Search results
Results from the WOW.Com Content Network
(the Fibonacci sequence) is formed by starting with 0 and 1 and then adding any two consecutive terms to obtain the next one: an implicit description (sequence A000045 in the OEIS). The sequence 0, 3, 8, 15, ... is formed according to the formula n 2 − 1 for the n th term: an explicit definition.
A bijection with the sums to n is to replace 1 with 0 and 2 with 11. The number of binary strings of length n without an even number of consecutive 0 s or 1 s is 2F n. For example, out of the 16 binary strings of length 4, there are 2F 4 = 6 without an even number of consecutive 0 s or 1 s—they are 0001, 0111, 0101, 1000, 1010, 1110. There is ...
1 1 1 0 110 1 100 1101100 1 1100100 110110011100100 1 110110001100100. In each iteration of this process, a 1 is placed at the end of the previous iteration's string, then this string is repeated in reverse order, replacing 0 by 1 and vice versa.
"subtract if possible, otherwise add": a(0) = 0; for n > 0, a(n) = a(n − 1) − n if that number is positive and not already in the sequence, otherwise a(n) = a(n − 1) + n, whether or not that number is already in the sequence.
In mathematics, the Fibonacci numbers form a sequence defined recursively by: = {= = + > That is, after two starting values, each number is the sum of the two preceding numbers.
Michael Stifel published the following method in 1544. [3] [4] Consider the sequence of mixed numbers,,,, … with = + +.To calculate a Pythagorean triple, take any term of this sequence and convert it to an improper fraction (for mixed number , the corresponding improper fraction is ).
In mathematics, the Jacobsthal numbers are an integer sequence named after the German mathematician Ernst Jacobsthal.Like the related Fibonacci numbers, they are a specific type of Lucas sequence (,) for which P = 1, and Q = −2 [1] —and are defined by a similar recurrence relation: in simple terms, the sequence starts with 0 and 1, then each following number is found by adding the number ...
For example, for p = 3 one has π 1 (3) = 8 which equals 3 2 − 1 = 8; for p = 7, one has π 1 (7) = 16, which properly divides 7 2 − 1 = 48. This analysis fails for p = 2 and p is a divisor of the squarefree part of k 2 + 4, since in these cases are zero divisors , so one must be careful in interpreting 1/2 or k 2 + 4 {\displaystyle {\sqrt ...