A024450 Sum of squares of the first n primes.

4, 13, 38, 87, 208, 377, 666, 1027, 1556, 2397, 3358, 4727, 6408, 8257, 10466, 13275, 16756, 20477, 24966, 30007, 35336, 41577, 48466, 56387, 65796, 75997, 86606, 98055, 109936, 122705, 138834, 155995, 174764, 194085, 216286, 239087, 263736, 290305, 318194

Offset: 1

Author

Clark Kimberling

Keywords

• nonn

Comments

It appears that the only square in this sequence is 4. Checked 10^11 terms. a(10^11) = 247754953701579144582110673365391267. - T. D. Noe, Sep 06 2005

a(2n-1) is divisible by 2, a(3n+1) is divisible by 3, a(4n-3) is divisible by 4, a(6n+1) is divisible by 6, a(8n-3) is divisible by 8, a(12n+1) is divisible by 12, a(24n-11) is divisible by 24. - Alexander Adamchuk, Jun 15 2006

The sequence is best looked at in base 12, with X for 10 and E for 11: 4, 11, 32, 73, 154, 275, 476, 717, X98, 1479, 1E3X, 289E, 3860, 4941, 6082, 7823, 9844, EX25, 12546, 15447, 18548, 20089, 2406X, 2876E, 320E0, 37E91, 42152, 488E3, 53754, 5E015, 68416, 76337, 85178, 94399, X51EX, E643E, 108760, 120001. Since the squares of all primes greater than 3 are always 1 mod 12, the sequence obeys the rule a(n) mod 12 = (n-1) mod 12 for n>=2. The rule gives a(2n-1) = (2n-2) mod 12 and so a(2n-1) must be divisible by 2. a(3n+1) = (3n) mod 12 so a(3n+1) is divisible by 3. The other rules are proved similarly. Remember: base 12 is a research tool! - Walter Kehowski, Jun 24 2006

For all primes p > 3, we have p^2 == 1 (mod m) for m dividing 24 (and only these m). Using a covering argument, it is not hard to show that all terms except a(24k+13) are nonsquares. Hence in the search for square a(n), only 1 out of every 24 terms needs to be checked. - T. D. Noe, Jan 23 2008

Links

• N. J. A. Sloane, Table of n, a(n) for n = 1..5000
• Carlos Rivera, Puzzle 128: Sum of consecutive squared primes a square, The Prime Puzzles & Problems Connection.
• Vladimir Shevelev, Asymptotics of sum of the first n primes with a remainder term
• Eric Weisstein's World of Mathematics, Prime Sums
• Eric Weisstein's World of Mathematics, Prime Zeta Function
• OEIS Wiki, Sums of powers of primes divisibility sequences

Crossrefs

Partial sums of A001248.

Cf. A024447, A024525, A098561, A098562.

Cf. A007504 (sum of the first n primes).

Programs

Haskell

a024450 n = a024450_list !! (n-1)
a024450_list = scanl1 (+) a001248_list
-- Reinhard Zumkeller, Jun 08 2015

Magma

[&+[NthPrime(k)^2: k in [1..n]]: n in [1..40]]; // Vincenzo Librandi, Oct 11 2018

Magma

[n le 1 select 4 else Self(n-1) + NthPrime(n)^2: n in [1..80]]; // G. C. Greubel, Jan 30 2025

Maple

A024450:=n->add(ithprime(i)^2, i=1..n); seq(A024450(n), n=1..100); # Wesley Ivan Hurt, Nov 09 2013

Mathematica

Table[ Sum[ Prime[k]^2, {k, 1, n} ], {n, 40} ]
Accumulate[Prime[Range[40]]^2] (* Harvey P. Dale, Apr 16 2013 *)

PARI

s=0;forprime(p=2,1e3,print1(s+=p^2", ")) \\ Charles R Greathouse IV, Jul 15 2011

PARI

a(n) = norml2(primes(n)); \\ Michel Marcus, Nov 26 2020

Python

from sympy import prime, primerange
def a(n): return sum(p*p for p in primerange(1, prime(n)+1))
print([a(n) for n in range(1, 40)]) # Michael S. Branicky, Apr 13 2021

Formula

a(n) = A007504(n)^2 - 2*A024447(n). - Alexander Adamchuk, Jun 15 2006

a(n) = Sum_{i=1..n} prime(i)^2. - Walter Kehowski, Jun 24 2006

a(n) = (1/3)*n^3*log(n)^2 + O(n^3*log(n)*log(log(n))). The proof is similar to proof for A007504(n) (see link of Shevelev). - Vladimir Shevelev, Aug 02 2013

a(n) = a(n-1) + prime(n)^2, with a(1) = 4. - G. C. Greubel, Jan 30 2025