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%I #10 Feb 13 2022 23:17:38
%S 1,1,1,2,2,1,3,5,3,1,5,10,8,4,1,8,20,17,11,5,1,13,38,35,24,14,6,1,21,
%T 71,68,50,31,17,7,1
%N Triangle read by rows, generated from Pascal's triangle.
%C The array P =
%C 1, 0, 0, 0, 0, 0, ...
%C 0, 1, 0, 0, 0, 0, ...
%C 0, 1, 1, 0, 0, 0, ...
%C 0, 0, 2, 1, 0, 0, ...
%C 0, 0, 1, 3, 1, 0, ...
%C 0, 0, 0, 3, 4, 1, ...
%C ...
%C ... as shown on page 107 of "A Primer for the Fibonacci Numbers".
%C The array A is composed of arithmetic sequences, as a matrix.
%C 1, 1, 1, 1, 1, ...
%C 1, 2, 3, 4, 5, ...
%C 1, 3, 5, 7, 9, ...
%C 1, 4, 7, 10, 13, ...
%C 1, 5, 9, 13, 17, ...
%C ...
%C Leftmost column = Fibonacci numbers, next column (1, 2, 5, 10, 20, ...) = Fibonacci numbers convolved with themselves.
%D V. E. Hoggatt, Jr., editor; "A Primer for the Fibonacci Numbers", 1963, p. 107.
%F Let P = an array with columns composed of Pascal's Triangle rows, offset, spaces filled in with zeros; A = an array composed of arithmetic sequences(n, k). Perform P * A and extract antidiagonals which become the rows of A106196.
%e The operation P * A generates the array:
%e 1, 1, 1, 1, 1, ...
%e 1, 2, 3, 4, 5, ...
%e 2, 5, 8, 11, 14, ...
%e 3, 10, 17, 24, 31, ...
%e 5, 20, 35, 50, 65, ...
%e ...
%e from which we extract antidiagonals, read by rows, become triangle A106196:
%e 1;
%e 1, 1;
%e 2, 2, 1;
%e 3, 5, 3, 1;
%e 5, 10, 8, 4, 1;
%e 8, 20, 17, 11, 5, 1;
%e 13, 38, 35, 24, 14, 6, 1;
%e 21, 71, 68, 50, 31, 17, 7, 1;
%e ...
%Y Cf. A052996, A007678, A106196.
%K nonn,tabl
%O 0,4
%A _Gary W. Adamson_, Apr 24 2005