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A154283 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). 18
1, 1, 4, 1, 1, 20, 48, 20, 1, 1, 72, 603, 1168, 603, 72, 1, 1, 232, 5158, 27664, 47290, 27664, 5158, 232, 1, 1, 716, 37257, 450048, 1822014, 2864328, 1822014, 450048, 37257, 716, 1, 1, 2172, 247236, 6030140, 49258935, 163809288, 242384856, 163809288, 49258935, 6030140, 247236, 2172, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
From Yahia Kahloune, Jan 30 2014: (Start)
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).
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).
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).
(End)
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
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..1600 (rows 1..40)
FORMULA
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).
From Yahia Kahloune, Jan 29 2014: (Start)
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).
binomial(n,2)^p = Sum_{k=0..2*p-2} T(p,k)*binomial(n+k,2*p). (End)
From Peter Bala, Dec 21 2019; (Start)
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).
The sequence of alternating row sums (unsigned) [1, 1, 2, 10, 104, 1816,...] appears to be A005799. (End)
EXAMPLE
Triangle begins:
1;
1, 4, 1;
1, 20, 48, 20, 1;
1, 72, 603, 1168, 603, 72, 1;
1, 232, 5158, 27664, 47290, 27664, 5158, 232, 1;
1, 716, 37257, 450048, 1822014, 2864328, 1822014, ...;
1, 2172, 247236, 6030140, 49258935, 163809288, 242384856, ...;
1, 6544, 1568215, 72338144, 1086859301, 6727188848, 19323413187, ...;
1, 19664, 9703890, 811888600, 21147576440, 225167210712, ... ;
...
The T(2,1) = 4 permutations of 1122 with 1 descent are 1212, 1221, 2112, 2211. - Andrew Howroyd, May 15 2020
MAPLE
A154283 := proc(n, k)
(1-x)^(2*n+1)*add( (l*(l+1)/2)^n*x^(l-1), l=0..k+1) ;
coeftayl(%, x=0, k) ;
end proc: # R. J. Mathar, Feb 01 2013
MATHEMATICA
p[x_, n_]= (1-x)^(2*n+1)*Sum[(k*(k+1)/2)^n*x^k, {k, 0, Infinity}]/x;
Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 10}]//Flatten
PROG
(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
(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
(SageMath)
def A154283(n, k): return sum((-1)^j*binomial(2*n+1, j)*binomial(k-j+2, 2)^n for j in (0..k))
flatten([[A154283(n, k) for k in (0..2*n-2)] for n in (1..12)]) # G. C. Greubel, Jun 13 2022
CROSSREFS
Row sums are A000680.
Similar triangles for e=1..6: A173018 (or A008292), this sequence, A174266, A236463, A237202, A237252.
Sequence in context: A176422 A156586 A181544 * A185946 A015113 A016519
KEYWORD
nonn,easy,tabf
AUTHOR
Roger L. Bagula, Jan 06 2009
EXTENSIONS
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).
Edited by Andrew Howroyd, May 09 2020
STATUS
approved

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Last modified July 17 09:00 EDT 2024. Contains 374363 sequences. (Running on oeis4.)