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2022 CCA Math Bonanza
TB4
Functions
Functions
Source:
April 28, 2022
function
Problem Statement
Let
f
(
x
)
f(x)
f
(
x
)
be a function such that
f
(
1
)
=
1234
f(1) = 1234
f
(
1
)
=
1234
,
f
(
2
)
=
1800
f(2)=1800
f
(
2
)
=
1800
, and
f
(
x
)
=
f
(
x
−
1
)
+
2
f
(
x
−
2
)
−
1
f(x) = f(x-1) + 2f(x-2)-1
f
(
x
)
=
f
(
x
−
1
)
+
2
f
(
x
−
2
)
−
1
for all integers
x
x
x
. Evaluate the number of divisors of
∑
i
=
1
2022
f
(
i
)
\sum_{i=1}^{2022}f(i)
i
=
1
∑
2022
f
(
i
)
2022 CCA Math Bonanza Tiebreaker Round #4
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