A152522 -id:A152522 - cp's OEIS frontend

A152523 Multiplicities of distinct terms in A152522.

1, 1, 1, 4, 1, 2, 4, 6, 6, 7, 6, 7, 4, 11, 3, 2, 5, 4, 1, 22, 11

Offset: 1

Vladimir Shevelev, Dec 06 2008

a152522(n) = my(P=prime(n)); forstep(k=6, 9e99, 2, forprime(p=3, P, if(isprime(k-p), next(2))); return(k));
lista(nn) = {my(last = a152522(1), newa, nb = 1); for (n=2, nn, newa = a152522(n); if (newa == last, nb++, print1(nb, ", "); last = newa; nb = 1););} \\ Michel Marcus, Dec 18 2018

a(15)-a(21) from Michel Marcus, Dec 18 2018

A152451 From every interval (2^(m-1), 2^m), we remove primes p for which 2^m-p is a prime; the sequence gives the remaining odd primes.

Original entry on oeis.org

3, 7, 17, 23, 31, 37, 43, 71, 73, 79, 83, 89, 101, 103, 107, 113, 127, 131, 137, 139, 151, 157, 163, 179, 181, 191, 193, 199, 211, 223, 229, 241, 257, 263, 269, 277, 281, 293, 307, 311, 317, 337, 347, 353, 359, 367, 379, 383, 389, 397, 401, 419, 421, 431, 443

Offset: 1

Vladimir Shevelev, Dec 04 2008, Dec 05 2008, Dec 08 2008, Dec 12 2008

nonn

Powers of 2 are not expressible as sums of two primes from this sequence.

Consider a strong Goldbach conjecture: every even number n >= 6 is a sum of two primes, the lesser of which is O((log(n))^2*log(log(n))) (cf. comment to A152522). The number of such representations for 2^k, trivially, is less than k^5 for k > k_0. Removing the maximal primes in every such representation of 2^k, k >= 3, we obtain an analog B of A152451 with the counting function H(x) = pi(x) - O((log(x))^5). Replacing in B the first N terms with N consecutive primes (with arbitrarily large N), we obtain a sequence which essentially is indistinguishable from the sequence of all primes with the help of the approximation of pi(x) by li(x), since, according to the well-known Littlewood result, the remainder term in the theorem of primes cannot be less than sqrt(x)logloglog(x)/log(x). But for this sequence we have infinitely many even numbers for which the considered strong Goldbach conjecture is wrong. Thus the conjecture is essentially unprovable.

Complement of A086081.

Cf. A152522.

lista(nn) = {forprime(p=3, nn, m = ceil(log(p)/log(2)); if (!isprime(2^m-p), print1(p, ", ")););} \\ Michel Marcus, Sep 12 2015; Jan 22 2023

If A(X) is the counting function for the terms a(n)<=x, then A(x) = x/log(x) + O(x*log(log(x))/(log(x))^2).

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.

A298736 a(n) = s(n) - prime(n+1)+3, where s(n) = smallest even number x > prime(n) such that the difference x-p is composite for all primes p <= prime(n).

Original entry on oeis.org

6, 10, 26, 90, 88, 84, 82, 200, 282, 280, 522, 518, 516, 512, 942, 936, 934, 928, 924, 922, 2566, 2562, 2556, 2548, 2544, 2542, 5268, 5266, 5262, 5248, 5244, 5238, 5236, 7280, 7278, 7272, 7266, 7262, 7256, 43356, 43354, 43344, 43342, 43338, 43336, 43324, 54024

Offset: 1

Felix Fröhlich, Jan 25 2018

The statement "a(n) >= 0 for n >= 1" is equivalent to Goldbach's conjecture (cf. Phong, Dongdong, 2004, Theorem (a)).

Records: 6, 10, 26, 90, 200, 282, 522, 942, 2566, 5268, 7280, 43356, 54024, ..., . - Robert G. Wilson v, Feb 28 2018

Robert G. Wilson v, Table of n, a(n) for n = 1..1205
Bui Minh Phong and Li Dongdong, Elementary problems which are equivalent to the Goldbach's Conjecture, Acta Academiae Paedagogicae Agriensis, Sectio Mathematicae 31 (2004) 33-37.
Wikipedia, Goldbach's conjecture
Index entries for sequences related to Goldbach conjecture

Cf. A000040, A082467, A152522, A242165.

N:= 100: # to get a(1)..a(N)
P:= [seq(ithprime(i),i=1..N+1)]:
s:= proc(n,k0) local k;
  for k from max(k0,P[n]+1) by 2 do
    if andmap(not(isprime), map(t -> k - t, P[1..n])) then return k
  fi
od
end proc:
K[1]:= 6: A[1]:= 6:
for n from 2 to N do
  K[n]:= s(n,K[n-1]);
  A[n]:= K[n]- P[n+1]+3;
od:
seq(A[n],n=1..N); # Robert Israel, Mar 01 2018

f[n_] := Block[{k, x = 2, q = Prime@ Range@ n}, x += Mod[x, 2]; While[k = 1; While[k < n +1 && CompositeQ[x - q[[k]]], k++]; k < n +1, z = x += 2]; x - Prime[n +1] +3]; Array[f, 47] (* Robert G. Wilson v, Feb 26 2018 *)

s(n) = my(p=prime(n), x); if(p==2, x=4, x=p+1); while(1, forprime(q=1, p, if(ispseudoprime(x-q), break, if(q==p, return(x)))); x=x+2)
a(n) = s(n)-prime(n+1)+3

a(n) = A152522(n)-A000040(n+1)+3 for n > 0.

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A152523 Multiplicities of distinct terms in A152522.

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A152451 From every interval (2^(m-1), 2^m), we remove primes p for which 2^m-p is a prime; the sequence gives the 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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A298736 a(n) = s(n) - prime(n+1)+3, where s(n) = smallest even number x > prime(n) such that the difference x-p is composite for all primes p <= prime(n).

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