a(1): {1}, therefore a(1) = 1;
a(6): {6, 4, 2, 1} or {6, 3, 2, 1}, therefore a(6) = 2;
a(12): {12, 8, 4, 2, 1}, {12, 6, 4, 2, 1} or {12, 6, 3, 2, 1}, therefore a(12) = 3;
a(14): {14, 12, 8, 4, 2, 1}, {14, 12, 6, 4, 2, 1}, {14, 12, 6, 3, 2, 1}, {14, 7, 6, 4, 2, 1} or {14, 7, 6, 3, 2, 1}, therefore a(14) = 5.
For n=15 we have five alternative paths from 15 to 1: {15, 10, 5, 4, 2, 1}, {15, 10, 8, 4, 2, 1}, {15, 12, 8, 4, 2, 1}, {15, 12, 6, 4, 2, 1}, {15, 12, 6, 3, 2, 1}, therefore a(15) = 5. These form a graph illustrated below:
15
/ \
/ \
10 12
/ \ / \
/ \ / \
5 8 6
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4 3
\ /
\ /
2
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1
(End)