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A003506
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Triangle of denominators in Leibniz's Harmonic Triangle a(n,k), n >= 1, 1 <= k <= n.
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65
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1, 2, 2, 3, 6, 3, 4, 12, 12, 4, 5, 20, 30, 20, 5, 6, 30, 60, 60, 30, 6, 7, 42, 105, 140, 105, 42, 7, 8, 56, 168, 280, 280, 168, 56, 8, 9, 72, 252, 504, 630, 504, 252, 72, 9, 10, 90, 360, 840, 1260, 1260, 840, 360, 90, 10, 11, 110, 495, 1320, 2310, 2772, 2310, 1320, 495, 110, 11
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OFFSET
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1,2
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COMMENTS
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Array 1/Beta(n,m) read by antidiagonals. - Michael Somos, Feb 05 2004
a(n,k) = total size of all of the elements of the family of k-size subsets of an n-element set. For example, a 2-element set, say, {1,2}, has 3 families of k-size subsets: one with 1 0-size element, one with 2 1-size elements and one with 1 2-size element; respectively, {{}}, {{1},{2}}, {{1,2}}. - Ross La Haye, Dec 31 2006
Second slice along the 1-2-plane in the cube a(m,n,o) = a(m-1,n,o) + a(m,n-1,o) + a(m,n,o-1) with a(1,0,0)=1 and a(m<>1=0,n>=0,0>=o)=0, for which the first slice is Pascal's triangle (slice read by antidiagonals). - Thomas Wieder, Aug 06 2006
Triangle, read by rows, given by [2,-1/2,1/2,0,0,0,0,0,0,...] DELTA [2,-1/2,1/2,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 07 2007
This sequence * [1/1, 1/2, 1/3, ...] = (1, 3, 7, 15, 31, ...). - Gary W. Adamson, Nov 14 2007
n-th row = coefficients of first derivative of corresponding Pascal's triangle row. Example: x^4 + 4x^3 + 6x^2 + 4x + 1 becomes (4, 12, 12, 4). - Gary W. Adamson, Dec 27 2007
Consider
1 1/2 1/3 1/4 1/5
-1/2 -1/6 -1/12 -1/20 -1/30
1/3 1/12 1/30 1/60 1/105
-1/4 -1/20 -1/60 -1/140 -1/280
1/5 1/30 1/105 1/280 1/630
This is an autosequence (the inverse binomial transform is the sequence signed) of the second kind: the main diagonal is 2 times the first upper diagonal.
Note that 2, 12, 60, ... = A005430(n+1), Apery numbers = 2*A002457(n). (End)
From Louis Conover (for the 9th grade G1c mathematics class at the Chengdu Confucius International School), Mar 02 2015: (Start)
The i-th order differences of n^-1 appear in the (i+1)th row.
1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, ...
1/2, 1/6, 1/12, 1/20, 1/30, 1/42, 1/56, 1/72, ...
1/3, 1/12, 1/30, 1/60, 1/105, 1/168, 1/252, 1/360, ...
1/4, 1/20, 1/60, 1/140, 1/280, 1/504, 1/840, 1/1320, ...
1/5, 1/30, 1/105, 1/280, 1/630, 1/1260, 1/2310, 1/3960, ...
1/6, 1/42, 1/168, 1/504, 1/1260, 1/2772, 1/5544, 1/12012, ...
(End)
T(n,k) is the number of edges of distance k from a fixed vertex in the n-dimensional hypercube. - Simon Burton, Nov 04 2022
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REFERENCES
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A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, see 130.
B. A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8. English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see p. 38.
G. Boole, A Treatise On The Calculus of Finite Differences, Dover, 1960, p. 26.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 83, Problem 25.
M. Elkadi and B. Mourrain, Symbolic-numeric methods for solving polynomial equations and applications, Chap 3. of A. Dickenstein and I. Z. Emiris, eds., Solving Polynomial Equations, Springer, 2005, pp. 126-168. See p. 152.
D. Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, 35.
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LINKS
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FORMULA
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a(n, 1) = 1/n; a(n, k) = a(n-1, k-1) - a(n, k-1) for k > 1.
Considering the integer values (rather than unit fractions): a(n, k) = k*C(n, k) = n*C(n-1, k-1) = a(n, k-1)*a(n-1, k-1)/(a(n, k-1) - a(n-1, k-1)) = a(n-1, k) + a(n-1, k-1)*k/(k-1) = (a(n-1, k) + a(n-1, k-1))*n/(n-1) = k*A007318(n, k) = n*A007318(n-1, k-1). Row sums of integers are n*2^(n-1) = A001787(n); row sums of the unit fractions are A003149(n-1)/A000142(n). - Henry Bottomley, Jul 22 2002
G.f.: x*y/(1-x-y*x)^2.
E.g.f.: x*y*exp(x+x*y). (End)
f(s,n) = Integral_{x=0..oo} exp(-s*x)*x^n dx = Gamma(n)/s^n; t(n,m) = f(s,n)/(f(s,n-m)*f(s,m)) = Gamma(n)/(Gamma(n - m)*Gamma(m); the powers of s cancel out. - Roger L. Bagula and Gary W. Adamson, Sep 14 2008
T(n,5) = T(n,n-4) = A174002(n-4) for n > 4.
T(2*n,n) = T(2*n,n+1) = A005430(n). (End)
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k) - 2*T(n-2,k-1) - T(n-2,k-2), T(1,1) = 1 and, for n > 1, T(n,k) = 0 if k <= 1 or if k > n. - Philippe Deléham, Mar 17 2012
T(n,k) = Sum_{i=1..k} i*binomial(k,i)*binomial(n+1-k,k+1-i). - Mircea Merca, Apr 11 2012
If we include a main diagonal of zeros so that the array is in the form
0
1 0
2 2 0
3 6 3 0
4 12 12 4 0
...
then we obtain the exponential Riordan array [x*exp(x),x], which factors as [x,x]*[exp(x),x] = A132440*A007318. This array is the infinitesimal generator for A116071. A signed version of the array is the infinitesimal generator for A215652. - Peter Bala, Sep 14 2012
a(n,k) = (n-1)!/((n-k)!(k-1)!) if k > n/2 and a(n,k) = (n-1)!/((n-k-1)!k!) otherwise. [Forms 'core' for Pascal's recurrence; gives common term of RHS of T(n,k) = T(n-1,k-1) + T(n-1,k)]. - Jon Perry, Oct 08 2013
Assuming offset 0: T(n, k) = FallingFactorial(n + 1, n) / (k! * (n - k)!). The counterpart using the rising factorial is A356546. - Peter Luschny, Aug 13 2022
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EXAMPLE
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The triangle begins:
1;
1/2, 1/2;
1/3, 1/6, 1/3;
1/4, 1/12, 1/12, 1/4;
1/5, 1/20, 1/30, 1/20, 1/5;
...
The triangle of denominators begins:
1
2 2
3 6 3
4 12 12 4
5 20 30 20 5
6 30 60 60 30 6
7 42 105 140 105 42 7
8 56 168 280 280 168 56 8
9 72 252 504 630 504 252 72 9
10 90 360 840 1260 1260 840 360 90 10
11 110 495 1320 2310 2772 2310 1320 495 110 11
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MAPLE
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with(combstruct):for n from 0 to 11 do seq(m*count(Combination(n), size=m), m = 1 .. n) od; # Zerinvary Lajos, Apr 09 2008
A003506 := (n, k) -> k*binomial(n, k):
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MATHEMATICA
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L[n_, 1] := 1/n; L[n_, m_] := L[n, m] = L[n - 1, m - 1] - L[n, m - 1]; Take[ Flatten[ Table[ 1 / L[n, m], {n, 1, 12}, {m, 1, n}]], 66]
t[n_, m_] = Gamma[n]/(Gamma[n - m]*Gamma[m]); Table[Table[t[n, m], {m, 1, n - 1}], {n, 2, 12}]; Flatten[%] (* Roger L. Bagula and Gary W. Adamson, Sep 14 2008 *)
Table[k*Binomial[n, k], {n, 1, 7}, {k, 1, n}] (* Peter Luschny, May 27 2011 *)
t[n_, k_] := Denominator[n!*k!/(n+k+1)!]; Table[t[n-k, k] , {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 28 2013 *)
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PROG
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(PARI) A(i, j)=if(i<1||j<1, 0, 1/subst(intformal(x^(i-1)*(1-x)^(j-1)), x, 1))
(PARI) A(i, j)=if(i<1||j<1, 0, 1/sum(k=0, i-1, (-1)^k*binomial(i-1, k)/(j+k)))
(PARI) {T(n, k) = (n + 1 - k) * binomial( n, k - 1)} /* Michael Somos, Feb 06 2011 */
(Haskell)
a003506 n k = a003506_tabl !! (n-1) !! (n-1)
a003506_row n = a003506_tabl !! (n-1)
a003506_tabl = scanl1 (\xs ys ->
zipWith (+) (zipWith (+) ([0] ++ xs) (xs ++ [0])) ys) a007318_tabl
a003506_list = concat a003506_tabl
(SageMath)
T_row = lambda n: (n*(x+1)^(n-1)).list()
# Assuming offset 0:
return falling_factorial(n+1, n)//(factorial(k)*factorial(n-k))
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CROSSREFS
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Columns k >= 1 (offset 1): A000027, A002378, A027480, A033488, A174002, A253946(n+4), ..., with sum of reciprocals: infinity, 1, 1/2, 1/3, 1/4, 1/5, ..., respectively. - Wolfdieter Lang, Jul 20 2022
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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