OFFSET
0,1
COMMENTS
The autosequence of first kind from (-1)^n/(n+1) is A189733.
For the second kind (the second increasing diagonal is (-1)^n/(n+1), half of the main one):
2, 1, 0, -1/2, -1/3, 1/6, 1/2, 5/12,
-1, -1, -1/2, 1/6, 1/2, 1/3, -1/12, -7/20,
0, 1/2, 2/3, 1/3, -1/6, -5/12, -4/15, 1/12,
1/2, 1/6, -1/3, -1/2, -1/4, 3/20, 7/20, 13/60,
-1/3, -1/2, -1/6, 1/4, 2/5, 1/5, -2/15, -3/10,
-1/6, 1/3, 5/12, 3/20, -1/5, -1/3, -1/6, 5/42,
1/2, 1/12, -4/15, -7/20, -2/15, 1/6, 2/7, 1/7,
-5/12, -7/20, -1/12, 13/60, 3/10, 5/42, -1/7, -1/4.
Main diagonal: (period 2:repeat 2, -1)/A026741(n+1).
Second (increasing) diagonal: (-1)^n / (n+1).
Fourth (increasing) diagonal: (-1)^(n+1)*A146535(n)/ a(n).
LINKS
OEIS Wiki, Autosequence.
Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-3,0,0,1).
FORMULA
a(3*n) = (3*n+1)*(3*n+2), a(3*n+1) = (n+1)*(3*n+2), a(3*n+2) = 3*(n+1)*(3*n+4).
G.f.: 2*(1+x+6*x^2+7*x^3+2*x^4+3*x^5+x^6)/(1-x^3)^3. - Jean-François Alcover, Nov 11 2016
a(n+2) = 2 * A306368(n) for n >= 0. - Joerg Arndt, Aug 25 2023
a(n) = (n+1) * A051176(n+2) for n >= 0. - Paul Curtz, Sep 13 2023
Sum_{n>=0} 1/a(n) = 1 + log(3) - Pi/(3*sqrt(3)). - Amiram Eldar, Sep 17 2023
MATHEMATICA
c = Table[1/9 (7 n + 7 n^2 + 2 n Cos[(2 n *Pi)/3] + 2 n^2 Cos[(2 n *Pi)/3] + 2 Sqrt[3] n Sin[(2 n *Pi)/3] + 2 Sqrt[3] n^2 Sin[(2 n *Pi/3]), {n, 1, 50}], (* Roger Bagula, Mar 25 2012*)
a[n_] := (n+1) * Numerator[(n+2)/3]; Array[a, 60, 0] (* Amiram Eldar, Sep 17 2023 *)
CROSSREFS
KEYWORD
nonn,frac,easy
AUTHOR
Paul Curtz, Sep 02 2011
STATUS
approved