|
%I #62 Jan 29 2022 12:19:24
%S 1,3,1,10,6,1,35,29,9,1,126,130,57,12,1,462,562,312,94,15,1,1716,2380,
%T 1578,608,140,18,1,6435,9949,7599,3525,1045,195,21,1,24310,41226,
%U 35401,19044,6835,1650,259,24,1,92378,169766,161052,97954,40963,12021,2450
%N A convolution triangle of numbers, generalizing Pascal's triangle A007318.
%C Replacing each '2' in the recurrence by '1' produces Pascal's triangle A007318(n-1,m-1). The columns appear as A001700, A008549, A045720, A045894, A035330, ...
%C Triangle T(n,k), 1 <= k <= n, given by (0, 3/1, 1/3, 5/3, 3/5, 7/5, 5/7, 9/7, 7/9, 11/9, 9/11, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Jan 28 2012
%C Riordan array (1, c(x)/sqrt(1-4x)) where c(x) = g.f. for Catalan numbers A000108, first column (k = 0) omitted. - _Philippe Deléham_, Jan 28 2012
%H Reinhard Zumkeller, <a href="/A035324/b035324.txt">Rows n = 1..120 of triangle, flattened</a>
%H Milan Janjić, <a href="https://www.emis.de/journals/JIS/VOL21/Janjic2/janjic103.html">Pascal Matrices and Restricted Words</a>, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
%H Wolfdieter Lang, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL3/LANG/lang.html">On generalizations of Stirling number triangles</a>, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
%H Wolfdieter Lang, <a href="/A035324/a035324.txt">First 10 rows.</a>
%F a(n+1, m) = 2*(2*n+m)*a(n, m)/(n+1) + m*a(n, m-1)/(n+1), n >= m >= 1; a(n, m) := 0, n<m; a(n, 0) := 0, a(1, 1)=1;
%F G.f. for column m: ((x*c(x)/sqrt(1-4*x))^m)/x, where c(x) = g.f. for Catalan numbers A000108.
%F a(n, m) =: s2(3; n, m).
%F With offset 0 (0 <= k <= n), T(n,k) = Sum_{j>=0} A039598(n,j)*binomial(j,k). - _Philippe Deléham_, Mar 30 2007
%F T(n+1,n) = 3*n = A008585(n).
%F T(n,k) = T(n-1,k-1) + 3*T(n-1,k) + Sum_{i>=0} T(n-1,k+1+i)*(-1)^i. - _Philippe Deléham_, Feb 23 2012
%F T(n,m) = Sum_{k=m..n} k*binomial(k-1,k-m)*2^(k-m)*binomial(2*n-k-1,n-k))/n. - _Vladimir Kruchinin_, Aug 07 2013
%e Triangle begins:
%e 1;
%e 3, 1;
%e 10, 6, 1;
%e 35, 29, 9, 1;
%e 126, 130, 57, 12, 1;
%e 462, 562, 312, 94, 15, 1;
%e Triangle (0, 3, 1/3, 5/3, 3/5, ...) DELTA (1,0,0,0,0,0, ...) has an additional first column (1,0,0,...).
%t a[n_, m_] /; n >= m >= 1 := a[n, m] = 2*(2*(n-1) + m)*(a[n-1, m]/n) + m*(a[n-1, m-1]/n); a[n_, m_] /; n < m = 0; a[n_, 0] = 0; a[1, 1] = 1; Flatten[ Table[ a[n, m], {n, 1, 10}, {m, 1, n}]] (* _Jean-François Alcover_, Feb 21 2012, from first formula *)
%o (Haskell)
%o a035324 n k = a035324_tabl !! (n-1) !! (k-1)
%o a035324_row n = a035324_tabl !! (n-1)
%o a035324_tabl = map snd $ iterate f (1, [1]) where
%o f (i, xs) = (i + 1, map (`div` (i + 1)) $
%o zipWith (+) ((map (* 2) $ zipWith (*) [2 * i + 1 ..] xs) ++ [0])
%o ([0] ++ zipWith (*) [2 ..] xs))
%o -- _Reinhard Zumkeller_, Jun 30 2013
%o (Sage)
%o @cached_function
%o def T(n, k):
%o if n == 0: return n^k
%o return sum(binomial(2*i-1, i)*T(n-1, k-i) for i in (1..k-n+1))
%o A035324 = lambda n,k: T(k, n)
%o for n in (1..8): print([A035324(n, k) for k in (1..n)]) # _Peter Luschny_, Aug 16 2016
%Y Cf. A000108, A007318, A039598.
%Y Row sums: A049027(n), n >= 1.
%Y Alternating row sums give A000108 (Catalan numbers).
%Y If offset 0 (n >= m >= 0): convolution triangle based on A001700 (central binomial coeffs. of odd order).
%K easy,nice,nonn,tabl
%O 1,2
%A _Wolfdieter Lang_
|
