Sequence related to figurate numbers

[Andrew Plewe]

(I'm assuming that the sum must have at least 2 terms after the initial 
one is dropped, otherwise we can trivially get everything except 1.)

Try looking just at column 2 plus column 3.  This gives, in rows 
(indexed by the increment):
5
8 10
11 13 15
14 16 18 20
17 19 21 23 25
20 22 24 26 28 30

As you can see, each row has an increment of 2, with alternating rows 
even and odd.  And from the point shown on, each row overlaps the 2nd 
previous row.  So the only values missing from this triangle are:

1,2,3,4,6,7,9,12.

We can verify that 9 can be represented (2+3+4); the others cannot.  (7 
= 3 + 4 and 12 = 3 + 4 + 5, but Andrew specified summation starting at 
the second column.)

Franklin T. Adams-Watters


-----Original Message-----
From: aplewe at sbcglobal.net

Consider all of the positive, constant-spaced integers:

1 2 3 4 5
1 3 5 7 9
2 4 6 8 10
1 4 7 10 13
2 5 8 11 14
3 6 9 12 15
etc.

The figurate numbers are the sums of rows starting with 1. The other 
rows
have various names, but are constructed similarly thus I'm including 
them
here. The question is are there any composite numbers greater than 6 
which
cannot be expressed as the sum of one of these rows minus the initial 
value?

    -Andrew Plewe-
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Yes, and I should have been a bit more specific. I'm mostly interested in
values with a representation of three or more terms. With just two terms you
can construct all integers (not just composites), or most of them anyway as
you've shown below. With three terms it gets more interesting; I realized a
few minutes ago that 8 would be in my sequence, and possibly other perfect
powers of two.


-----Original Message-----

(I'm assuming that the sum must have at least 2 terms after the initial
one is dropped, otherwise we can trivially get everything except 1.)

Try looking just at column 2 plus column 3.  This gives, in rows
(indexed by the increment):
5
8 10
11 13 15
14 16 18 20
17 19 21 23 25
20 22 24 26 28 30

As you can see, each row has an increment of 2, with alternating rows
even and odd.  And from the point shown on, each row overlaps the 2nd
previous row.  So the only values missing from this triangle are:

1,2,3,4,6,7,9,12.

We can verify that 9 can be represented (2+3+4); the others cannot.  (7
= 3 + 4 and 12 = 3 + 4 + 5, but Andrew specified summation starting at
the second column.)

Franklin T. Adams-Watters





Here's one way of examining the numbers with odd divisors. Any number with
an odd divisor can be mapped to a row in the array as follows:

(7*5 = 35)
7..5..3..1
7..6..5..4
7..7..7..7
7..8..9..10
7..9..11.13

(11 * 7 = 77)
11.8..5..2
11.9..7..5
11.10.9..8
11.11.11.11
11.12.13.14
11.13.15.17
11.14.17.20

(12 * 5 = 60)
12.10..8..6..4..2
12.11.10..9..8..7
12.12.12.12.12.12
12.13.14.15.16.17
12.14.16.18.20.22
etc.

I'm fairly confident that all numbers with an odd divisor can be reduced to
the sum of one of the rows of the array. However, it's not immediately
obvious that they can be reduced to the sum of a row minus the initial
value. Perhaps there's another bijection that maps numbers with odd divisors
to such a representation?