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A105438 Triangle, row sums = (Fibonacci numbers - 2).

%I A105438 #17 May 04 2024 14:53:00
%S A105438 1,2,1,3,2,1,4,4,2,1,5,6,5,2,1,6,9,8,6,2,1,7,12,14,10,7,2,1,8,16,20,
%T A105438 20,12,8,2,1,9,20,30,30,27,14,9,2,1,10,25,40,50,42,35,16,10,2,1,11,30,
%U A105438 55,70,77,56,44,18,11,2,1
%N A105438 Triangle, row sums = (Fibonacci numbers - 2).
%C A105438 Row sums = 1, 3, 6, 11, 19, 32, 53...(Fibonacci numbers - 2; starting with F(4)) The first few rows of the triangle are:
%C A105438 Row sums = (Fibonacci numbers - 2; starting 1, 3, 6...).
%C A105438 Column 1 = A002620; Column 2 = A006918; Column 3 = A096338.
%C A105438 Inverse array is A105522. - _Paul Barry_, Apr 11 2005
%C A105438 Diagonal sums are A027383(n). - _Philippe Deléham_, Jan 16 2014
%H A105438 T. Mansour and A. O. Munagi, <a href="https://doi.org/10.1216/rmj-2012-42-4-1313">Alternating subsets modulo m</a>, Rocky Mt. J. Math. 42, No. 4, 1313-1325 (2012), eq. (2)
%F A105438 By columns (k = 0, 1, 2...); use partial sum operator on (bin(n, k) numbers repeated).
%F A105438 T(n,k) = Sum_{j=0..n-k} C((j+2k)/2, k)*(1+(-1)^j)+C((j-1+2k)/2, k)*(1-(-1)^j)/2; Riordan array (1/(1-x)^2, x/(1-x^2)). - _Paul Barry_, Apr 11 2005
%F A105438 T(n,k) = T(n-1,k)+T(n-1,k-1)+T(n-2,k)-T(n-2,k-1)-T(n-3,k), T(0,0)=1, T(1,0)=2, T(1,1)= 1, T(n,k)= 0 if k<0 or if k>n. - _Philippe Deléham_, Jan 16 2014
%e A105438 Column 2: 1, 2, 5, 8, 14, 20, 30...is generated by using the partial sum operator on 1, 1, 3, 3, 6, 6, 10, 10...
%e A105438 The first few rows of the triangle are:
%e A105438   1;
%e A105438   2, 1;
%e A105438   3, 2, 1;
%e A105438   4, 4, 2, 1;
%e A105438   5, 6, 5, 2, 1;
%e A105438   6, 9, 8, 6, 2, 1;
%e A105438   7, 12, 14, 10, 7, 2, 1;
%e A105438   8, 16, 20, 20, 12, 8, 2, 1;
%e A105438   9, 20, 30, 30, 27, 14, 9, 2, 1;
%e A105438   10, 25, 40, 50, 42, 35, 16, 10, 2, 1;
%e A105438   ...
%Y A105438 Cf. A001911, A002620, A006918, A096338, A105108.
%K A105438 nonn,tabl,easy
%O A105438 0,2
%A A105438 _Gary W. Adamson_, Apr 09 2005