OFFSET
1,2
COMMENTS
k > 1 never ends in 1, 3, 6 or 8 (that is, k*(k+1) does not end in 2). - Lekraj Beedassy, Jul 09 2004
k > 1 can never be congruent to (1 or 3) mod 5, because if it were, then k^2 + (k+1)^2 would be divisible by 5. In other words, for k > 1, this sequence cannot contain any values in A047219. This means that we can immediately discard 40% of all possible k. - Dmitry Kamenetsky, Sep 02 2008
The number of primes of this type is roughly twice the number of primes of the form k^2+1 (A005574), and bigger than the number of primes (A000040) up to same k. For example: pi_{k^2+1}(10^8) = 3954181 < pi(10^8) = pi_{k}(10^8) = 5761455 < pi_{k^2+(k+1)^2}(10^8) = 7755330. - Hermann Stamm-Wilbrandt, Dec 22 2025
LINKS
T. D. Noe and Zak Seidov, Table of n, a(n) for n = 1..10000
FORMULA
a(n) = (A002731(n)-1)/2.
a(n) = (sqrt(2*A027862(n)-1)-1)/2. - Zak Seidov, Jul 22 2013
a(n) = floor(sqrt(A027862(n)/2)). - Rémi Guillaume, Apr 02 2025
MATHEMATICA
Select[Range[250], PrimeQ[#^2+(#+1)^2]&] (* Harvey P. Dale, Dec 31 2017 *)
PROG
(Magma) [n: n in [0..1000] |IsPrime(n^2 + (n+1)^2)]; // Vincenzo Librandi, Nov 19 2010
(Haskell)
a027861 n = a027861_list !! (n-1)
a027861_list = filter ((== 1) . a010051 . a001844) [0..]
-- Reinhard Zumkeller, Jul 13 2014
(PARI) is(n)=isprime(n^2 + (n+1)^2) \\ Charles R Greathouse IV, Apr 28 2015
CROSSREFS
Complement of A012132.
KEYWORD
nonn,easy
AUTHOR
STATUS
approved
