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A247665
a(1)=2; thereafter a(n) is the smallest number >= 2 not yet used which is compatible with the condition that a(n) is relatively prime to the next n terms.
27
2, 3, 4, 5, 7, 9, 8, 11, 13, 17, 19, 23, 15, 29, 14, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 25, 27, 79, 83, 16, 49, 89, 97, 101, 103, 107, 109, 113, 121, 127, 131, 137, 139, 149, 151, 157, 163, 167, 169, 173, 179, 181, 191, 85, 193, 57, 197, 199, 211, 223
OFFSET
1,1
COMMENTS
It appears that a(k) is even iff k = 2^i-1 (cf. A248379). It also appears that all powers of 2 occur in the sequence. (Amarnath Murthy)
The indices of even terms and their values are [1, 2], [3, 4], [7, 8], [15, 14], [31, 16], [63, 32], [127, 64], [255, 128], [511, 122], ...
Will the numbers 6, 10, 21, 22, ... ever occur? 12, 18, 20, ... are also missing, but if 6 never appears then neither will 12, etc.
A related question: are all terms deficient? - Peter Munn, Jul 20 2017
It appears that the missing numbers are 6, 10, 12, 18, 20, 21, 22, 24, 26, 28, 30, 33, 34, 35, 36, 38, 39, 40, 42, ..., but since there is no proof that any one of these is really missing, this sequence cannot yet be added to the OEIS. - N. J. A. Sloane, May 18 2022
REFERENCES
Amarnath Murthy, Email to N. J. A. Sloane, Oct 05 2014.
LINKS
Russ Cox, Go program (can compute 5 million terms in about 5 minutes)
EXAMPLE
a(1) = 2 must be rel. prime to a(2), so a(2)=3.
a(2) = 3 must be rel. prime to a(3) and a(4), so we can take them to be 4 and 5.
a(3) = 4 must be rel. prime to a(5), a(6), so we must take them to be 7,9.
a(4) = 5 must be rel. prime to a(7), a(8), so we must take them to be 8,11.
At each step after the first, we must choose two new numbers, and we must make sure that not only are they rel. prime to a(n), they are also rel. prime to all a(i), i>n, that have been already chosen.
PROG
(PARI) m=100; v=vector(m); u=vectorsmall(100*m); for(n=1, m, for(i=2, 10^9, if(!u[i], for(j=(n+1)\2, n-1, if(gcd(v[j], i)>1, next(2))); v[n]=i; u[i]=1; break))); v \\ Jens Kruse Andersen, Oct 08 2014
(Haskell)
a247665 n = a247665_list !! (n-1)
a247665_list = 2 : 3 : f [3] [4..] where
f (x:xs) zs = ys ++ f (xs ++ ys) (zs \\ ys) where
ys = [v, head [w | w <- vs, gcd v w == 1]]
(v:vs) = filter (\u -> gcd u x == 1 && all ((== 1) . (gcd u)) xs) zs
-- Reinhard Zumkeller, Oct 09 2014
(Sage)
# s is the starting point (2 in A247665).
def gen(s):
sequence = [s]
available = list(range(2, 2*s))
available.pop(available.index(s))
yield s
while True:
available.extend(range(available[-1]+1, next_prime(available[-1])+1))
for i, e in enumerate(available):
if all(gcd(e, sequence[j])==1 for j in range(-len(sequence)//2, 0)):
available.pop(i)
sequence.append(e)
yield(e)
break
g = gen(2)
[next(g) for i in range(40)] # (gets first 40 terms of A247665)
# Nadia Heninger, Oct 28 2014
(Python)
from itertools import count, islice
from math import gcd
from collections import deque
def A247665_gen(): # generator of terms
aset, aqueue, c, f = {2}, deque([2]), 3, True
yield 2
while True:
for m in count(c):
if m not in aset and all(gcd(m, a) == 1 for a in aqueue):
yield m
aset.add(m)
aqueue.append(m)
if f: aqueue.popleft()
f = not f
while c in aset:
c += 1
break
A247665_list = list(islice(A247665_gen(), 50)) # Chai Wah Wu, May 19 2022
CROSSREFS
Indices of primes and prime powers: A248387, A248918.
Lengths of runs of primes: A249033.
A090252 = similar to A247665 but start with a(1)=1. A249559 starts with a(1)=3.
A249064 is a different generalization.
A064413 is another similar sequence.
Sequence in context: A105362 A361299 A115928 * A117331 A080612 A039261
KEYWORD
nonn,look
AUTHOR
N. J. A. Sloane, Oct 06 2014 and Oct 08 2014
EXTENSIONS
More terms from Jens Kruse Andersen, Oct 06 2014
Further terms from Russ Cox, Oct 08 2014
Added condition a(n) >= 2 to definition. - N. J. A. Sloane, May 16 2022
STATUS
approved