A100810 -id:A100810 - cp's OEIS frontend

A100821 a(n) = 1 if prime(n) + 2 = prime(n+1), otherwise 0.

0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0

Offset: 1

Giovanni Teofilatto, Jan 06 2005

Same as A062301 except for starting point.

a(n)=1 iff prime(n) is the smaller of a pair of twin primes, else a(n)=0. This sequence can be derived from the sequence b(n)=1 iff n and n+2 are both prime, else b(n)=0. This latter sequence has as its inverse Moebius transform the sequence c(n) = the number of distinct factors of n which are the smaller of a pair of twin primes. For example, c(15)=2 because 15 is divisible by 3 and 5, each of which is the smaller of a pair of twin primes. - Jonathan Vos Post, Jan 07 2005

Table[If[Prime[n] + 2 == Prime[n + 1], 1, 0], {n, 120}] (* Ray Chandler, Jan 09 2005 *)

a(n) = A062301(n+1) = 1 - A100810(n).

Corrected and extended by Ray Chandler, Jan 09 2005

A106002 a(n)=1 if there is a number of the form 6k+3 such that prime(n) < 6k+3 < prime(n+1), otherwise 0.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1

Offset: 1

Giovanni Teofilatto, Apr 29 2005

Except for first two primes and twin primes, there is always at least one number of the form 6k+3 between two successive primes.

			a(3)=0 because between prime(3)=5 and prime(4)=7 there are no numbers of the form 6k+3;
a(4)=1 because between prime(4)=7 and prime(5)=11 there is 9=6*1+3.

Karl-Heinz Hofmann, Table of n, a(n) for n = 1..10000

Same as A100810 after first term.

Table[If[Prime[n]<6Ceiling[Prime[n]/6]+3James C. McMahon, Jan 29 2024 *)

a(n) = my(p=prime(n)); for(k=p+1, nextprime(p+1)-1, if (!((k-3) % 6), return(1))); \\ Michel Marcus, Jan 30 2024

from sympy import sieve
def A106002(n):
    for comp in range(sieve[n]+1, sieve[n+1]):
        if (comp-3) % 6 == 0: return 1
    return 0 # Karl-Heinz Hofmann, Jan 30 2024

Edited by Ray Chandler, Oct 17 2006

A105470 a(n)=1 if there is number of the form 6k+3 with prime(n) <= 6k+3 <= prime(n+1), otherwise 0.

Original entry on oeis.org

1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1

Offset: 1

Giovanni Teofilatto, May 02 2005

Except for the first pair of primes and for twin primes there is always at least one number of the form 6n+3 between two successive primes.

			a(3)=0 because between prime(3) and prime(4) there are no numbers of the form 6k+3;
a(4)=1 because between prime(4) and prime(5) there is one number of the form 6k+3: 9.

Cf. A100810, A106002.

f[n_] := Count[Table[Mod[k, 6], {k, Prime[n], Prime[n + 1]}], 3];Table[If[f[n] == 0, 0, 1], {n, 120}] (* Ray Chandler, Oct 17 2006 *)
Join[{1,1},If[Last[#]-First[#]==2,0,1]&/@Partition[Prime[Range[ 3,200]],2,1]] (* Harvey P. Dale, Nov 27 2013 *)

Edited by Ray Chandler, Oct 17 2006

cp's OEIS Frontend

A100821 a(n) = 1 if prime(n) + 2 = prime(n+1), otherwise 0.

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A106002 a(n)=1 if there is a number of the form 6k+3 such that prime(n) < 6k+3 < prime(n+1), otherwise 0.

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A105470 a(n)=1 if there is number of the form 6k+3 with prime(n) <= 6k+3 <= prime(n+1), otherwise 0.

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