Derive a formula for the number of partitions of n into parts that are odd and bigger than 1; e.g. a(12)=5 cases: 3+3+3+3, 5+7, 7+5, 3+9, 9+3.
Verify your formula by evaluating a(14).
(In reply to
re: P is for what? by Charlie)
Actually I was referring to the formula given at http://mathworld.wolfram.com/PadovanSequence.html
which has 3 terms based on the roots of <img src="http://mathworld.wolfram.com/images/equations/PadovanSequence/NumberedEquation3.gif" class="numberedequation" alt=" x^3+x^21=0. " border="0" height="17" width="91">
which reminds me of the closed form given as (6) at
http://mathworld.wolfram.com/FibonacciNumber.html
which has two terms based on the roots of <img src="http://mathworld.wolfram.com/images/equations/FibonacciNumber/NumberedEquation4.gif" class="numberedequation" alt=" x^2x1=0, " border="0" height="17" width="85">
But you are right there are are other ways of generating Fibonacci numbers that have analogues for Padovan numbers.

Posted by Jer
on 20130125 09:18:28 