1,10

Right border = A000009; row sums = A000009 with offset 1.

Sum of n-th row terms = rightmost term in next row.

The INVERTi transform of A000009 starting with offset 1 = (1, 0, 1, -1, -1,

1, -2, 2, -2, 2, -3, 3, -3, 4, -5, 5, -5, 6,...); i.e. A000700 signed = left border.

A000700 is derived from parity changes of A000041 as follows: Given A000041: (1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135,...). Write down the parity starting (1, 1, 0, 1, 1, 1, 1, 1...) then add "1" starting in the next

string of A000041 with a change in parity. Since the next 4 terms of A000041

are (22, 30, 42, 56...) we denote these by (...2, 2, 2, 2...). The next three p(n)

terms are 77, 101, 135, so these are (...3, 3, 3,...) in A000700.

The signed version of A000700 as indicated: (alternate signs starting with

A000700(3): (+-+...) = the INVERTi transform of A000009.

Let M = triangle by columns: A000700 (signed, starting 1, 0, 1, -1, 1, -1, 1, -2,...)

in every column and P = an infinite lower triangular matrix with

A000009 (1, 1, 1, 2, 2, 3, 4, 5, 6,...) as the right border and the rest zeros.

A146061 = M * P

First few rows of the triangle =

1;

0, 1;

1, 0, 1;

-1, 1, 0, 2;

1, -1, 1, 0, 2;

-1, 1, -1, 2, 0, 3;

1, -1, 1, -2, 2, 0, 4;

-2, 1, -1, 2, -2, 3, 0, 5;

2, -2, 1, -2, 2, -3, 4, 0, 6;

-2, 2, -2, 2, -2, 3, -4, 5, 0, 8;

2, -2, 2, -4, 2, -3, 4, -5, 6, 0, 10;

-3, 2, -2, 4, -4, 3, -4, 5, -6, 8, 0, 12;

3, -3, 2, -4, 4, -6, 4, -5, 6, -8, 10, 0, 15;

-3, 3, -3, 4, -4, 6, -8, 5, -6, 8, -10, 12, 0, 18;

4, -3, 3, -6, 4, -6, 8, -10, 6, -8, 10, -12, 15, 0, 22;

-5, 4, -3, 6, -6, 6, -8, 10, -12, 8, -10, 12, -15, 18, 0, 27;

5, -5, 4, -6, 6, -9, 8, -10, 12, -16, 10, -12, 15, -18, 22, 0, 32;

...

A000009, Cf. A000700

Adjacent sequences: A146058 A146059 A146060 this_sequence A146062 A146063 A146064

Sequence in context: A026609 A090340 A117162 this_sequence A135936 A109707 A064272

tabl,sign

Gary W. Adamson (qntmpkt(AT)yahoo.com), Oct 26 2008