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%I #63 Jun 22 2023 12:22:22
%S 1,1,1,1,2,2,1,3,6,5,1,4,12,20,14,1,5,20,50,70,42,1,6,30,100,210,252,
%T 132,1,7,42,175,490,882,924,429,1,8,56,280,980,2352,3696,3432,1430,1,
%U 9,72,420,1764,5292,11088,15444,12870,4862,1,10,90,600,2940,10584,27720
%N Triangle read by rows, T(n,k) = C(n,k)*C(2*k,k)/(k+1), n >= 0, 0 <= k <= n.
%C A Catalan scaled binomial matrix.
%C From _Philippe Deléham_, Sep 01 2005: (Start)
%C Table U(n,k), k >= 0, n >= 0, read by antidiagonals, begins:
%C row k = 0: 1, 1, 2, 5, 14, ... is A000108
%C row k = 1: 1, 2, 6, 20, 70, ... is A000984
%C row k = 2: 1, 3, 12, 50, 280, ... is A007854
%C row k = 3: 1, 4, 20, 104, 548, ... is A076035
%C row k = 4: 1, 5, 30, 185, 1150, ... is A076036
%C G.f. for row k: 1/(1-(k+1)*x*C(x)) where C(x) is the g.f. = for Catalan numbers A000108.
%C U(n,k) = Sum_{j=0..n} A106566(n,j)*(k+1)^j. (End)
%C This sequence gives the coefficients (increasing powers of x) of the Jensen polynomials for the Catalan sequence A000108 of degree n and shift 0. For the definition of Jensen polynomials for a sequence see a comment in A094436. - _Wolfdieter Lang_, Jun 25 2019
%H Indranil Ghosh, <a href="/A098474/b098474.txt">Rows 0..125, flattened</a>
%F G.f.: 2/(1-x+(1-x-4*x*y)^(1/2)). - _Vladeta Jovovic_, Sep 11 2004
%F E.g.f.: exp(x*(1+2*y))*(BesselI(0, 2*x*y)-BesselI(1, 2*x*y)). - _Vladeta Jovovic_, Sep 11 2004
%F G.f.: 1/(1-x-xy/(1-xy/(1-x-xy/(1-xy/(1-x-xy/(1-xy/(1-x-xy/(1-xy/(1-... (continued fraction). - _Paul Barry_, Feb 11 2009
%F Sum_{k=0..n} T(n,k)*x^(n-k) = A126930(n), A005043(n), A000108(n), A007317(n+1), A064613(n), A104455(n) for x = -2, -1, 0, 1, 2, 3 respectively. - _Philippe Deléham_, Dec 12 2009
%F T(n,k) = (-1)^k*Catalan(k)*Pochhammer(-n,k)/k!. - _Peter Luschny_, Feb 05 2015
%F O.g.f.: [1 - sqrt(1-4tx/(1-x))]/(2tx) = 1 + (1+t) x + (1+2t+2t^2) x^2 + (1+3t+6t^2+5t^3) x^3 + ... , generating the polynomials of this entry, reverse of A124644. See A011973 for a derivation and the inverse o.g.f., connected to the Fibonacci, Chebyshev, and Motzkin polynomials. See also A267633. - _Tom Copeland_, Jan 25 2016
%F From _Peter Bala_, Jun 13 2016: (Start)
%F The o.g.f. F(x,t) = ( 1 - sqrt(1 - 4*t*x/(1 - x)) )/(2*t*x) satisfies the partial differential equation d/dx(x*(1 - x)*F) - x*t*(1 + 4*t)*dF/dt - 2*x*t*F = 1. This gives a recurrence for the row polynomials: (n + 2)*R(n+1,t) = t*(1 + 4*t)*R'(n,t) + (2*t + n + 2)*R(n,t), where the prime ' indicates differentiation with respect to t.
%F Equivalently, setting Q(n,t) = t^(n+2)*R(n,-t)/(1 - 4*t)^(n + 3/2) we have t^2*d/dt(Q(n,t)) = (n + 2)*Q(n+1,t).
%F This leads to the following expansions:
%F Q(0,t) = (1/2)*Sum_{k >= 1} k*binomial(2*k,k)*t^(k+1)
%F Q(1,t) = (1/2)*Sum_{k >= 1} k*(k+1)/2!*binomial(2*k,k)*t^(k+2)
%F Q(2,t) = (1/2)*Sum_{k >= 1} k*(k+1)*(k+2)/3!*binomial(2*k,k) *t^(k+3) and so on. (End)
%F Sum_{k=0..n} T(n,k)*x^k = A007317(n+1), A162326(n+1), A337167(n) for x = 1, 2, 3 respectively. - _Sergii Voloshyn_, Mar 31 2022
%e Rows begin:
%e 1;
%e 1, 1;
%e 1, 2, 2;
%e 1, 3, 6, 5;
%e 1, 4, 12, 20, 14;
%e 1, 5, 20, 50, 70, 42;
%e 1, 6, 30, 100, 210, 252, 132;
%e ...
%e Row 3: t*(1 - 3*t + 6*t^2 - 5*t^3)/(1 - 4*t)^(9/2) = 1/2*Sum_{k >= 1} k*(k+1)*(k+2)*(k+3)/4!*binomial(2*k,k)*t^k. - _Peter Bala_, Jun 13 2016
%p p := proc(n) option remember; if n = 0 then 1 else normal((x*(1 + 4*x)*diff(p(n-1, x), x) + (2*x + n + 1)*p(n-1, x))/(n + 1)) fi end:
%p row := n -> local k; seq(coeff(p(n), x, k), k = 0..n):
%p for n from 0 to 6 do row(n) od; # _Peter Luschny_, Jun 21 2023
%t Table[Binomial[n, k] Binomial[2 k, k]/(k + 1), {n, 0, 10}, {k, 0, n}] // Flatten (* or *)
%t Table[(-1)^k*CatalanNumber[k] Pochhammer[-n, k]/k!, {n, 0, 10}, {k, 0, n}] // Flatten (* _Michael De Vlieger_, Feb 17 2017 *)
%o (Sage)
%o def A098474(n,k):
%o return (-1)^k*catalan_number(k)*rising_factorial(-n,k)/factorial(k)
%o for n in range(7): [A098474(n,k) for k in (0..n)] # _Peter Luschny_, Feb 05 2015
%o (Python)
%o from functools import cache
%o @cache
%o def A098474row(n: int) -> list[int]:
%o if n == 0: return [1]
%o a = A098474row(n - 1) + [0]
%o row = [0] * (n + 1)
%o row[0] = 1; row[1] = n
%o for k in range(2, n + 1):
%o row[k] = (a[k] * (n + k + 1) + a[k - 1] * (4 * k - 2)) // (n + 1)
%o return row # _Peter Luschny_, Jun 22 2023
%Y Row sums are A007317.
%Y Antidiagonal sums are A090344.
%Y Principal diagonal is A000108.
%Y Mirror image of A124644.
%Y Cf. A000984, A007854, A011973, A076035, A076036, A098443, A098473, A106566, A267633.
%K easy,nonn,tabl
%O 0,5
%A _Paul Barry_, Sep 09 2004
%E New name using a formula of Paul Barry by _Peter Luschny_, Feb 05 2015
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