Let $b_k(n)$ be the number of ways to reach the $k-th$ pole in $n$ flights. Then obviously $b_k(n) = 0$ if $k$ and $n$ have the same parity, and $b_k(n) = b_{k-1}(n - 1) + b_{k+1}(n - 1)$ for $1 < k < 8,$ $b_1(n) = b_2(n - 1),$ $b_8(n) = b_7(n - 1).$ Clearly $a(n) = b_8(2n + 1).$ Then $a(n) - 7a(n - 1) = b_8(2n + 1) - 7b_8(2n - 1) = b_7(2n) - 7b_8(2n - 1) = b_6(2n - 1) + b_8(2n - 1) - 7b_8(2n - 1) = b_6(2n - 1) - 6b_8(2n - 1) = b_5(2n - 2) - 5b_7(2n - 2).$ Let us extend our attention to $a(n - 2)$ and further reduce the number of flights by unscrupulous use of $(*):$ $a(n) - 7a(n - 1) + 15a(n - 2) = b_5(2n - 2) - 5b_7(2n - 2) + 15b_8(2n - 3) = b_4(2n - 3) - 4b_6(2n - 3) + 10b_8(2n - 3) = b_3(2n - 4) - 3b_5(2n - 4) + 6b_8(2n - 4).$ And on, and on in the same vein: $a(n) - 7a(n - 1) + 15a(n - 2) - 10a(n - 3) = b_3(2n - 4) - 3b_5(2n - 4) + 6b_8(2n - 4) - 10b_8(2n - 5) = b_2(2n - 5) - 2b_4(2n - 5) + 3b_6(2n - 5) - 4b_8(2n - 5) = b_1(2n - 6) - b_3(2n - 6) + b_5(2n - 6) - b_7(2n - 6).$ Turning finally to the case of $2n - 7$ flights, we get $a(n) - 7a(n - 1) + 15a(n - 2) - 10a(n - 3) + a(n - 4) = b_1(2n - 6) - b_3(2n - 6) + b_5(2n - 6) - b_7(2n - 6) - b_8(2n - 7) = 0.$