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A098033
Parity of p*(p+1)/2 for n-th prime p.
2
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
OFFSET
1,1
COMMENTS
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.
Analogous to the prime race (mod 3). - Robert G. Wilson v, Sep 17 2004
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
LINKS
FORMULA
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
EXAMPLE
a(1) = parity of (2*(2+1)/2 = 3) = 1 (odd).
MAPLE
seq((ithprime(n) mod 4) mod 3, n = 2..105] # Gary Detlefs, Oct 27 2011
MATHEMATICA
Table[ Mod[ Prime[n](Prime[n] + 1)/2, 2], {n, 105}] (* Robert G. Wilson v, Sep 17 2004 *)
Mod[(#(#+1))/2, 2]&/@Prime[Range[110]] (* Harvey P. Dale, Mar 29 2015 *)
PROG
(PARI) a(n)=prime(n)%4<3 \\ Charles R Greathouse IV, Oct 27 2011
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Jeremy Gardiner, Sep 10 2004
EXTENSIONS
More terms from Robert G. Wilson v, Sep 17 2004
STATUS
approved