A130330 Triangle read by rows, the matrix product A130321 * A000012, both taken as infinite lower triangular matrices.

1, 3, 1, 7, 3, 1, 15, 7, 3, 1, 31, 15, 7, 3, 1, 63, 31, 15, 7, 3, 1, 127, 63, 31, 15, 7, 3, 1, 255, 127, 63, 31, 15, 7, 3, 1, 511, 255, 127, 63, 31, 15, 7, 3, 1, 1023, 511, 255, 127, 63, 31, 15, 7, 3, 1, 2047, 1023, 511, 255, 127, 63, 31, 15, 7, 3, 1

Offset: 0

Author

Gary W. Adamson, May 24 2007

Keywords

• nonn
• easy
• tabl

Comments

Row sums are A000295: (1, 4, 11, 26, 57, 120, 247, ...), the Eulerian numbers.

T(n,k) is the number of length n+1 binary words containing at least two 1's such that the first 1 is preceded by exactly (k-1) 0's. T(3,2) = 3 because we have: 0101, 0110, 0111. - Geoffrey Critzer, Dec 31 2013

Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array

/I_k 0\

\ 0 M/

having the k x k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite matrix product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to A110441. - Peter Bala, Jul 22 2014

From Wolfdieter Lang, Oct 28 2019:(Start)

This triangle gives the solution of the following problem. Iterate the function f(x) = (x - 1)/2 to obtain f^{[k]}(x) = (x - (2^(k+1) - 1))/2^(k+1), for k >= 0. Find the positive integer x values for which the iterations stay integer and reach 1. Only odd integers x qualify, and the answer is x = x(n) = 2*T(n, 0) = 2*(2^(n+1) - 1), with the iterations T(n,0), ..., T(n,n) = 1.

This iteration is motivated by a problem posed by Johann Peter Hebel (1760 - 1826) in "Zweites Rechnungsexempel" from 1804, with the solution x = 31 corresponding to row n = 3 [15 7 3 1]. The egg selling woman started with 31 = T(4, 0) eggs and after four customers obtained, one after the other, always a number of eggs which was one half of the woman's remaining number of eggs plus 1/2 (selling only whole eggs, of course) she had one egg left. See the link and reference. [For Hebel's first problem see a comment in A000225.]

(End)

Examples

			First few rows of the triangle T(n, k):
n\k     0    1    2    3   4   5   6  7  8  9 10 11 12 ...
0:      1
1:      3    1
2:      7    3    1
3      15    7    3    1
4:     31   15    7    3   1
5:     63   31   15    7   3   1
6:    127   63   31   15   7   3   1
7:    255  127   63   31  15   7   3  1
8:    511  255  127   63  31  15   7  3  1
9:   1023  511  255  127  63  31  15  7  3  1
10:  2047 1023  511  255 127  63  31 15  7  3  1
11:  4095 2047 1023  511 255 127  63 31 15  7  3  1
12:  8191 4095 2047 1023 511 255 127 63 31 15  7  3  1
... reformatted and extended. - _Wolfdieter Lang_, Oct 28 2019
		

References

• Johann Peter Hebel, Gesammelte Werke in sechs Bänden, Herausgeber: Jan Knopf, Franz Littmann und Hansgeorg Schmidt-Bergmann unter Mitarbeit von Ester Stern, Wallstein Verlag, 2019. Band 3, S. 36-37, Solution, S. 40-41. See also the link below.

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..11324 [first 150 rows; offset shifted by _Georg Fischer_, Oct 29 2019]
• Johann Peter Hebel, Zweites Rechnungsexempel., 1804; Solution: Auflösung des zweiten Rechnungsexempels. , 1805.

Crossrefs

Cf. A130321, A000012, A000225. A110441.

Programs

Haskell

a130330 n k = a130330_row n !! (k-1)
a130330_row n = a130330_tabl !! (n-1)
a130330_tabl = iterate (\xs -> (2 * head xs + 1) : xs) [1]
-- Reinhard Zumkeller, Mar 31 2012

Mathematica

nn=12;a=1/(1- x);b=1/(1-2x);Map[Select[#,#>0&]&,Drop[CoefficientList[Series[a x^2 b/(1-y x),{x,0,nn}],{x,y}],2]]//Grid  (* Geoffrey Critzer, Dec 31 2013 *)

Formula

A130321 * A000012 as infinite lower triangular matrices, where A130321 = (1; 2,1; 4,2,1; ...) and A000012 = (1; 1,1; 1,1,1; ...).

In every column k with offset n = k: 2^(m+1) - 1 = A000225(m+1) = (1, 3, 7, 15, ...), for m >= 0.

G.f.: 1/((1-y*x)*(1-x)*(1-2x)). - Geoffrey Critzer, Dec 31 2013

T(n, k) = 2^((n - k) + 1) - 1, n >= 0, k = 0..n. - Wolfdieter Lang, Oct 28 2019

Extensions

More terms from Geoffrey Critzer, Dec 31 2013

Edited by Wolfdieter Lang, Oct 28 2019