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%I #10 Dec 05 2017 08:25:26
%S 1,1,1,1,2,1,1,3,4,1,1,4,11,7,1,1,5,26,32,12,1,1,6,57,122,92,20,1,1,7,
%T 120,423,582,252,33,1,1,8,247,1389,3333,2598,681,54,1,1,9,502,4414,
%U 18054,24117,11451,1815,88,1,1,10,1013,13744,94684,210990,172980,49566
%N Triangle read by rows, based on a simple Fibonacci recursion rule.
%C Subdiagonal is A000071(n+3). Row sums of inverse are 0^n.
%C Row sums are given by A135934 - _Emanuele Munarini_, Dec 05 2017
%F Number triangle T(n, k)=T(n-1, k-1)+F(k+1)*T(n-1, k) where F(n)=A000045(n); Column k has g.f. x^k/Product(1-F(j+1)x, j, 0, k).
%e Triangle begins
%e 1....1....2....3....5....8...13....F(k+1)
%e 1
%e 1....1
%e 1....2....1
%e 1....3....4....1
%e 1....4...11....7....1
%e 1....5...26...32...12....1
%e 1....6...57..122...92...20....1
%e For example, T(6,3)=122=26+3*32=T(5,2)+F(4)*T(5,3)
%t (* To generate the triangle *)
%t Grid[RecurrenceTable[{F[n,k] == F[n-1,k-1] + Fibonacci[k+1] F[n-1,k], F[0,k] == KroneckerDelta[k]}, F, {n,0,10}, {k,0,10}]] (* _Emanuele Munarini_, Dec 05 2017 *)
%Y Cf. A111577, A111578, A111579, A008277, A039755, A135934.
%K easy,nonn,tabl
%O 0,5
%A _Gary W. Adamson_, Aug 14 2005
%E Edited by _Paul Barry_, Nov 14 2005
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