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A111669 Triangle read by rows, based on a simple Fibonacci recursion rule.

%I A111669 #30 May 25 2024 09:05:31
%S A111669 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 A111669 120,423,582,252,33,1,1,8,247,1389,3333,2598,681,54,1,1,9,502,4414,
%U A111669 18054,24117,11451,1815,88,1,1,10,1013,13744,94684,210990,172980,49566,4807,143,1
%N A111669 Triangle read by rows, based on a simple Fibonacci recursion rule.
%C A111669 Subdiagonal is A000071(n+3). Row sums of inverse are 0^n.
%C A111669 Row sums are given by A135934. - _Emanuele Munarini_, Dec 05 2017
%H A111669 Michel Marcus, <a href="/A111669/b111669.txt">Rows n = 0..20, flattened</a>
%F A111669 T(n, k) = T(n-1, k-1) + F(k+1)*T(n-1, k) where F(n)=A000045(n).
%F A111669 Column k has g.f. x^k/Product_{j=0..k} (1 - F(j+1)*x).
%e A111669 Triangle begins
%e A111669   1....1....2....3....5....8...13....F(k+1)
%e A111669   1
%e A111669   1....1
%e A111669   1....2....1
%e A111669   1....3....4....1
%e A111669   1....4...11....7....1
%e A111669   1....5...26...32...12....1
%e A111669   1....6...57..122...92...20....1
%e A111669 For example, T(6,3) = 122 = 26 + 3*32 = T(5,2) + F(4)*T(5,3).
%t A111669 (* To generate the triangle *)
%t A111669 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 *)
%o A111669 (PARI) T(n, k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, T(n-1, k-1) + fibonacci(k+1)*T(n-1, k))); \\ _Michel Marcus_, May 25 2024
%Y A111669 Cf. A111577, A111578, A111579, A008277, A039755, A135934.
%K A111669 easy,nonn,tabl
%O A111669 0,5
%A A111669 _Gary W. Adamson_, Aug 14 2005
%E A111669 Edited by _Paul Barry_, Nov 14 2005