A131980 A coefficient tree from the list partition transform relating A000129, A000142, A000165, A110327, and A110330.
1, 2, 6, 2, 24, 24, 120, 240, 24, 720, 2400, 720, 5040, 25200, 15120, 720, 40320, 282240, 282240, 40320, 362880, 3386880, 5080320, 1451520, 40320, 3628800, 43545600, 91445760, 43545600, 3628800, 39916800, 598752000, 1676505600, 1197504000, 199584000, 3628800
Offset: 0
Examples
Triangle begins as: 1; 2; 6, 2; 24, 24; 120, 240, 24; 720, 2400, 720; 5040, 25200, 15120, 720; 40320, 282240, 282240, 40320; 362880, 3386880, 5080320, 1451520, 40320; 3628800, 43545600, 91445760, 43545600, 3628800;
Links
- M. V. Koutras, Eulerian numbers associated with sequences of polynomials, The Fibonacci Quarterly, 32 (1994), 44-57.
Crossrefs
Cf. A228955.
Programs
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GAP
Flat(List([0..10], n-> List([0..Int(n/2)], k-> Factorial(n)*Binomial(n+1, 2*k+1) ))); # G. C. Greubel, Dec 30 2019
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Magma
[Factorial(n)*Binomial(n+1, 2*k+1): k in [0..Floor(n/2)], n in [0..10]]; // G. C. Greubel, Dec 30 2019
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Maple
for n from 0 to 10 do seq( n!*binomial(n+1,2*k+1), k = 0..floor(n/2) ) end do; # Peter Bala, Sep 09 2013
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Mathematica
Table[n!*Binomial[n+1, 2*k+1], {n,0,10}, {k,0,Floor[n/2]}]//Flatten (* G. C. Greubel, Dec 30 2019 *)
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PARI
T(n,k) = n!*binomial(n+1, 2*k+1); for(n=0,10, for(k=0, n\2, print1(T(n,k), ", "))) \\ G. C. Greubel, Dec 30 2019
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Sage
[[factorial(n)*binomial(n+1, 2*k+1) for k in (0..floor(n/2))] for n in (0..10)] # G. C. Greubel, Dec 30 2019
Formula
E.g.f. for the polynomials b(.,t), introduced above, is 1 - 2x - (t-1) * x^2; therefore e.g.f. for the polynomials a(.,t), which are the row polynomials of this array, is 1 / ( 1 - 2x - (t-1) * x^2 ) = (t-1) / ( t - ( 1 + x*(t-1) )^2 ).
Also, a(n,t) = (1 - t*u^2)^(n+1) (D_u)^n [ 1 / (1 - t*u^2) ] with eval. at u = 1/t. Compare A076743.
a(n,t) = n!*Sum_{k>=0} binomial(n+1,2k+1) * t^k = n!*Sum_{k>=0} A034867(n,k) * t^k.
Additional relations are given by formulas in A133314.
From Peter Bala, Sep 09 2013: (Start)
Recurrence equation: T(n+1,k) = (n+2 +2*k)T(n,k) + (n +2 -2*k)T(n,k-1).
Let P(n,x) = x*(x + 1)*...*(x + n - 1) denote the n-th rising factorial.
T(n,k) = Sum_{j=0..k+1} (-1)^(k+1-j)*binomial(n+1,k+1-j)*P(n,2*j) for n >= 1.
The row polynomial a(n,t) satisfies t*a(n,t)/(1 - t)^(n+1) = Sum_{j>=1} P(n,2*j)*t^j. For example, for n = 3 we have t*(24 + 24*t)/(1 - t)^4 = 2*3*4*t + (4*5*6)*t^2 + (6*7*8)*t^3 + ..., while for n = 4 we have t*(120 + 240*t + 24*t^2)/(1 - t)^5 = (2*3*4*5)*t + (4*5*6*7)*t^2 + (6*7*8*9)*t^3 + .... (End)
Extensions
Removed erroneous and duplicate statements. - Tom Copeland, Dec 03 2013
Comments