1,1

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.

A109210(n)=A109212(n)-A109211(n)+1.

Position of the final digit in the subsequence gives A109211 and length of susequence gives A109212.

Sequence in context: A108705 A113933 A127871 this_sequence A056707 A069256 A120069

Adjacent sequences: A109207 A109208 A109209 this_sequence A109211 A109212 A109213

base,nonn

Zak Seidov (zakseidov(AT)yahoo.com), Jun 22 2005