lpetrich
Contributor
I'll try to determine its general rule.
11, 2(12), 2(13)+(22), 2(14)+2(23), 2(15)+2(24)+(33), 2(16)+2(25)+2(34) over 1,2,3,4,5,6, ...
In other words, for positive-integer k, is there a set of ak all
\( \frac1n \sum_{k=1}^{n} \, a_k a_{n-k+1} \)
are equal?
Let each one's value be Sn, and let us find the solution by folding the values together by multiplying by tn and adding:
\( \sum_n n S_n t^n = A(t)^2 \)
where
\( A(t) = \sum_n a_n t^n = \sqrt{ \sum_n n S_n t^n } \)
If all the Sn's equal S0, then
\( \sum_n n S_n t^n = S_0 \sum_n n t^n = \frac{S_0}{(1 - t)^2} \)
That gives A(t) = a0/(1 - t) or an = a0 -- all equal.
11, 2(12), 2(13)+(22), 2(14)+2(23), 2(15)+2(24)+(33), 2(16)+2(25)+2(34) over 1,2,3,4,5,6, ...
In other words, for positive-integer k, is there a set of ak all
\( \frac1n \sum_{k=1}^{n} \, a_k a_{n-k+1} \)
are equal?
Let each one's value be Sn, and let us find the solution by folding the values together by multiplying by tn and adding:
\( \sum_n n S_n t^n = A(t)^2 \)
where
\( A(t) = \sum_n a_n t^n = \sqrt{ \sum_n n S_n t^n } \)
If all the Sn's equal S0, then
\( \sum_n n S_n t^n = S_0 \sum_n n t^n = \frac{S_0}{(1 - t)^2} \)
That gives A(t) = a0/(1 - t) or an = a0 -- all equal.