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A162312
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Triangular array, inverse of 2*P - I, where P is Pascal's triangle and I is the identity matrix.
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4
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1, -2, 1, 6, -4, 1, -26, 18, -6, 1, 150, -104, 36, -8, 1, -1082, 750, -260, 60, -10, 1, 9366, -6492, 2250, -520, 90, -12, 1, -94586, 65562, -22722, 5250, -910, 126, -14, 1, 1091670, -756688, 262248, -60592, 10500, -1456, 168, -16, 1, -14174522, 9825030
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OFFSET
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0,2
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COMMENTS
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We make a few remarks about the general array M(a) := (I - a*P)^-1, where a <> 1, and its connection with weighted sums of powers of positive integers. The present case corresponds to -M(2).
The array M(a) begins
/
| 1/(1-a)
| a/(1-a)^2............... 1/(1-a)
| (a+a^2)/(1-a)^3......... 2*a/(1-a)^2........ 1/(1-a)
| (a+4*a^2+a^3)/(1-a)^4... 3*(a+a^2)/(1-a)^3.. 3*a/(1-a)^2... 1/(1-a)
| ...
\
In the first column the numerator polynomials are the Eulerian polynomials A_n(a). See A008292.
The e.g.f. for this array is
(1)... exp(x*t)/(1-a*exp(t)) = 1/(1-a) + [a/(1-a)^2 + x/(1-a)]*t
+ [(a+a^2)/(1-a)^3 + 2*a*x/(1-a)^2 + x^2/(1-a)]*t^2/2! + ....
The row generating polynomials P_m(x) of the array M(a), which, of course, depend on a, have properties similar to those of the Bernoulli polynomials. They form an Appell sequence and may be expressed in terms of the Eulerian polynomials as
(2)... P_m(x) = sum {k=0..m} binomial(m,k) * A_k(a) / (1-a)^(k+1) * x^(m-k).
As a Newton series we have
(3)... P_m(x) = 1/(1-a)*sum {j = 0..m} sum {k = j..m}(a/(1-a))^j * k! * Stirling2(m,k) * binomial(x,k-j).
The proof of this result in the particular case a = -1 given in [Roman, p. 100] can be easily generalized to a proof of (3).
A result equivalent to (3) is
(4)... P_m(x) = 1/(1-a)*sum {j = 0..m} sum {k = 0..j} (a/(1-a))^j * (-1)^(j-k) * comb(j,k) * (x + k)^m,
which in turn leads to the infinite series expansion
(5)... P_m(x) = sum {k = 0..inf} a^k * (x + k)^m,
provided |a| < 1. See [Nelsen].
The polynomials P_m(x) satisfy the difference equation
(6)... P_m(x) - a*P_m(x + 1) = x^m (recall a <> 1),
which leads easily to the evaluation of the weighted sums of powers of integers
(7)... sum {k = 0..n-1} a^k * k^m = P_m(0) - a^n * P_m(n).
for m = 0,1,2,... and a <> 1.
More generally we have
(8)... sum {k = 0..n-1} a^k * (x + k)^m = P_m(x) - a^n * P_m(x + n).
for m = 0,1,2,... and a <> 1.
In the remaining case a = 1 the sums are evaluated in terms of the Bernoulli polynomials.
The most well-studied case is when a = -1. The row polynomials of the array M(-1) are then one half of the Euler polynomials E_m(x), which may be used to evaluate the alternating sums of powers of integers
(9)... 2*sum {k = 1..n-1} (-1)^k * k^m = E_m(0) - (-1)^n * E_m(n).
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REFERENCES
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S. Roman, The Umbral Calculus, Dover Publications.
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LINKS
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R. B. Nelsen, Problem E3062, Amer. Math. Monthly, Vol. 94, No. 4 (Apr., 1987), 376-377.
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FORMULA
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TABLE ENTRIES
(1)... T(n,k) = (-1)^(n+k) * binomial(n,k) * A000629(n-k).
(2)... T(n,k) = (-1)^(n+k) * binomial(n,k) * sum {j = 0..n} j! * Stirling2(n-k+1,j+1).
GENERATING FUNCTION
(3)... exp(x*t)/(2*exp(t)-1) = 1 + (-2+x)*t + (6-4*x+x^2)*t^2/2!
+ ....
PROPERTIES OF ROW POLYNOMIALS
The row generating polynomials R_n(x) form an Appell sequence. The first few values are R_0(x) = 1, R_1(x) = x-2, R_2(x) = x^2-4*x+6 and R_3(x) = x^3-6*x^2+18*x-26.
They may be recursively computed by means of
(4)... R_n(x) = x^n - 2*sum {k = 0..n-1} binomial(n,k) * R_k(x).
Explicit formulas are
(5)... R_n(x) = sum {j = 0..n} sum {k = j..n} (-2)^j * k! * Stirling2(n,k) * binomial(x,k-j),
(6)... R_n(x) = (-1)^n * sum {j = 0..n} sum {k = j..n} k! * Stirling2(n,k) * binomial(-x+1,k-j),
and
(7)... R_n(x) = sum {j = 0..n} sum {k = 0..j} 2^j * (-1)^k * comb(j,k) * (x + k)^n.
Other expansions include
(8)... R_n(x) = sum {k = 0..n} binomial(n,k) * (-1)^k * A000670(k) * (x-1)^(n-k),
(9)... R_n(x) = sum {k = 0..n} binomial(n,k) * (-1/2)^k * A080253(k) * (x-1/2)^(n-k)
and
(10)... R_n(x) = sum {k = 0..n} binomial(n,k) * (-1)^k * A007047(k) * (x+1)^(n-k).
SUMS OF POWERS OF INTEGERS
The row polynomials satisfy the difference equation
(11)... 2*R_n(x+1) - R_n(x) = x^n,
and so may be used to evaluate the weighted sum of powers of integers
(12)... sum {k = 0..n-1} 2^k * k^m = 2^n*R_m(n) - R_m(0).
For example, m = 3 gives
(13)... sum {k = 0..n-1} 2^k * k^3 = 2^n*(n^3-6*n^2+18*n-26) + 26.
More generally we have
(14)... sum {k = 0..n-1} 2^k * (x + k)^m = 2^n * R_m(x + n) - R_m(x).
RELATIONS WITH OTHER SEQUENCES
(15)... Row sums [1,-1,3,-13,75,...] = (-1)^n*A000670(n).
(16)... Alt. row sums [1,-3,11,-51,299,...] = (-1)^n * A007047(n).
(17)... Column 0: (-1)^n * A000629(n).
(18)... (-2)^n * R_n(1/2) = A080253(n).
(19)... R_n(1-x) = (-1)^n * P_n(x),
where P_n(x) are the row generating polynomials of A154921.
This provides the connection between (12) and the result
(20)... sum {k = 0..n-1} (1/2)^k * k^m = 2*P_m(0) - (1/2)^(n-1) * P_m(n).
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EXAMPLE
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Triangle begins
====================================================
n\k|.....0......1......2......3......4......5......6
====================================================
0..|.....1
1..|....-2......1
2..|.....6.....-4......1
3..|...-26.....18.....-6......1
4..|...150...-104.....36.....-8......1
5..|.-1082....750...-260.....60....-10......1
6..|..9366..-6492...2250...-520.....90....-12......1
...
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MAPLE
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with(combinat):
T := (n, k) -> (-1)^(n+k)*binomial(n, k)
*add( j!*stirling2(n-k+1, j+1), j = 0..n):
for n from 0 to 9 do
seq(T(n, k), k = 0..n);
end do;
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MATHEMATICA
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Table[(-1)^(n+k) Binomial[n, k] PolyLog[k-n, 1/2], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 14 2019 *)
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PROG
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(PARI) matrix(10, 10, n, k, 2*binomial(n-1, k-1) - (n==k))^(-1) \\ Michel Marcus, Jul 12 2018
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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