A116981 Number of distinct representations of 8n^3 as the sum of two primes.
1, 5, 13, 11, 28, 53, 50, 53, 135, 106, 116, 253, 165, 229, 568, 244, 313, 656, 381, 575, 1123, 600, 612, 1297, 956, 871, 1735, 1130, 1102, 3025, 1288, 1314, 3169, 1620, 2671, 3582, 1954, 2149, 4729, 3064, 2513, 6244, 2822, 3276, 8242, 3450, 3590, 7305, 4598, 5402, 9028, 4825, 4809, 9886, 7552, 6446
Offset: 1
Examples
a(1) = 1 because (2*1)^3 = 8 = 3 + 5 uniquely. a(2) = 5 because (2*2)^3 = 64 = 3 + 61 = 5 + 59 = 11 + 53 = 17 + 47 = 23 + 41. a(3) = 13 because (2*3)^3 = 216 = 5 + 211 = 17 + 199 = 19 + 197 = 23 + 193 = 37 + 179 = 43 + 173 = 53 + 163 = 59 + 157 = 67 + 149 = 79 + 137 = 89 + 127 = 103 + 113 = 107 + 109.
References
- H. Halberstam and H. E. Richert, "Sieve methods", Academic Press, London, New York, San Francisco, 1974.
Programs
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Maple
a:=proc(n) local ct,j: ct:=0: for j from 1 to prevprime((2*n)^3) do if isprime((2*n)^3-ithprime(j))=true then ct:=ct+1 else ct:=ct fi od: ct/2: end: seq(a(n),n=1..40); # execution takes hours - Emeric Deutsch, Apr 17 2006 # Faster alternative N:= 100: # to get a(1)..a(N) V:= Vector(8*N^3,datatype=float[8]): P:= select(isprime, [2,seq(i,i=3..8*N^3,2)]): V[P]:= 1: C:= SignalProcessing:-Convolution(V,V): seq(round(C[8*n^3-1])/2,n=1..N); # Robert Israel, Jan 24 2018
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Mathematica
Table[Count[IntegerPartitions[8*n^3,{2}],?(AllTrue[#,PrimeQ]&)],{n,60}] (* The program uses the AllTrue function from Mathematica version 10 *) (*The program will take a long time to run. *) (* _Harvey P. Dale, Oct 11 2019 *)
Formula
Extensions
Corrected and extended by Emeric Deutsch, Apr 17 2006
More terms from Robert Israel, Jan 24 2018