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A117301
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a(n) = prime(n+3)*prime(n) - prime(n+1)*prime(n+2).
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10
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-1, -2, -12, -24, -12, -24, 56, -78, -48, 42, -184, -24, 152, 46, -260, -48, 102, -304, 110, 126, -60, 276, -250, -630, -24, -12, -24, 1272, -72, -1156, -294, 476, -24, -676, 580, -374, -60, 286, -740, 644, -24, -1206, -12, 1520, 1942, -1880
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OFFSET
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1,2
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COMMENTS
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The number of negative values in this sequence appears to be consistently larger than the number of positive values. The following list gives the number of positive terms among the first n terms divided by the number of negative terms among the first n terms for various n.
n ratio
10^2 0.51515151515...
10^3 0.70940170940...
10^4 0.80212650928...
10^5 0.83826908582...
10^6 0.86339454584...
Cino Hilliard conjectures that this ratio converges and that there are infinitely many elements in the sequence whose absolute value is 12.
It appears that the positions of negative multiples of 12 are given by A064026(n+1) for n >= 1. If so, then Hilliard's conjecture is true, and a further conjecture is that if k >= 2 then there are infinitely many multiples of -12*k in this sequence. - Clark Kimberling, Jan 01 2014
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LINKS
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FORMULA
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EXAMPLE
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a(4) = prime(4)*prime(7) - prime(5)*prime(6) = 7*17 - 11*13 = -24.
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MATHEMATICA
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Table[Prime[n]*Prime[n + 3] - Prime[n + 1]Prime[n + 2], {n, 1, 100}] (* Stefan Steinerberger, Jun 27 2007 *)
(* The following program is significantly faster: *)
(First[#]Last[#]-#[[2]]#[[3]])&/@Partition[Prime[Range[50]], 4, 1] (* Harvey P. Dale, May 08 2011 *)
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PROG
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(PARI) det2cont(n) = {local(m, p, x, D); m=0; p=0; for(x=1, n, D=prime(x)*prime(x+3)-prime(x+1)*prime(x+2); if(D<0, m++, p++); print1(D", ") ); print(); print("neg= "m); print("pos= "p); print("pos/neg = "p/m+.) }
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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