%I #94 Aug 08 2022 17:40:15

%S 0,2,3,5,7,8,10,12,13,15,17,18,20,22,23,25,27,28,30,32,33,35,37,38,40,

%T 42,43,45,47,48,50,52,53,55,57,58,60,62,63,65,67,68,70,72,73,75,77,78,

%U 80,82,83,85,87,88,90,92,93,95,97,98,100,102,103,105,107

%N Numbers that are congruent to {0, 2, 3} mod 5.

%C Row sum of a triangle where the top value is 2 and every elementary triangle or triple is required to have the values 1,2,2 (see link below). Compare with A008854 where the triple contains 1,2,2 with 1 at the top. - _Craig Knecht_, Oct 18 2015

%C Also, numbers k such that k*(k^2+1)/5 is a nonnegative integer. - _Bruno Berselli_, Jan 16 2016

%C Conjecture: Apart from 0, the sequence gives the values for c/6, such that an infinite number of primes, p, result in both p^2-c and p^2+c being positive primes, except when c is a square. When c is square solutions exist for c (both within and outside of the a(n) set), but occur at only a single prime p. See A274609. Other c values with only one prime providing a solution occur when p^2-c=3. See A274610. The only remaining c values with single p solutions are: c=2 (with p=3) and c=6 (with p=5). - _Richard R. Forberg_, Jun 26 2016

%C See A047363 for case of p^3 +- c. See A005097 and A177735 for observations on the general case p^q +- c. - _Richard R. Forberg_, Aug 11 2016

%H Vincenzo Librandi, <a href="/A047222/b047222.txt">Table of n, a(n) for n = 1..1000</a>

%H Craig Knecht, <a href="/A047222/a047222_1.jpg">Row sum for the 1,2,2 triangle with 2 at the top</a>.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,1,-1).

%F From _R. J. Mathar_, Oct 18 2008: (Start)

%F G.f.: x^2*(2 + x + 2*x^2)/((1 - x)^2*(1 + x + x^2)).

%F a(n) = A028738(n-2), 1 < n < 16. (End)

%F a(n) = floor((5*n-4)/3). - _Gary Detlefs_, Oct 28 2011

%F a(n) = 2*n - 2 - floor(n/3). - _Wesley Ivan Hurt_, Nov 07 2013

%F From _Wesley Ivan Hurt_, Jun 14 2016: (Start)

%F a(n) = a(n-1) + a(n-3) - a(n-4) for n>4.

%F a(n) = (15*n-15-3*cos(2*n*Pi/3)-sqrt(3)*sin(2*n*Pi/3))/9.

%F a(3k) = 5k-2, a(3k-1) = 5k-3, a(3k-2) = 5k-5. (End)

%F a(n) = n - 1 + floor((2n-1)/3). - _Wesley Ivan Hurt_, Dec 27 2016

%F Sum_{n>=2} (-1)^n/a(n) = arccoth(3/sqrt(5))/sqrt(5) - log(2)/5. - _Amiram Eldar_, Dec 10 2021

%F From _Peter Bala_, Aug 04 2022: (Start)

%F a(n) = a(floor(n/2)) + a(1 + ceiling(n/2)) for n >= 4 with a(1) = 0, a(2) = 2 and a(3) = 3.

%F a(2*n) = a(n) + a(n+1); a(2*n+1) = a(n) + a(n+2). Cf. A008854 and A042965. (End)

%p A047222 := n -> 2*n - 2 - iquo(n,3): seq(A047222(n), n=1..100); # _Wesley Ivan Hurt_, Nov 07 2013

%t Floor[(Range[5, 305, 5] - 4)/3] (* _Vladimir Joseph Stephan Orlovsky_, Jan 26 2012 *)

%t Flatten[Table[5n + {0, 2, 3}, {n, 0, 19}]] (* _Alonso del Arte_, Nov 07 2013 *)

%t LinearRecurrence[{1, 0, 1, -1}, {0, 2, 3, 5}, 100] (* _Vincenzo Librandi_, Jun 15 2016 *)

%o (PARI) a(n)=(5*n-4)\3 \\ _Charles R Greathouse IV_, Oct 28 2011

%o (PARI) concat(0, Vec(x^2*(2+x+2*x^2)/((1-x)^2*(1+x+x^2)) + O(x^100))) \\ _Altug Alkan_, Oct 26 2015

%o (Magma) [n : n in [0..150] | n mod 5 in [0, 2, 3]]; // _Wesley Ivan Hurt_, Jun 14 2016

%Y Cf. A008854, A028738, A042965, A047363, A005097.

%K nonn,easy

%O 1,2

%A _N. J. A. Sloane_