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A022543 Number of distinct 'failure tables' for a string of length n. 1
1, 2, 4, 9, 20, 47, 110, 263, 630, 1525, 3701, 9039, 22140, 54460, 134339, 332439, 824735, 2051307, 5113298, 12773067, 31968041, 80152901, 201297338, 506324357, 1275385911, 3216901194, 8124150323, 20541362001, 51994801119, 131747424892 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
REFERENCES
Knuth-Morris-Pratt pattern matching algorithm.
LINKS
Dennis Moore, W. F. Smyth and Dianne Miller, Counting distinct strings, Algorithmica, Vol. 23 (1999), 1-13.
EXAMPLE
For example, a string of length 3 can have one of the following 4 'failure tables': 012, 001, 010, 000.
PROG
(C++)
// check(p, n) returns true if and only if there exists a string of length n that have provided failure table (assuming that p[0] == -1).
bool check(int *p, int n) {
static int a[64];
for (int i = 0; i <= n; i++)
a[i] = i;
for (int i = 1, k = 0; i <= n; i++) {
for (; k >= p[i]; k = p[k]);
if (++k != p[i])
return false;
if (k)
a[i] = a[k];
}
for (int i = 1, k = 0; i <= n; i++, k++)
for (; k >= p[i]; k = p[k])
if (k + 1 < i && a[k + 1] == a[i])
return false;
return true;
}
// count(n) returns number of different failure tables for string of length n.
long long count(int n, int i = 1) {
static int p[64] = {-1};
if (!check(p, i - 1))
return 0;
if (i > n)
return 1;
long long result = 0;
for (p[i] = 0; p[i] <= p[i - 1] + 1; p[i]++)
result += count(n, i + 1);
return result;
} // Pavel Irzhavski, Feb 25 2014
CROSSREFS
Sequence in context: A039808 A138164 A130802 * A307557 A036618 A349014
KEYWORD
nonn,nice
AUTHOR
Dianne Miller (millerdm(AT)mcmaster.ca)
EXTENSIONS
a(19) and beyond from Pavel Irzhavski, Feb 25 2014
STATUS
approved

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Last modified April 23 06:04 EDT 2024. Contains 371906 sequences. (Running on oeis4.)