A392982 Run lengths of consecutive increases of prime(n)/n where prime(n) is the n-th prime.

1, 4, 2, 3, 3, 4, 3, 6, 2, 3, 2, 2, 3, 3, 2, 1, 1, 3, 1, 1, 2, 5, 2, 1, 3, 1, 1, 4, 1, 5, 3, 3, 2, 2, 3, 1, 1, 3, 2, 3, 6, 5, 3, 1, 3, 1, 3, 2, 9, 3, 2, 4, 2, 1, 1, 3, 1, 1, 2, 1, 1, 2, 4, 4, 6, 2, 2, 3, 2, 3, 1, 1, 7, 4, 1, 3, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 5

Offset

1,2

Comments

Conjecture: For every positive integer k there exists m with a(m) = k.

Links

Table of n, a(n) for n=1..93.

Formula

a(n) = A079418(n) - A079418(n-1) for n>1.

Example

The first eight values of prime(n)/n are: 2/1 = 2, 3/2 = 1.5, 5/3 = 1.666..., 7/4 = 1.75, 11/5 = 2.2, 13/6 = 2.166..., 17/7 = 2.428..., 19/8 = 2.375.
Since 2 > 1.5, the first run has length 1, so a(1) = 1.
Since 1.5 < 1.666... < 1.75 < 2.2 and then 2.2 > 2.166..., the next run has length 4, so a(2) = 4.
Since 2.166... < 2.428... and then 2.428... > 2.375, the third run has length 2, so a(3) = 2.

Mathematica

s={1}; p=1; Do[c=1; Until[Prime[p+1]/(p+1)<Prime[p]/p, c++; p++]; AppendTo[s, c-1], {i, 92}]; s (* James C. McMahon, Mar 16 2026 *)

Prog

(PARI) lista(nn) = my(v=vector(nn, n, prime(n)/n), nb=1, list=List()); for (n=2, nn, if (v[n] > v[n-1], nb++, listput(list, nb); nb = 1); ); Vec(list); \\ Michel Marcus, Mar 10 2026

Crossrefs

First differences of A079418.

Sequence in context: A136626 A079623 A177864 * A090112 A226087 A118945

Adjacent sequences: A392979 A392980 A392981 * A392984 A392985 A392986

Keyword

nonn

Author

Philip Jameson, Mar 07 2026

Status

approved