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A130330
Triangle read by rows, the matrix product A130321 * A000012, both taken as infinite lower triangular matrices.
5
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
0,2
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)
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.
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
EXAMPLE
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
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,easy,tabl
AUTHOR
Gary W. Adamson, May 24 2007
EXTENSIONS
More terms from Geoffrey Critzer, Dec 31 2013
Edited by Wolfdieter Lang, Oct 28 2019
STATUS
approved