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COMMENTS
| Take decimal expansion of Pi, s= 3,1,4,1,5,9,2,6,5,3,5,8,9,7,9,3,2,3,8,4,6,2,6,4,3,3,8,3,2,7,9,5,0,2,8,8,4,1. Starting with the first term 3 add terms consecutively until we get sum = 100 or larger. If sum = 100 then a(1)=3. It is impossible to get 100 this way. Then we try starting with second digit, 1. It happens that the subsequence from i=2 to 21: s1={1,4,1,5,9,2,6,5,3,5,8,9,7,9,3,2,3,8,4,6} has a sum = 100, hence a(1)=2. Next similar subsequence with sum = 100 has i=16-37: s2={3,2,3,8,4,6,2,6,4,3,3,8,3,2,7,9,5,0,2,8,8,4}, hence a(2)=16, etc.
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