A098084 -id:A098084 - cp's OEIS frontend

A098085 Primes P(n)+P(n+1)+b(n) = least prime >= P(n)+P(n+1), P(i)=i-th prime, b(n) given in A098084.

5, 11, 13, 19, 29, 31, 37, 43, 53, 61, 71, 79, 89, 97, 101, 113, 127, 131, 139, 149, 157, 163, 173, 191, 199, 211, 211, 223, 223, 241, 263, 269, 277, 293, 307, 311, 331, 331, 347, 353, 367, 373, 389, 397, 397, 419, 439, 457, 457, 463, 479, 487, 499, 509, 521

Offset: 1

Pierre CAMI, Sep 13 2004

			P(1) + P(2) = 2 + 3 = 5; least prime >= 5 = 5.
P(2) + P(3) = 3 + 5 = 8; least prime > 8 = 11.
P(3) + P(4) = 5 + 7 = 12; least prime > 12 = 13.

f[n_] := Block[{k = 0, p = Prime[n] + Prime[n + 1]}, While[ !PrimeQ[p + k], k++ ]; p + k]; Table[ f[n], {n, 55}] (* Robert G. Wilson v, Sep 24 2004 *)

More terms from Robert G. Wilson v, Sep 25 2004

A366274 a(n) is the least k such that prime(n+1+k) >= prime(n)+prime(n+1).

Original entry on oeis.org

1, 2, 2, 3, 4, 4, 4, 5, 6, 7, 8, 9, 10, 10, 10, 13, 13, 13, 14, 14, 15, 15, 16, 18, 20, 20, 19, 19, 18, 22, 24, 24, 25, 27, 27, 27, 29, 28, 29, 30, 31, 31, 33, 33, 32, 34, 37, 39, 38, 39, 40, 40, 41, 42, 42, 43, 42, 43, 43, 43

Offset: 1

Patrick Butler, Oct 05 2023

nonn

a(n) is the number of primes between prime(n) and prime(n) + prime(n+1).

Conjecture: for n >= 3, a(n) < n.

			For n = 5 prime(n) = 11. prime(5) + prime(6) = 11+13=24.  The 4th prime after 13 is 29 which is the next prime after 13 greater than or equal to 24. So a(5) = 4.

Patrick Butler, Table of n, a(n) for n = 1..10000

Cf. A000720, A001043, A076273, A098084, A098085.

R:= 1: pn:= 2: pn1:= 3: p:=5: m:= 4: pp:= 7:
for n from 2 to 100 do
  pn:= pn1; pn1:= nextprime(pn1);
  while pp <= pn + pn1 do m:= m+1; pp:= nextprime(pp); od;
  R:= R, m-n-1;
od:
R; # Robert Israel, Oct 31 2023

A366274[n_]:=PrimePi[Prime[n]+Prime[n+1]-1]-n;Array[A366274,100] (* Paolo Xausa, Dec 09 2023 *)

a(n) = my(k=1, q=prime(n)+prime(n+1)); while(prime(n+k) < q, k++); k; \\ Michel Marcus, Oct 06 2023

m=0
#list here is a list of prime numbers A000040.
def a(n):
    global list
    sum= list[n]+list[n+1]
    i=n+2
    while True:
        if(list[i]>=sum):
            m=i
            break
        i=i+1
    k = m-(n+1)
    return k
#
#calculate the terms of the sequence a(n).
seq = []
for n in range(0,firstN):
   seq.append(a(n))

a(n) = A000720(A001043(n)-1)-n = A000720(A076273(n+1))-n. - Paolo Xausa, Dec 09 2023

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A098085 Primes P(n)+P(n+1)+b(n) = least prime >= P(n)+P(n+1), P(i)=i-th prime, b(n) given in A098084.

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A366274 a(n) is the least k such that prime(n+1+k) >= prime(n)+prime(n+1).

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