A162313 Triangular array P*(2*I - P^2)^-1, where P is Pascal's triangle A007318 and I is the identity matrix.

1, 3, 1, 17, 6, 1, 147, 51, 9, 1, 1697, 588, 102, 12, 1, 24483, 8485, 1470, 170, 15, 1, 423857, 146898, 25455, 2940, 255, 18, 1, 8560947, 2966999, 514143, 59395, 5145, 357, 21, 1, 197613377, 68487576, 11867996, 1371048, 118790, 8232, 476, 24, 1

Offset: 0

Author

Peter Bala, Jul 01 2009

Keywords

• easy
• nonn
• tabl

Comments

Unsigned inverse of A162315.

The row generating polynomials of this triangle may be used to evaluate the weighted sums of powers of odd numbers

(1)... 1^m + 2*3^m + 4*5^m + ... + 2^n*(2*n+1)^m

and also the sums

(2)... 1^m + (1/2)*3^m + (1/4)*5^m + ... + (1/2)^n*(2*n+1)^m.

See the Formula section below.

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.

The array M(a) begins

/

| a/(1-a)

| (a^2+a)/(1-a)^2 ................. a/(1-a)

| (a^3+6*a^2+a)/(1-a)^3 ........... 2*(a^2+a)/(1-a)^2 ... a/(1-a)

(a^4+23*a^3+23*a^2+a)/(1-a)^4 ...

| .

\ .

In the first column we recognize the numerator polynomials as the Eulerian polynomials of type B. See A060187.

The e.g.f. for this array is

(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! + ....

The row polynomials P_m(x), which depend on a, satisfy the difference equation

(4)... P_m(x) - a*P_m(x+2) = a*(x+1)^m.

for m >= 1.

The first few values are

P_0(x) = a/(1-a), P_1(x) = a*x/(1-a) + (a^2+a)/(1-a)^2 and

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.

Using (4) leads to the evaluations of the weighted sums of powers of even and odd positive integers:

(5)... Sum_{k = 1..n} a^k*(2*k)^m = P_m(1) - a^n*P_m(2*n+1)

and

(6)... Sum_{k = 1..n} a^k*(2*k-1)^m = P_m(0) - a^n*P_m(2*n),

with m = 0,1,2,... and a <> 1.

In the remaining case a = 1 the sums are evaluated in terms of the Bernoulli polynomials.

Examples

			Triangle begins
  n\k|.......0.......1......2......3......4......5......6
  =======================================================
  0..|.......1
  1..|.......3.......1
  2..|......17.......6......1
  3..|.....147......51......9......1
  4..|....1697.....588....102.....12......1
  5..|...24483....8485...1470....170.....15......1
  6..|..423857..146898..25455...2940....255.....18......1
  ...
		

Crossrefs

A000629, A007318, A060187, A080253 (column 0), A154921, A162312, A162314 (row sums), A162315 (unsigned inverse).

Programs

Mathematica

m = 8;
P = Table[Binomial[n, k], {n, 0, m}, {k, 0, m}];
T = P.Inverse[2 IdentityMatrix[m+1] - P.P];
Table[T[[n+1, k+1]], {n, 0, m}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 02 2019 *)

Formula

TABLE ENTRIES

(1)... T(n,k) = binomial(n,k)*A080253(n-k).

GENERATING FUNCTION

(2)... exp((x+1)*t)/(2-exp(2*t)) = 1 + (3+x)*t + (17+6*x+x^2)*t^2/2!

+ ....

The e.g.f. can also be written as

(3)... exp(x*t)*G(t), where G(t) = exp(t)/(2-exp(2*t)) is the e.g.f. for A080253.

ROW GENERATING POLYNOMIALS

The row generating polynomials R_n(x) form an Appell sequence. The first few values are R_0(x) = 1, R_1(x) = 3 + x, R_2(x) = 17 + 6*x + x^2 and R_3(x) = 147 + 51*x + 9*x^2 + x^3.

The row polynomials may be recursively computed by means of

(4)... R_n(x) = (x+1)^n + Sum_{k=0..n-1} 2^(n-k)*binomial(n,k)*R_k(x).

An explicit formula is

(5)... R_n(x) = Sum_{j = 0..n} Sum_{k = 0..j} (-1)^(j-k)*binomial(j,k)*(x+2*k+1)^n.

There is also a representation as an infinite series

(6)... R_n(x) = (1/2)*Sum_{k >= 0} (1/2)^k*(x+2*k+1)^n.

SUMS OF POWERS OF INTEGERS

The row polynomials satisfy the difference equation

(7)... 2*R_n(x) - R_n(x+2) = (x+1)^n,

and hence may be used to evaluate the weighted sums of powers of odd integers

(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)

as well as

(9)... Sum_{k=0..n-1} 2^k*(2*k+1)^m = (-1)^m*(2^n*R_m(-2*n)-R_m(0)).

For example, m = 2 gives

(10)... Sum_{k=0..n-1} (1/2)^k*(2*k+1)^2 = 34-2^(1-n)*(4*n^2+12*n+17)

and

(11)... Sum_{k = 0..n-1} 2^k*(2*k+1)^2 = 2^n*(4*n^2 - 12*n + 17)-17.

RELATIONS WITH OTHER SEQUENCES

(12)... Row sums = [1,4,24,208,2400,...] = 2^n*A000629(n) = A162314(n).

(13)... Column 0 = [1,3,17,147,1697,...] = A080253.

The identity

(14)... R_n(2*x-1) = 2^n*P_n(x),

where P_n(x) are the row generating polynomials of A154921, provides a surprising connection between (6) and the result

(15)... Sum_{k = 0..n-1} (1/2)^k*k^m = 2*P_m(0) - (1/2)^(n-1)*P_m(n).