A225854 Frequency of prime numbers between consecutive partial sums of primes.
1, 2, 1, 3, 2, 4, 3, 5, 4, 6, 6, 8, 6, 9, 6, 9, 10, 10, 8, 12, 12, 11, 12, 12, 15, 14, 14, 14, 14, 17, 17, 16, 17, 19, 19, 22, 16, 24, 21, 20, 20, 20, 28, 22, 26, 21, 24, 28, 23, 31, 23, 30, 28, 28, 32, 28, 31, 30, 27, 36, 29, 32, 31, 39, 33, 38, 36, 36, 37
Offset: 1
Keywords
Examples
List the numbers with an increment of 1 beginning at n=1, and stop when the number of numbers reaches a prime, in this case the list would be {1,2} since its size is 2. Find the number of primes in that interval and add it to the sequence. In this case, there is 1 prime in the list. Continue counting from the last number in the previous list and apply the same rules, the next list will be {3,4,5} of size 3 and contains 2 prime numbers. The list after that will be {6,7,8,9,10} of size 5 and contains 1 prime number.
Links
- Giovanni Resta, Table of n, a(n) for n = 1..10000
Crossrefs
Cf. A014284.
Programs
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Mathematica
numberOfLines = 100; (*How many elements desired in the sequence*) a = {0}; distribution = {}; last = 0; For[j = 1, j <= numberOfLines, j++, frequency = 0; b = {}; For[i = 1, i <= Prime[j], i++, b = Append[b, last + i]; If[PrimeQ[b[[i]]], frequency += 1];];last += Prime[j]; distribution = Append[distribution, frequency];]; Print["Distribution = ", distribution]; ListPlot[distribution]; (*original program*) seq[n_] := Block[{a=0, b=2, p=2, v}, Table[v = PrimePi@b-PrimePi@a; p = NextPrime@p; a = b; b += p; v, {n}]]; seq[100] (* faster version, Giovanni Resta, May 18 2013 *)
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