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CHMMC problems
2014 CHMMC (Fall)
4
2014 CHMMC Tiebreaker 4 - f(i, j, k) = f(i - 1, j + k, 2i - 1)
2014 CHMMC Tiebreaker 4 - f(i, j, k) = f(i - 1, j + k, 2i - 1)
Source:
March 1, 2024
algebra
CHMMC
Problem Statement
If
f
(
i
,
j
,
k
)
=
f
(
i
−
1
,
j
+
k
,
2
i
−
1
)
f(i, j, k) = f(i - 1, j + k , 2i - 1)
f
(
i
,
j
,
k
)
=
f
(
i
−
1
,
j
+
k
,
2
i
−
1
)
and
f
(
0
,
j
,
k
)
=
j
+
k
f(0, j, k) = j + k
f
(
0
,
j
,
k
)
=
j
+
k
, evaluate
f
(
n
,
0
,
0
)
f(n, 0, 0)
f
(
n
,
0
,
0
)
.
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