In recreational mathematics, van Eck's sequence is an integer sequencedefined recursively as follows. Let a0 = 0. Then, for n ≥ 0, if there exists an m < n such that am = an, take the largest such m and set an+1 = n − m; otherwise an+1 = 0. Thus the first occurrence of an integer in the sequence is followed by a 0, and the second and subsequent occurrences are followed by the size of the gap between the two most recent occurrences.
The first few terms of the sequence are(OEIS: A181391):
- 0, 0, 1, 0, 2, 0, 2, 2, 1, 6, 0, 5, 0, 2, 6, 5, 4, 0, 5 .. [1]
- Multiple choice questions follow each reading passage and guide students to identify the sequence of what they've read. Perfect for skill reinforcement and assessment! Ideal for individual students, large groups, and as a center activity. Reading Level 2.0-3.5 Interest Level 2-12.
- Apr 22, 2014 I believe they can be ignored too. The thing I would check is the sequence value against the max value of the column they populate. If the sequence is lower you could run into duplicate value exception when inserting into the table using that sequence until the sequence generates numbers larger than the max value of the column.
- 2) 0, 3, 8, 15, 24n term 1 0 2 3 3 8 4 15 5 24 6 Think of another sequence that this one is near. How about this sequence: n term 1 1 2 4 3 9 4 16 5 25 6 36 Each term of our sequence is 1 less than this sequence and this sequence is easily recognized as the sequence of the squares of n or n².
The sequence was named by Neil Sloane after Jan Ritsema van Eck, who contributed it to the On-Line Encyclopedia of Integer Sequences in 2010.
Nov 28, 2016 The explicit formula for the following geometric sequence which is 0.2, -0.06, 0.018, -0.0054, 0.00162. Would be this: an = 0.2(-0.3)^n-1. When we say geometric sequence, this is the sequence of numbers wherein each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.
Properties[edit]
It is known that the sequence contains infinitely many zeros and that it is unbounded.[1]
It is conjectured, but not proved, that the sequence contains every positive integer, and that every pair of non-negative integers apart from (1,1) and (n,n+1) appears as consecutive terms in the sequence.[1]
Variations[edit]
The sequence OEIS: A181391 is defined with a0 = 0. This can be changed such that the sequence starts with any integer.
For Example:
With a0 = 1, OEIS: A171911:
1, 0, 0, 1, 3, 0, 3, 2, 0, 3, 3, 1, 8, 0, 5, 0, 2, 9, 0, 3 ..[2]
With a0 = 2, OEIS: A171912:
2, 0, 0, 1, 0, 2, 5, 0, 3, 0, 2, 5, 5, 1, 10, 0, 6, 0, 2, 8 ..[3]
In fact, OEIS has eight other entries, from A171911 to A171918, corresponding to the separate sequences generated with a0 = 1 to 8.
References[edit]
- ^ abcvan Eck's sequence (A181391) at the On-Line Encyclopedia of Integer Sequences
- ^'A171911 - OEIS'. oeis.org. Retrieved 2019-06-17.
- ^'A171912 - OEIS'. oeis.org. Retrieved 2019-06-17.
External links[edit]
- Brady Haran and N. J. A. Sloane, Don't Know (the Van Eck Sequence) on YouTube
Retrieved from 'https://en.wikipedia.org/w/index.php?title=Van_Eck%27s_sequence&oldid=926591207'
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
Sequence 2.0.3 1
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
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 2019-ncov
2: | | O | O O . | O O . O . . . | Movavi video editor plus 2020 v20.2.1 crack.
3: | | | O | O O O . |
4: | | | | O |
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
Sequence 20l
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