A251600 -id:A251600 - cp's OEIS frontend

A098037 Number of prime divisors, counted with multiplicity, of the sum of two consecutive primes.

1, 3, 3, 3, 4, 3, 4, 3, 3, 4, 3, 3, 4, 4, 4, 5, 5, 7, 3, 6, 4, 5, 3, 3, 4, 4, 4, 6, 3, 6, 3, 3, 4, 7, 5, 4, 7, 4, 4, 6, 6, 4, 8, 4, 5, 3, 3, 5, 5, 4, 4, 7, 4, 3, 5, 4, 6, 3, 4, 4, 8, 6, 3, 6, 5, 7, 3, 5, 5, 5, 4, 4, 4, 5, 3, 3, 3, 4, 6, 5, 6, 4, 8, 4, 5, 3, 3, 5, 5, 4, 3, 4, 3, 5, 3, 4, 3, 5, 5, 7, 6, 7, 3, 5, 4

Offset: 1

Cino Hilliard, Sep 10 2004

Clearly sum of two consecutive primes prime(x) and prime(x+1) has more than 2 prime divisors for all x > 1.

			Prime(2) + prime(3) = 2*2*2, 3 factors, the second term in the sequence.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A071215, A251600 (greedy inverse).

A098037 := proc(n)
    ithprime(n)+ithprime(n+1) ;
    numtheory[bigomega](%) ;
end proc:
seq(A098037(n),n=1..40) ; # R. J. Mathar, Jan 20 2025

PrimeOmega[Total[#]]&/@Partition[Prime[Range[110]],2,1] (* Harvey P. Dale, Jun 14 2011 *)

b(n) = for(x=1,n,y1=(prime(x)+prime(x+1));print1(bigomega(y1)","))

a(n) = A001222(A001043(n)). - Michel Marcus, Feb 15 2014

Definition corrected by Andrew S. Plewe, Apr 08 2007

A251609 Least k such that prime(k) + prime(k+1) contains n distinct prime divisors.

Original entry on oeis.org

1, 3, 6, 27, 276, 1755, 24865, 646029, 7946521, 195711271, 4129119136, 198635909763, 6351380968517, 322641218722443, 11068897188590241, 501741852481602261, 24367382928343066431, 1292304206793356882286

Offset: 1

Michel Lagneau, Dec 05 2014

			a(1) = 1 because prime(1) + prime(2) = 2 + 3 = 5, which is a prime power and so has one distinct prime divisor; the other prime indices yielding a prime power are 2, 18, 564,...(A071352) since prime(2) + prime(3) = 3 + 5 = 2^3, prime(18) + prime(19) = 61 + 67 = 2^7, prime(564) + prime(565)= 4093 + 4099 = 2^13,...

Cf. A071215, A001221, A071352, A230518, A251600.

N:= 10^6: # to use primes <= N
Primes:= select(isprime, [2,seq(2*i+1,i=1..(N-1)/2)]):
for i from 1 to nops(Primes)-1 do
  f:= nops(numtheory:-factorset(Primes[i]+Primes[i+1]));
  if not assigned(A[f]) then A[f]:= i fi
od:
seq(A[j],j=1..max(indices(A))); # Robert Israel, Dec 05 2014

lst={};Do[k=1;While[Length[FactorInteger[Prime[k]+Prime[k+1]]]!=n,k++];AppendTo[lst,k],{n,1,5}];lst

a(n) = A000720(A230518(n)). - Amiram Eldar, Feb 17 2019

a(10)-a(18) from Amiram Eldar, Feb 17 2019

A288507 Least number k such that both prime(k+1) -/+ prime(k) are products of n prime factors (counting multiplicity).

Original entry on oeis.org

24, 319, 738, 57360, 1077529, 116552943

Offset: 3

Zak Seidov, Jun 10 2017

Prime(k) + prime(k+1) cannot be semiprime, so the offset is 3.

For n=3 to 8, all terms k happen to satisfy prime(k+1) - prime(k) = 2^n. - Michel Marcus, Jul 24 2017

			n = 8: k = 116552943, p = prime(k) = 2394261637, q = prime(k+1) = 2394261893; both q-p = 2^8 and  p+q = 2*3^2*5*7^3*155119 are 8-almost primes (A046310).

Cf. A001043, A001223, A046310, A251600.

a(n) = my(k = 1, p = 2, q = nextprime(p+1)); while((bigomega(p+q)!= n) || (bigomega(p-q)!= n), k++; p = q; q = nextprime(p+1)); k; \\ Michel Marcus, Jul 24 2017

from sympy import factorint, nextprime
def A288507(n):
    k, p, q = 1, 2, 3
    while True:
        if sum(factorint(q-p).values()) == n and sum(factorint(q+p).values()) == n:
            return k
        k += 1
        p, q = q, nextprime(q) # Chai Wah Wu, Jul 23 2017

cp's OEIS Frontend

A098037 Number of prime divisors, counted with multiplicity, of the sum of two consecutive primes.

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A251609 Least k such that prime(k) + prime(k+1) contains n distinct prime divisors.

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Maple

Mathematica

Formula

Extensions

A288507 Least number k such that both prime(k+1) -/+ prime(k) are products of n prime factors (counting multiplicity).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Python