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A267413
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Dropping any binary digit gives a prime number.
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1
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6, 7, 11, 15, 35, 39, 63, 135, 255, 999, 2175, 8223, 16383, 57735, 131075, 131079, 262143, 524295, 1048575, 536870919, 1073735679, 2147483655, 4294967295, 17179770879, 4260641103903, 4611686018427387903, 4720069647059686260735, 1237940039285380274899124223
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OFFSET
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1,1
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COMMENTS
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This is the binary analog of A034895. The sequence contains mostly numbers with very few binary digit runs (BDR, A005811). Those with one BDR are of the type 2^k-1, such that 2^(k-1)-1 is a Mersenne prime (A000668). Vice versa, if M is any Mersenne prime, then 2*M+1 is a term. The number 6 is the only term with an even number of BDRs. There are many terms with 3 BDRs. The first term with 5 BDRs is 57735. The next terms with at least 5 BDRs (if they exist at all) are larger than 10^10. So far, I could test that a(24) > 10^10.
For n >= 2, a(n) == 3 (mod 4).
2^k+3 is in the sequence if 2^(k-1)+1 and 2^(k-1)+3 are primes, i.e., 2^(k-1)+1 is in the intersection of A019434 and A001359. The only known terms of the sequence in this class are 7, 11, 35, 131075.
2^k+7 is in the sequence if 2^(k-1)+3 and 2^(k-1)+7 are primes: thus 2^(k-1)+3 is in A057733 and 2^(k-1)+7 is in A104066. Terms of the sequence in this class include 15, 39, 135, 131079, 524295, 536870919, 2147483655 (but no more for k <= 2000).
(End)
For n > 1, a(n) = 2p+1 for some prime p. - Chai Wah Wu, Aug 27 2021
There are no more terms with an odd number of binary digits: from any number having an odd number of binary digits, one can always drop a digit and obtain a multiple of 3. Numbers of the form 2^k+3 (k even and k > 2) cannot be terms because 2^(k-1)+1 is a multiple of 3.
(End)
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LINKS
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EXAMPLE
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Decimal and binary forms of the known terms:
1 6 110
2 7 111
3 11 1011
4 15 1111
5 35 100011
6 39 100111
7 63 111111
8 135 10000111
9 255 11111111
10 999 1111100111
11 2175 100001111111
12 8223 10000000011111
13 16383 11111111111111
14 57735 1110000110000111 <--- (a binary palindrome
15 131075 100000000000000011 with 5 digit runs)
16 131079 100000000000000111
17 262143 111111111111111111
18 524295 10000000000000000111
19 1048575 11111111111111111111
20 536870919 100000000000000000000000000111
21 1073735679 111111111111111110011111111111
22 2147483655 10000000000000000000000000000111
23 4294967295 11111111111111111111111111111111
24 17179770879 1111111111111111100111111111111111
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MAPLE
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filter:= proc(n) local B, k, y;
if not isprime(floor(n/2)) then return false fi;
B:= convert(n, base, 2);
for k from 2 to nops(B) do
if B[k] <> B[k-1] then
y:= n mod 2^(k-1);
if not isprime((y+n-B[k]*2^(k-1))/2) then return false fi
fi
od;
true
end proc:
select(filter, [6, seq(i, i=7..10^6, 4)]); # Robert Israel, Jan 14 2016
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MATHEMATICA
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Select[Range[2^20], AllTrue[Function[w, Map[FromDigits[#, 2] &@ Drop[w, {#}] &, Range@ Length@ w]]@ IntegerDigits[#, 2], PrimeQ] &] (* Michael De Vlieger, Jan 16 2016, Version 10 *)
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PROG
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(PARI) DroppingAnyDigitGivesAPrime(N, b) = {
\\ Property-testing function; returns 1 if true for N, 0 otherwise
\\ Works with any base b. Here used with b=2.
my(k=b, m); if(N<b, return(0));
while(N>=(k\b), m=(N\k)*(k\b)+(N%(k\b));
if ((m<2)||(!isprime(m)), return(0)); k*=b);
return(1);
}
(Python)
from sympy import isprime
def ok(n):
if n < 7 or n%4 != 3: return n == 6
b = bin(n)[2:]
return all(isprime(int(b[:i]+b[i+1:], 2)) for i in range(len(b)))
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CROSSREFS
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KEYWORD
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nonn,base,more,hard
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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