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
Keywords
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
Programs
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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 (m
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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]
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