A106831 Define a triangle in which the entries are of the form +-1/(b!c!d!e!...), where the order of the factorials is important; read the triangle by rows and record and expand the denominators.

2, 6, 4, 24, 12, 12, 8, 120, 48, 36, 24, 48, 24, 24, 16, 720, 240, 144, 96, 144, 72, 72, 48, 240, 96, 72, 48, 96, 48, 48, 32, 5040, 1440, 720, 480, 576, 288, 288, 192, 720, 288, 216, 144, 288, 144, 144, 96, 1440, 480, 288, 192, 288, 144, 144, 96, 480, 192, 144, 96, 192

Offset: 0

Author

N. J. A. Sloane, May 22 2005

Keywords

• nonn
• tabf
• frac
• easy
• nice

Comments

Row n has 2^n terms. Row 0 is +1/2!. An entry +-1/b!c!d!... has two children, a left child -+1/(a+1)!b!c!... and a right child +-1/2!b!c!d!...

Let S_n = sum of entries in row n of the triangle. Then for n > 0, n!S_{n-1} is the Bernoulli number B_n.

Examples

			Woon's "Bernoulli Tree" begins like this (see also the given Wikipedia-link). This sequence gives the values of the denominators:
                                     +1
                                    ────
                                     2!
                 -1                 /  \                  +1
                ──── ............../    \.............. ─────
                 3!                                      2!2!
        +1        .         -1                 -1         .         +1
       ────      / \       ────               ────       / \      ──────
        4! ...../   \..... 2!3!               3!2! ...../   \.... 2!2!2!
       / \                 / \                 / \                 / \
      /   \               /   \               /   \               /   \
     /     \             /     \             /     \             /     \
    -1      +1         +1       -1         +1      -1          -1       +1
   ────    ────       ────     ──────     ────   ──────      ──────  ────────
    5!     2!4!       3!3!     2!2!3!     4!2!   2!3!2!      3!2!2!  2!2!2!2!
etc.
		

Links

• Reinhard Zumkeller, Rows n = 0..13 of table, flattened
• Wikipedia, Bernoulli number: A binary tree representation
• S. C. Woon, A tree for generating Bernoulli numbers, Math. Mag., 70 (1997), 51-56.
• Index entries for sequences related to Bernoulli numbers.

Crossrefs

Cf. A242179 (numerators), A050925, A050932, A000142.

Cf. A001013, A060054, A075180, A164555, A027642, A246660.

Cf. A323505 (mirror image), and also A005940, A283477, A322827 for other similar trees.

Programs

Haskell

a106831 n k = a106831_tabf !! n !! n
a106831_row n = a106831_tabf !! n
a106831_tabf = map (map (\(, , left, right) -> left * right)) $
   iterate (concatMap (\(x, f, left, right) -> let f' = f * x in
   [(x + 1, f', f', right), (3, 2, 2, left * right)])) [(3, 2, 2, 1)]
-- Reinhard Zumkeller, May 05 2014

Maple

Contribution from Peter Luschny, Jun 12 2009: (Start)
The routine computes the triangle row by row and gives the numbers with their sign.
Thus A(1)=[2]; A(2)=[ -6,4]; A(3)=[24,-12,-12,8]; etc.
A := proc(n) local k, i, j, m, W, T; k := 2;
W := array(0..2^n); W[1] := [1,`if`(n=0,1,2)];
for i from 1 to n-1 do for m from k by 2 to 2*k-1 do
T := W[iquo(m,2)]; W[m] := [ -T[1],T[2]+1,seq(T[j],j=3..nops(T))];
W[m+1] := [T[1],2,seq(T[j],j=2..nops(T))]; od; k := 2*k; od;
seq(W[i][1]*mul(W[i][j]!,j=2..nops(W[i])),i=iquo(k,2)..k-1) end:
seq(print(A(i)),i=1..5); (End)

Mathematica

a [n_] := Module[{k, i, j, m, w, t}, k = 2; w = Array[0&, 2^n]; w[[1]] := {1, If[n == 0, 1, 2]}; For[i = 1, i <= n-1, i++, For[m = k, m <= 2*k-1 , m = m+2, t = w[[Quotient[m, 2]]]; w[[m]] = {-t[[1]], t[[2]]+1, Sequence @@ Table[t[[j]], {j, 3, Length[t]}]}; w[[m+1]] = {t[[1]], 2, Sequence @@ Table[t[[j]], {j, 2, Length[t]}]}]; k = 2*k]; Table[w[[i, 1]]*Product[w[[i, j]]!, {j, 2, Length[w[[i]]]}], {i, Quotient[k, 2], k-1}]]; Table[a[i] , {i, 1, 6}] // Flatten // Abs (* Jean-François Alcover, Dec 20 2013, translated from Maple *)

PARI

A106831off1(n) = if(!n,1, my(rl=1, m=1); while(n,if(!(n%2), rl++, m *= ((1+rl)!); rl=1); n >>= 1); (m));
A106831(n) = A106831off1(1+n); \\ Antti Karttunen, Jan 16 2019

PARI

A001511(n) = (1+valuation(n,2));
A106831r1(n) = if(!n,1,if(n%2, 2*A106831r1((n-1)/2), (1+A001511(n))*A106831r1(n/2))); \\ Implements the given recurrence.
A106831(n) = A106831r1(1+n); \\ Antti Karttunen, Jan 16 2019

Formula

From Antti Karttunen, Jan 16 2019: (Start)

If sequence is shifted one term to the right, then the following recurrence works:

a(0) = 1; and for n > 0, a(2n) = (1+A001511(2n))*a(n), a(2n+1) = 2*a(n).

(End)

Extensions

More terms from Franklin T. Adams-Watters, Apr 28 2006

Example section reillustrated by Antti Karttunen, Jan 16 2019