A107711 - OEIS

A107711

Triangle read by rows: T(0,0)=1, T(n,m) = binomial(n,m) * gcd(n,m)/n.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 5, 10, 5, 1, 1, 1, 1, 3, 5, 5, 3, 1, 1, 1, 1, 7, 7, 35, 7, 7, 1, 1, 1, 1, 4, 28, 14, 14, 28, 4, 1, 1, 1, 1, 9, 12, 42, 126, 42, 12, 9, 1, 1, 1, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 1, 1, 11, 55, 165, 66, 462, 66, 165, 55, 11, 1, 1

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OFFSET

0,13

COMMENTS

T(0,0) is an indeterminate, but 1 seems a logical value to assign it. T(n,0) = T(n,1) = T(n,n-1) = T(n,n) = 1.

T(2n,n) = A001700(n-1) (n>=1). - Emeric Deutsch, Jun 13 2005

LINKS

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened

Wolfdieter Lang, On Collatz' Words, Sequences and Trees, arXiv:1404.2710 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.11.7.

FORMULA

From Wolfdieter Lang, Feb 28 2014 (Start)

T(n, m) = T(n-1,m)*(n-1)*gcd(n,m)/((n-m)*gcd(n-1,m)), n > m >= 1, T(n, 0) = 1, T(n, n) = 1, otherwise 0.

T(n, m) = binomial(n-1,m-1)*gcd(n,m)/m for n >= m >= 1, T(n,0) = 1, otherwise 0 (from iteration of the preceding recurrence).

T(n, m) = T(n-1, m-1)*(n-1)*gcd(n,m)/(m*gcd(n-1,m-1)) for n >= m >= 2, T(n, 0) = 1, T(n, 1) = 0, otherwise 0 (from the preceding formula).

T(2*n, n) = A001700(n-1) (n>=1) (see the Emeric Deutsch comment above), T(2*n, n-1) = A234040(n), T(2*n+1,n) = A000108(n), n >= 0 (Catalan numbers).

Column sequences: T(n+2, 2) = A026741(n+1), T(n+3, 3) = A234041(n), T(n+4, 4) = A208950(n+2), T(n+5, 5) = A234042, n >= 0. (End)

EXAMPLE

T(6,2)=5 because binomial(6,2)*gcd(6,2)/6 = 15*2/6 = 5.

The triangle T(n,m) begins:

n\m 0 1 2 3 4 5 6 7 8 9 10...

0: 1

1: 1 1

2: 1 1 1

3: 1 1 1 1

4: 1 1 3 1 1

5: 1 1 2 2 1 1

6: 1 1 5 10 5 1 1

7: 1 1 3 5 5 3 1 1

8: 1 1 7 7 35 7 7 1 1

9: 1 1 4 28 14 14 28 4 1 1

10: 1 1 9 12 42 126 42 12 9 1 1

n\m 0 1 2 3 4 5 6 7 8 9 10...

... reformatted - Wolfdieter Lang, Feb 23 2014

MAPLE

a:=proc(n, k) if n=0 and k=0 then 1 elif k<=n then binomial(n, k)*gcd(n, k)/n else 0 fi end: for n from 0 to 13 do seq(a(n, k), k=0..n) od; # yields sequence in triangular form. - Emeric Deutsch, Jun 13 2005

MATHEMATICA

T[0, 0] = 1; T[n_, m_] := Binomial[n, m] * GCD[n, m]/n;

Table[T[n, m], {n, 1, 13}, {m, 1, n}] // Flatten (* Jean-François Alcover, Nov 16 2017 *)

PROG

(Haskell)

a107711 n k = a107711_tabl !! n !! k

a107711_row n = a107711_tabl !! n

a107711_tabl = [1] : zipWith (map . flip div) [1..]

(tail $ zipWith (zipWith (*)) a007318_tabl a109004_tabl)

-- Reinhard Zumkeller, Feb 28 2014

CROSSREFS

Row sums A159554.