A282100 - OEIS

A282100

a(n) is the least number of successive prime numbers modulo 100 required such that their sum is at least 100.

9, 4, 3, 2, 2, 2, 2, 3, 6, 3, 2, 2, 2, 2, 2, 5, 3, 2, 2, 2, 2, 6, 3, 2, 2, 2, 3, 4, 3, 2, 2, 2, 2, 6, 2, 2, 2, 2, 7, 3, 2, 2, 2, 6, 3, 2, 2, 2, 5, 3, 2, 2, 2, 5, 3, 2, 2, 2, 6, 3, 2, 2, 2, 2, 5, 2, 2, 2, 6, 3, 2, 2, 2, 7, 2, 2, 5, 3, 2, 2

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OFFSET

1,1

COMMENTS

I conjecture that the limit (a(1) + ... + a(n))/n exists and is equal to e = 2.718281828459045235360287471....

On the contrary, I conjecture that the limit is smaller, around 2.704, and that furthermore the limit is rational. - Charles R Greathouse IV, Feb 06 2017

If the first conjecture is true, then the prime numbers are distributed randomly. If the second is true, we conclude that the prime numbers are not so random. - Dimitris Valianatos, Feb 08 2017

First appearances of new values: a(1) = 9, a(2) = 4, a(3) = 3, a(4) = 2, a(9) = 6, a(16) = 5, a(39) = 7, a(197) = 8, a(260614) = 10, a(76605811) = 11, a(2070246794) = 12, a(20564734002) = 13, a(1162175131698) = 14, .... - Charles R Greathouse IV, Mar 06 2017

LINKS

Table of n, a(n) for n=1..80.

EXAMPLE

a(1)=9 because 9 "prime numbers mod 100" are required so that the sum is >= 100 (2+3+5+7+11+13+17+19+23 = 100).

a(2)=4 because the next 4 "prime numbers mod 100" 29+31+37+41 = 138 >= 100.

MATHEMATICA

j = 1; Differences@ Join[{1}, Table[k = j; While[Total@ Mod[#, 100] &@ Prime@ Range[j, k] < 100, k++]; Prime@ k; j = k + 1, {n, 120}]] (* Michael De Vlieger, Feb 07 2017 *)

PROG

(PARI) {

s1=0; k=0; a=2; m=2000;

forprime(n=a, m,

d=n%100;

s1+=d; k++;

if(s1>=100,

print1(k", ");

s1=0; k=0;

)

)}

CROSSREFS