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Triangle read by rows, antidiagonals of an array generated from INVERT transforms of variants of (1, 2, 3, ...).
2

%I #9 Mar 30 2024 23:08:41

%S 1,1,2,1,2,5,1,2,2,16,1,2,2,5,45,1,2,2,2,12,125,1,2,2,2,5,24,341,1,2,

%T 2,2,2,12,48,918,1,2,2,2,2,7,18,97,2453,1,2,2,2,2,2,16,28,195,6515

%N Triangle read by rows, antidiagonals of an array generated from INVERT transforms of variants of (1, 2, 3, ...).

%C Row sums = A001906, the even-indexed Fibonacci numbers starting (1, 3, 8, 21, ...).

%F Given S(x) = (1 + 2x + 3x^2 + ...), where (1, 2, 3, ...) = the INVERTi transform of (1, 3, 8, 21, 55, ...); k-th row of the array = INVERT transform of S(x^k). Take finite differences of array columns starting from the topmost "1"; becoming rows of the triangle.

%e First few rows of the array:

%e 1, 3, 8, 21, 55, 144, 377, 987, 2584, ...

%e 1, 1, 3, 5, 10, 19, 36, 69, 131, ...

%e 1, 1, 1, 3, 5, 7, 12, 21, 34, ...

%e 1, 1, 1, 1, 3, 5, 7, 9, 16, ...

%e 1, 1, 1, 1, 1, 3, 5, 7, 9, ...

%e 1, 1, 1, 1, 1, 1, 3, 5, 7, ...

%e ...

%e Taking finite differences from the bottom to top starting with the last "1" we obtain triangle A175011:

%e 1;

%e 1, 2;

%e 1, 2, 5;

%e 1, 2, 2, 16;

%e 1, 2, 2, 5, 45;

%e 1, 2, 2, 2, 12, 125;

%e 1, 2, 2, 2, 5, 24, 341;

%e 1, 2, 2, 2, 2, 12, 48, 918;

%e 1, 2, 2, 2, 2, 7, 18, 97, 2453;

%e 1, 2, 2, 2, 2, 2, 16, 28, 195, 6515;

%e ...

%Y Cf. A001906.

%K nonn,tabl

%O 1,3

%A _Gary W. Adamson_, Apr 03 2010