

A156695


Odd numbers that are not of the form p + 2^a + 2^b, a, b > 0, p prime.


51



1, 3, 5, 6495105, 848629545, 1117175145, 2544265305, 3147056235, 3366991695, 3472109835, 3621922845, 3861518805, 4447794915, 4848148485, 5415281745, 5693877405, 6804302445, 7525056375, 7602256605, 9055691835, 9217432215
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OFFSET

1,2


COMMENTS

Crocker shows that this sequence is infinite.
All members above 5 found so far (up to 2.5 * 10^11) are divisible by 255 = 3 * 5 * 17, and many are divisible by 257. I conjecture that all members of this sequence greater than 5 are divisible by 255. This implies that all odd numbers (greater than 7) are the sum of a prime and at most three positive powers of two.
Pan shows that, for every c > 1, a(n) << x^c. More specifically, there are constants C,D > 0 such that there are at least Dx/exp(C log x log log log log x/log log log x) members of this sequence up to x.  Charles R Greathouse IV, Apr 11 2016
All terms > 5 are numbers k > 3 such that k  2^n is a de Polignac number (A006285) for every n > 0 with 2^n < k. Are there numbers K such that K  2^n is a Riesel number (A101036) for every n > 0? If so, K  2^n  2^m is composite for every pair m,n > 0, by the dual Riesel conjecture.  Thomas Ordowski, Jan 06 2024


LINKS



EXAMPLE

Prime factorization of terms:
F_0 = 3, F_1 = 5, F_2 = 17, F_3 = 257 are Fermat numbers (cf. A000215)
6495105 = 3 * 5 * 17 * 25471
848629545 = 3 * 5 * 17 * 461 * 7219
1117175145 = 3 * 5 * 17 * 257 * 17047
2544265305 = 3^2 * 5 * 17 * 257 * 12941
3147056235 = 3^2 * 5 * 17 * 257 * 16007
3366991695 = 3 * 5 * 17 * 83 * 257 * 619
3472109835 = 3 * 5 * 17 * 257 * 52981
3621922845 = 3 * 5 * 17^2 * 257 * 3251
3861518805 = 3^3 * 5 * 17 * 257 * 6547
4447794915 = 3^3 * 5 * 17 * 257 * 7541
4848148485 = 3^4 * 5 * 17 * 704161
5415281745 = 3 * 5 * 17 * 21236399
5693877405 = 3^2 * 5 * 17 * 257 * 28961
6804302445 = 3^2 * 5 * 17 * 53 * 257 * 653
7525056375 = 3^2 * 5^3 * 17 * 257 * 1531
7602256605 = 3 * 5 * 17 * 257 * 311 * 373
9055691835 = 3 * 5 * 17 * 257 * 138181
9217432215 = 3^2 * 5 * 17 * 173 * 257 * 271


PROG

(PARI) is(n)=if(n%2==0, return(0)); for(a=1, log(n)\log(2), for(b=1, a, if(isprime(n2^a2^b), return(0)))); 1 \\ Charles R Greathouse IV, Nov 27 2013
(Python)
from itertools import count, islice
from sympy import isprime
def A156695_gen(startvalue=1): # generator of terms >= startvalue
for n in count(max(startvalue+(startvalue&1^1), 1), 2):
l = n.bit_length()1
for a in range(l, 0, 1):
c = n(1<<a)
for b in range(min(a, l1), 0, 1):
if isprime(c(1<<b)):
break
else:
continue
break
else:
yield n


CROSSREFS



KEYWORD

nonn,hard,nice


AUTHOR



EXTENSIONS



STATUS

approved



