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%I A162312
%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 9366,6492,2250,520,90,12,1,94586,65562,22722,5250,910,126,14,1,
%U A162312 1091670,756688,262248,60592,10500,1456,168,16,1,14174522,9825030
%V A162312 1,-2,1,6,-4,1,-26,18,-6,1,150,-104,36,-8,1,-1082,750,-260,60,-10,1,
%W A162312 9366,-6492,2250,-520,90,-12,1,-94586,65562,-22722,5250,-910,126,-14,1,
%X A162312 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,
%C A162312 where a <> 1, and its connection with weighted sums of powers of
%C A162312 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
%C A162312 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,
%C A162312 of course, depend on a, have properties similar to those of the
%C A162312 Bernoulli polynomials. They form an Appell sequence and may be
%C A162312 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!
%C A162312 *Stirling2(m,k)*binomial(x,k-j).
%C A162312 The proof of this result in the particular case a = -1 given in
%C A162312 [Roman, p.100] can be easily generalised 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
%C A162312 *(-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
%C A162312 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
%C A162312 Bernoulli polynomials.
%C A162312 The most well-studied case is when a = -1. The row polynomials of the
%C A162312 array M(-1) are then one half of the Euler polynomials E_m(x), which
%C A162312 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 G. F. C. De Bruyn, Formulas for a + a^2*2^p + a^3*3^p + ... + a^n*n^p. 
               Fibonacci Quart. 33 (1995), no. 2, 98-103.
%D A162312 R. B. Nelsen, Problem E3062: Amer. Math. Monthly, Vol. 94, No. 4 (Apr., 
               1987), 376-377.
%D A162312 S. Roman, The Umbral Calculus, Dover Publications.
%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!
%F A162312 *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
%F A162312 first few values are R_0(x) = 1, R_1(x) = x-2, R_3(x) = x^2-4*x+6
%F A162312 and R_4(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)
%F A162312 * binomial(x,k-j),
%F A162312 (6)... R_n(x) = (-1)^n*sum {j = 0..n} sum {k = j..n} k!*Stirling2(n,k)
%F A162312 *binomial(-x+1,k-j),
%F A162312 as well as
%F A162312 (7)... R_n(x) = sum {j = 0..n} sum {k = 0..j} 2^j*(-1)^k*comb(j,k)
%F A162312 *(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)
%F A162312 *(x-1)^(n-k),
%F A162312 (9)... R_n(x) = sum {k = 0..n} binomial(n,k)*(-1/2)^k*A080253(k)
%F A162312 *(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)
%F A162312 *(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;
%Y A162312 A000629, A000670, A007047, A008292, A080253, A154921.
%Y A162312 Sequence in context: A132159 A112356 A135885 this_sequence A141715 A098697 
               A021466
%Y A162312 Adjacent sequences: A162309 A162310 A162311 this_sequence A162313 A162314 
               A162315
%K A162312 easy,sign,tabl
%O A162312 0,2
%A A162312 Peter Bala (pbala(AT)talktalk.net), Jul 01 2009

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