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A098033
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Parity of p*(p+1)/2 for n-th prime p.
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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
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OFFSET
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1,1
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COMMENTS
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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 the next prime(n) natural numbers and the sum of the next n primes; A006254 = numbers n such that 2n-1 is prime; A067076 = numbers n such that 2n+3 is a prime.
See also A089253 = 2n-5 is a prime.
For n > 1, if A000040(n) == 1 (mod 4), then a(n) = 1, otherwise a(n)=0, so (for n>1) also a(n) = number of representations of A000040(n) as a difference of hexagonal numbers (A000384) (cf. [Nyblom, p. 262]). - L. Edson Jeffery, Feb 16 2013
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LINKS
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FORMULA
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a(n) = parity of p*(p+1)/2 for n-th prime p.
a(n) = 1 - A100672(n), n > 1. - Steven G. Johnson (stevenj(AT)math.mit.edu), Sep 18 2008
For n > 1, a(n) = (prime(n) mod 4) mod 3. - Gary Detlefs, Oct 27 2011
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EXAMPLE
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a(1) = parity of (2*(2+1)/2 = 3) = 1 (odd).
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MAPLE
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seq((ithprime(n) mod 4) mod 3, n = 2..105] # Gary Detlefs, Oct 27 2011
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MATHEMATICA
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Mod[(#(#+1))/2, 2]&/@Prime[Range[110]] (* Harvey P. Dale, Mar 29 2015 *)
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PROG
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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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