A283371 Maximum number of pairs of primes (p, q) such that p < q = < prime (n) and q - p = constant.

0, 1, 1, 2, 2, 3, 3, 4, 5, 6, 6, 7, 7, 8, 9, 10, 11, 11, 12, 12, 13, 14, 14, 15, 15, 15, 16, 17, 18, 19, 19, 19, 20, 21, 21, 21, 22, 22, 23, 23, 24, 25, 25, 26, 27, 27, 28, 29, 30, 31, 31, 31, 32, 32, 33, 34, 35, 36, 36, 37, 37, 38, 39, 40, 41, 41, 41, 42, 43, 43, 43, 43, 44, 44, 45, 46, 47, 48, 48, 49, 50, 50

Offset

1,4

Comments

Maximum number of different ways of expressing a positive number as a difference of two distinct primes less than or equal to prime(n).

Is there any n such that a(n+1) - a(n) > 1?

What is the asymptotic behavior of a(n)?

To answer the first question: for all n, either a(n+1) = a(n) or a(n+1) = a(n) + 1. - Charles R Greathouse IV, Mar 06 2017

Links

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Formula

a(n) >> n/log n. In particular, lim inf a(n) * (log n)/n >= 1/2. - Charles R Greathouse IV, Mar 06 2017

Example

a(1)=0 because there are no two distinct primes less than or equal to prime(1)=2.
a(2)=1 because there are only two distinct primes less than or equal to prime(2)=3, and then there is only one positive difference among them: 3 - 2 = 1.
a(3)=1 because the three pairs of distinct primes less than or equal to prime(3)=5, i.e., (2,3), (3,5), and (2,5), produce different positive differences: 3 - 2 = 1, 5 - 3 = 2, and 5 - 2 = 3.
a(4)=2 because among all pairs of distinct primes taken from the first four primes, 2, 3, 5, and 7, there are two pairs with same positive difference, i.e., 7 - 5 = 5 - 3 = 2.
a(6)=3 because among all pairs of distinct primes taken from the first six primes, 2, 3, 5, 7, 11, and 13, there are at most three pairs with the same positive difference, i.e., 13 - 11 = 7 - 5 = 5 - 3 = 2.

Mathematica

a[n_]:=Module[{fp, fps, fpst, fpstts, fpsttst},
fp=Prime[Range[n]];
fps=Subsets[fp, {2}];
fpst=Table[Abs@(fps[[j]][[2]]-fps[[j]][[1]]), {j, 1, Length[fps]}];
fpstts=fpst//Tally;
If[n<2, 0, fpsttst=fpstts//Transpose; fpsttst[[2]]//Max]//Return];
Table[a[n], {n, 1, 120}]

Prog

(PARI) first(n)=my(v=vector(n), P=primes(n), H=vectorsmall((P[#P]-P[2])/2, i, 0)); v[2]=1; for(n=3, #P, for(i=2, n-1, H[(P[n]-P[i])/2]++); v[n]=vecmax(H)); v \\ Charles R Greathouse IV, Mar 06 2017

Crossrefs

Cf. A283302, A030173.

Sequence in context: A197432 A255573 A062298 * A116579 A156253 A346693

Adjacent sequences: A283368 A283369 A283370 * A283372 A283373 A283374

Keyword

nonn

Author

Andres Cicuttin, Mar 06 2017

Status

approved