cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A047222 Numbers that are congruent to {0, 2, 3} mod 5.

Original entry on oeis.org

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, 42, 43, 45, 47, 48, 50, 52, 53, 55, 57, 58, 60, 62, 63, 65, 67, 68, 70, 72, 73, 75, 77, 78, 80, 82, 83, 85, 87, 88, 90, 92, 93, 95, 97, 98, 100, 102, 103, 105, 107
Offset: 1

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

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Formula

From R. J. Mathar, Oct 18 2008: (Start)
G.f.: x^2*(2 + x + 2*x^2)/((1 - x)^2*(1 + x + x^2)).
a(n) = A028738(n-2), 1 < n < 16. (End)
a(n) = floor((5*n-4)/3). - Gary Detlefs, Oct 28 2011
a(n) = 2*n - 2 - floor(n/3). - Wesley Ivan Hurt, Nov 07 2013
From Wesley Ivan Hurt, Jun 14 2016: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4) for n>4.
a(n) = (15*n-15-3*cos(2*n*Pi/3)-sqrt(3)*sin(2*n*Pi/3))/9.
a(3k) = 5k-2, a(3k-1) = 5k-3, a(3k-2) = 5k-5. (End)
a(n) = n - 1 + floor((2n-1)/3). - Wesley Ivan Hurt, Dec 27 2016
Sum_{n>=2} (-1)^n/a(n) = arccoth(3/sqrt(5))/sqrt(5) - log(2)/5. - Amiram Eldar, Dec 10 2021
From Peter Bala, Aug 04 2022: (Start)
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.
a(2*n) = a(n) + a(n+1); a(2*n+1) = a(n) + a(n+2). Cf. A008854 and A042965. (End)