A109578 -id:A109578 - cp's OEIS frontend

A096500 Let f(n) = smallest prime > n; a(n) = f(n+1) - f(n).

1, 2, 0, 2, 0, 4, 0, 0, 0, 2, 0, 4, 0, 0, 0, 2, 0, 4, 0, 0, 0, 6, 0, 0, 0, 0, 0, 2, 0, 6, 0, 0, 0, 0, 0, 4, 0, 0, 0, 2, 0, 4, 0, 0, 0, 6, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 2, 0, 6, 0, 0, 0, 0, 0, 4, 0, 0, 0, 2, 0, 6, 0, 0, 0, 0, 0, 4, 0, 0, 0, 6, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 2, 0, 4, 0, 0, 0

Offset: 1

Labos Elemer, Jul 09 2004

nonn

Antti Karttunen, Table of n, a(n) for n = 1..30000
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..100000

Cf. A000720, A001223, A010051, A096501, A175851.
First differences of A151800.
Cf. also A109578.

seq(nextprime(n+1)-nextprime(n),n=1..250); # Muniru A Asiru, Jan 03 2019

Mathematica
```
<Michael De Vlieger, Jan 03 2019 *)
```

A151800(n) = nextprime(1+n);
A096500(n) = (A151800(1+n)-A151800(n)); \\ Antti Karttunen, Jan 03 2019

From Antti Karttunen, Jan 03 2019: (Start)
a(n) = A151800(n+1) - A151800(n).
a(n) = A010051(1+n) * A001223(A000720(1+n)).
(End)

A109579 Sum([pi(j+1)-pi(j)][prime(j+1)-prime(j)],j=1..n), where pi(k) is the number of prime numbers less than or equal to k and prime(k) is the k-th prime.

Original entry on oeis.org

1, 3, 3, 7, 7, 11, 11, 11, 11, 13, 13, 17, 17, 17, 17, 23, 23, 29, 29, 29, 29, 33, 33, 33, 33, 33, 33, 35, 35, 49, 49, 49, 49, 49, 49, 55, 55, 55, 55, 61, 61, 71, 71, 71, 71, 83, 83, 83, 83, 83, 83, 85, 85, 85, 85, 85, 85, 91, 91, 93, 93, 93, 93, 93, 93, 107, 107, 107, 107, 111

Offset: 1

Roger L. Bagula, Jun 29 2005

nonn

			a(2)=3 because pi(1)=0,p(2)=1,p(3)=2,prime(1)=2,prime(2)=3,prime(3)=5 and so a(2)=(1-0)(3-2)+(2-1)(5-3)=1+2=3.

Cf. A001223, A010051.

Partial sums of A109578.

with(numtheory): a:=n->sum((pi(k+1)-pi(k))*(ithprime(k+1)-ithprime(k)),k=1..n): seq(a(n),n=1..70);

an = Table[(PrimePi[n + 1] - PrimePi[n])*(Prime[n + 1] - Prime[n]), {n, 1, 200}] a[0] = 0; a[n_] := a[n] = a[n - 1] + an[[n]] aa = Table[a[n], {n, 0, Length[an]}]

cp's OEIS Frontend

A096500 Let f(n) = smallest prime > n; a(n) = f(n+1) - f(n).

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