| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Start with 1 2 3 4 5 6 7 8 9 Cross out every 2nd number 1 3 5 7 9 Form partial sums 1 4 9 16 25 squares Start with 1 2 3 4 5 6 7 8 9 10 11 Cross out every 3rd number 1 2 4 5 7 8 10 11 Form partial sums 1 3 7 12 19 27 37 48 Cross out every 2nd number 1 7 19 37 Form partial sums 1 8 27 64 cubes Start with 1 2 3 4 5 6 7 8 9 10 11 12 13 Cross out every 4th number 1 2 3 5 6 7 9 10 11 13 Form partial sums 1 3 6 11 17 24 33 43 54 67 Cross out every 3rd number 1 3 11 17 33 43 67 Form partial sums 1 4 15 32 65 108 175 Cross out every 2nd number 1 15 65 175 Form partial sums 1 16 81 256 4th powers! Does this pattern continue? Reference --------- C. J. Smyth, Curioser and Curioser. New James Cook Mathematical Notes. Issue 36 (Feb 1985) page 4115.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 271.1 | SPRITE::OSMAN | Wed May 01 1985 09:03 | 12 | ||
This seems obvious to me. The difference of two "partial sum"s is a finite difference. It's well known that if you take a sequence of numbers whose first level finite differences are constant, the sequence is linear, i.e. proportional to FIRST powers. ... and that if second level finite differences are constant, the sequence is quadratic, i.e. proportional to SQUARES etc. to CUBES, QUARTICS. /Eric | |||||
| 271.2 | HARE::STAN | Wed May 01 1985 13:16 | 5 | ||
This may provide some motivation about why it works; but doesn't make the result obvious. Why isn't the result a constant times the nth powers? (or even a quadratic times the nth powers). Also, deleting every kth element in the middle of the procedure means that you're not working with strict partial sums. | |||||
| 271.3 | LATOUR::JMUNZER | Sun Jun 02 1985 08:29 | 92 | ||
These delightful patterns can be interpreted (and understood) as a novel
method of producing n-tuples.
Suppose you want 3-tuples.
Start with {aaa}.
One by one, go through the a's, change them to b's, keeping the
old and the new 3-tuples. Now you have 8 3-tuples.
One by one, go through the a's again, changing them to c's, and
keeping the old and the new 3-tuples. 27.
etc.
A possible way to diagram the process is below. The spirit is to
list the candidates for change, with capitalization indicating
the "a" which may be changed at that step. Roughly, if the "a"
changes, you move down (and land above the hyphens); if it doesn't,
you move right (to below the hyphens).
Aaa aAa aaA aaa Aaa aAa aaA aaa Aaa aAa aaA aaa
bAa abA aab cAa acA aac dAa adA aad
--- --- --- --- --- --- --- ---
baA aba Aab caA aca Aac daA ada
baa Aba aAb caa Aca aAc daa
bAa abA aab cAa acA aac
baA aba Aab caA aca
baa Aba aAb caa
bAa abA aab
baA aba
baa
bbA abb ccA acc ddA add
bab cAb cac dAc dad
--- cbA acb dcA adc
bba bcA abc cdA acd
--- bac dAb cad
Abb --- dbA adb
bAb cca bdA abd
bbA cab --- bad
cba Acc ---
bca cAc dda
abb Acb dac
bab Abc dca
bba bAc cda
ccA dab
cAb dba
etc. etc.
bbb ccc ddd
ccb ddc
cbc dcd
bcc cdd
cbb ddb
bcb dbd
bbc bdd
--- dcc
bbb cdc
dcb
dbc
etc.
Count the items in each little list:
1 1 1 1 1 1 1 1 1 1 1 1
1 2 3 4 5 6 7 8 9
1 3 7 12 19 27
1 8 27
This is, in essence, the cubes pattern of the problem.
Patterns for other powers are diagrams of the n-tuple procedure for other
n's.
Trying to tie this n-tuple procedure to the problem:
[1] "Crossing out every ---th number" corresponds to the holes
in the procedure diagram, which cause only the correct number
of a's to be examined for change.
[2] Forming partial sums" corresponds to the collection of
candidates for change (above & below the hyphens).
[3] Squares, cubes, etc. are the numbers of 2-tuples, 3-tuples,
etc. And what a great way to use a linear-looking thing
to produce very unlinear results.
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