 2003

Status

approved