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A007504
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Sum of the first n primes.
(Formerly M1370)
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463
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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
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OFFSET
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0,2
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COMMENTS
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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]
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
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
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REFERENCES
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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).
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LINKS
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FORMULA
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a(n) ~ n^2 * log(n) / 2. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 24 2001 (see Bach & Shallit 1996)
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
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MAPLE
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s1:=[2]; for n from 2 to 1000 do s1:=[op(s1), s1[n-1]+ithprime(n)]; od: s1;
add(ithprime(i), i=1..n) ;
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MATHEMATICA
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Accumulate[Prime[Range[100]]] (* Zak Seidov, Apr 10 2011 *)
primeRunSum = 0; Table[primeRunSum = primeRunSum + Prime[k], {k, 100}] (* Zak Seidov, Apr 16 2011 *)
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PROG
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(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)
(Haskell)
a007504 n = a007504_list !! n
a007504_list = scanl (+) 0 a000040_list
(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
(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))
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CROSSREFS
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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).
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KEYWORD
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nonn,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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