A007504 Sum of the first n primes.

0, 2, 5, 10, 17, 28, 41, 58, 77, 100, 129, 160, 197, 238, 281, 328, 381, 440, 501, 568, 639, 712, 791, 874, 963, 1060, 1161, 1264, 1371, 1480, 1593, 1720, 1851, 1988, 2127, 2276, 2427, 2584, 2747, 2914, 3087, 3266, 3447, 3638, 3831, 4028, 4227, 4438, 4661, 4888

Offset: 0

Author

N. J. A. Sloane, Robert G. Wilson v

Keywords

• nonn
• nice

Comments

It appears that a(n)^2 - a(n-1)^2 = A034960(n). - Gary Detlefs, Dec 20 2011

This is true. Proof: By definition we have A034960(n) = Sum_{k = (a(n-1)+1)..a(n)} (2*k-1). Since Sum_{k = 1..n} (2*k-1) = n^2, it follows A034960(n) = a(n)^2 - a(n-1)^2, for n > 1. - Hieronymus Fischer, Sep 27 2012 [formulas above adjusted to changed offset of A034960 - Hieronymus Fischer, Oct 14 2012]

Row sums of the triangle in A037126. - Reinhard Zumkeller, Oct 01 2012

Ramanujan noticed the apparent identity between the prime parts partition numbers A000607 and the expansion of Sum_{k >= 0} x^a(k)/((1-x)...(1-x^k)), cf. A046676. See A192541 for the difference between the two. - M. F. Hasler, Mar 05 2014

For n > 0: row 1 in A254858. - Reinhard Zumkeller, Feb 08 2015

a(n) is the smallest number that can be partitioned into n distinct primes. - Alonso del Arte, May 30 2017

For a(n) < m < a(n+1), n > 0, at least one m is a perfect square.

Proof: For n = 1, 2, ..., 6, the proposition is clear. For n > 6, a(n) < ((prime(n) - 1)/2)^2, set (k - 1)^2 <= a(n) < k^2 < ((prime(n) + 1)/2)^2, then k^2 < (k - 1)^2 + prime(n) <= a(n) + prime(n) = a(n+1), so m = k^2 is this perfect square. - Jinyuan Wang, Oct 04 2018

For n >= 5 we have a(n) < ((prime(n)+1)/2)^2. This can be shown by noting that ((prime(n)+1)/2)^2 - ((prime(n-1)+1)/2)^2 - prime(n) = (prime(n)+prime(n-1))*(prime(n)-prime(n-1)-2)/4 >= 0. - Jianing Song, Nov 13 2022

Washington gives an oscillation formula for |a(n) - pi(n^2)|, see links. - Charles R Greathouse IV, Dec 07 2022

References

• E. Bach and J. Shallit, §2.7 in Algorithmic Number Theory, Vol. 1: Efficient Algorithms, MIT Press, Cambridge, MA, 1996.
• H. L. Nelson, "Prime Sums", J. Rec. Math., 14 (1981), 205-206.
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• R. J. Mathar, Table of n, a(n) for n = 0..100000
• C. Axler, On a Sequence involving Prime Numbers, J. Int. Seq. 18 (2015) # 15.7.6.
• Christian Axler, New bounds for the sum of the first n prime numbers, arXiv:1606.06874 [math.NT], 2016.
• P. Hecht, Post-Quantum Cryptography: S_381 Cyclic Subgroup of High Order, International Journal of Advanced Engineering Research and Science (IJAERS, 2017) Vol. 4, Issue 6, 78-86.
• R. J. Mathar, Table of 100000n, a(100000n) for n = 1..10000
• Romeo Meštrović, Curious conjectures on the distribution of primes among the sums of the first 2n primes, arXiv:1804.04198 [math.NT], 2018.
• Vladimir Shevelev, Asymptotics of sum of the first n primes with a remainder term
• Nilotpal Kanti Sinha, On the asymptotic expansion of the sum of the first n primes, arXiv:1011.1667 [math.NT], 2010-2015.
• Lawrence C. Washington, Sums of Powers of Primes II, arXiv preprint (2022). arXiv:2209.12845 [math.NT]
• Eric Weisstein's World of Mathematics, Prime Sums
• OEIS Wiki, Sums of powers of primes divisibility sequences

Crossrefs

Cf. A000041, A034386, A111287, A013916, A013918 (primes), A045345, A050247, A050248, A068873, A073619, A034387, A014148, A014150, A178138, A254784, A254858.

See A122989 for the value of Sum_{n >= 1} 1/a(n).

Cf. A008472, A002110, A038107.

Programs

GAP

P:=Filtered([1..250],IsPrime);;
a:=Concatenation([0],List([1..Length(P)],i->Sum([1..i],k->P[k]))); # Muniru A Asiru, Oct 07 2018

Haskell

a007504 n = a007504_list !! n
a007504_list = scanl (+) 0 a000040_list
-- Reinhard Zumkeller, Oct 01 2014, Oct 03 2011

Magma

[0] cat [&+[ NthPrime(k): k in [1..n]]: n in [1..50]]; // Bruno Berselli, Apr 11 2011 (adapted by Vincenzo Librandi, Nov 27 2015 after Hasler's change on Mar 05 2014)

Maple

s1:=[2]; for n from 2 to 1000 do s1:=[op(s1),s1[n-1]+ithprime(n)]; od: s1;
A007504 := proc(n)
    add(ithprime(i), i=1..n) ;
end proc: # R. J. Mathar, Sep 20 2015

Mathematica

Accumulate[Prime[Range[100]]] (* Zak Seidov, Apr 10 2011 *)
primeRunSum = 0; Table[primeRunSum = primeRunSum + Prime[k], {k, 100}] (* Zak Seidov, Apr 16 2011 *)

PARI

A007504(n) = sum(k=1,n,prime(k)) \\ Michael B. Porter, Feb 26 2010

PARI

a(n) = vecsum(primes(n)); \\ Michel Marcus, Feb 06 2021

Python

from itertools import accumulate, count, islice
from sympy import prime
def A007504_gen(): return accumulate(prime(n) if n > 0 else 0 for n in count(0))
A007504_list = list(islice(A007504_gen(),20)) # Chai Wah Wu, Feb 23 2022

Formula

a(n) ~ n^2 * log(n) / 2. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 24 2001 (see Bach & Shallit 1996)

a(n) = A014284(n+1) - 1. - Jaroslav Krizek, Aug 19 2009

a(n+1) - a(n) = A000040(n+1). - Jaroslav Krizek, Aug 19 2009

a(A051838(n)) = A002110(A051838(n)) / A116536(n). - Reinhard Zumkeller, Oct 03 2011

a(n) = min(A068873(n), A073619(n)) for n > 1. - Jonathan Sondow, Jul 10 2012

a(n) = A033286(n) - A152535(n). - Omar E. Pol, Aug 09 2012

For n >= 3, a(n) >= (n-1)^2 * (log(n-1) - 1/2)/2 and a(n) <= n*(n+1)*(log(n) + log(log(n))+ 1)/2. Thus a(n) = n^2 * log(n) / 2 + O(n^2*log(log(n))). It is more precise than in Fares's comment. - Vladimir Shevelev, Aug 01 2013

a(n) = (n^2/2)*(log n + log log n - 3/2 + (log log n - 3)/log n + (2 (log log n)^2 - 14 log log n + 27)/(4 log^2 n) + O((log log n/log n)^3)) [Sinha]. - Charles R Greathouse IV, Jun 11 2015

G.f: (x*b(x))/(1-x), where b(x) is the g.f. of A000040. - Mario C. Enriquez, Dec 10 2016

a(n) = A008472(A002110(n)), for n > 0. - Michel Marcus, Jul 16 2020

Extensions

More terms from Stefan Steinerberger, Apr 11 2006

a(0) = 0 prepended by M. F. Hasler, Mar 05 2014