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%I #38 Feb 20 2022 14:57:12
%S 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,
%T 255,127,63,31,15,7,3,1,511,255,127,63,31,15,7,3,1,1023,511,255,127,
%U 63,31,15,7,3,1,2047,1023,511,255,127,63,31,15,7,3,1
%N Triangle read by rows, the matrix product A130321 * A000012, both taken as infinite lower triangular matrices.
%C Row sums are A000295: (1, 4, 11, 26, 57, 120, 247, ...), the Eulerian numbers.
%C 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
%C Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
%C /I_k 0\
%C \ 0 M/
%C 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
%C From _Wolfdieter Lang_, Oct 28 2019:(Start)
%C 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.
%C 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.]
%C (End)
%D 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.
%H Reinhard Zumkeller, <a href="/A130330/b130330.txt">Table of n, a(n) for n = 0..11324</a> [first 150 rows; offset shifted by _Georg Fischer_, Oct 29 2019]
%H Johann Peter Hebel, <a href="https://gutenberg.spiegel.de/buch/schatzkastlein-des-rheinischen-hausfreundes-8818/15">Zweites Rechnungsexempel.</a>, 1804; Solution: <a href="https://gutenberg.spiegel.de/buch/schatzkastlein-des-rheinischen-hausfreundes-8818/17">Auflösung des zweiten Rechnungsexempels. </a>, 1805.
%F A130321 * A000012 as infinite lower triangular matrices, where A130321 = (1; 2,1; 4,2,1; ...) and A000012 = (1; 1,1; 1,1,1; ...).
%F In every column k with offset n = k: 2^(m+1) - 1 = A000225(m+1) = (1, 3, 7, 15, ...), for m >= 0.
%F G.f.: 1/((1-y*x)*(1-x)*(1-2x)). - _Geoffrey Critzer_, Dec 31 2013
%F T(n, k) = 2^((n - k) + 1) - 1, n >= 0, k = 0..n. - _Wolfdieter Lang_, Oct 28 2019
%e First few rows of the triangle T(n, k):
%e n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 ...
%e 0: 1
%e 1: 3 1
%e 2: 7 3 1
%e 3 15 7 3 1
%e 4: 31 15 7 3 1
%e 5: 63 31 15 7 3 1
%e 6: 127 63 31 15 7 3 1
%e 7: 255 127 63 31 15 7 3 1
%e 8: 511 255 127 63 31 15 7 3 1
%e 9: 1023 511 255 127 63 31 15 7 3 1
%e 10: 2047 1023 511 255 127 63 31 15 7 3 1
%e 11: 4095 2047 1023 511 255 127 63 31 15 7 3 1
%e 12: 8191 4095 2047 1023 511 255 127 63 31 15 7 3 1
%e ... reformatted and extended. - _Wolfdieter Lang_, Oct 28 2019
%t 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 *)
%o (Haskell)
%o a130330 n k = a130330_row n !! (k-1)
%o a130330_row n = a130330_tabl !! (n-1)
%o a130330_tabl = iterate (\xs -> (2 * head xs + 1) : xs) [1]
%o -- _Reinhard Zumkeller_, Mar 31 2012
%Y Cf. A130321, A000012, A000225. A110441.
%K nonn,easy,tabl
%O 0,2
%A _Gary W. Adamson_, May 24 2007
%E More terms from _Geoffrey Critzer_, Dec 31 2013
%E Edited by _Wolfdieter Lang_, Oct 28 2019
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