cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A309121 a(n) gives the number of primes in the interval I_j = [(j^2 + 3*j - 2)/2, j*(j + 5)/2] = [A034856(j), A095998(j)], for j >= 1.

Original entry on oeis.org

2, 2, 1, 2, 2, 2, 2, 2, 3, 3, 2, 3, 4, 2, 3, 4, 3, 4, 3, 4, 5, 4, 4, 3, 5, 5, 4, 6, 5, 5, 3, 5, 7, 7, 4, 5, 7, 4, 7, 6, 6, 6, 7, 7, 8, 5, 6, 6, 11, 4, 5, 9, 8, 8, 9, 7, 8, 7, 8, 7, 9, 7, 11, 6, 9, 9, 11, 9, 7, 7, 11, 11, 10, 9, 8, 9, 7, 11, 9, 12, 9, 12, 11, 11, 10, 10, 10, 12, 11, 13, 9, 10, 11, 12
Offset: 1

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Author

Wolfdieter Lang, Jul 13 2019

Keywords

Comments

Conjecture: a(n) >= 1. The length of interval I_n is n+2.
These intervals are considered in A307213.

Examples

			The intervals are I_1 = [1, 2, 3], I_2 = [4, 5, 6, 7], ...
		

Crossrefs

Programs

  • Magma
    [#PrimesInInterval(Binomial(j+1,2)+j-1,Binomial(j+1,2)+2*j):j in [1..94]]; // Marius A. Burtea, Jul 13 2019

A307213 Divide the natural numbers into sets of successive sizes 3,4,5,6,7,...,, starting with {1,2,3}. Cycle through each set until you reach a prime; if the prime was the n-th element in its set, jump to the n-th element of the next set.

Original entry on oeis.org

1, 2, 5, 9, 10, 11, 16, 17, 23, 30, 31, 39, 40, 41, 50, 51, 52, 43, 53, 64, 65, 66, 67, 79, 92, 93, 94, 95, 96, 97, 111, 112, 113, 128, 129, 130, 131, 147, 148, 149, 166, 167, 185, 186, 187, 169, 170, 171, 172, 173, 192, 193, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 244, 245, 246, 247, 248, 249, 250, 229
Offset: 1

Views

Author

Christopher Cormier, Mar 28 2019

Keywords

Comments

"Cycle" in the definition, means that if no prime is found, go back to the start of the set.
If a set does not contain a prime, the sequence goes into an infinite loop, but it is conjectured that this does not happen since the sets are of increasing length.
The sets (rather intervals) are I_j = [(j^2 + 3*j - 2)/2, j*(j + 5)/2] =[A034856(j), A095998(j)], for j >= 1. For the number of primes in these intervals see A309121. - Wolfdieter Lang, Jul 13 2019

Examples

			The first set is {1,2,3}. We look at 1 then 2. 2 is prime, and it is the second number in the set. The next set is {4,5,6,7}. So we jump to the second element, 5. 5 is also prime, so we jump to the second element of the next set, {8,9,10,11,12}, which is 9, etc. If we reach the end of a set without reaching a prime, we loop back to the first element, which is the only way for a(n) < a(n-1) to happen.
		

Crossrefs

Programs

Extensions

Edited by N. J. A. Sloane, Jul 13 2019
Showing 1-2 of 2 results.