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%I #9 Nov 02 2019 11:41:14
%S 1,3,1,17,6,1,147,51,9,1,1697,588,102,12,1,24483,8485,1470,170,15,1,
%T 423857,146898,25455,2940,255,18,1,8560947,2966999,514143,59395,5145,
%U 357,21,1,197613377,68487576,11867996,1371048,118790,8232,476,24,1
%N Triangular array P*(2*I - P^2)^-1, where P is Pascal's triangle A007318 and I is the identity matrix.
%C Unsigned inverse of A162315.
%C The row generating polynomials of this triangle may be used to evaluate the weighted sums of powers of odd numbers
%C (1)... 1^m + 2*3^m + 4*5^m + ... + 2^n*(2*n+1)^m
%C and also the sums
%C (2)... 1^m + (1/2)*3^m + (1/4)*5^m + ... + (1/2)^n*(2*n+1)^m.
%C See the Formula section below.
%C We make a few remarks about the general array M(a) := a*P*(I-a*P^2)^-1, where a <> 1, and its connection with weighted sums of powers of odd numbers. The present case corresponds to a = 1/2. Compare with the remarks in A162312.
%C The array M(a) begins
%C /
%C | a/(1-a)
%C | (a^2+a)/(1-a)^2 ................. a/(1-a)
%C | (a^3+6*a^2+a)/(1-a)^3 ........... 2*(a^2+a)/(1-a)^2 ... a/(1-a)
%C (a^4+23*a^3+23*a^2+a)/(1-a)^4 ...
%C | .
%C \ .
%C In the first column we recognize the numerator polynomials as the Eulerian polynomials of type B. See A060187.
%C The e.g.f. for this array is
%C (3)... a*exp((x+1)*t)/(1-a*exp(2*t)) = a/(1-a) +[(a^2+a)/(1-a)^2 + a/(1-a)*x]*t + [(a^3+6*a^2+a)/(1-a)^3 + 2*(a^2+a)*x/(1-a)^2 + a/(1-a)*x^2]*t^2/2! + ....
%C The row polynomials P_m(x), which depend on a, satisfy the difference equation
%C (4)... P_m(x) - a*P_m(x+2) = a*(x+1)^m.
%C for m >= 1.
%C The first few values are
%C P_0(x) = a/(1-a), P_1(x) = a*x/(1-a) + (a^2+a)/(1-a)^2 and
%C P_2(x) = a*x^2/(1-a) + 2*(a^2+a)*x/(1-a)^2 + (a^3+6*a^2+a)/(1-a)^3.
%C Using (4) leads to the evaluations of the weighted sums of powers of even and odd positive integers:
%C (5)... sum {k = 1..n} a^k*(2*k)^m = P_m(1) - a^n*P_m(2*n+1)
%C and
%C (6)... sum {k = 1..n} a^k*(2*k-1)^m = P_m(0) - a^n*P_m(2*n),
%C with m = 0,1,2,... and a <> 1.
%C In the remaining case a = 1 the sums are evaluated in terms of the Bernoulli polynomials.
%F TABLE ENTRIES
%F (1)... T(n,k) = binomial(n,k)*A080253(n-k).
%F GENERATING FUNCTION
%F (2)... exp((x+1)*t)/(2-exp(2*t)) = 1 + (3+x)*t + (17+6*x+x^2)*t^2/2!
%F + ....
%F The e.g.f. can also be written as
%F (3)... exp(x*t)*G(t), where G(t) = exp(t)/(2-exp(2*t)) is the e.g.f. for A080253.
%F ROW GENERATING POLYNOMIALS
%F The row generating polynomials R_n(x) form an Appell sequence. The first few values are
%F R_0(x) = 1, R_1(x) = 3 + x, R_2(x) = 17 + 6*x + x^2 and
%F R_3(x) = 147 + 51*x + 9*x^2 + x^3.
%F The row polynomials may be recursively computed by means of
%F (4)... R_n(x) = (x+1)^n + sum {k=0..n-1} 2^(n-k)*binomial(n,k)*R_k(x).
%F An explicit formula is
%F (5)... R_n(x) = sum {j = 0..n} sum {k = 0..j} (-1)^(j-k)*binomial(j,k)*(x+2*k+1)^n).
%F There is also a representation as an infinite series
%F (6)... R_n(x) = 1/2*sum {k = 0..inf}(1/2)^k*(x+2*k+1)^n.
%F SUMS OF POWERS OF INTEGERS
%F The row polynomials satisfy the difference equation
%F (7)... 2*R_n(x) - R_n(x+2) = (x+1)^n,
%F and hence may be used to evaluate the weighted sums of powers of odd integers
%F (8)... sum {k=0..n-1} (1/2)^k*(2*k+1)^m = 2*R_m(0)-1/2^(n-1)*R_m(2*n)
%F as well as
%F (9)... sum {k=0..n-1} 2^k*(2*k+1)^m = (-1)^m*(2^n*R_m(-2*n)-R_m(0)).
%F For example, m = 2 gives
%F (10)... sum {k=0..n-1} (1/2)^k*(2*k+1)^2 = 34-2^(1-n)*(4*n^2+12*n+17)
%F and
%F (11)... sum {k = 0..n-1} 2^k*(2*k+1)^2 = 2^n*(4*n^2 - 12*n + 17)-17.
%F RELATIONS WITH OTHER SEQUENCES
%F (12)... Row sums = [1,4,24,208,2400,...] = 2^n*A000629(n) = A162314(n).
%F (13)... Column 0 = [1,3,17,147,1697,...] = A080253.
%F The identity
%F (14)... R_n(2*x-1) = 2^n*P_n(x),
%F where P_n(x) are the row generating polynomials of A154921, provides a surprising connection between (6) and the result
%F (15)... sum {k = 0..n-1} (1/2)^k*k^m = 2*P_m(0) - (1/2)^(n-1)*P_m(n).
%e Triangle begins
%e n\k|.......0.......1......2......3......4......5......6
%e =======================================================
%e 0..|.......1
%e 1..|.......3.......1
%e 2..|......17.......6......1
%e 3..|.....147......51......9......1
%e 4..|....1697.....588....102.....12......1
%e 5..|...24483....8485...1470....170.....15......1
%e 6..|..423857..146898..25455...2940....255.....18......1
%e ...
%t m = 8;
%t P = Table[Binomial[n, k], {n, 0, m}, {k, 0, m}];
%t T = P.Inverse[2 IdentityMatrix[m+1] - P.P];
%t Table[T[[n+1, k+1]], {n, 0, m}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Nov 02 2019 *)
%Y A000629, A007318, A060187, A080253 (column 0), A154921, A162312, A162314 (row sums), A162315 (unsigned inverse).
%K easy,nonn,tabl
%O 0,2
%A _Peter Bala_, Jul 01 2009
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