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A113631
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Number of distinct representations of (2n)^2 as the sum of two primes.
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3
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0, 1, 2, 4, 5, 6, 11, 9, 8, 20, 14, 14, 26, 17, 18, 48, 22, 22, 49, 28, 36, 69, 33, 37, 68, 47, 43, 83, 49, 47, 125, 50, 53, 118, 56, 94, 126, 63, 63, 153, 98, 71, 186, 79, 94, 230, 89, 91, 197, 127, 127, 215, 112, 105, 220, 172, 147
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OFFSET
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0,3
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COMMENTS
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From Halberstam and Richert: A045917(2n)<(8+0(1))*c(n)*n/log(n)^2 where c(n)=prod(p>2,(1-1/(p-1)^2))*prod(p|n,p>2,(p-1)/(p-2)). Hence a(n) = A045917(2n) < (8+0(1))*c(2n)*2n/log(2n)^2 where c(k)=prod(p>2,(1-1/(p-1)^2))*prod(p|k,p>2,(p-1)/(p-2)). See also: A045917 From Goldbach problem: number of decompositions of 2n into unordered sums of two primes. A016742 Even squares: (2n)^2.
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REFERENCES
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R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd ed., 1994, p. 99.
H. Halberstam and H. E. Richert, 1974, "Sieve methods", Academic press, London, New York, San Francisco.
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LINKS
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FORMULA
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a(n) = A045917(2n). a(n) = #{p(i) + p(j) = A016742(n) for p(k) = A000040(k) and i >= j}. a(n) = #{p(i) + p(j) = (2*n)^2 for p(k) = A000040(k) and i >= j}.
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EXAMPLE
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a(1) = 1 because (2*1)^2 = 4 = 2 + 2 uniquely.
a(2) = 2 because (2*2)^2 = 16 = 3 + 13 = 5 + 11.
a(3) = 4 because (2*3)^2 = 36 = 5 + 31 = 7 + 29 = 13 + 23 = 17 + 19.
a(4) = 5 because (2*4)^2 = 64 = 3 + 61 = 5 + 59 = 11 + 53 = 17 + 47 = 23 + 41.
a(5) = 6 because (2*5)^2 = 100 = 3 + 97 = 11 + 89 = 17 + 83 = 29 + 71 = 41 + 59 = 47 + 53.
a(6) = 11 because (2*6)^2 = 144 = 5 + 139 = 7 + 137 = 13 + 131 = 17 + 127 = 31 + 113 = 37 + 107 = 41 + 103 = 43 + 101 = 47 + 97 = 61 + 83 = 71 + 73.
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MAPLE
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g:=sum(sum(x^(ithprime(i)+ithprime(j)), i=1..j), j=1..1500): gser:=series(g, x=0, 12560): 0, seq(coeff(gser, x^(4*n^2)), n=1..56); # Emeric Deutsch, Apr 03 2006
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PROG
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(Magma) [#RestrictedPartitions(4*n^2, 2, {p:p in PrimesUpTo(20000)}):n in [0..56] ] // Marius A. Burtea, Jan 19 2019
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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