- Above 3 2, write 3 above 2 3, and so on. This sequence is bounded below and bounded above, so it is bounded (if you are asked for it, the GLB is 0 and the LUB is 3 2). This sequence is not monotonic. The sequence in Example (4) is the same as −1,1,−1,1,1,1,−1,. This sequence is bounded.
- CPHmodels 3.2 is a protein homology modeling server. The template recognition is based on profile-profile alignment guided by secondary structure and exposure predictions. New in version 3.2: An improvement in the alignment algorithm in case of remote homology modeling where a structure dependant gap penalty has been introduced.
Identify the Sequence 2/3, 2/9, 2/27, 2/81, This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by gives the next term. In other words,. Geometric Sequence: This is the form of a geometric sequence.
Example5.1.1What sequence is represented by the generating series (3 + 8x^2 + x^3 + frac{x^5}{7} + 100x^6 + cdotstext{?})
Solution
We just read off the coefficients of each (x^n) term. So (a_0 = 3) since the coefficient of (x^0) is 3 ((x^0 = 1) so this is the constant term). What is (a_1text{?}) It is NOT 8, since 8 is the coefficient of (x^2text{,}) so 8 is the term (a_2) of the sequence. To find (a_1) we need to look for the coefficient of (x^1) which in this case is 0. So (a_1 = 0text{.}) Continuing, we have (a_2 = 8text{,}) (a_3 = 1text{,}) (a_4 = 0text{,}) and (a_5 = frac{1}{7}text{.}) So we have the sequence
begin{equation*}3, 0, 8, 1, frac{1}{7}, 100, ldotsend{equation*}Note that when discussing generating functions, we always start our sequence with (a_0text{.})
The binary weight of n is also called Hamming weight of n.
a(n) is also the largest integer such that 2^a(n) divides binomial(2n, n) = A000984(n). - Benoit Cloitre, Mar 27 2002
To construct the sequence, start with 0 and use the rule: If k >= 0 and a(0), a(1), .., a(2^k-1) are the 2^k first terms, then the next 2^k terms are a(0) + 1, a(1) + 1, .., a(2^k-1) + 1. - Benoit Cloitre, Jan 30 2003
An example of a fractal sequence. That is, if you omit every other number in the sequence, you get the original sequence. And of course this can be repeated. So if you form the sequence a(0 * 2^n), a(1 * 2^n), a(2 * 2^n), a(3 * 2^n), .. (for any integer n > 0), you get the original sequence. - Christopher.Hills(AT)sepura.co.uk, May 14 2003
The n-th row of Pascal's triangle has 2^k distinct odd binomial coefficients where k = a(n) - 1. - Lekraj Beedassy, May 15 2003
Fixed point of the morphism 0 -> 01, 1 -> 12, 2 -> 23, 3 -> 34, 4 -> 45, etc., starting from a(0) = 0. - Robert G. Wilson v, Jan 24 2006
a(n) = number of times n appears among the mystery calculator sequences: A005408, A042964, A047566, A115419, A115420, A115421. - Jeremy Gardiner, Jan 25 2006
a(n) = number of solutions of the Diophantine equation 2^m*k + 2^(m-1) + i = n, where m >= 1, k >= 0, 0 <= i < 2^(m-1); a(5) = 2 because only (m, k, i) = (1, 2, 0) [2^1*2 + 2^0 + 0 = 5] and (m, k, i) = (3, 0, 1) [2^3*0 + 2^2 + 1 = 5] are solutions. - Hieronymus Fischer, Jan 31 2006
The first appearance of k, k >= 0, is at a(2^k-1). - Robert G. Wilson v, Jul 27 2006
Sequence is given by T^(infinity)(0) where T is the operator transforming any word w = w(1)w(2)..w(m) into T(w) = w(1)(w(1)+1)w(2)(w(2)+1)..w(m)(w(m)+1). I.e., T(0) = 01, T(01) = 0112, T(0112) = 01121223. - Benoit Cloitre, Mar 04 2009
For n >= 2, the minimal k for which a(k(2^n-1)) is not multiple of n is 2^n + 3. - Vladimir Shevelev, Jun 05 2009
Triangle inequality: a(k+m) <= a(k) + a(m). Equality holds if and only if C(k+m, m) is odd. - Vladimir Shevelev, Jul 19 2009
U5z8 Z 5u 4
The number of occurrences of value k in the first 2^n terms of the sequence is equal to binomial(n, k), and also equal to the sum of the first n - k + 1 terms of column k in the array A071919. Example with k = 2, n = 7: there are 21 = binomial(7,2) = 1 + 2 + 3 + 4 + 5 + 6 2's in a(0) to a(2^7-1). - Brent Spillner (spillner(AT)acm.org), Sep 01 2010, simplified by R. J. Mathar, Jan 13 2017
Let m be the number of parts in the listing of the compositions of n as lists of parts in lexicographic order, a(k) = n - length(composition(k)) for all k < 2^n and all n (see example); A007895 gives the equivalent for compositions into odd parts. - Joerg Arndt, Nov 09 2012
From Daniel Forgues, Mar 13 2015: (Start)
Just tally up row k (binary weight equal k) from 0 to 2^n - 1 to get the binomial coefficient C(n,k). (See A007318.)
0 1 3 7 15
0: O | . | . . | . . . . | . . . . . . . . |
1: | O | O . | O . . . | O . . . . . . . |
Sequence 2 0 3 Multiplication
2: | | O | O O . | O O . O . . . |
Sequence 0 1 2 3 4
3: | | | O | O O O . |
4: | | | | O | Vidconvert 1 4 5.
Due to its fractal nature, the sequence is quite interesting to listen to.
(End)
The binary weight of n is a particular case of the digit sum (base b) of n. - Daniel Forgues, Mar 13 2015
The mean of the first n terms is 1 less than the mean of [a(n+1),..,a(2n)], which is also the mean of [a(n+2),..,a(2n+1)]. - Christian Perfect, Apr 02 2015
Imazing 2 0 0 b1 download free. a(n) is also the largest part of the integer partition having viabin number n. The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. 'Viabin' is coined from 'via binary'. For example, consider the integer partition [2, 2, 2, 1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20. - Emeric Deutsch, Jul 20 2017
Sequence 2 0 3 23
a(n) is also known as the population count of the binary representation of n. - Chai Wah Wu, May 19 2020
