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A024450
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a(n) = sum of squares of the first n primes.
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36
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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; internal format)
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OFFSET
| 1,1
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COMMENTS
| It appears that the only square in this sequence is 4. Checked 10^11 terms. a(10^11)=247754953701579144582110673365391267. - T. D. Noe (noe(AT)sspectra.com), 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 (alex(AT)kolmogorov.com), 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 (wkehowski(AT)cox.net), Jun 24 2006
For all primes p > 3, we have p^2 = 1 (mod m) for m=2,3,4,6,8,12,24 (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 (noe(AT)sspectra.com), Jan 23 2008
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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
Eric Weisstein's World of Mathematics, Prime Sums
Eric Weisstein's World of Mathematics, Prime Zeta Function
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FORMULA
| a(n) = A007504(n)^2 - 2*A024447(n). - Alexander Adamchuk (alex(AT)kolmogorov.com), Jun 15 2006
a(n) = sum(prime(i)^2,i=1..n) - Walter Kehowski (wkehowski(AT)cox.net), Jun 24 2006
a(n) ~ 0.40 n^3 log(n)^2 - T. D. Noe (noe(AT)sspectra.com), Jan 23 2008
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MATHEMATICA
| Table[ Sum[ Prime[k]^2, {k, 1, n} ], {n, 1, 40} ]
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PROG
| (PARI) s=0; forprime(p=2, 1e3, print1(s+=p^2", ")) \\ Charles R Greathouse IV, Jul 15 2011
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CROSSREFS
| Partial sums of A001248. Cf. A098561, A098562.
Cf. A007504 (sum of the first n primes).
Sequence in context: A155344 A155418 A155235 * A047094 A145128 A089092
Adjacent sequences: A024447 A024448 A024449 * A024451 A024452 A024453
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KEYWORD
| nonn
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AUTHOR
| Clark Kimberling (ck6(AT)evansville.edu)
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