A078587 -id:A078587 - cp's OEIS frontend

A277583 Goldbach's problem extended to squares of prime gaps (>=2): smallest integer >= ((A078587(n) - A078496(n))^2)/n for n >= 4.

1, 4, 1, 10, 5, 2, 4, 14, 1, 12, 3, 2, 3, 9, 1, 31, 2, 1, 15, 7, 5, 6, 2, 3, 12, 20, 1, 19, 11, 2, 2, 5, 3, 4, 9, 1, 1, 15, 1, 54, 1, 1, 20, 4, 3, 12, 1, 6, 7, 3, 4, 11, 1, 2, 16, 10, 1, 22, 6, 2, 1, 3, 2, 3, 14, 1, 1, 9, 1, 2, 13, 1, 1, 2, 2, 17, 5, 1, 11, 28, 2, 7, 1, 10, 4, 15

Offset: 4

Juri-Stepan Gerasimov, Oct 22 2016

nonn

Where A078587(n) + A078496(n) = 2n and A078587(n) < A078496(n).

			a(5) = 4 because ((A078587(5) - A078496(5))^2)/5 = ((3 - 7)^2)/5 < 4, where 3 (prime) = 7 (prime) = 2*5;
a(6) = 1 because ((A078587(6) - A078496(6))^2)/6 = ((5 - 7)^2)/6 < 1, where 5 (prime) + 7 (prime) = 2*6;
a(7) = 10 because ((A078587(7) - A078496(7))^2)/7 = ((3 - 11)^2)/7 < 10, where 3 (prime) + 11 (prime) = 2*7.

Cf. A002375, A078496, A078587, A277581 (Goldbach's problem extended to squares of prime gaps >= 0).

Table[k = 1; While[k < ((Last@ # - First@ #)^2)/n, k++] &@ Block[{p = n + 1, q}, q = 2 n - p; While[q > 0 && Nand[PrimeQ@ p, PrimeQ@ q], p++; q--]; {p, q}]; k, {n, 4, 89}] (* or *)
Table[Ceiling[4 (n - #)^2/n] &@ Block[{p = n + 1, q}, q = 2 n - p; While[q > 0 && Nand[PrimeQ@ p, PrimeQ@ q], p++; q--]; p], {n, 4, 89}] (* Michael De Vlieger, Oct 26 2016, after T. D. Noe at A078587 and Michel Marcus PARI *)

maxp(n) = {my(p = precprime(n-1)); while(!isprime(2*n-p), p = precprime(p-1)); p;}
a(n) = ceil(4*(n - maxp(n))^2/n); \\ Michel Marcus, Oct 22 2016

A082467 Least k>0 such that n-k and n+k are both primes.

Original entry on oeis.org

1, 2, 1, 4, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 12, 3, 2, 9, 6, 5, 6, 3, 4, 9, 12, 1, 12, 9, 4, 3, 6, 5, 6, 9, 2, 3, 12, 1, 24, 3, 2, 15, 6, 5, 12, 3, 8, 9, 6, 7, 12, 3, 4, 15, 12, 1, 18, 9, 4, 3, 6, 5, 6, 15, 2, 3, 12, 1, 6, 15, 4, 3, 6, 5, 18, 9, 2, 15, 24, 5, 12, 3, 14, 9, 18, 7, 12, 9, 4, 15, 6, 7, 30, 9

Offset: 4

Benoit Cloitre, Apr 27 2003

nonn

The existence of k>0 for all n >= 4 is equivalent to the strong Goldbach Conjecture that every even number >= 8 is the sum of two distinct primes.
n and k are coprime, because otherwise n + k would be composite. So the rational sequence r(n) = a(n)/n = k/n is injective. - Jason Kimberley, Sep 21 2011
Because there are arbitrarily many composites from m!+2 to m!+m, there are also arbitrarily large a(n) but they increase very slowly. The twin prime conjecture implies that infinitely many a(n) are 1. - Juhani Heino, Apr 09 2020

			n=10: k=3 because 10-3 and 10+3 are both prime and 3 is the smallest k such that n +/- k are both prime.

Klaus Brockhaus, Table of n, a(n) for n = 4..5000
OEIS (Plot 2), log_10(A082467(n)/n) vs n
J. S. Kimberley, A082467

Cf. A087695, A087696, A087697, A087678, A087679, A087680, A087681, A087682, A087683, A087711.
Cf. A129301 (records), A129302 (where records occur).
Cf. A047160 (allows k=0).
Cf. A078611 (subset for prime n).

A082467 := func; [A082467(n):n in [4..98]]; // Jason Kimberley, Sep 03 2011

A082467 := proc(n) local k; k := 1+irem(n,2);
while n > k do if isprime(n-k) then if isprime(n+k)
then RETURN(k) fi fi; k := k+2 od; print("Goldbach erred!") end:
seq(A082467(i),i=4..90); # Peter Luschny, Sep 21 2011

f[n_] := Block[{k}, If[OddQ[n], k = 2, k = 1]; While[ !PrimeQ[n - k] || !PrimeQ[n + k], k += 2]; k]; Table[ f[n], {n, 4, 98}] (* Robert G. Wilson v, Mar 28 2005 *)

a(n)=if(n<0,0,k=1; while(isprime(n-k)*isprime(n+k) == 0,k++); k)

A078496(n)-a(n) = A078587(n)+a(n) = n.

Entries checked by Klaus Brockhaus, Apr 08 2007

A078496 Smallest prime p such that p>n and 2*n-p is also prime.

Original entry on oeis.org

5, 7, 7, 11, 11, 11, 13, 17, 13, 19, 17, 17, 19, 23, 19, 31, 23, 23, 31, 29, 29, 31, 29, 31, 37, 41, 31, 43, 41, 37, 37, 41, 41, 43, 47, 41, 43, 53, 43, 67, 47, 47, 61, 53, 53, 61, 53, 59, 61, 59, 61, 67, 59, 61, 73, 71, 61, 79, 71, 67, 67, 71, 71, 73, 83, 71, 73, 83, 73, 79

Offset: 4

Serhat Sevki Dincer (sevki(AT)ug.bilkent.edu.tr), Nov 26 2002

nonn

Suggested by Goldbach Conjecture.

Values of q from A143697. This follows from the factorization n^2-k^2 = (n-k)(n+k).

			a(11)=17.

P. CAMI, Table of n, a(n) for n = 4..60000

Cf. A143697, A078587.
a(n) = 2n - A078587(n).
Cf. A082467

Table[p=n+1; q=2n-p; While[q>0&&!(PrimeQ[p]&&PrimeQ[q]), p++; q-- ]; p, {n, 4, 100}]

a(n) = {my(p=nextprime(n+1)); while(!isprime(2*n-p), p = nextprime(p+1)); p;} \\ Michel Marcus, Oct 22 2016

n>3 integer; a(n)=min{p: p>n; p, 2*n-p are primes}.

Edited by N. J. A. Sloane, Jan 24 2009 at the suggestion of R. J. Mathar and T. D. Noe.

A063748 Greatest x that is a solution to x-phi(x)=n or zero if there is no solution, where phi(x) is Euler's totient function.

Original entry on oeis.org

4, 9, 8, 25, 10, 49, 16, 27, 0, 121, 22, 169, 26, 55, 32, 289, 34, 361, 38, 85, 30, 529, 46, 133, 0, 187, 52, 841, 58, 961, 64, 253, 0, 323, 68, 1369, 74, 391, 76, 1681, 82, 1849, 86, 493, 70, 2209, 94, 589, 0, 667, 0, 2809, 106, 703, 104, 697, 0, 3481, 118, 3721, 122

Offset: 2

Labos Elemer, Aug 13 2001

nonn

See A051953 for x-phi(x), the cototient function. Note that a(n)=0 for n in A005278. Also note that n=1 has an infinite number of solutions. If n is prime, then a(n)=n^2. If n is even, then a(n)<=2n. In particular, if n=p+1 for a prime p, then a(n)=2n-2. Also, if n=2^k, then a(n)=2n. If n>9 is odd and composite, then a(n)=pq, with p>q odd primes with p+q=n+1 and p-q minimal. We can take p=A078496((n+1)/2) and q=A078587((n+1)/2).

			For n=15, the solutions are x=39 and x=55, so a(15)=55. Note that 55=5*11 and 5+11=n+1.

T. D. Noe, Table of n, a(n) for n=2..1000

Cf. A000010, A051953.

Cf. A063507 (least solution to x-phi(x)=n), A063740 (number of solutions to x-phi(x)=n).

nn=10^4; lim=Floor[Sqrt[nn]]; mx=Table[0,{lim}]; Do[c=n-EulerPhi[n]; If[0T. D. Noe *)
Table[Module[{k = n^2}, While[And[k - EulerPhi@ k != n, k > 0], k--];
k], {n, 2, 62}] (* Michael De Vlieger, Mar 17 2017 *)

a(n)=Max{x : A051953(x)=n} if the inverse set is not empty; a(n)=0 if no inverse exists.

Corrected and edited by T. D. Noe, Oct 30 2006

A143697 Least square k^2 such that n^2-k^2 = p*q with p and q odd primes and p= 4.

Original entry on oeis.org

1, 4, 1, 16, 9, 4, 9, 36, 1, 36, 9, 4, 9, 36, 1, 144, 9, 4, 81, 36, 25, 36, 9, 16, 81, 144, 1, 144, 81, 16, 9, 36, 25, 36, 81, 4, 9, 144, 1, 576, 9, 4, 225, 36, 25, 144, 9, 64, 81, 36, 49, 144, 9, 16, 225, 144, 1, 324, 81, 16, 9, 36, 25, 36, 225, 4, 9, 144, 1, 36, 225

Offset: 4

Pierre CAMI, Aug 29 2008

nonn

The product p*q is the sum of p consecutive odd numbers with 2*n-1 the greatest.
For n=4 p*q=3*5=15, 15=7+5+3
For n=5 p*q=3*7=21, 21=9+7+5
For n=6 p*q=5*7=35, 35=11+9+7+5+3
For n=7 p*q=3*11=33, 33=13+11+9
k^2 is the sum of the k first consecutive odd numbers p=n-k and q=n+k.
Assuming a strong version of the Goldbach conjecture, every term exists and we have a(n)=A082467(n)^2, p(n)=A078587(n) and q(n)=A078496(n). [T. D. Noe, Jan 22 2009]

			4*4-1=3*5 p=3 q=5
5*5-4=3*7 p=3 q=7
6*6-1=5*7 p=5 q=7
7*7-16=3*11 p=3 q=11

Pierre CAMI, Table of n, a(n) for n = 4..60000

Cf. A078587, A078496.

a(n) = {for (k=1, n-1, my(x=n^2-k^2); if ((omega(x)==2) && (bigomega(x)==2) && (x%2), return(k^2);););} \\ Michel Marcus, Sep 23 2019

A167485 Smallest positive integer m such that n can be expressed as the sum of an initial subsequence of the divisors of m, or 0 if no such m exists.

Original entry on oeis.org

1, 1, 0, 2, 3, 0, 5, 4, 7, 15, 12, 21, 6, 9, 13, 8, 12, 30, 10, 42, 19, 18, 20, 57, 14, 36, 46, 30, 12, 102, 29, 16, 21, 42, 62, 84, 22, 36, 37, 18, 27, 63, 20, 50, 43, 66, 52, 129, 33, 75, 40, 78, 48, 220, 34, 36, 28, 49, 60, 265, 24, 132, 61, 32, 56, 117, 54, 100, 67, 90, 84

Offset: 0

Franklin T. Adams-Watters, Nov 04 2009

It appears that 2 and 5 are the only zeros in this sequence. This would follow from a slightly stronger version of the Goldbach conjecture: every even integer > 22 can be expressed as the sum of two primes p and q, with 5 < p < q < 5p. Then odd numbers can be obtained for pq and even numbers for 5pq.
Is a(n) = o(n)? - Arkadiusz Wesolowski, Nov 09 2013
The above question has been posed by Erdős. See Guy. - Stefano Spezia, Sep 25 2024
a(A000203(n)) <= n. Since A000203(n)/n can be arbitrarily large, that shows that lim inf_{n -> oo} a(n)/n = 0.  - Robert Israel, Sep 26 2024

			The divisors of 15 are 1,3,5,15, with cumulative sums 1,4,9,24. Since this is the smallest number where 9 occurs in the sums, a(9) = 15.

R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section B2.

Robert Israel, Table of n, a(n) for n = 0..20000 (n = 0 .. 1000 from Michel Marcus)
Robert Israel, Plot of a(n)/n for n = 1 .. 10^6

Cf. A000203, A001065, A078587, A051444.

N:= 100: # for a(0) .. a(N)
count:= 1: V:= Array(0..N): V[0]:= 1:
for m from 1 while count < N-1 do
  L:= ListTools:-PartialSums(sort(convert(numtheory:-divisors(m),list)));
  for x in L do
    if x > N then break fi;
    if V[x] = 0 then V[x]:= m; count:= count+1 fi;
od od:
convert(V,list); # Robert Israel, Sep 26 2024

{u=vector(100); for(n=1,1000,ds=divisors(n);s=0; for(k=1,#ds,s+=ds[k];if(s>#u,break);if(!u[s],u[s]=n))); u}

A305883 Triangle read by rows: row n lists the pairs (p, q) such that p, q are primes, p+q=2*n and p < q.

Original entry on oeis.org

3, 5, 3, 7, 5, 7, 3, 11, 3, 13, 5, 11, 5, 13, 7, 11, 3, 17, 7, 13, 3, 19, 5, 17, 5, 19, 7, 17, 11, 13, 3, 23, 7, 19, 5, 23, 11, 17, 7, 23, 11, 19, 13, 17, 3, 29, 13, 19, 3, 31, 5, 29, 11, 23, 5, 31, 7, 29, 13, 23, 17, 19, 7, 31, 3, 37, 11, 29, 17, 23, 5, 37, 11, 31

Offset: 4

Seiichi Manyama, Jun 13 2018

			  n  | (p,q)
  ---+----------------------------
   4 | (3,  5);
   5 | (3,  7);
   6 | (5,  7);
   7 | (3, 11);
   8 | (3, 13), (5, 11);
   9 | (5, 13), (7, 11);
  10 | (3, 17), (7, 13);
  11 | (3, 19), (5, 17);
  12 | (5, 19), (7, 17), (11, 13);

Seiichi Manyama, Rows n = 4..374, flattened
Eric Weisstein's World of Mathematics, Goldbach Partition
Wikipedia, Goldbach's conjecture

Cf. A002373, A020481, A061357 (the size of row n), A078496, A078587.

row[n_] := Select[Table[{p, 2 n - p}, {p, Prime[Range[PrimePi[n]]]}], Less @@ # && AllTrue[#, PrimeQ]&] // Union;
Table[row[n], {n, 4, 25}] // Flatten (* Jean-François Alcover, Jun 16 2018 *)

A234649 Difference between the first members of the widest and the narrowest prime pair having an arithmetic mean of n.

Original entry on oeis.org

2, 2, 4, 2, 6, 4, 6, 6, 10, 8, 12, 0, 14, 14, 10, 14, 14, 16, 18, 16, 16, 12, 22, 16, 20, 24, 24, 26, 26, 28, 26, 32, 30, 26, 36, 16, 36, 36, 28, 36, 36, 18, 44, 38, 40, 44, 42, 40, 50, 48, 40, 42, 52, 30, 42, 46, 42, 56, 56, 58, 48, 60, 64, 56, 66, 60, 48, 60, 70, 68, 68, 54, 68, 74, 60, 56

Offset: 8

Ralf Stephan, Dec 29 2013

nonn

The widest prime pair with a mean of n is (A002373(n),A020482(n)) and the narrowest is (A078587(n),A078496(n)).
Existence of a(n) for all n depends on A061357(n) > 0.
Even numbers missing in the subsequence with n<10^5 are 34,62,82,88,112,116,118,122,130,140,152...
a(n) = 0 for n=4,5,6,7,19 because A061357(n) = 1.

			The prime pairs with an arithmetic mean of 18 are (17,19), (13,23), (7,29), and (5,31), so a(18) = 17-5 = 31-19 = 12. The only pair with mean of 19 is (7,31) so a(19) = 0.

Ralf Stephan, Table of n, a(n) for n = 8..100000

Cf. A045917.

a(n)=mi=0;ma=0;forprime(p=3,n-1,if(isprime(2*n-p),if(!mi,mi=2*n-p);ma=2*n-p));if(!ma,-1,mi-ma)

a(n) = A078587(n) - A002373(n) = A078496(n) - A020482(n).

cp's OEIS Frontend

A277583 Goldbach's problem extended to squares of prime gaps (>=2): smallest integer >= ((A078587(n) - A078496(n))^2)/n for n >= 4.

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A082467 Least k>0 such that n-k and n+k are both primes.

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A078496 Smallest prime p such that p>n and 2*n-p is also prime.

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A063748 Greatest x that is a solution to x-phi(x)=n or zero if there is no solution, where phi(x) is Euler's totient function.

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A143697 Least square k^2 such that n^2-k^2 = p*q with p and q odd primes and p= 4.

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A167485 Smallest positive integer m such that n can be expressed as the sum of an initial subsequence of the divisors of m, or 0 if no such m exists.

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A305883 Triangle read by rows: row n lists the pairs (p, q) such that p, q are primes, p+q=2*n and p < q.

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