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A075888
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Difference of successive primes squared divided by 24, (prime(n+1)^2-prime(n)^2)/24, n >= 3.
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3
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1, 3, 2, 5, 3, 7, 13, 5, 17, 13, 7, 15, 25, 28, 10, 32, 23, 12, 38, 27, 43, 62, 33, 17, 35, 18, 37, 140, 43, 67, 23, 120, 25, 77, 80, 55, 85, 88, 30, 155, 32, 65, 33, 205, 217, 75, 38, 77, 118, 40, 205, 127, 130, 133, 45, 137, 93, 47, 240, 350, 103, 52, 105, 378, 167, 285
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OFFSET
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3,2
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COMMENTS
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For n>=3, prime(n+1)^2-prime(n)^2 is always divisible by 24.
It follows from the previous comment that for n>=3, prime(n)= sqrt(5^2 + k*24) where integer k>= 0 Then it follows from above that for n>=3, ((prime(n))^2 - 1)/24 always gives integral values - see A024702 [From Alexander R. Povolotsky, Sep 20 2008]
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LINKS
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Table of n, a(n) for n=3..68.
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FORMULA
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a(n)=(prime(n+1)^2-prime(n)^2)/24, n >= 3.
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EXAMPLE
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a(4)=3 because (p(5)^2-p(4)^2)/24=(11^2-7^2)/24=3.
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MATHEMATICA
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(#[[2]]-#[[1]])/24&/@(Partition[Prime[Range[3, 70]], 2, 1]^2) (* Harvey P. Dale, Apr 06 2013 *)
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PROG
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(PARI) j=[]; for(n=3, 300, if(((floor((((prime(n+1))^2)-((prime(n))^2))/24))==(ceil(((((prime(n+1))^2)-((prime(n))^2))/24)))), j=concat(j, ((((prime(n+1))^2) - ((prime(n))^2))/24)), j=concat(j, -1))); j [From Alexander R. Povolotsky, Sep 08 2008]
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CROSSREFS
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Cf. A024702 [From Alexander R. Povolotsky, Sep 20 2008]
Sequence in context: A111618 A107128 A086670 * A075889 A181771 A045766
Adjacent sequences: A075885 A075886 A075887 * A075889 A075890 A075891
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KEYWORD
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easy,nonn
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AUTHOR
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Zak Seidov, Oct 17 2002
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STATUS
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approved
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