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The total number of presents the subject of the song receives is 364. A partridge in a pear tree for 12 days, 2 turtle doves for 11 days etc etc.
To the Maths literate on the forum (and I know there are a few), does this illustrate any deeper rules in Maths or is it just a novelty? Is there for example, a simple way of working out the total of such series?
Ta.
Post edited at 20:31
In reply to DaveHK:
Shirley anyone would start with fibonacci, then move on looking for the obvious.
Shirley anyone would start with fibonacci, then move on looking for the obvious.
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In reply to BannedUser27:
Fibonacci might be in there somewhere but is the number of presents on each individual day not 1,3,6,10...?
> Shirley anyone would start with fibonacci, then move on looking for the obvious.
Fibonacci might be in there somewhere but is the number of presents on each individual day not 1,3,6,10...?
Post edited at 20:46
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In reply to DaveHK:
Never mind the maths. I think it's extremely unlikely a partridge would fly into a pear tree and therefore, what was it doing there? Was it nailed to a branch or something? Just saying.
Never mind the maths. I think it's extremely unlikely a partridge would fly into a pear tree and therefore, what was it doing there? Was it nailed to a branch or something? Just saying.
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In reply to DaveHK:
I dunno, the math is going to be in one sequence or another. It's kind of mixes things up a bit, with the pagan/ Christian mix up. If I was serious about finding a sequence I'd start in base 16 or do a modern translation and use hexadecimal. Ho Ho Ho
I dunno, the math is going to be in one sequence or another. It's kind of mixes things up a bit, with the pagan/ Christian mix up. If I was serious about finding a sequence I'd start in base 16 or do a modern translation and use hexadecimal. Ho Ho Ho
In reply to Lusk:
Is the cumulative total not better shown in 3d?
> What 2d shape is a Christmas tree....?
Is the cumulative total not better shown in 3d?
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In reply to BannedUser27:
Go on then.
> If I was serious about finding a sequence I'd start in base 16 or do a modern translation and use hexadecimal. Ho Ho Ho
Go on then.
Post edited at 20:52
In reply to DaveHK:
Ok. The rule is that co-incidences are not that uncommon.
> You're meant to tell me.
Ok. The rule is that co-incidences are not that uncommon.
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In reply to DaveHK:
Well start with the basics 12 days of Christmas, 12 months in the year, 12 tribes of Judah. 12 signs of the zodiac, etc. A common sequence. So start at least one base higher no?
Well start with the basics 12 days of Christmas, 12 months in the year, 12 tribes of Judah. 12 signs of the zodiac, etc. A common sequence. So start at least one base higher no?
Post edited at 20:59
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In reply to DaveHK:
By using a spreadsheet and some intuition I came up with this - it seems to work but I don't know why.
Total=(n/2 +1)*(n+1)*(n/3) where n is number of days.
By using a spreadsheet and some intuition I came up with this - it seems to work but I don't know why.
Total=(n/2 +1)*(n+1)*(n/3) where n is number of days.
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In reply to keith-ratcliffe:
That's the spirit!
> By using a spreadsheet and some intuition I came up with this - it seems to work but I don't know why.
> Total=(n/2 +1)*(n+1)*(n/3) where n is number of days.
That's the spirit!
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In reply to DaveHK:
The number of presents each day are triangular numbers: 1, 1+2=3, 1+2+3=6 etc. so the total number is the sum of the first 12 triangular numbers: 1+3+6+10+15+21+28+36+45+55+66+78=364
The general formula for the sum of the first n triangular numbers is n(n+1)(n+2)/6
This is a bit technical but standard stuff to prove, but you can check that putting n=12 gives (12x13x14)/6=364
Anyway, if Christmas had n days, the total number of presents would be n(n+1)(n+2)/6
So if Christmas lasts all year (not far from the truth!), there should be (365x366x367)/6=8171255 presents.
EDIT: Some simple algebra shows that my formula is equivalent to keiths's (though I've no idea how he got it from a spread sheet!)
> The total number of presents the subject of the song receives is 364. A partridge in a pear tree for 12 days, 2 turtle doves for 11 days etc etc.
> To the Maths literate on the forum (and I know there are a few), does this illustrate any deeper rules in Maths or is it just a novelty? Is there for example, a simple way of working out the total of such series?
The number of presents each day are triangular numbers: 1, 1+2=3, 1+2+3=6 etc. so the total number is the sum of the first 12 triangular numbers: 1+3+6+10+15+21+28+36+45+55+66+78=364
The general formula for the sum of the first n triangular numbers is n(n+1)(n+2)/6
This is a bit technical but standard stuff to prove, but you can check that putting n=12 gives (12x13x14)/6=364
Anyway, if Christmas had n days, the total number of presents would be n(n+1)(n+2)/6
So if Christmas lasts all year (not far from the truth!), there should be (365x366x367)/6=8171255 presents.
EDIT: Some simple algebra shows that my formula is equivalent to keiths's (though I've no idea how he got it from a spread sheet!)
Post edited at 21:45
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In reply to keith-ratcliffe:
I get n^3/6 + n^2/2 + n/3 which looks like the same thing multiplied out.
But I do know why
A bit of algebra to sum series on Sum(m = 1 to n) m(n - (m - 1)) where n is the number of days
I get n^3/6 + n^2/2 + n/3 which looks like the same thing multiplied out.
But I do know why
A bit of algebra to sum series on Sum(m = 1 to n) m(n - (m - 1)) where n is the number of days
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In reply to Robert Durran:
About as technical as scratching ones arse. What about the partridge?
> The number of presents each day are triangular numbers: 1, 1+2=3, 1+2+3=6 etc. so the total number is the sum of the first 12 triangular numbers: 1+3+6+10+15+21+28+36+45+55+66+78=364
> The general formula for the sum of the first n triangular numbers is n(n+1)(n+2)/6
> This is a bit technical but standard stuff to prove, but you can check that putting n=12 gives (12x13x14)/6=364
> Anyway, if Christmas had n days, the total number of presents would be n(n+1)(n+2)/6
> So if Christmas lasts all year (not far from the truth!), there should be (365x366x367)/6=8171255 presents.
> EDIT: Some simple algebra shows that my formula is equivalent to keiths's (though I've no idea how he got it from a spread sheet!)
About as technical as scratching ones arse. What about the partridge?
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In reply to Robert Durran:
I set the spreadsheet up to just work out the summation but then noticed that the result 364 was one less than days in the year which is 7*52 (days & weeks) or 7*4*13. I just chanced that this was made up of 12/2-1, 12/3, 12+1. I tested it and it worked.
edit I can now see that this is the same as Tom in Edinburgh & Roberts formula.
I set the spreadsheet up to just work out the summation but then noticed that the result 364 was one less than days in the year which is 7*52 (days & weeks) or 7*4*13. I just chanced that this was made up of 12/2-1, 12/3, 12+1. I tested it and it worked.
edit I can now see that this is the same as Tom in Edinburgh & Roberts formula.
Post edited at 22:01
In reply to Oceanrower:
One. The pear tree is owned by a farmer.
> Just to confuse things, is a partridge in a pear tree one gift or two?
One. The pear tree is owned by a farmer.
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In reply to DaveHK:
Thanks Robert and Keith. I'm only a Geography teacher but think I might go off piste with this on the last day of term, after all the government keeps telling us we're all responsible for numeracy.
Thanks Robert and Keith. I'm only a Geography teacher but think I might go off piste with this on the last day of term, after all the government keeps telling us we're all responsible for numeracy.
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In reply to marsbar:
I like the link with Pascal's triangle:
n(n+1)(n+2)/6
= (n+2)!/(n-1)!3!
= (n+2)C3
Giving 14C3 for 12 days.
I like the link with Pascal's triangle:
n(n+1)(n+2)/6
= (n+2)!/(n-1)!3!
= (n+2)C3
Giving 14C3 for 12 days.
In reply to keith-ratcliffe:
Very lucky; the link to the number of days in a year is, of course, entirely coincidental!
> I set the spreadsheet up to just work out the summation but then noticed that the result 364 was one less than days in the year which is 7*52 (days & weeks) or 7*4*13. I just chanced that this was made up of 12/2-1, 12/3, 12+1. I tested it and it worked.
Very lucky; the link to the number of days in a year is, of course, entirely coincidental!
In reply to keith-ratcliffe:
Indeed. The sum of triangular numbers as required appears as a case of the "hockey stick addition"
I'd not seen the powers of 11 one before. Cool.
I'd also not seen before the link between Pascal's Triangle and the standard formula
Sum[n(n+1)(n+2).....(n+r)]
=[1/(r+2)]n(n+1)(n+2)......(n+r+1)
of which the Presents is just the case r=1 (give or take a factor of 2)
> That third link is awesome - the elegance of maths sometimes astounds me.
Indeed. The sum of triangular numbers as required appears as a case of the "hockey stick addition"
I'd not seen the powers of 11 one before. Cool.
I'd also not seen before the link between Pascal's Triangle and the standard formula
Sum[n(n+1)(n+2).....(n+r)]
=[1/(r+2)]n(n+1)(n+2)......(n+r+1)
of which the Presents is just the case r=1 (give or take a factor of 2)
Post edited at 22:34
In reply to keith-ratcliffe:
Hockey stick addition, brilliant
> That third link is awesome - the elegance of maths sometimes astounds me.
Hockey stick addition, brilliant
In reply to Robert Durran:
The days/weeks was indeed chance recognition of a known pattern but it got me thinking about the origins of our calendars - which civilisation first used the term weeks in a year and identified it as 52? The moons cycle is an obvious phenomena but doesn't easily lead to weeks & a year.
The days/weeks was indeed chance recognition of a known pattern but it got me thinking about the origins of our calendars - which civilisation first used the term weeks in a year and identified it as 52? The moons cycle is an obvious phenomena but doesn't easily lead to weeks & a year.
In reply to Lusk:
Indeed it seems the brainwashed who wanna brainwash can't either add up or to 365 plus 1 in 4 and also see the fable aspect to the rhyme. Maybe they are just like dim innit. Bottomless pow brah bottomless po!
Indeed it seems the brainwashed who wanna brainwash can't either add up or to 365 plus 1 in 4 and also see the fable aspect to the rhyme. Maybe they are just like dim innit. Bottomless pow brah bottomless po!
In reply to BannedUser27:
?
> Indeed it seems the brainwashed who wanna brainwash can't either add up or to 365 plus 1 in 4 and also see the fable aspect to the rhyme. Maybe they are just like dim innit. Bottomless pow brah bottomless po!
?
In reply to BannedUser27:
It's an early example of David Bowie song writing, chuck a load of things in a box and randomly pick them out and write a song about it.
It's an early example of David Bowie song writing, chuck a load of things in a box and randomly pick them out and write a song about it.
In reply to BannedUser27:
Eh?
> Indeed it seems the brainwashed who wanna brainwash can't either add up or to 365 plus 1 in 4 and also see the fable aspect to the rhyme. Maybe they are just like dim innit. Bottomless pow brah bottomless po!
Eh?
In reply to DaveHK:
Maybe, when you free the heal and go steep and deep ask one of the folk who cross the line with you what they think.
> ?
Maybe, when you free the heal and go steep and deep ask one of the folk who cross the line with you what they think.
In reply to BannedUser27:
Is that you mum? Best stay off the sherry.
> Maybe, when you free the heal and go steep and deep ask one of the folk who cross the line with you what they think.
Is that you mum? Best stay off the sherry.
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In reply to BannedUser27:
Something to do with skiing? On drugs?
Anything interesting to contribute?
> Maybe, when you free the heal and go steep and deep ask one of the folk who cross the line with you what they think.
Something to do with skiing? On drugs?
Anything interesting to contribute?
In reply to BannedUser27:
Jimmy Page, get a grip!
> I'm not sure about him with that Lori Maddox episode.
Jimmy Page, get a grip!
In reply to Robert Durran:
Are you really this stupid, read the thread, who dropped the first ski reference, not me. We're talking about occult numerology here. Try to keep up!
Are you really this stupid, read the thread, who dropped the first ski reference, not me. We're talking about occult numerology here. Try to keep up!
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In reply to DaveHK:
The thread sent me off in a different direction - this site allows you to search all the digits in the decimal part of Pi - there are 200 million to have a look at.
http://www.angio.net/pi/
So my question is - is there such a thing as a 'Pi-whack' - a bit like a Googlewhack - a set of digits that only appear once in the decimal part of Pi?
Anyone up for the challenge?
The thread sent me off in a different direction - this site allows you to search all the digits in the decimal part of Pi - there are 200 million to have a look at.
http://www.angio.net/pi/
So my question is - is there such a thing as a 'Pi-whack' - a bit like a Googlewhack - a set of digits that only appear once in the decimal part of Pi?
Anyone up for the challenge?
1
In reply to Lusk:
Was that not started by Bob Dylan?
vimeo.com/72540087
> It's an early example of David Bowie song writing, chuck a load of things in a box and randomly pick them out and write a song about it.
Was that not started by Bob Dylan?
vimeo.com/72540087
Post edited at 21:57
In reply to Lusk:
It's Christmas tree shaped. I know this coz my 6yr old drew one and says that's what shape it is.
> What 2d shape is a Christmas tree....?
It's Christmas tree shaped. I know this coz my 6yr old drew one and says that's what shape it is.
In reply to DaveHK:
it occurred to me today that the total number of presents (364) is 4 times the sum of the first 6 square numbers. Explain that one! Spotting this has the advantage that the general formula for a Christmas with n days can be easily found from the standard formula for summing square numbers.
it occurred to me today that the total number of presents (364) is 4 times the sum of the first 6 square numbers. Explain that one! Spotting this has the advantage that the general formula for a Christmas with n days can be easily found from the standard formula for summing square numbers.
Post edited at 23:58
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In reply to DaveHK:
One further nice bit of maths from this:
The answer is the sum of the first 12 triangular numbers.
It is also 12x1 + 11x2 + 10x3 + .........2x11 + 1x12
Generalising to n days and equating (using the standard formula r(r+1)/2 for the rth triangular number) gives:
Sum{r(r+1)/2} = Sum{(n+1-r)r}
This leads with a little algebra to the standard formula for summing square numbers:
Sum{r^2} = n(n+1)(2n+1)/6
I have always found it a bit unsatisfactory to have to resort to induction to prove this (especially since my pupils are expected to know the formula before meeting induction!)
In fact a similar method with a bit of tweaking gets the formula for summing cubes from the formulae for summing squares and for summing whole numbers (triangular numbers). It is possible to continue to fourth powers and so on.
One further nice bit of maths from this:
The answer is the sum of the first 12 triangular numbers.
It is also 12x1 + 11x2 + 10x3 + .........2x11 + 1x12
Generalising to n days and equating (using the standard formula r(r+1)/2 for the rth triangular number) gives:
Sum{r(r+1)/2} = Sum{(n+1-r)r}
This leads with a little algebra to the standard formula for summing square numbers:
Sum{r^2} = n(n+1)(2n+1)/6
I have always found it a bit unsatisfactory to have to resort to induction to prove this (especially since my pupils are expected to know the formula before meeting induction!)
In fact a similar method with a bit of tweaking gets the formula for summing cubes from the formulae for summing squares and for summing whole numbers (triangular numbers). It is possible to continue to fourth powers and so on.
2
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