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 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 (list; graph; refs; listen; history; text; internal format)
 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(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 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]^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 !! (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 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 STATUS approved

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Last modified January 17 19:19 EST 2021. Contains 340247 sequences. (Running on oeis4.)