[math-fun] Sums of digits of powers of 2
Let E(n) and V(n) be the expected value and variance, respectively, of the sum of n independent draws from {0,1,2,...,9}; we can easily write E(n) and V(n) in closed form, and thereby extend to E(x) and V(x) where x is any positive real number. Let S(n) be the sum of the digits of 2^n, and let c = log_10 2, so that 2^n has about cn digits. What if anything is known (empirically or rigorously) about the distribution and autocorrelations of the sequence D_n whose nth term is (S(n)-E(cn))/sqrt(V(cn))? If that's obscure, just think of it as the sum of the digits of 2^n, adjusted to make it look like a Gaussian of mean 0 and variance 1. (The "D" is for discrepancy.) Jim Propp
S(n) is the sequence http://oeis.org/A001370 where it is asserted that S(n) ~ ne/2 Maple calculates the first 5000 terms in just over 9 seconds -- so experiments should be easy. On Wed, Apr 17, 2013 at 3:46 PM, James Propp <jamespropp@gmail.com> wrote:
Let E(n) and V(n) be the expected value and variance, respectively, of the sum of n independent draws from {0,1,2,...,9}; we can easily write E(n) and V(n) in closed form, and thereby extend to E(x) and V(x) where x is any positive real number.
Let S(n) be the sum of the digits of 2^n, and let c = log_10 2, so that 2^n has about cn digits.
What if anything is known (empirically or rigorously) about the distribution and autocorrelations of the sequence D_n whose nth term is (S(n)-E(cn))/sqrt(V(cn))?
If that's obscure, just think of it as the sum of the digits of 2^n, adjusted to make it look like a Gaussian of mean 0 and variance 1. (The "D" is for discrepancy.)
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
How can S(n) grow like ne/2? Surely it must grow like n times the log of 2 to the base 10 (the number of digits) times 9/2 (the average of 0,1,...,9). Jim Propp On Wednesday, April 17, 2013, W. Edwin Clark <wclark@mail.usf.edu> wrote:
S(n) is the sequence http://oeis.org/A001370 where it is asserted that S(n) ~ ne/2 Maple calculates the first 5000 terms in just over 9 seconds -- so experiments should be easy.
On Wed, Apr 17, 2013 at 3:46 PM, James Propp <jamespropp@gmail.com> wrote:
Let E(n) and V(n) be the expected value and variance, respectively, of the sum of n independent draws from {0,1,2,...,9}; we can easily write E(n) and V(n) in closed form, and thereby extend to E(x) and V(x) where x is any positive real number.
Let S(n) be the sum of the digits of 2^n, and let c = log_10 2, so that 2^n has about cn digits.
What if anything is known (empirically or rigorously) about the distribution and autocorrelations of the sequence D_n whose nth term is (S(n)-E(cn))/sqrt(V(cn))?
If that's obscure, just think of it as the sum of the digits of 2^n, adjusted to make it look like a Gaussian of mean 0 and variance 1. (The "D" is for discrepancy.)
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Shouldn't that be 4.5 log_10(2) n? Admittedly, 1.354635 is close to 1.359141. Rich ----- Quoting "W. Edwin Clark" <wclark@mail.usf.edu>:
S(n) is the sequence http://oeis.org/A001370 where it is asserted that S(n) ~ ne/2 Maple calculates the first 5000 terms in just over 9 seconds -- so experiments should be easy.
On Wed, Apr 17, 2013 at 3:46 PM, James Propp <jamespropp@gmail.com> wrote:
Let E(n) and V(n) be the expected value and variance, respectively, of the sum of n independent draws from {0,1,2,...,9}; we can easily write E(n) and V(n) in closed form, and thereby extend to E(x) and V(x) where x is any positive real number.
Let S(n) be the sum of the digits of 2^n, and let c = log_10 2, so that 2^n has about cn digits.
What if anything is known (empirically or rigorously) about the distribution and autocorrelations of the sequence D_n whose nth term is (S(n)-E(cn))/sqrt(V(cn))?
If that's obscure, just think of it as the sum of the digits of 2^n, adjusted to make it look like a Gaussian of mean 0 and variance 1. (The "D" is for discrepancy.)
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I would say you are right -- unless the author of that assertion knows something about the distribution of the digits of 2^n that I don't know. Has anything been proved about the distribution of the digits of 2^n? For example some powers of 2 have all even digits: http://oeis.org/A068994 :-) On Wed, Apr 17, 2013 at 5:48 PM, <rcs@xmission.com> wrote:
Shouldn't that be 4.5 log_10(2) n? Admittedly, 1.354635 is close to 1.359141.
Rich
----- Quoting "W. Edwin Clark" <wclark@mail.usf.edu>:
S(n) is the sequence http://oeis.org/A001370 where it is asserted that
S(n) ~ ne/2 Maple calculates the first 5000 terms in just over 9 seconds -- so experiments should be easy.
On Wed, Apr 17, 2013 at 3:46 PM, James Propp <jamespropp@gmail.com> wrote:
Let E(n) and V(n) be the expected value and variance, respectively, of
the sum of n independent draws from {0,1,2,...,9}; we can easily write E(n) and V(n) in closed form, and thereby extend to E(x) and V(x) where x is any positive real number.
Let S(n) be the sum of the digits of 2^n, and let c = log_10 2, so that 2^n has about cn digits.
What if anything is known (empirically or rigorously) about the distribution and autocorrelations of the sequence D_n whose nth term is (S(n)-E(cn))/sqrt(V(cn))?
If that's obscure, just think of it as the sum of the digits of 2^n, adjusted to make it look like a Gaussian of mean 0 and variance 1. (The "D" is for discrepancy.)
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I think so, too. It would be very surprising otherwise. Charles Greathouse Analyst/Programmer Case Western Reserve University On Wed, Apr 17, 2013 at 5:48 PM, <rcs@xmission.com> wrote:
Shouldn't that be 4.5 log_10(2) n? Admittedly, 1.354635 is close to 1.359141.
Rich
-----
Quoting "W. Edwin Clark" <wclark@mail.usf.edu>:
S(n) is the sequence http://oeis.org/A001370 where it is asserted that
S(n) ~ ne/2 Maple calculates the first 5000 terms in just over 9 seconds -- so experiments should be easy.
On Wed, Apr 17, 2013 at 3:46 PM, James Propp <jamespropp@gmail.com> wrote:
Let E(n) and V(n) be the expected value and variance, respectively, of
the sum of n independent draws from {0,1,2,...,9}; we can easily write E(n) and V(n) in closed form, and thereby extend to E(x) and V(x) where x is any positive real number.
Let S(n) be the sum of the digits of 2^n, and let c = log_10 2, so that 2^n has about cn digits.
What if anything is known (empirically or rigorously) about the distribution and autocorrelations of the sequence D_n whose nth term is (S(n)-E(cn))/sqrt(V(cn))?
If that's obscure, just think of it as the sum of the digits of 2^n, adjusted to make it look like a Gaussian of mean 0 and variance 1. (The "D" is for discrepancy.)
Jim Propp ______________________________**_________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/**cgi-bin/mailman/listinfo/math-**fun<http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun>
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Let's take a (light-hearted) look at some empirical evidence. I calculated (Sum-of-Digits of 2^n)/n for a handful of random, increasingly-large n: 9*log10(2)/2 1.354634980 e/2 1.359140914 n SD(2^n)/n 48811 1.359140358 54622 1.359141005 95411 1.359140980 159152 1.359140947 506737 1.359140935 Yes of course, I'm being facetious. Back in 2006, I did this with powers of 5 and the number pi: http://oeis.org/A119666 On Apr 17, 2013, at 5:48 PM, rcs@xmission.com wrote:
Shouldn't that be 4.5 log_10(2) n? Admittedly, 1.354635 is close to 1.359141.
Quoting "W. Edwin Clark" <wclark@mail.usf.edu>:
S(n) is the sequence http://oeis.org/A001370 where it is asserted that S(n) ~ ne/2
participants (5)
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James Propp -
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W. Edwin Clark