A309701 Primes with record Manhattan distance from origin. When starting rightwards in a grid, turn left after a prime number. If not, walk straight on.

2, 3, 11, 29, 59, 97, 151, 193, 211, 223, 239, 281, 307, 311, 331, 337, 479, 541, 593, 613, 631, 641, 659, 877, 881, 907, 911, 997, 1409, 1861, 1907, 2267, 2281, 2287, 2309, 2311, 2503, 2579, 2609, 2617, 2657, 2671, 2677, 3671, 3691, 3697, 3727, 3761, 3767, 3793, 3797, 4201, 4327, 4357, 4391, 4397, 4507, 4721, 4751, 4909

Offset: 1

Author

Pieter Post, Aug 13 2019

Keywords

• nonn

Comments

This sequence differs from A309755 where the Euclidean distance is used.

Examples

			Grid of the first 34 steps. 0 represents (0,0).
xx  xx   xx   31   30    29
xx  xx   xx   32   xx    28
xx  xx   xx   33   xx    27
xx  xx   xx   34   xx    26
xx 5/17 4/16 3/15  14   13/25
x 0/6/18  1    2   xx   12/24
xx 7/19 8/20 9/21 10/22 11/23
2 (2,0) is 2 steps away from the origin, 3 (2,1) has a distance of 3. Next record distance is 11 (4,-1), distance 5. Next is 29 (4,5), distance 9.
		

Links

• Pieter Post, Table of n, a(n) for n = 1..166

Crossrefs

Cf. A000040, A309755.

Programs

Mathematica

step[n_] := Switch[n, 0, {1,0}, 1, {0,1}, 2, {-1,0}, 3, {0,-1}]; r = {0,0}; q = 0; s={}; rm=0; Do[p = NextPrime[q]; r += step[Mod[n, 4]] * (p-q); r1 = Total @ Abs @ r; If[r1 > rm, rm = r1; AppendTo[s, p]]; q = p, {n, 0, 3000}]; s (* Amiram Eldar, Aug 15 2019 *)

PARI

upto(n) = {my(pos = [0, 0], rotateLeft = [0, -1; 1, 0], step = [1, 0], recordDistance = 0, q = 0, res = List(), i = 0); forprime(p = 2, n, pos += (p - q) * step; step *= rotateLeft; if(abs(pos[1]) + abs(pos[2]) > recordDistance, i++; recordDistance = abs(pos[1]) + abs(pos[2]); listput(res, p)); q = p); res} \\ David A. Corneth, Aug 15 2019

Python

def prime(z):
    isPrime=True
    for y in range(2,int(z**0.5)+1) :
        if z%y==0:
            isPrime=False
            break
    return isPrime
m,n,g,h=[],[],[1,0,-1,0],[0,1,0,-1]
for c in range (2,10000):
    if prime(c)==True:
        m.append(c)
ca,cb,cc=2,0,0
for j in range(2,10000):
    if j in m:
        cc=cc+1
        cd,ce=g[cc%4],h[cc%4]
    ca,cb=ca+cd,cb+ce
    n.append([j+1,ca,cb,abs(ca)+abs(cb)])
v=2
for j in n:
    if j[3]>v and j[0] in m:
        print (j)
        v=j[3]