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Triangle read by rows, generated from Pascal's triangle.
1

%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