

A109210


In the decimal expansion of Pi, positions of first digit of subsequences with total sum = 100.


2



2, 16, 30, 31, 39, 42, 45, 48, 53, 54, 66, 67, 68, 69, 71, 74, 94, 100, 103, 110, 111, 116, 119, 124, 126, 128, 131, 139, 140, 142, 145, 147, 148, 150, 152, 157, 160, 161, 162, 165, 166, 172, 174, 176, 180, 183, 184, 186, 191, 199, 201, 208, 209, 213, 230, 235
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OFFSET

1,1


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. 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 to 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.


LINKS

Table of n, a(n) for n=1..56.


FORMULA

a(n) = A109211(n)  A109212(n) + 1. [corrected by Jinyuan Wang, Mar 22 2020]


PROG

(PARI) default(realprecision, 101000); v=digits(floor(Pi*10^10^5));
lista(nn) = {my(x=21, s=103); for(k=2, nn, s=v[k1]; if(s==100, print1(k, ", "), while(s<100, s+=v[x++]); if(s==100, print1(k, ", ")))); } \\ Jinyuan Wang, Mar 22 2020


CROSSREFS

Cf. A000796.
Cf. A109211 (position of the final digit), A109212 (length of subsequence).
Sequence in context: A127871 A242933 A321308 * A056707 A069256 A279034
Adjacent sequences: A109207 A109208 A109209 * A109211 A109212 A109213


KEYWORD

nonn,base,less


AUTHOR

Zak Seidov, Jun 22 2005


STATUS

approved



