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Search: id:A098033
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| A098033 |
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Parity of p(p+1)/2 for n-th prime p. |
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+0
2
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| 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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The following sequences (possibly with a different offset for first term) all appear to have the same parity: A034953 = triangular numbers with prime indices; A054269 = length of period of continued fraction for sqrt(p), p prime; A082749 = difference between the sum of next prime(n) natural numbers and the sum of next n primes; A006254 = numbers n such that 2n-1 is prime; A067076 = numbers n such that 2n+3 is a prime.
Analogous to the prime race (mod 3). - Robert G. Wilson v Sep 17 2004.
See also A089253 = 2n-5 is a prime.
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FORMULA
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a(n) = parity of (p(p+1)/2) for n-th prime p(n)
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EXAMPLE
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a(1) = parity of (2(2+1)/2 = 3) = 1 (odd).
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MATHEMATICA
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Table[ Mod[ Prime[n](Prime[n] + 1)/2, 2], {n, 105}] (from Robert G. Wilson v Sep 17 2004)
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CROSSREFS
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Cf. A034953, A054269, A082749, A006254, A067076.
equal to 1 minus A100672 [From Steven G. Johnson (stevenj(AT)math.mit.edu), Sep 18 2008]
Sequence in context: A145273 A120522 A157423 this_sequence A135022 A071982 A157749
Adjacent sequences: A098030 A098031 A098032 this_sequence A098034 A098035 A098036
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KEYWORD
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easy,nonn
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AUTHOR
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Jeremy Gardiner (jeremy.gardiner(AT)btinternet.com), Sep 10 2004
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EXTENSIONS
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More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Sep 17 2004
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