A152451 -id:A152451 - cp's OEIS frontend

A156284 From every interval (2^(m-1), 2^m), m >= 3, we remove primes p for which 2^m-p is a prime that was not removed for smaller values of m; the sequence gives all remaining odd primes.

3, 7, 11, 17, 19, 23, 31, 37, 43, 59, 67, 71, 73, 79, 83, 89, 101, 103, 107, 113, 127, 131, 137, 139, 151, 157, 163, 179, 181, 191, 193, 199, 211, 223, 227, 229, 241, 251, 257, 263, 269

Offset: 1

Vladimir Shevelev, Feb 07 2009

nonn

Powers of 2 are not expressible as sums of two primes from this sequence. This is attained by a more economical algorithm than that for construction of A152451. If A(x) is the counting function for the terms a(n) <= x, then A(x) = pi(x) - O(x/(log^2(x)). It is known that the approximation of pi(x) by x/log(x) gives the remainder term as, at best, O(x/log^2(x)). Therefore beginning our process from m >= M (with arbitrarily large M), we obtain a sequence which essentially is indistinguishable from the sequence of all odd primes with the help of the approximation of pi(x) by x/log(x). Hence it is in principle impossible to prove the binary Goldbach conjecture by such an approximation of pi(x).

Cf. A002375, A152451, A156537.

A241922 Smallest k^2>=0 such that n-k^2 is semiprime, or a(n)=2 if there is no such k^2.

Original entry on oeis.org

2, 2, 2, 0, 1, 0, 1, 4, 0, 0, 1, 2, 4, 0, 0, 1, 2, 4, 4, 16, 0, 0, 1, 9, 0, 0, 1, 2, 4, 4, 9, 2, 0, 0, 0, 1, 4, 0, 0, 1, 16, 4, 4, 9, 36, 0, 1, 9, 0, 1, 0, 1, 4, 16, 0, 1, 0, 0, 1, 9, 4, 0, 1, 9, 0, 1, 9, 64, 0, 1, 9, 2, 4, 0, 1, 25, 0, 1, 64, 25, 4, 0, 1, 49, 0, 0, 0, 1, 4, 4, 0, 1, 0, 0, 0, 1, 4, 4, 4, 9, 16

Offset: 1

Vladimir Shevelev, May 01 2014

nonn

If n = m^2, m>=2, then the condition {a(n) differs from 2} is equivalent to the Goldbach binary conjecture. Indeed, if m^2 - k^2 is semiprime, then (m-k)*(m+k) = p*q, where p<=q are primes. Here we consider two possible cases. 1) m-k=1, m+k=p*q and 2) m-k=p, m+k=q. But in the first case k=m-1>m-p, i.e., more than k in the second case. In view of the minimality k, we only have to consider case 2). In this case we have m-/+k both are primes p<=q (with equality in case k=0) and thus 2*m = p + q. Conversely, let the Goldbach conjecture be true. Then for a perfect square n>=4, we have 2*sqrt(n)=p+q (p<=q are both primes). Thus n=((p+q)/2)^2 and n-((p-q)/2)^2=p*q is semiprime. Hence a(n) is a square not exceeding ((p-q)/2)^2.
Note that a(n)=2 for 1,2,3,12,17,28,32,72,...
All these numbers are in A100570. Thus the Goldbach binary conjecture is true if and only if A100570 does not contain perfect squares.
The largest term found in the first 2^28 terms is a(106956964) = 369^2 = 136161. This further encourages one to believe that Goldbach's binary conjecture holds true. - Daniel Mikhail, Nov 23 2020

Peter J. C. Moses, Table of n, a(n) for n = 1..1000
Daniel Mikhail, Lists of up to the first 15 integers that are a squared distance, k^2, away from a semiprime for all k's found between [5..2^28]

Cf. A000290, A001358, A100570, A152522, A152451, A156537.

a(n) = {my(lim = if (issquare(n), sqrtint(n)-1, sqrtint(n))); for (k=0, lim, if (bigomega(n-k^2) == 2, return (k^2));); return (2);} \\ Michel Marcus, Nov 26 2020

a(A001358(n)) = 0.

A241927 Smallest k^2>=1 such that n-k^2 is semiprime p*q in Fermi-Dirac arithmetic (A176525) with additional requirement that, if n is a square, then p and q are of the same parity; or a(n)=2 if there is no such k^2.

Original entry on oeis.org

2, 2, 2, 2, 2, 2, 1, 2, 1, 4, 1, 4, 1, 4, 1, 1, 9, 4, 1, 2, 1, 1, 1, 4, 4, 4, 1, 1, 1, 4, 4, 4, 1, 1, 1, 1, 1, 4, 1, 1, 9, 4, 4, 9, 1, 1, 1, 4, 4, 4, 1, 1, 1, 4, 4, 1, 9, 1, 1, 9, 4, 4, 1, 1, 1, 1, 4, 4, 1, 1, 9, 4, 4, 9, 1, 1, 1, 1, 4, 4, 4, 25, 1, 4, 9, 1, 1, 1, 4, 4, 4, 1, 1, 1, 1, 1, 4, 4, 1, 1, 1, 4, 4, 4, 25

Offset: 1

Vladimir Shevelev, May 02 2014

nonn

A semiprime in Fermi-Dirac arithmetic is a product of two distinct terms of A050376, or, equivalently, an infinitary semiprime. The conjecture that every even number>=4 is a sum of two A050376 terms is a weaker form of the Goldbach conjecture; as such, it is natural to refer to it as a Goldbach conjecture in Fermi-Dirac arithmetic (FDGC).
Let us prove that the condition {a(m^2) differs from 2} is equivalent to the FDGC.
Indeed, from the FDGC for a perfect square n>=4, we have 2*sqrt(n)=p+q (pA050376 terms of the same parity). Thus n=((p+q)/2)^2 and n-((p-q)/2)^2=p*q is Fermi-Dirac semiprime. Hence, a(n)>=1 is a square not exceeding ((p-q)/2)^2. Thus the condition {a(m^2) differs from 2} is necessary for the truth of the FDGC.
Let us prove that the condition {a(m^2) differs from 2} is also sufficient. Indeed, a(m^2)-k^2 = p*q, where, say, pA050376, and p,q are of the same parity. If p,q are primes, then the proof repeats one in A241922. Let, e.g., p=s^2A050376). Consider two principal cases: 1) m-k = s, m+k = s*q; 2) m-k = s^2, m+k = q. In 1) k=m-s, in 2) k=m-s^2. In view of the minimality of k, we should accept 2) and thus m-k=p, m+k=q. So, 2*m=p+q as the FDGC requires.
The sequence of numbers n for which a(n)=2 begins 1, 2, 3, 4, 5, 6, 8, 20, ... (A241947).

			a(17)=9, since 9 is the smallest square such that 17-9 = 8 = 2*4 is a Fermi-Dirac semiprime.

V. S. Shevelev, Multiplicative functions in the Fermi-Dirac arithmetic, Izvestia Vuzov of the North-Caucasus region, Nature sciences 4 (1996), 28-43 (in Russian; MR 2000f: 11097, pp. 3912-3913).

Peter J. C. Moses, Table of n, a(n) for n = 1..10000

Cf. A000290, A001358, A050376, A176525, A100570, A152522, A152451, A156537, A241922.

A152871 Irregular table with first row containing the single term 3; in the n-th row, n>=2, we list in increasing order those d=2^(n+1)-a, for each term a in all the preceding rows, such that d is prime.

Original entry on oeis.org

3, 5, 11, 13, 19, 29, 53, 59, 61, 67, 109, 197, 227, 251, 499, 509, 773, 797, 827, 971, 1013, 1019, 1021, 1549, 1987, 2029, 3083, 3299, 3323, 4091, 4093, 4099, 6163, 8179, 15413, 16187, 16381, 28669, 30781, 31219, 32707, 32749, 50123, 62213, 64709, 64763

Offset: 1

Vladimir Shevelev, Dec 14 2008

Since primes above the n-th row are <2^n, primes in the n-th row are >2^(n+1)-2^n=2^n. Thus in different rows primes are different.

Questions: 1) Is every row nonempty? 2) Is the sequence infinite (an infinite number of nonempty rows)?

			1: 3;
2: 5;
3: 11, 13;
4: 19, 29;
5: 53, 59, 61;
6: 67, 109;
7: 197, 227, 251;
8: 499, 509;
9: 773, 797, 827, 971, 1013, 1019, 1021;

Jason Kimberley, Table of n, a(n) for n = 1..1016

The n-th row has length A152872(n).

Cf. A152451.

A152871and2 :=
    function(N)
        A := [[3]]; C := [1];
        for n in [2..N] do
            C[n] := 0;
            A[n] := [];
            for a in Reverse(&cat A) do
                d := 2^(n+1) - a;
                if
                    IsPrime(d)
                then
                    Append(~A[n],d);
                    C[n] +:= 1;
                end if;
            end for;
        end for;
        return A, C;
    end function;
A152871and2(20);

Heavily edited by Jason Kimberley, Feb 12 2013

A152872 The length of the n-th row in A152871.

Original entry on oeis.org

1, 1, 2, 2, 3, 2, 3, 2, 7, 3, 5, 3, 3, 5, 6, 7, 8, 9, 9, 6, 8, 9, 5, 10, 4, 10, 7, 11, 7, 11, 8, 11, 19, 18, 10, 12, 17, 19, 15, 7, 20, 24, 15, 15, 14, 16, 16, 26, 18, 19, 22, 22, 21, 23, 13, 29, 25, 22, 21, 23, 23, 30, 22, 21, 18, 47, 31, 41, 23, 23, 28, 36, 17, 38, 31, 48, 28, 46, 24, 46

Offset: 1

Vladimir Shevelev, Dec 14 2008

Jason Kimberley, Table of n, a(n) for n = 1..170

Cf. A152451.

Magma
```
use code in A152871.
```

Renamed and extended by Jason Kimberley, Feb 12 2013

A158848 Prime numbers p where 2^k-p is prime, with k>6 and minimal.

Original entry on oeis.org

67, 97, 109, 149, 167, 173, 197, 227, 233, 239, 251, 271, 283, 313, 331, 349, 373, 409, 433, 439, 499, 509, 521, 557, 563, 593, 641, 677, 743, 761, 773, 797, 827, 857, 887, 911, 941, 953, 971, 977, 983, 1013, 1019, 1021, 1039, 1051, 1129, 1171, 1237, 1279, 1291, 1297, 1321

Offset: 1

Vladimir Shevelev, Mar 28 2009

nonn

These are the primes removed according to algorithm of A156284 beginning with m=7.

Powers 2^m, m>=7, are not expressible as sums of two primes which are not in the sequence.

Bill McEachen, Table of n, a(n) for n = 1..10000

Cf. A152451, A156284, A158756, A158759, A158846, A158847, A086081.

isok(p) = if (isprime(p), my(k=ceil(log(p)/log(2))); (k >= 7) && isprime(2^k-p)); \\ Michel Marcus, Jan 22 2023

a(n) = A086081(n+11). - Bill McEachen, Jan 22 2023

Missing terms 773, 797, 827, 857 added by Michel Marcus, Jan 23 2023

New name from Bill McEachen, Jan 22 2023

A152460 Primes p such that there exist positive integer k and prime q with p > q and 3^k = p + 2q or 3^k = q + 2p.

Original entry on oeis.org

3, 5, 11, 13, 17, 23, 29, 31, 37, 43, 47, 59, 67, 71, 97, 101, 103, 107, 109, 113, 137, 149, 157, 181, 197, 229, 233, 239, 251, 263, 269, 271, 281, 283, 307, 311, 313, 331, 347, 349, 353, 359, 367, 383, 431, 467, 503, 523, 563, 571, 587, 607, 643, 647, 683, 691

Offset: 1

Vladimir Shevelev, Dec 05 2008, Dec 12 2008

nonn

a(n) is the greater of primes (p,q) in representations of a power of 3 in Lemoine-Levy's form p+2q (see A046927)

If 3^n=p+2q, then 3^(n-1)<=max(p,q)<3^n. Therefore the sets of greater primes for different powers of 3 do not intersect.

			27=5+2*11=13+2*7=17+2*5=23+2*2, so that 11,13,17 and 23 are in the sequence.

Cf. A103151, A086081, A152451

aa(n)={my(v=[]); forprime(p=2,n\2,q=n-p*2; if(isprime(q),v=concat(v,(max(p,q))))); vecsort(v,,8)};
for(n=2, 7, v=aa(3^n); for(i=1,#v,print1(v[i], ", ")))

If A(x) is the counting function of a(n)<=x, then A(x)=O(xloglogx/(logx)^2).

Program and editing by Charles R Greathouse IV, Nov 02 2009

A242247 Maximal k <= n^2 + 1 such that every Goldbach representation of 2*k = p+q contains at least one prime from the set {prime(1), prime(2), ..., prime(n)}, or a(n)=0 if there is no such k.

Original entry on oeis.org

2, 4, 8, 10, 14, 22, 22, 28, 32, 38, 46, 49, 49, 58, 58, 68, 74, 74, 82, 82, 87, 94, 94, 98, 112, 116, 121, 128, 136, 146, 146, 146, 155, 155, 164, 166, 184, 184, 184, 200, 206, 206, 221, 221, 224, 238, 244, 265, 265, 268, 268, 268, 286, 286, 286, 286, 344

Offset: 1

Vladimir Shevelev, May 09 2014

nonn

The restriction a(n) <= n^2 + 1 allows one to make the sequence computable for n >= 1 and, at the same time, to somewhat agree with heuristic arguments for large n.
The sequence is based on a conjecture stronger than Goldbach's Conjecture: for arbitrarily large N there exists a number m(N) such that, for k > m(N), the number of unordered Goldbach representations (A002375) of 2*k is greater than N.
Heuristic arguments would imply that m(N) ~ N*log^2(2*N). Then, conjecturally, for n >= 3, a(n) < n*log^2(2*n).
The existence of a(n) for arbitrary n says that, if we remove from the sequence of primes an arbitrarily large number M of the first terms, the Goldbach representations remain for all sufficiently large even numbers.

			Let n=3. Then the set is {2,3,5}. The Goldbach representations of 2*k=16 are 3+13 and 5+11. Each of them contains a prime from {2,3,5}. So a(3) >= 8. Since, by definition, a(3) <= 10, consider also 2*k=18 and 20. We have 18=7+11, 20=7+13. These representations contain none of the primes 2,3,5. Therefore a(3)=8.

Cf. A002375, A152451, A156284, A156537.

More terms from Peter J. C. Moses, May 10 2014

cp's OEIS Frontend

A156284 From every interval (2^(m-1), 2^m), m >= 3, we remove primes p for which 2^m-p is a prime that was not removed for smaller values of m; the sequence gives all remaining odd primes.

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A241922 Smallest k^2>=0 such that n-k^2 is semiprime, or a(n)=2 if there is no such k^2.

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A241927 Smallest k^2>=1 such that n-k^2 is semiprime p*q in Fermi-Dirac arithmetic (A176525) with additional requirement that, if n is a square, then p and q are of the same parity; or a(n)=2 if there is no such k^2.

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A152871 Irregular table with first row containing the single term 3; in the n-th row, n>=2, we list in increasing order those d=2^(n+1)-a, for each term a in all the preceding rows, such that d is prime.

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A152872 The length of the n-th row in A152871.

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Magma

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A158848 Prime numbers p where 2^k-p is prime, with k>6 and minimal.

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A152460 Primes p such that there exist positive integer k and prime q with p > q and 3^k = p + 2q or 3^k = q + 2p.

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A242247 Maximal k <= n^2 + 1 such that every Goldbach representation of 2*k = p+q contains at least one prime from the set {prime(1), prime(2), ..., prime(n)}, or a(n)=0 if there is no such k.

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