

A024450


Sum of squares of the first n primes.


131



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
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OFFSET

1,1


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(2n1) is divisible by 2, a(3n+1) is divisible by 3, a(4n3) is divisible by 4, a(6n+1) is divisible by 6, a(8n3) is divisible by 8, a(12n+1) is divisible by 12, a(24n11) 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 = (n1) mod 12 for n>=2. The rule gives a(2n1) = (2n2) mod 12 and so a(2n1) 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


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


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 *)


PROG

(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
(Haskell)
a024450 n = a024450_list !! (n1)
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


CROSSREFS

Partial sums of A001248. Cf. A098561, A098562.
Cf. A007504 (sum of the first n primes).
Sequence in context: A155344 A155418 A155235 * A047094 A145128 A277974
Adjacent sequences: A024447 A024448 A024449 * A024451 A024452 A024453


KEYWORD

nonn


AUTHOR

Clark Kimberling


STATUS

approved



