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%I #55 Apr 05 2019 09:26:42
%S 0,1,1,7,13,151,421,6847,25453,532231,2473141,63206287,352444093,
%T 10645162711,69251478661,2413453999327,17943523153933,708721089607591,
%U 5927841361456981,261679010699505967,2431910546406522973,118654880542567722871,1212989379862721528101
%N "Stirling-Bernoulli transform" of Fibonacci numbers.
%C From _Paul Curtz_, Oct 11 2013: (Start)
%C Differences table:
%C 0, 1, 1, 7, 13, 151, 421, 6847, ...
%C 1, 0, 6, 6, 138, 270, 6426, ...
%C -1, 6, 0, 132, 132, 6156, ...
%C 7, -6, 132, 0, 6024, ...
%C -13, 138, -132, 6024, ...
%C 151, -270, 6156, ...
%C -421, 6426, ...
%C 6847, ... .
%C a(n) is an autosequence of first kind: the inverse binomial transform is the sequence signed, the main diagonal is A000004=0's.
%C The "Stirling-Bernoulli transform" applied to an autosequence of first kind is an autosequence of first kind.
%C Now consider the Akiyama-Tanigawa transform or algorithm applied to A000045(n):
%C 0, 1, 1, 2, 3, 5, 8, ...
%C -1, 0, -3, -4, -10, -18, ... = -A006490
%C -1, 6, 3, 24, 40, ...
%C -7, 6, -63, -64, ...
%C -13, 138, 3, ...
%C -151, 270, ...
%C -421, ... .
%C Hence -a(n). The Akiyama-Tanigawa algorithm applied to an autosequence of first kind is an autosequence of first kind.
%C a(n+5) - a(n+1) = 150, 420, 6840, ... is divisible by 30.
%C For an autosequence of the second kind, the inverse binomial transform is the sequence signed with the main diagonal double of the first upper diagonal.
%C The Akiyama-Tanigawa algorithm applied to an autosequence leads to an autosequence of the same kind. Example: the A-T algorithm applied to the autosequence of second kind 1/n leads to the autosequence of the second kind A164555(n)/A027642(n).
%C Note that a2(n) = 2*a1(n+1) - a1(n) applied to the autosequence of the first kind a1(n) is a corresponding autosequence of the second kind. (End)
%H Alois P. Heinz, <a href="/A050946/b050946.txt">Table of n, a(n) for n = 0..447</a>
%H C. J. Pita Ruiz V., <a href="https://cs.uwaterloo.ca/journals/JIS/VOL16/Pita/pita19.html">Some Number Arrays Related to Pascal and Lucas Triangles</a>, J. Int. Seq. 16 (2013) #13.5.7
%F O.g.f.: Sum_{n>=1} Fibonacci(n) * n! * x^n / Product_{k=1..n} (1+k*x). - _Paul D. Hanna_, Jul 20 2011
%F A100872(n)=a(2*n) and A100868(n)=a(2*n-1).
%F From _Paul Barry_, Apr 20 2005: (Start)
%F E.g.f.: exp(x)*(1-exp(x))/(1-3*exp(x)+exp((2*x))).
%F a(n) = Sum_{k=0..n} (-1)^(n-k)*S2(n, k)*k!*Fibonacci(k). [corrected by _Ilya Gutkovskiy_, Apr 04 2019] (End)
%F a(n) ~ c * n! / (log((3+sqrt(5))/2))^(n+1), where c = 1/sqrt(5) if n is even and c = 1 if n is odd. - _Vaclav Kotesovec_, Aug 13 2013
%F a(n) = -1 * Sum_{k = 0..n} A163626(n,k)*A000045(k). - _Philippe Deléham_, May 29 2015
%p with(combinat):
%p a:= n-> add((-1)^(k+1) *k! *stirling2(n+1, k+1)*fibonacci(k), k=0..n):
%p seq(a(n), n=0..30); # _Alois P. Heinz_, May 17 2013
%t CoefficientList[Series[E^x*(1-E^x)/(1-3*E^x+E^(2*x)), {x, 0, 20}], x]* Range[0, 20]! (* _Vaclav Kotesovec_, Aug 13 2013 *)
%t t[0, k_] := Fibonacci[k]; t[n_, k_] := t[n, k] = (k+1)*(t[n-1, k] - t[n-1, k+1]); a[n_] := t[n, 0] // Abs; Table[a[n], {n, 0, 22}] (* _Jean-François Alcover_, Oct 22 2013, after _Paul Curtz_ *)
%o (PARI) {a(n)=polcoeff(sum(m=0, n, fibonacci(m)*m!*x^m/prod(k=1, m, 1+k*x+x*O(x^n))), n)} /* _Paul D. Hanna_, Jul 20 2011 */
%Y Cf. A000045, A051782, A105796, A163626.
%K nonn
%O 0,4
%A _N. J. A. Sloane_, Jan 02 2000
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