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%I #69 Jun 13 2022 02:39:03
%S 1,1,4,1,1,20,48,20,1,1,72,603,1168,603,72,1,1,232,5158,27664,47290,
%T 27664,5158,232,1,1,716,37257,450048,1822014,2864328,1822014,450048,
%U 37257,716,1,1,2172,247236,6030140,49258935,163809288,242384856,163809288,49258935,6030140,247236,2172,1
%N Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(2*n+1,i) * binomial(k+2-i,2)^n, 0 <= k <= 2*(n-1).
%C From _Yahia Kahloune_, Jan 30 2014: (Start)
%C In general, let b(k,e,p) = Sum_{i=0..k} (-1)^i*binomial(e*p+1,i)*binomial(k+e-i,e)^p. Then T(n,k) = b(k,2,n).
%C With these coefficients we can calculate: Sum_{i=1..n} binomial(i+e-1,e)^p = Sum_{k=0..e*(p-1)} b(k,e,p)*binomial(n+e+k,e*p+k).
%C For example, A085438(n) = Sum_{i=1..n} binomial(1+i,2)^3 = T(3,0)*binomial(2+n,7) + T(3,1)*binomial(3+n,7) + T(3,2)*binomial(4+n,7) + T(3,3)*binomial(5+n,7) + T(3,4)*binomial(6+n,7) = (1/5040)*(90*n^7 + 630*n^6 + 1638*n^5 + 1890*n^4 + 840*n^3 - 48*n).
%C (End)
%C T(n,k) is the number of permutations of 2 indistinguishable copies of 1..n with exactly k descents. A descent is a pair of adjacent elements with the second element less than the first. - _Andrew Howroyd_, May 06 2020
%H Andrew Howroyd, <a href="/A154283/b154283.txt">Table of n, a(n) for n = 1..1600</a> (rows 1..40)
%H H. Prodinger, <a href="https://arxiv.org/abs/1102.5186">On Touchard's continued fraction and extensions: combinatorics-free, self-contained proofs </a>, arXiv:1102.5186 [math.CO], 2011.
%F T(n,k) = (-1) times coefficient of x^k in (x-1)^(2*n+1) * Sum_{k>=0} (k*(k+1)/2)^n *x^(k-1).
%F From _Yahia Kahloune_, Jan 29 2014: (Start)
%F Sum_{i=1..n} binomial(1+i,2)^p = Sum_{k=0..2*p-2} T(p,k)*binomial(n+2+k,2*p+1).
%F binomial(n,2)^p = Sum_{k=0..2*p-2} T(p,k)*binomial(n+k,2*p). (End)
%F From _Peter Bala_, Dec 21 2019; (Start)
%F E.g.f. as a continued fraction: (1-x)/(1-x + ( 1-exp((1-x)^2*t))*x/(1-x + (1-exp(2*(1-x)^2*t))*x/(1-x + (1-exp(3*(1-x)^2*t))*x/(1-x + ... )))) = 1 + x*t + x*(x^2 + 4*x + 1)*t^2/2! + x*(x^4 + 20*x^3 + 48*x^2 + 20*x + 1)*t^3/3! + ... (use Prodinger equation 1.1).
%F The sequence of alternating row sums (unsigned) [1, 1, 2, 10, 104, 1816,...] appears to be A005799. (End)
%e Triangle begins:
%e 1;
%e 1, 4, 1;
%e 1, 20, 48, 20, 1;
%e 1, 72, 603, 1168, 603, 72, 1;
%e 1, 232, 5158, 27664, 47290, 27664, 5158, 232, 1;
%e 1, 716, 37257, 450048, 1822014, 2864328, 1822014, ...;
%e 1, 2172, 247236, 6030140, 49258935, 163809288, 242384856, ...;
%e 1, 6544, 1568215, 72338144, 1086859301, 6727188848, 19323413187, ...;
%e 1, 19664, 9703890, 811888600, 21147576440, 225167210712, ... ;
%e ...
%e The T(2,1) = 4 permutations of 1122 with 1 descent are 1212, 1221, 2112, 2211. - _Andrew Howroyd_, May 15 2020
%p A154283 := proc(n,k)
%p (1-x)^(2*n+1)*add( (l*(l+1)/2)^n*x^(l-1),l=0..k+1) ;
%p coeftayl(%,x=0,k) ;
%p end proc: # _R. J. Mathar_, Feb 01 2013
%t p[x_, n_]= (1-x)^(2*n+1)*Sum[(k*(k+1)/2)^n*x^k, {k, 0, Infinity}]/x;
%t Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n,10}]//Flatten
%o (PARI) T(n,k)={sum(i=0, k, (-1)^i*binomial(2*n+1, i)*binomial(k+2-i, 2)^n)} \\ _Andrew Howroyd_, May 09 2020
%o (Magma) [(&+[(-1)^j*Binomial(2*n+1,j)*Binomial(k-j+2,2)^n: j in [0..k]]): k in [0..2*n-2], n in [1..12]]; // _G. C. Greubel_, Jun 13 2022
%o (SageMath)
%o def A154283(n,k): return sum((-1)^j*binomial(2*n+1, j)*binomial(k-j+2, 2)^n for j in (0..k))
%o flatten([[A154283(n,k) for k in (0..2*n-2)] for n in (1..12)]) # _G. C. Greubel_, Jun 13 2022
%Y Columns k=0..9 are A000012, A061981, A151624, A151625, A151626, A151627, A151628, A151629, A151630, A151631.
%Y Row sums are A000680.
%Y Similar triangles for e=1..6: A173018 (or A008292), this sequence, A174266, A236463, A237202, A237252.
%Y Cf. A000680, A005799, A085438, A334781.
%K nonn,easy,tabf
%O 1,3
%A _Roger L. Bagula_, Jan 06 2009
%E Edited by _N. J. A. Sloane_, Jan 30 2014 following suggestions from _Yahia Kahloune_ (among other things, the signs of all terms have been reversed).
%E Edited by _Andrew Howroyd_, May 09 2020
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