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2024 JHMT HS
14
Infinite Product of Trinomials
Infinite Product of Trinomials
Source:
September 8, 2024
algebra
2024
Problem Statement
Let
N
13
N_{13}
N
13
be the answer to problem 13, and let
k
=
1
N
13
+
6
k = \tfrac{1}{N_{13} + 6}
k
=
N
13
+
6
1
. Compute the infinite product
(
1
−
k
+
k
2
)
(
1
−
k
3
+
k
6
)
(
1
−
k
9
+
k
18
)
(
1
−
k
27
+
k
54
)
⋯
,
(1 - k + k^2)(1 - k^3 + k^6)(1 - k^9 + k^{18})(1 - k^{27} + k^{54})\cdots,
(
1
−
k
+
k
2
)
(
1
−
k
3
+
k
6
)
(
1
−
k
9
+
k
18
)
(
1
−
k
27
+
k
54
)
⋯
,
where the factors take the form
(
1
−
k
3
a
+
k
2
⋅
3
a
)
(1 - k^{3^a} + k^{2\cdot 3^a})
(
1
−
k
3
a
+
k
2
⋅
3
a
)
for all nonnegative integers
a
a
a
.
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