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%I #49 Feb 15 2022 13:04:49
%S 1,1,1,2,1,1,3,2,1,1,5,3,2,1,1,8,5,3,2,1,1,13,8,5,3,2,1,1,21,13,8,5,3,
%T 2,1,1,34,21,13,8,5,3,2,1,1,55,34,21,13,8,5,3,2,1,1,89,55,34,21,13,8,
%U 5,3,2,1,1,144,89,55,34,21,13,8,5,3,2,1,1,233,144,89,55,34,21,13,8,5,3,2,1
%N Triangle read by rows: row n contains first n nonzero Fibonacci numbers in decreasing order.
%C Sum of n-th row = F(n+2) - 1; sequence A000071 starting (1, 2, 4, 7, 12, 20, ...).
%C Riordan array (1/(1-x-x^2),x). - _Philippe Deléham_, Apr 23 2009 [with offset 0]
%C Sequence B is called a reverse reluctant sequence of sequence A, if B is triangle array read by rows: row number k lists first k elements of the sequence A in reverse order. Sequence A104762 is the reverse reluctant sequence of Fibonacci numbers (A000045), except 0. - _Boris Putievskiy_, Dec 13 2012
%H Harvey P. Dale, <a href="/A104762/b104762.txt">Table of n, a(n) for n = 1..1000</a>
%H Boris Putievskiy, <a href="http://arxiv.org/abs/1212.2732">Transformations Integer Sequences And Pairing Functions</a>, arXiv:1212.2732 [math.CO], 2012.
%F In every column, (1, 1, 2, 3, 5, ...); the nonzero Fibonacci numbers, A000045.
%F a(n,k) = A000045(n-k+1). - _R. J. Mathar_, Jun 23 2006
%F a(n) = A000045(m), where m = (t*t+3*t+4)/2-n, t = floor((-1+sqrt(8*n-7))/2). - _Boris Putievskiy_, Dec 13 2012
%F Let P denote Pascal's triangle. Then P*A104762*P^(-1 ) = A121461. - _Peter Bala_, Apr 11 2013
%F a(n,k) = |round[(r^n)*(s^k)/sqrt(5)|, where r = golden ratio = (1+ sqrt(5))/2, s = (1 - sqrt(5))/2, 1 < = k <= n-1, n > = 2. - _Clark Kimberling_, May 01 2016
%F G.f. of triangle: G(x,y) = x*y/((1-x-x^2)*(1-x*y)). - _Robert Israel_, May 01 2016
%e First few rows of the triangle:
%e 1;
%e 1, 1;
%e 2, 1, 1;
%e 3, 2, 1, 1;
%e 5, 3, 2, 1, 1;
%e 8, 5, 3, 2, 1, 1;
%e ...
%e From _Philippe Deléham_, Oct 07 2014: (Start)
%e Production matrix begins:
%e 1, 1;
%e 1, 0, 1;
%e 0, 0, 0, 1;
%e 0, 0, 0, 0, 1;
%e 0, 0, 0, 0, 0, 1;
%e 0, 0, 0, 0, 0, 0, 1;
%e 0, 0, 0, 0, 0, 0, 0, 1;
%e ... (End)
%p seq(seq(combinat:-fibonacci(n-i),i=0..n-1),n=1..20); # _Robert Israel_, May 01 2016
%t r = N[(1 + Sqrt[5])/2, 100]; s = N[(1 - Sqrt[5])/2, 100];
%t t = Table[Abs[Round[(r^n)*(s^k)/Sqrt[5]]], {n, 2, 15}, {k, 1, n - 1}]
%t Flatten[t]
%t TableForm[t]
%t (* _Clark Kimberling_, May 01 2016 *)
%t Table[Reverse[Fibonacci[Range[n]]],{n,15}]//Flatten (* _Harvey P. Dale_, Jan 28 2019 *)
%Y Cf. A000045, A000071, A271355 (analogous Lucas triangle).
%Y Companion triangle A104763, Fibonacci sequence in each row starting from the left. A121461.
%K nonn,easy,tabl
%O 1,4
%A _Gary W. Adamson_, Mar 23 2005, Mar 05 2007
%E Edited by _N. J. A. Sloane_ at the suggestion of _Philippe Deléham_, Jun 11 2007
%E More terms from _Philippe Deléham_, Apr 21 2009
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