A247824 Least positive integer m such that m + n divides prime(m) + prime(n).

1, 5, 5, 5, 2, 2, 38, 16, 40, 12, 13, 1, 11, 1, 11, 4, 35, 38, 35, 35, 38, 35, 35, 36, 31, 31, 33, 33, 36, 36, 25, 25, 2, 25, 4, 3, 4, 6, 6, 8, 222, 8, 95, 223, 99, 98, 95, 88, 222, 94, 93, 94, 95, 92, 226, 88, 83, 92, 225, 92

Offset: 1

Author

Zhi-Wei Sun, Sep 24 2014

Keywords

• nonn

Comments

Conjecture: a(n) exists for any n > 0. Moreover, a(n) < n*(n-1) for all n > 2. - Zhi-Wei Sun, Sep 25 2014

A247869(n) = (prime(a(n)) + prime(n)) / (a(n) + n). - Reinhard Zumkeller, Sep 27 2014

I have verified the conjecture for n up to 10^5, and noted that max{a(n): n=1..10^5} = a(79276) = 3141281384 > 3*10^9. - Zhi-Wei Sun, Oct 08 2014

I would like to offer 500 US dollars as the prize for the first proof of the above conjecture. - Zhi-Wei Sun, Feb 24 2018

Chang Zhang (a student of Nanjing Univ.) has verified the conjecture for n up to 4*10^5. For example, a(337647) = 21342496785. - Zhi-Wei Sun, Jun 22 2020

Examples

			a(2) = 5 since 5 + 2 = 7 divides prime(5) + prime(2) = 11 + 3 = 14.
a(10409) = 69804276 since 69804276 + 10409 = 69814685 divides prime(10409) + prime(69804276) = 109481 + 1396184219 = 1396293700 = 20*69814685.
a(35980) = 180302246 since 35980 + 180302246 = 180338226 divides prime(35980) + prime(180302246) = 427727 + 3786675019 = 3787102746 = 21*180338226.
a(79276) = 3141281384 since 79276 + 3141281384 = 3141360660 divides prime(79276) + prime(3141281384) = 1010431 + 75391645409 = 75392655840 = 24*3141360660.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..100000
• Zhi-Wei Sun, A new theorem on the prime-counting function, arXiv:1409.5685, 2014.
• Zhi-Wei Sun, m+n divides prime(m)+prime(n) for some n>0, a message to Number Theory List, Sept. 27, 2014.
• Zhi-Wei Sun, A new theorem on the prime-counting function, Ramanujan J. 42(2017), no.1, 59-67.

Crossrefs

Cf. A000040, A247600, A247793.

Programs

Haskell

import Data.List (genericIndex)
a247824 n = genericIndex a247824_list (n - 1)
a247824_list = f ips where
   f ((x, p) : xps) = head
     [y | (y, q) <- ips, (p + q) `mod` (x + y) == 0] : f xps
   ips = zip [1..] a000040_list
-- Reinhard Zumkeller, Sep 27 2014

Mathematica

Do[m=1;Label[aa];If[Mod[Prime[m]+Prime[n],m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]
lpi[n_]:=Module[{k=1,p=Prime[n]},While[!Divisible[p+Prime[k],k+n], k++]; k]; Array[lpi,60] (* Harvey P. Dale, Apr 23 2015 *)

PARI

a(n) = {m = 1; while ((prime(m) + prime(n)) % (m + n), m++); m;} \\ Michel Marcus, Sep 25 2014

PARI

a(n)=my(p=prime(n),m); forprime(q=2,, if((p+q)%(n+m++)==0, return(m))) \\ Charles R Greathouse IV, Sep 25 2014