Re: [math-fun] More fun with sequences embedded in decimals
Yes, in fact you cracked it. What I thought was notable about this solution was: - It arises naturally from a common length conversion. - We know that 0.3937 = 0.3937 1/2.54 = 0.3937007874... Given that the numbers themselves are so close, it would stand to reason that their geometric mean would be near their arithmetic mean, so that sqrt(0.3937 * (1/2.54)) .= (0.3937+(1/2.54))/2 = 0.3937003937... Both accurate to the decimals shown. This gives us a square root with a pretty repeated start block, which you observed and reproduced in another example. - But then there is the nice common factor between the block values 3937 = 31*127 254 = 2*127 This arises from the unusual square factor in the prime factorization: 999998 = 2*31*127^2 This has the peculiar side benefit that the quotient of the repeating blocks is a terminating decimal: 3937/254 = 15.5 Going back to our geometric mean identity, we get sqrt(0.3937 * (1/2.54)) = sqrt(0.155) = 0.3937003937... Which gives us a pretty square root of a terminating decimal. Multiplying by 20 on both sides then gives sqrt(62) = 7.874 007874 ... I will guess that small integer square roots with pretty blocks like this are rare in any base. - And an aside about the Mersenne primes You can't miss that 31 and 127 are Mersenne primes, which at first might seem incidental. However, consider that 2^20 = 1048576 .= 10^6 is close to a power of 10. Obviously this is well known, since we call 2^10 bytes a kilobyte, 2^20 bytes a megabyte, etc. We can break this up to 2 * 2^5 * (2^7)^2 = 1048576 .= 10^6 And if we tweak a couple of these factors slightly, we shouldn't affect the product too much, so lets tweak 2^5 and 2^7 down a notch: 2 * (2^5-1) * (2^7-1)^2 .= 999998 .= 10^6 So perhaps it is now less surprising that Mersenne primes might multiply to a value near 10^6, since Mersenne primes are slightly smaller than powers of 2 and 2^20 is slightly larger than 10^6. But of course, this is a hindsight analysis, since I knew the factorization to begin with. On 2/5/2012 3:09 AM, Robert Munafo wrote:
I don't see how the Mersenne factors matter.
The first two of David's properties, 1/2.54 = 00.3937 007874... and 1/39.37=00.0254 000508... are explained by their product being 99.9998 = (10^N-2) / 10^K, as I suggested before.
The other of David's properties, regarding the geometric mean, also comes from (10^N-2) / 10^K, as you can see here:
1/sqrt(99.9998) = .100000100000150000250000437500787501443752681255027353246111 1/sqrt(99.998) = .100001000015000250004375078751443776813002743871274179225808 1/sqrt(99.98) = .100010001500250043757876444018175282935398352327750882087668 1/sqrt(99.8) = .100100150250438288946436286857920839193514369563735147583946
These all have the same pattern, related to the series expansion of sqrt(1/(1-2x)), which you can get by "Series[sqrt(1/(1-2*x)),{x,0,8}]" in Mathematica or Wolfram Alpha. The coefficients are: 1, 1, 3/2, 5/2, 35/8, 63/8, 231/16, 429/16, 6435/128, ... The numerators are https://oeis.org/A001790, which are also the numerators of the sqrt(1/(1-x)) series expansion because only the denominators change.
Also, for fun, what if I had decided to use a system that was the harmonic mean between avoirdupois and metric. How many inches in a centimeter?
On Sun, Feb 5, 2012 at 10:20, David Wilson <davidwwilson@comcast.net> wrote:
What I thought was notable about this solution was: - It arises naturally from a common length conversion. [...] [and]
sqrt(62) = 7.874 007874 ...
I will guess that small integer square roots with pretty blocks like this are rare in any base.
Well, of course there's 2*7^2 = 98, which gives us the pretty digit pattern: sqrt(2) = 1.4 14 ... More on this later, plus a mnemonic for the coefficients of Series[sqrt(1/(1-2x))] In base 5, because 111^2*3963+2 = 5^11, the square root of 3963*5 = 19815_10 = 1113230_5 = 1030.34033 00103034033 00132103324... where the digit-groups are 11 digits long. In base 6, because 46*13^2+2 = 6^5, the square root of 46*6 = 276_10 = 1140_6 = 24.34 02434 04053... where the digit-groups are 5 digits long. In base 7, because 977423*17^2+2 = 7^10, the square root of 977423_10 = 11210426_7 = 2611.43464 0261143464 0421565362... where the digit-groups are 10 digits long. In each of these examples, the third and 4th pretty digit-groups are 3/2 and 5/2 times the first two digit groups. For example in the familiar square root of 2, the first 8 digits (in groups of 2) are 14, 14, 21, 35 and 14*1.5 = 21, 14*5/2=35. (You can check the other pretty digit examples above, but remember I showed each in its native base and in odd bases 1/2 is an infinite "repeating decimal". So 261143464 times 3/2 in base 7 is 421565362.3333... where the "0.3333..." is 1/2 in base 7. These 3's overrun the following digit-group which hides the 5/2 term.) So there are 4 examples (-: The digits of the square root of 2, taken in pairs and divided by 14, are one way of remembering the coefficients of the series expansion of sqrt(1/(1-2x)), if you ever find yourself in the need and are stuck with just a pocket calculator. (-: The "62" ought to be 61.25, but it's overrun by the leading "1" of the following term which is 14*63/8=110.25) (And yes, I think these are cool, thank you David. I'll add a bit about it to my numbers pages.) - - - Related to this (in my mind at least) is the Catalan numbers and (1-sqrt(1-4x))/2... Folks who experiment with nice-looking digits might eventually play around with things like: 1.001^2 = 1.002001 1.002001^2 = 1.004006004001 How about a sequence of digits where you can square it and get almost the same sequence? 0.00010001000a000b000c...^2 = 0.00000001000a000b000c... where the two sides differ only in one digit. This can be worked out iteratively: A_0 = 0 A_1 = A_0^2 + .0001 = .0001 A_2 = A_1^2 + .0001 = .00010001 A_3 = A_2^2 + .0001 = .000100010002... A_4 = A_3^2 + .0001 = .0001000100020005... A_5 = A_4^2 + .0001 = .00010001000200050014... and so on. In the limit you get (1-sqrt(1-0.0004))/2 = .000100010002000500140042013204291430... which has the same digits as the easier-to-remember 50-sqrt(49*51) = .0100010002000500140042013204291430... On 2/5/2012 3:09 AM, Robert Munafo wrote:
The other of David's properties, regarding the geometric mean, also comes from (10^N-2) / 10^K, as you can see here:
1/sqrt(99.9998) = .**10000010000015000025000043750... [...] These all have the same pattern, related to the series expansion of sqrt(1/(1-2x)), which you can get by "Series[sqrt(1/(1-2*x)),{x,0,**8}]" in Mathematica or Wolfram Alpha. The coefficients are: 1, 1, 3/2, 5/2, 35/8, 63/8, 231/16, 429/16, 6435/128, ... The numerators are https://oeis.org/A001790, which are also the numerators of the sqrt(1/(1-x)) series expansion because only the denominators change.
-- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com
Hello, back in the old days, I used a trick like that to find the algebraic generating function of a finite sequence. The trick was the following, 1) find if the sequence satisfies a linear polynomial recurrence. 2) Extend the sequence several hundred terms. 3) Evaluate the sequence at a very small point like 1/10000 4) combine a series of evaluations of the sequence at 1/10001, 1/10002, 1/10003, 5) Use Newton interpolation formula to find the algebraic generating function. That method was based on the fact that a number like sqrt(51)/14 = 0.5101020306102035... shows the first terms of a known sequence. And many examples I saw here in the past days. Here is the result of this (an article of 1993), http://arxiv.org/ftp/arxiv/papers/0912/0912.0072.pdf Best regards, ps : well the article is in french. Simon Plouffe Le 06/02/2012 03:13, Robert Munafo a écrit :
On Sun, Feb 5, 2012 at 10:20, David Wilson<davidwwilson@comcast.net> wrote:
What I thought was notable about this solution was: - It arises naturally from a common length conversion. [...] [and]
sqrt(62) = 7.874 007874 ...
I will guess that small integer square roots with pretty blocks like this are rare in any base.
Well, of course there's 2*7^2 = 98, which gives us the pretty digit pattern:
sqrt(2) = 1.4 14 ...
More on this later, plus a mnemonic for the coefficients of Series[sqrt(1/(1-2x))]
In base 5, because 111^2*3963+2 = 5^11, the square root of 3963*5 = 19815_10 = 1113230_5 = 1030.34033 00103034033 00132103324... where the digit-groups are 11 digits long.
In base 6, because 46*13^2+2 = 6^5, the square root of 46*6 = 276_10 = 1140_6 = 24.34 02434 04053... where the digit-groups are 5 digits long.
In base 7, because 977423*17^2+2 = 7^10, the square root of 977423_10 = 11210426_7 = 2611.43464 0261143464 0421565362... where the digit-groups are 10 digits long.
In each of these examples, the third and 4th pretty digit-groups are 3/2 and 5/2 times the first two digit groups. For example in the familiar square root of 2, the first 8 digits (in groups of 2) are 14, 14, 21, 35 and 14*1.5 = 21, 14*5/2=35.
(You can check the other pretty digit examples above, but remember I showed each in its native base and in odd bases 1/2 is an infinite "repeating decimal". So 261143464 times 3/2 in base 7 is 421565362.3333... where the "0.3333..." is 1/2 in base 7. These 3's overrun the following digit-group which hides the 5/2 term.)
So there are 4 examples (-:
The digits of the square root of 2, taken in pairs and divided by 14, are one way of remembering the coefficients of the series expansion of sqrt(1/(1-2x)), if you ever find yourself in the need and are stuck with just a pocket calculator. (-: The "62" ought to be 61.25, but it's overrun by the leading "1" of the following term which is 14*63/8=110.25)
(And yes, I think these are cool, thank you David. I'll add a bit about it to my numbers pages.)
- - -
Related to this (in my mind at least) is the Catalan numbers and (1-sqrt(1-4x))/2...
Folks who experiment with nice-looking digits might eventually play around with things like:
1.001^2 = 1.002001
1.002001^2 = 1.004006004001
How about a sequence of digits where you can square it and get almost the same sequence?
0.00010001000a000b000c...^2 = 0.00000001000a000b000c...
where the two sides differ only in one digit. This can be worked out iteratively:
A_0 = 0
A_1 = A_0^2 + .0001 = .0001
A_2 = A_1^2 + .0001 = .00010001
A_3 = A_2^2 + .0001 = .000100010002...
A_4 = A_3^2 + .0001 = .0001000100020005...
A_5 = A_4^2 + .0001 = .00010001000200050014...
and so on. In the limit you get
(1-sqrt(1-0.0004))/2 = .000100010002000500140042013204291430...
which has the same digits as the easier-to-remember
50-sqrt(49*51) = .0100010002000500140042013204291430...
On 2/5/2012 3:09 AM, Robert Munafo wrote:
The other of David's properties, regarding the geometric mean, also comes from (10^N-2) / 10^K, as you can see here:
1/sqrt(99.9998) = .**10000010000015000025000043750... [...] These all have the same pattern, related to the series expansion of sqrt(1/(1-2x)), which you can get by "Series[sqrt(1/(1-2*x)),{x,0,**8}]" in Mathematica or Wolfram Alpha. The coefficients are: 1, 1, 3/2, 5/2, 35/8, 63/8, 231/16, 429/16, 6435/128, ... The numerators are https://oeis.org/A001790, which are also the numerators of the sqrt(1/(1-x)) series expansion because only the denominators change.
Also, many of the formulas found where added to the OEIS, in 2009, 2011, 18 years later, some people rediscovered the weel and posted their finding but this is another story. Simon Plouffe Le 06/02/2012 03:13, Robert Munafo a écrit :
On Sun, Feb 5, 2012 at 10:20, David Wilson<davidwwilson@comcast.net> wrote:
What I thought was notable about this solution was: - It arises naturally from a common length conversion. [...] [and]
sqrt(62) = 7.874 007874 ...
I will guess that small integer square roots with pretty blocks like this are rare in any base.
Well, of course there's 2*7^2 = 98, which gives us the pretty digit pattern:
sqrt(2) = 1.4 14 ...
More on this later, plus a mnemonic for the coefficients of Series[sqrt(1/(1-2x))]
In base 5, because 111^2*3963+2 = 5^11, the square root of 3963*5 = 19815_10 = 1113230_5 = 1030.34033 00103034033 00132103324... where the digit-groups are 11 digits long.
In base 6, because 46*13^2+2 = 6^5, the square root of 46*6 = 276_10 = 1140_6 = 24.34 02434 04053... where the digit-groups are 5 digits long.
In base 7, because 977423*17^2+2 = 7^10, the square root of 977423_10 = 11210426_7 = 2611.43464 0261143464 0421565362... where the digit-groups are 10 digits long.
In each of these examples, the third and 4th pretty digit-groups are 3/2 and 5/2 times the first two digit groups. For example in the familiar square root of 2, the first 8 digits (in groups of 2) are 14, 14, 21, 35 and 14*1.5 = 21, 14*5/2=35.
(You can check the other pretty digit examples above, but remember I showed each in its native base and in odd bases 1/2 is an infinite "repeating decimal". So 261143464 times 3/2 in base 7 is 421565362.3333... where the "0.3333..." is 1/2 in base 7. These 3's overrun the following digit-group which hides the 5/2 term.)
So there are 4 examples (-:
The digits of the square root of 2, taken in pairs and divided by 14, are one way of remembering the coefficients of the series expansion of sqrt(1/(1-2x)), if you ever find yourself in the need and are stuck with just a pocket calculator. (-: The "62" ought to be 61.25, but it's overrun by the leading "1" of the following term which is 14*63/8=110.25)
(And yes, I think these are cool, thank you David. I'll add a bit about it to my numbers pages.)
- - -
Related to this (in my mind at least) is the Catalan numbers and (1-sqrt(1-4x))/2...
Folks who experiment with nice-looking digits might eventually play around with things like:
1.001^2 = 1.002001
1.002001^2 = 1.004006004001
How about a sequence of digits where you can square it and get almost the same sequence?
0.00010001000a000b000c...^2 = 0.00000001000a000b000c...
where the two sides differ only in one digit. This can be worked out iteratively:
A_0 = 0
A_1 = A_0^2 + .0001 = .0001
A_2 = A_1^2 + .0001 = .00010001
A_3 = A_2^2 + .0001 = .000100010002...
A_4 = A_3^2 + .0001 = .0001000100020005...
A_5 = A_4^2 + .0001 = .00010001000200050014...
and so on. In the limit you get
(1-sqrt(1-0.0004))/2 = .000100010002000500140042013204291430...
which has the same digits as the easier-to-remember
50-sqrt(49*51) = .0100010002000500140042013204291430...
On 2/5/2012 3:09 AM, Robert Munafo wrote:
The other of David's properties, regarding the geometric mean, also comes from (10^N-2) / 10^K, as you can see here:
1/sqrt(99.9998) = .**10000010000015000025000043750... [...] These all have the same pattern, related to the series expansion of sqrt(1/(1-2x)), which you can get by "Series[sqrt(1/(1-2*x)),{x,0,**8}]" in Mathematica or Wolfram Alpha. The coefficients are: 1, 1, 3/2, 5/2, 35/8, 63/8, 231/16, 429/16, 6435/128, ... The numerators are https://oeis.org/A001790, which are also the numerators of the sqrt(1/(1-x)) series expansion because only the denominators change.
participants (3)
-
David Wilson -
Robert Munafo -
Simon Plouffe