A248831 Prime sieve of the square root of 2.

1, 4, 1, 6, 309, 6, 8078, 1875, 69480, 66, 7, 32, 8462, 388, 87, 2764, 27350, 846, 12, 24, 6055850, 4, 999358, 41322266, 27505, 9995050, 527820605, 470, 5, 716059, 453459686, 285, 86408, 5, 87, 4508, 627995, 8968, 396546, 808, 640620, 5, 50, 502, 599, 2

Offset: 1

Author

Jared Kish, Oct 15 2014

Keywords

• nonn
• base

Comments

Algorithm: see A245770 (prime sieve of Pi).

Links

• Manfred Scheucher, Table of n, a(n) for n = 1..431
• Wikipedia, Square root of 2
• Manfred Scheucher, Sage Script (Note: should also run in pure python with a few modifications)

Crossrefs

Cf. A002193 (square root of 2), A245770 (prime sieve of Pi), A248804 (prime sieve of e).

Programs

Python

def sqroot(a, digits):
    a = a * (10**(2*digits))
    x_prev = 0
    x_next = 1 * (10**digits)
    while x_prev != x_next:
        x_prev = x_next
        x_next = (x_prev + (a // x_prev)) >> 1
    return x_next
def primes(n):
    sieve = [True] * n
    for i in range(3,int(n**0.5)+1,2):
        if sieve[i]:
            sieve[i*i::2*i]=[False]*((n-i*i-1)/(2*i)+1)
    return [2] + [i for i in range(3,n,2) if sieve[i]]
a = sqroot(2,300)#300 digits is arbitrary - lengthen for more digits
b = primes(10000000)#make sure to scan primes up to longest term in sequence
y = str(a)
for x in b:
    if str(x) in y:
        y = y.replace(str(x)," ",1)#replace first occurrence only
while "  " in y:
    y = y.replace("  "," ")#replace chains of spaces with a single space
z = y.split(" ")#split terms into a list
z = filter(None, z)#remove null terms
f = map(int,z)#convert to integers
print(f)
# David Consiglio, Jr., Jan 03 2015

Extensions

More terms from David Consiglio, Jr., Dec 02 2014, Jan 03 2015

More terms from Manfred Scheucher, May 25 2015