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Triangle generated from INVERT transforms of variants of A080995.
1

%I #12 Nov 12 2017 03:58:23

%S 1,1,1,1,1,1,1,1,1,2,1,1,1,1,3,1,1,1,1,2,5,1,1,1,1,1,4,6,1,1,1,1,1,2,

%T 6,9,1,1,1,1,1,1,4,8,12,1,1,1,1,1,1,2,6,12,16,1,1,1,1,1,1,1,4,8,19,18,

%U 1,1,1,1,1,1,1,2,6,11,28,23

%N Triangle generated from INVERT transforms of variants of A080995.

%C Row sums = A000041 starting with offset 1: (1, 1, 2, 3, 5, 7, 11, 15, ...).

%C The INVERTi transform of A000041 starting with offset 1 follows from the definition of the INVERT transform, given 1/p(x) = A010815.

%F Given the INVERTi transform of the partition numbers starting with offset 1 = a signed variant of A080995 such that Q = (1, 1, 0, 0, -1, 0, -1, 0, 0, 0, 0, 1, ...).

%F Construct an array in which k-th row (k=1,2,3,...) = the INVERT transform of Q(x^k), i.e., where polcoeff Q(x) is interleaved with 0,1,2,3,... zeros.

%F Take finite differences of the array terms starting with the last "1" going from the bottom to top, becoming rows of triangle A175010.

%e First few rows of the array:

%e 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101,

%e 1, 1, 2, 3, 4, 6, 9, 13, 18, 26, 38, 54, 76,

%e 1, 1, 1, 2, 3, 4, 5, 7, 10, 14, 19, 26, 35,

%e 1, 1, 1, 1, 2, 3, 4, 5, 6, 8, 11, 15, 20,

%e 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 9, 12,

%e 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 8,

%e ...

%e Taking finite differences from the bottom starting with the top "1", we obtain rows of the triangle:

%e 1;

%e 1, 1;

%e 1, 1, 1;

%e 1, 1, 1, 2;

%e 1, 1, 1, 1, 3;

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

%e 1, 1, 1, 1, 1, 4, 6;

%e 1, 1, 1, 1, 1, 2, 6, 9;

%e 1, 1, 1, 1, 1, 1, 4, 8, 12;

%e 1, 1, 1, 1, 1, 1, 2, 6, 12, 16;

%e 1, 1, 1, 1, 1, 1, 1, 4, 8, 19, 18;

%e 1, 1, 1, 1, 1, 1, 1, 2, 6, 11, 28, 23;

%e 1, 1, 1, 1, 1, 1, 1, 1, 4, 8, 15, 41, 25;

%e 1, 1, 1, 1, 1, 1, 1, 1, 2, 6, 10, 22, 61, 26;

%e ...

%e Example: Row 2 = INVERT transform of Q(x^2), (i.e., Q(x) interleaved with one zero between terms).

%Y Cf. A000041, A080995, A010815.

%K nonn,tabl

%O 1,10

%A _Gary W. Adamson_, Apr 03 2010