A189731 a(n) = numerator of B(0,n) where B(n,n) = 0, B(n-1,n) = 1/n, and B(m,n) = B(m-1,n+1) - B(m-1,n).

0, 1, 1, 3, 2, 17, 4, 23, 25, 61, 18, 107, 40, 421, 1363, 1103, 210, 5777, 492, 7563, 24475, 19801, 2786, 103681, 33552, 135721, 146401, 355323, 39650, 1860497, 97108, 2435423, 2627065, 6376021, 20633238, 11128427, 1459960, 43701901

Offset: 0

Author

Paul Curtz, Apr 26 2011

Keywords

• nonn
• easy

Comments

Square array B(m,n) begins:

0, 1/1, 1/1, 3/2, 2/1, 17/6, ...

1/1, 0, 1/2, 1/2, 5/6, 7/6, ...

-1/1, 1/2, 0, 1/3, 1/3, 7/12, ...

3/2, -1/2, 1/3, 0, 1/4, 1/4, ...

-2/1, 5/6, -1/3, 1/4, 0, 1/5, ...

17/6, -7/6, 7/12, -1/4, 1/5, 0, ...

The inverse binomial transform of B(0,n) gives B(n,0) and thus it is an eigensequence in the sense that it remains the same (up to a sign) under inverse binomial transform.

The bisection of B(0,n) (odd part) gives A175385/A175386, and thus a(2*n+1) = A175385(n+1).

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A000204, A242926 (denominators).

Cf. A174341, A177690, A181722.

Programs

Maple

B:= proc(m, n) option remember;
      if m=n then 0
    elif n=m+1 then 1/n
    elif n>m then B(m, n-1) +B(m+1, n-1)
    else B(m-1, n+1) -B(m-1, n)
      fi
    end:
a:= n-> numer(B(0, n)):
seq(a(n), n=0..50);  # Alois P. Heinz, Apr 29 2011

Mathematica

Rest[Numerator[Abs[CoefficientList[Normal[Series[Log[1 - x^2/(1 + x)], {x, 0, 40}]], x]]]] (* Vaclav Kotesovec, Jul 07 2020 *)
Table[Numerator[(LucasL[n]-1)/n],{n,1,38}] (* Artur Jasinski, Oct 21 2022 *)

Formula

Numerator of (A000204(n) - 1)/n. - Artur Jasinski, Oct 21 2022