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A290691
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Triangle read by rows: infinite braid made of periodically-colored yarns in which the crossing of two adjacent yarns occurs when two color 0's are side by side (see comments).
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0
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1, 0, 2, 1, 1, 3, 0, 0, 2, 4, 2, 1, 1, 3, 5, 1, 0, 0, 2, 4, 6, 0, 3, 1, 1, 3, 5, 7, 2, 2, 0, 0, 2, 4, 6, 8, 1, 1, 4, 1, 1, 3, 5, 7, 9, 0, 0, 3, 0, 0, 2, 4, 6, 8, 10, 3, 2, 2, 5, 1, 1, 3, 5, 7, 9, 11, 2, 1, 1, 4, 0, 0, 2, 4, 6, 8, 10, 12, 1, 0, 0, 3, 6, 1, 1
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OFFSET
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1,3
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COMMENTS
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Construction: row numbers start at n = 3; column numbers run from k = 1 to k = n - 2. For all y >= 2, a yarn called "yarn y" is made of the repeated (y - 1, y - 2, ..., 1, 0) sequence. Pin it (for now vertically) at coordinates (y + 1, y - 1). Progress by increasing n: for a given row n, if two adjacent yarns show side by side 0's, then cross them at this point.
Properties: n is a prime iff there is no 0 in row n. n is a square iff there is an isolated 0 in row n column 1. If there is a crossing between yarn y1 and yarn y2 in row n, then n = y1 * y2.
Alternative definition: row n is the list of (-n) mod (y) sorted in ascending order of abs(y - n / y), for all y candidate divisor of n, y between 2 and (n - 1) inclusive.
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LINKS
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FORMULA
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T(n,k) = (-n) mod y(n,k), with y(n,k) the yarn going through (n,k); ambiguity at a crossing doesn't matter, both mod yielding 0.
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EXAMPLE
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Array begins:
1
0 2
1 1 3
0 0 2 4
2 1 1 3 5
1 0 0 2 4 6
0 3 1 1 3 5 7
2 2 0 0 2 4 6 8
1 1 4 1 1 3 5 7 9
...
Viewed as a braid (pairs of adjacent zeros being replaced by crossings):
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ------> k
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3 1
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4 0 2
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5 1 1 3
\ / |
6 0 2 4
/ \ | |
7 2 1 1 3 5
| \ / | |
8 1 0 2 4 6
| / \ | | |
9 0 3 1 1 3 5 7
| | \ / | | |
10 2 2 0 2 4 6 8
| | / \ | | | |
11 1 1 4 1 1 3 5 7 9
\ / | \ / | | | |
12 0 3 0 2 4 6 8 10
/ \ | / \ | | | | |
13 3 2 2 5 1 1 3 5 7 9 11
| | | | \ / | | | | |
14 2 1 1 4 0 2 4 6 8 10 12
| \ / | / \ | | | | | |
15 1 0 3 6 1 1 3 5 7 9 11 13
| / \ | | \ / | | | | | |
16 0 4 2 2 5 0 2 4 6 8 10 12 14
| | | | | / \ | | | | | | |
17 3 3 1 1 4 7 1 1 3 5 7 9 11 13 15
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V ...
n
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MATHEMATICA
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Ev[E_] := Module[{x, dx}, x = First[E]; dx = Last[E]; If[x == 0 && dx < 0, {-dx, -dx}, {x + dx, dx}]]
EvL[n_, L_] := Module[{LL}, LL = Ev /@ L; LL = Sort[LL]; LL = Append[LL, {n - 1, -1/n}]; LL]
It[nStart_, nEnd_, LStart_] := Module[{n, LL}, For[n = nStart; LL = LStart, n <= nEnd, n++, LL = EvL[n, LL]]; LL]
Encours[n_] := It[2, n, {}]
Countdown[x_, dx_] := If[dx > 0, (Ceiling[x] - x)/dx, (Floor[x] - x)/dx]
A[n_] := Drop[Apply[Countdown, #] & /@ Encours[n], -1]
Table[A[n], {n, 2, 25}] // Flatten
(* or *)
(Last /@ # &) /@ Sort /@ Table[{Abs[k - n/k], Mod[-n, k]}, {n, 3, 20}, {k, 2, n - 1}] // Flatten
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PROG
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(C)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define NMAX 40
struct cell { int f; int v; };
struct line { struct cell t[NMAX]; };
void display(struct line *T) { int n, k; for (n = 3; n <= NMAX; n ++) { for (k = 1; k < n - 1; k ++) { printf("%2d, ", T[n].t[k].v, T[n].t[k].f); } printf("\n"); } }
void swap(int *a, int *b) { int x; x = *a; *a = *b; *b = x; }
void fill(struct line *T)
{
int n, k;
for (n = 3; n <= NMAX; n ++)
{
for (k = 1; k < n - 2; k ++)
{
T[n].t[k].v = T[n - 1].t[k].v - 1;
T[n].t[k].f = T[n - 1].t[k].f;
}
T[n].t[n - 2].v = n - 2;
T[n].t[n - 2].f = n - 1;
for (k = 1; k < n - 2; k ++)
{
if ((T[n].t[k].v == 0) && (T[n].t[k + 1].v == 0))
{
swap(&T[n].t[k].f, &T[n].t[k + 1].f);
}
}
for (k = 1; k < n - 2; k ++)
{
if (T[n].t[k].v == -1)
{
T[n].t[k].v += T[n].t[k].f;
}
}
}
}
int main() { struct line T[NMAX + 1]; memset(T, 0x0, sizeof(T)); fill(T); display(T); }
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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