cp's OEIS Frontend

A162312 Triangular array, inverse of 2*P - I, where P is Pascal's triangle and I is the identity matrix.

%I A162312 #14 Jan 05 2025 19:51:39
%S A162312 1,-2,1,6,-4,1,-26,18,-6,1,150,-104,36,-8,1,-1082,750,-260,60,-10,1,
%T A162312 9366,-6492,2250,-520,90,-12,1,-94586,65562,-22722,5250,-910,126,-14,
%U A162312 1,1091670,-756688,262248,-60592,10500,-1456,168,-16,1,-14174522,9825030
%N A162312 Triangular array, inverse of 2*P - I, where P is Pascal's triangle and I is the identity matrix.
%C A162312 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).
%C A162312 The array M(a) begins
%C A162312 /
%C A162312 | 1/(1-a)
%C A162312 | a/(1-a)^2............... 1/(1-a)
%C A162312 | (a+a^2)/(1-a)^3......... 2*a/(1-a)^2........ 1/(1-a)
%C A162312 | (a+4*a^2+a^3)/(1-a)^4... 3*(a+a^2)/(1-a)^3.. 3*a/(1-a)^2... 1/(1-a)
%C A162312 | ...
%C A162312 \
%C A162312 In the first column the numerator polynomials are the Eulerian polynomials A_n(a). See A008292.
%C A162312 The e.g.f. for this array is
%C A162312 (1)... exp(x*t)/(1-a*exp(t)) = 1/(1-a) + [a/(1-a)^2 + x/(1-a)]*t
%C A162312 + [(a+a^2)/(1-a)^3 + 2*a*x/(1-a)^2 + x^2/(1-a)]*t^2/2! + ....
%C A162312 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
%C A162312 (2)... P_m(x) = sum {k=0..m} binomial(m,k) * A_k(a) / (1-a)^(k+1) * x^(m-k).
%C A162312 As a Newton series we have
%C A162312 (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).
%C A162312 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).
%C A162312 A result equivalent to (3) is
%C A162312 (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,
%C A162312 which in turn leads to the infinite series expansion
%C A162312 (5)... P_m(x) = sum {k = 0..inf} a^k * (x + k)^m,
%C A162312 provided |a| < 1. See [Nelsen].
%C A162312 The polynomials P_m(x) satisfy the difference equation
%C A162312 (6)... P_m(x) - a*P_m(x + 1) = x^m (recall a <> 1),
%C A162312 which leads easily to the evaluation of the weighted sums of powers of integers
%C A162312 (7)... sum {k = 0..n-1} a^k * k^m = P_m(0) - a^n * P_m(n).
%C A162312 for m = 0,1,2,... and a <> 1.
%C A162312 More generally we have
%C A162312 (8)... sum {k = 0..n-1} a^k * (x + k)^m = P_m(x) - a^n * P_m(x + n).
%C A162312 for m = 0,1,2,... and a <> 1.
%C A162312 In the remaining case a = 1 the sums are evaluated in terms of the Bernoulli polynomials.
%C A162312 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
%C A162312 (9)... 2*sum {k = 1..n-1} (-1)^k * k^m = E_m(0) - (-1)^n * E_m(n).
%D A162312 S. Roman, The Umbral Calculus, Dover Publications.
%H A162312 G. F. C. De Bruyn, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/33-2/debruyn.pdf">Formulas for a + a^2*2^p + a^3*3^p + ... + a^n*n^p</a>, Fibonacci Quart. 33 (1995), no. 2, 98-103.
%H A162312 R. B. Nelsen, <a href="https://www.jstor.org/stable/2323104">Problem E3062</a>, Amer. Math. Monthly, Vol. 94, No. 4 (Apr., 1987), 376-377.
%F A162312 TABLE ENTRIES
%F A162312 (1)... T(n,k) = (-1)^(n+k) * binomial(n,k) * A000629(n-k).
%F A162312 (2)... T(n,k) = (-1)^(n+k) * binomial(n,k) * sum {j = 0..n} j! * Stirling2(n-k+1,j+1).
%F A162312 GENERATING FUNCTION
%F A162312 (3)... exp(x*t)/(2*exp(t)-1) = 1 + (-2+x)*t + (6-4*x+x^2)*t^2/2!
%F A162312 + ....
%F A162312 PROPERTIES OF ROW POLYNOMIALS
%F A162312 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.
%F A162312 They may be recursively computed by means of
%F A162312 (4)... R_n(x) = x^n - 2*sum {k = 0..n-1} binomial(n,k) * R_k(x).
%F A162312 Explicit formulas are
%F A162312 (5)... R_n(x) = sum {j = 0..n} sum {k = j..n} (-2)^j * k! * Stirling2(n,k) * binomial(x,k-j),
%F A162312 (6)... R_n(x) = (-1)^n * sum {j = 0..n} sum {k = j..n} k! * Stirling2(n,k) * binomial(-x+1,k-j),
%F A162312 and
%F A162312 (7)... R_n(x) = sum {j = 0..n} sum {k = 0..j} 2^j * (-1)^k * comb(j,k) * (x + k)^n.
%F A162312 Other expansions include
%F A162312 (8)... R_n(x) = sum {k = 0..n} binomial(n,k) * (-1)^k * A000670(k) * (x-1)^(n-k),
%F A162312 (9)... R_n(x) = sum {k = 0..n} binomial(n,k) * (-1/2)^k * A080253(k) * (x-1/2)^(n-k)
%F A162312 and
%F A162312 (10)... R_n(x) = sum {k = 0..n} binomial(n,k) * (-1)^k * A007047(k) * (x+1)^(n-k).
%F A162312 SUMS OF POWERS OF INTEGERS
%F A162312 The row polynomials satisfy the difference equation
%F A162312 (11)... 2*R_n(x+1) - R_n(x) = x^n,
%F A162312 and so may be used to evaluate the weighted sum of powers of integers
%F A162312 (12)... sum {k = 0..n-1} 2^k * k^m = 2^n*R_m(n) - R_m(0).
%F A162312 For example, m = 3 gives
%F A162312 (13)... sum {k = 0..n-1} 2^k * k^3 = 2^n*(n^3-6*n^2+18*n-26) + 26.
%F A162312 More generally we have
%F A162312 (14)... sum {k = 0..n-1} 2^k * (x + k)^m = 2^n * R_m(x + n) - R_m(x).
%F A162312 RELATIONS WITH OTHER SEQUENCES
%F A162312 (15)... Row sums [1,-1,3,-13,75,...] = (-1)^n*A000670(n).
%F A162312 (16)... Alt. row sums [1,-3,11,-51,299,...] = (-1)^n * A007047(n).
%F A162312 (17)... Column 0: (-1)^n * A000629(n).
%F A162312 (18)... (-2)^n * R_n(1/2) = A080253(n).
%F A162312 (19)... R_n(1-x) = (-1)^n * P_n(x),
%F A162312 where P_n(x) are the row generating polynomials of A154921.
%F A162312 This provides the connection between (12) and the result
%F A162312 (20)... sum {k = 0..n-1} (1/2)^k * k^m = 2*P_m(0) - (1/2)^(n-1) * P_m(n).
%e A162312 Triangle begins
%e A162312 ====================================================
%e A162312 n\k|.....0......1......2......3......4......5......6
%e A162312 ====================================================
%e A162312 0..|.....1
%e A162312 1..|....-2......1
%e A162312 2..|.....6.....-4......1
%e A162312 3..|...-26.....18.....-6......1
%e A162312 4..|...150...-104.....36.....-8......1
%e A162312 5..|.-1082....750...-260.....60....-10......1
%e A162312 6..|..9366..-6492...2250...-520.....90....-12......1
%e A162312 ...
%p A162312 #A162312
%p A162312 with(combinat):
%p A162312 T := (n,k) -> (-1)^(n+k)*binomial(n,k)
%p A162312 *add( j!*stirling2(n-k+1,j+1),j = 0..n):
%p A162312 for n from 0 to 9 do
%p A162312 seq(T(n,k), k = 0..n);
%p A162312 end do;
%t A162312 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 *)
%o A162312 (PARI) matrix(10, 10, n, k, 2*binomial(n-1,k-1) - (n==k))^(-1) \\ _Michel Marcus_, Jul 12 2018
%Y A162312 Cf. A000629, A000670, A007047, A008292, A080253, A154921.
%K A162312 easy,sign,tabl
%O A162312 0,2
%A A162312 _Peter Bala_, Jul 01 2009
%E A162312 Typo corrected by _Peter Bala_, Nov 05 2010