A078496 -id:A078496 - cp's OEIS frontend

A083060 a(n) is the number of natural numbers k such that A078496(k) = prime(n), where prime(n) denotes the n-th prime.

0, 0, 1, 2, 3, 2, 3, 3, 3, 3, 5, 3, 5, 4, 3, 4, 3, 6, 4, 5, 4, 4, 5, 4, 6, 4, 4, 5, 4, 4, 7, 4, 5, 7, 5, 6, 5, 5, 7, 4, 5, 6, 4, 5, 5, 5, 5, 6, 8, 4, 7, 4, 5, 6, 5, 4, 3, 8, 8, 5, 5, 5, 7, 7, 5, 5, 9, 4, 7, 8, 8, 6, 7, 4, 5, 7, 4, 7, 7, 6, 8, 7, 5, 5, 6, 7, 6

Offset: 1

Serhat Sevki Dincer (sevki(AT)ug.bilkent.edu.tr), Apr 18 2003

			a(3)=1 since there is only k=4 for A078496(k) = prime(3) = 5.
a(8)=3 since there is only k=13,16,18 for A078496(k) = prime(8) = 19.

Iain Fox, Table of n, a(n) for n = 1..10000

Cf. A078496.

A078496(n) = {my(p=nextprime(n+1)); while(!isprime(2*n-p), p = nextprime(p+1)); p; }
first(n) = { my(res=vector(n)); for(x=4, prime(n), my(r=primepi(A078496(x))); if(r <= n, res[r]++)); res; } \\ Iain Fox, Nov 23 2017

For n > 2, a(n) = card({k: k > 3; A078496(k) = prime(n)}).

a(1) and a(2) prepended by Iain Fox, Nov 23 2017

More terms from Iain Fox, Nov 23 2017

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

Original entry on oeis.org

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

A083059 a(n) is the number of natural numbers k such that A078496(k)=n.

Original entry on oeis.org

1, 0, 2, 0, 0, 0, 3, 0, 2, 0, 0, 0, 3, 0, 3, 0, 0, 0, 3, 0, 0, 0, 0, 0, 3, 0, 5, 0, 0, 0, 0, 0, 3, 0, 0, 0, 5, 0, 4, 0, 0, 0, 3, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 3, 0, 6, 0, 0, 0, 0, 0, 4, 0, 0, 0, 5, 0, 4, 0, 0, 0, 0, 0, 4, 0, 0, 0, 5, 0, 0, 0, 0, 0, 4, 0

Offset: 5

Serhat Sevki Dincer (sevki(AT)ug.bilkent.edu.tr), Apr 18 2003

			a(19)=3 since there are only k=13,16,18 for which A078496(k)=19

Cf. A078496.

b(n)={forprime(p=nextprime(n+1), 2*n, if(isprime(2*n-p), return(p))); 0}
a(n)={if(isprime(n), sum(i=n\2+1, n-1, b(i)==n), 0)} \\ Andrew Howroyd, Nov 06 2019

a(n) = #{k: k>3 integer; A078496(k)=n}.

Terms a(24) and beyond from Andrew Howroyd, Nov 06 2019

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

A078587 Largest prime p such that p

Original entry on oeis.org

3, 3, 5, 3, 5, 7, 7, 5, 11, 7, 11, 13, 13, 11, 17, 7, 17, 19, 13, 17, 19, 19, 23, 23, 19, 17, 29, 19, 23, 29, 31, 29, 31, 31, 29, 37, 37, 29, 41, 19, 41, 43, 31, 41, 43, 37, 47, 43, 43, 47, 47, 43, 53, 53, 43, 47, 59, 43, 53, 59, 61, 59, 61, 61, 53, 67, 67, 59, 71, 67, 59, 71

Offset: 4

T. D. Noe, Dec 02 2002

Suggested by Goldbach Conjecture.

Also, values of p from A143697. This follows from the factorization n^2-k^2 = (n-k)(n+k). - T. D. Noe, Jan 22 2009

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

Cf. A143697, A078496.

Cf. A082467.

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

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

a(n) = 2n - A078496(n)

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

A078497 The member r of a triple of primes (p,q,r) in arithmetic progression which sum to 3*prime(n) = A001748(n) = p + q + r.

Original entry on oeis.org

7, 11, 17, 19, 23, 31, 29, 41, 43, 43, 53, 67, 53, 59, 71, 79, 73, 83, 79, 97, 107, 107, 127, 113, 109, 113, 139, 137, 151, 149, 167, 151, 167, 163, 163, 199, 197, 179, 191, 199, 233, 223, 227, 241, 223, 283, 257, 277, 239, 251, 271, 263, 263, 269, 281, 313

Offset: 3

Serhat Sevki Dincer (sevki(AT)ug.bilkent.edu.tr), Nov 27 2002

nonn

In case more than one triple of primes p, q=p+d and r=p+2*d exists, we take r=a(n) from the triple with the smallest d. This shows the difference from A092940, which would take the maximum r over all triples. - R. J. Mathar, May 19 2007

			a(1) = 7 because 3+5+7 = 15;
a(2) = 11 because 3+7+11 = 21;
a(3) = 17 because 5+11+17= 33.

Cf. A078496, A078498, A001748, A092940, A071681, A078611.

A078497 := proc(n) local p3, i,d,r,p; p3 := ithprime(n) ; i := n+1 ; while true do r := ithprime(i) ; d := r-p3 ; p := p3-d ; if isprime(p) then RETURN(r) ; fi ; i := i+1 ; od ; RETURN(-1) ; end: for n from 3 to 60 do printf("%d, ",A078497(n)) ; od ; # R. J. Mathar, May 19 2007

f[n_] := Block[{p = Prime[n], k}, k = p + 1; While[ !PrimeQ[k] || !PrimeQ[2p - k], k++ ]; k]; Table[ f[n], {n, 3, 60}]

Edited and extended by Robert G. Wilson v, Nov 29 2002

Further edited by N. J. A. Sloane, Jul 03 2008 at the suggestion of R. J. Mathar

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

A078498 Let q(n) be the prime defined in A078497; sequence gives (q(n)-prime(n))/6.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 2, 1, 2, 4, 1, 1, 2, 3, 1, 2, 1, 3, 4, 3, 5, 2, 1, 1, 5, 4, 4, 3, 5, 2, 3, 2, 1, 6, 5, 1, 2, 3, 7, 5, 5, 7, 2, 10, 5, 8, 1, 2, 5, 2, 1, 1, 2, 7, 1, 2, 9, 4, 4, 7, 6, 6, 3, 5, 6, 3, 1, 7, 5, 1, 5, 6, 5, 4, 3, 2, 5, 2, 2, 4, 3, 4, 3, 14, 3, 4, 4, 2, 9, 2, 7, 9, 8, 7, 4, 13

Offset: 5

Serhat Sevki Dincer (sevki(AT)ug.bilkent.edu.tr), Nov 27 2002

			a(6)=1, a(25)=5.

Cf. A078496, A078497.

Table[p = Prime[i]; j = 0; While[j++; df = 6*j; ! ((PrimeQ[p + df]) && (PrimeQ[p - df]))]; j, {i, 5, 100}]

For n>4 a(n)=( min{p : p>prime(n), p and 2*prime(n)-p are primes} - prime(n) ) / 6.

More terms from Pab Ter (pabrlos(AT)yahoo.com), May 27 2004

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

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 *)

cp's OEIS Frontend

A083060 a(n) is the number of natural numbers k such that A078496(k) = prime(n), where prime(n) denotes the n-th prime.

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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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A083059 a(n) is the number of natural numbers k such that A078496(k)=n.

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

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A078587 Largest prime p such that p

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A078497 The member r of a triple of primes (p,q,r) in arithmetic progression which sum to 3*prime(n) = A001748(n) = p + q + r.

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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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