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A282100 a(n) is the least number of successive prime numbers modulo 100 required such that their sum is at least 100. 0
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 (list; graph; refs; listen; history; text; internal format)
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
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
Sequence in context: A050016 A033329 A097326 * A199965 A021110 A010540
KEYWORD
nonn
AUTHOR
Dimitris Valianatos, Feb 06 2017
STATUS
approved

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Last modified April 24 00:30 EDT 2024. Contains 371917 sequences. (Running on oeis4.)