>>492

まづ
　b[0] = d[0] = 0,　a[1] = c[1] = 0,
と漸化式から
　a[奇数] = b[偶数] = c[奇数] = d[偶数] = 0,
が分かる。

次に、漸化式
　a[n+1] = d[n]/2,
　d[n+1] = (2/3)c[n],
　c[n+1] = b[n],
から、
　d[n] = 2a[n+1],
　c[n] = 3a[n+2],
　b[n] = 3a[n+3],
これを b[n+1] = a[n] + (1/3)c[n] に入れて　
　3a[n+4] = a[n] + a[n+2]
a[n]の初期値は
　a[0] = 1, a[1] = a[2] = a[3] = 0,
これを解く。