A309755 Primes with record Euclidean distance from the origin. When starting rightwards in a grid, turn left after a prime number, if not walk straight on.

2, 3, 11, 29, 59, 97, 149, 151, 191, 193, 211, 223, 239, 263, 281, 307, 311, 331, 337, 593, 613, 631, 641, 653, 659, 853, 857, 877, 881, 907, 911, 967, 971, 991, 997, 1801, 1811, 1847, 1861, 1901, 1907, 2251, 2267, 2281, 2287, 2309, 2311, 2657, 2671, 2677, 3163, 3167, 3187, 3191, 3299, 3319, 3343, 3691, 3697, 3719, 3727

Offset: 1

Author

Pieter Post, Aug 15 2019

Keywords

• nonn

Comments

Cells can contain more than one number.

This sequence differs from A309701, where the Manhattan distance is taken.

Examples

			Grid of the first 34 steps. 0 (second cell in sixth row) 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 two steps away from the origin, 3 (2,1) is at a distance of sqrt(5). Next record distance is 11 (4,-1), at distance sqrt(17). Next is 29 (4,5), at distance sqrt(41).
		

Links

• Pieter Post, Table of n, a(n) for n = 1..182

Crossrefs

Cf. A000040, A309701.

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 @ (r^2); If[r1 > rm, rm = r1; AppendTo[s, p]]; q = p, {n, 0, 3000}]; s (* Amiram Eldar, Aug 15 2019 *)

PARI

z=0; d=1; m=0; for (n=1, 3727, z+=d; if (isprime(n), d*=I; if (mRémy Sigrist, 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]
z=10000
for c in range (2,z):
    if prime(c)==True:
        m.append(c)
ca,cb,cc=2,0,0
for j in range(2,z):
    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,((ca)**2+(cb)**2)**(0.5)])
    #print (j+1,ca,cb)
v=2
for j in n:
    if j[3]>v and j[0] in m:
        print (j)
        v=j[3]