Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 15"
(Created page with "==Problem== For any finite sequence of positive integers <math>A=(a_1,a_2,\cdots,a_n)</math>, let <math>f(A)</math> be the sequence of the differences between consecutive term...") |
m |
||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | For any finite sequence of positive integers <math>A=(a_1,a_2,\cdots,a_n)</math>, let <math>f(A)</math> be the sequence of the differences between consecutive terms of <math>A</math>. i.e. <math>f(A)=(a_2-a_1,a_3-a_2,\cdots,a_n-a_{n-1})</math>. Let <math>F^k(A)</math> denote <math>F</math> applied <math>k</math> times to <math>A</math>. If all of the sequences <math>A, f(A), f^2(A),\cdots, f^{n-2}(A)</math> are strictly increasing and the only term of <math>f^{ | + | For any finite sequence of positive integers <math>A=(a_1,a_2,\cdots,a_n)</math>, let <math>f(A)</math> be the sequence of the differences between consecutive terms of <math>A</math>. i.e. <math>f(A)=(a_2-a_1,a_3-a_2,\cdots,a_n-a_{n-1})</math>. Let <math>F^k(A)</math> denote <math>F</math> applied <math>k</math> times to <math>A</math>. If all of the sequences <math>A, f(A), f^2(A),\cdots, f^{n-2}(A)</math> are strictly increasing and the only term of <math>f^{n-1}(A)</math> is <math>1</math>, we call the sequence <math>A</math> <math>\textit{superpositive}</math>. How many sequences <math>A</math> with at least two terms and no terms exceeding <math>18</math> are <math>\textit{superpositive}</math>? |
==Solution== | ==Solution== | ||
{{Solution}} | {{Solution}} |
Revision as of 20:36, 29 April 2025
Problem
For any finite sequence of positive integers , let
be the sequence of the differences between consecutive terms of
. i.e.
. Let
denote
applied
times to
. If all of the sequences
are strictly increasing and the only term of
is
, we call the sequence
. How many sequences
with at least two terms and no terms exceeding
are
?
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.