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%I #38 Mar 15 2024 02:24:42
%S 1,1,1,1,2,1,1,3,4,1,1,4,10,6,1,1,5,21,22,8,1,1,6,40,64,38,10,1,1,7,
%T 72,163,140,58,12,1,1,8,125,382,442,256,82,14,1,1,9,212,846,1259,954,
%U 420,110,16,1,1,10,354,1800,3334,3166,1794,640,142,18,1,1,11,585,3719,8366,9657,6754,3074,924,178,20,1
%N Triangle T(n,k) read by rows. T(n,1)=1; T(n,k) = Sum_{i=1..k-1} ( T(n-i,k-1) + T(n-i,k) ), k>1.
%C Are there closed forms for diagonals and columns?
%C With the definition of the array, I note that the sequence (phi(k)) (phi(k)= g.f. of the column number k) is given by the recurrence relation: phi(k+1)=phi(k)*(1-z^k)/(1-2*z+z^(k+1)). The consequence is: the sequence number k+1 column is the convolution of the k-one and a "-acci like" sequence whose g.f. is given by (1-z^k)/(1-2*z+z^(k+1)). E.g., the 2-column is the convolution of the 1-column and the sequence 1, 2, 3, 5, ... classical Fibonacci sequence without the first 1. The 3-column is the convolution of the 2-column and 1, 2, 4, 7, 13, ... tribonacci like-sequence (exactly: A000073 without beginning 0, 0, 1). - _Richard Choulet_, Feb 19 2010
%C Relation to metallic means:
%C T(n,1)=1, k>1: T(n,k) = Sum_{i=1..k-1} T(n-i,k-1) + 0*Sum_{i=1..k-1} T(n-i,k)
%C has antidiagonal sums for which the limiting ratio tends to the golden ratio, A001622.
%C T(n,1)=1, k>1: T(n,k) = Sum_{i=1..k-1} T(n-i,k-1) + 1*Sum_{i=1..k-1} T(n-i,k)
%C has antidiagonal sums for which the limiting ratio tends to the silver ratio, A014176.
%C T(n,1)=1, k>1: T(n,k) = Sum_{i=1..k-1} T(n-i,k-1) + 2*Sum_{i=1..k-1} T(n-i,k)
%C has antidiagonal sums for which the limiting ratio tends to the bronze ratio, A098316.
%C A similar point can be made about variations of the Pascal triangle.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Metallic_mean">Metallic mean</a>
%F The g.f of the number k column is phi(k)(z) = (1/(1-z))*Product_{i=1..k-1}(1-z^i)/(1-2*z+z^(i+1)). - _Richard Choulet_, Feb 19 2010
%e Table begins:
%e n/k| 1 2 3 4 5 6 7 8 9 10 11
%e ---+-----------------------------------------------------
%e 1 | 1
%e 2 | 1 1
%e 3 | 1 2 1
%e 4 | 1 3 4 1
%e 5 | 1 4 10 6 1
%e 6 | 1 5 21 22 8 1
%e 7 | 1 6 40 64 38 10 1
%e 8 | 1 7 72 163 140 58 12 1
%e 9 | 1 8 125 382 442 256 82 14 1
%e 10 | 1 9 212 846 1259 954 420 110 16 1
%e 11 | 1 10 354 1800 3334 3166 1794 640 142 18 1
%e Example: T(8,4) = 163 because it is the sum of the numbers:
%e 10 6
%e 21 22
%e 40 64
%e For k=1, we obtain phi(k)(z)=1/(1-z) which is clear; for k=2, we obtain phi(k)(z)=1/(1-z)^2. For k=3, we obtain phi(3)(z)=(1+z)/((1-2*z+z^3)*(1-z)); this is A001891 without the beginning zero. - _Richard Choulet_, Feb 19 2010
%p A175105 := proc(n,k) if k =1 then 1; elif k > n or k< 1 then 0 ; else add(procname(n-i,k-1)+procname(n-i,k),i=1..k-1) ; end if; end proc; # _R. J. Mathar_, Feb 16 2011
%t T[_, 1] = 1;
%t T[n_, k_] /; 1<k<=n := T[n, k] = Sum[T[n-i, k-1]+T[n-i, k], {i, 1, k-1}];
%t T[_, _] = 0;
%t Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Dec 19 2019 *)
%o (Excel) =if(column()=1;1;if(row()>=column();sum(indirect(address(row()-column()+1;column()-1;4)&":"&address(row()-column()+column()-1;column()-1;4);4))+sum(indirect(address(row()-column()+1;column();4)&":"&address(row()-column()+column()-1;column();4);4));0)) ' _Mats Granvik_, Mar 28 2010
%Y Cf. A172119, A051731, A001891 (column k=3), A176084 (row sums).
%Y (1-((-1)^T(n, k)))/2 = T(n, k) mod 2 = A051731.
%Y Cf. A179807=antidiagonal sums. A179748 has simpler recurrence.
%K nonn,tabl
%O 1,5
%A _Mats Granvik_, Feb 10 2010
%E Corrected and edited by _Mats Granvik_, Jul 28 2010, Dec 09 2010
%E Choulet formulas indices shifted (to adapt to the new column index) by _R. J. Mathar_, Dec 13 2010
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