By calculation

a 1=a 1，a2= 1+a 1，a3=2-a 1，a4=7-a 1 (S4= 10)

a5=a 1，a6=9+a 1，a7=2-a 1，a8= 15-a 1 (S8-S4=26)

a9=a 1，a 10= 17+a 1，a 1 1=2-a 1，a 12 = 23-a 1(s 12-S8 = 42)

Since S4= 10, S8-S4=26, and S 12-S8=42, the arithmetic progression of S4k-S4(k- 1) takes 10 as the first term, and the tolerance is 16.

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Special solution: If a 1= 1, a2=2. By calculating A3 = 1, A4 = 6, A5 = 1, A6 = 10 ..., we can guess that the odd-numbered items are all 1 and all items are. So S60 = 30 *1+(A2+A4+A6+... A60) = 30+30 * A2+1/2 * 30 * (30-1) * 4 =1830.