A341765 - OEIS

A341765

Consider gaps between successive odd primes from 3 up to prime(n+2). Let k1 be number of gaps congruent to 2 (mod 6) and let k2 be number of gaps congruent to 4 (mod 6). Then a(n) = k1 - k2.

1, 2, 1, 2, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 2

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

Theorem A: for all n, a(n) belongs to the set: {1,2}, for proof see A342156.

The indices n for which numbers of 1's and 2's in this sequence are equal are 2, 4, 6, 10, 12, 20, 36, 46, 48 and no other up to n=10^6.

LINKS

Table of n, a(n) for n=1..104.

FORMULA

a(n) = 3 - A039701(n+2). - Andrey Zabolotskiy, Nov 04 2024

MAPLE

a(1)=1 because prime(2+1)-prime(2)=5-3=2 then the gap 2 is congruent to 2 mod 6, then k1=1 and k2=0 so k1 - k2 = 1.

MATHEMATICA

k1 = 0; k2 = 0; cc = {}; Do[

gap = Prime[n + 1] - Prime[n];

If[Mod[gap/2, 3] == 1, k1 = k1 + 1,

If[Mod[gap/2, 3] == 2, k2 = k2 + 1]]; AppendTo[cc, k1 - k2];

If[k1 - k2 == 1, , If[k1 - k2 == 2, , Print[{n, k1 - k2}]]], {n, 2,

105}]; cc

PROG

(PARI) a(n) = {my(vp = vector(n+1, k, prime(k+1)), dp = vector(#vp-1, k, (vp[k+1] - vp[k])/2)); my(s=0); for (k=1, #dp, if ((dp[k]%3)==1, s++); if ((dp[k]%3) == 2, s--)); s; } \\ Michel Marcus, Feb 27 2021

CROSSREFS

Cf. A000230, A001223, A028334, A321856, A341952, A342156, A039701.

Sequence in context: A081129 A022934 A107450 * A104248 A358359 A249973

Adjacent sequences: A341762 A341763 A341764 * A341766 A341767 A341768

KEYWORD

nonn,changed

AUTHOR

Artur Jasinski, Feb 19 2021

EXTENSIONS

Name edited by Andrey Zabolotskiy, Nov 04 2024

STATUS

approved