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%I #72 Jul 14 2020 23:25:54
%S 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,
%T 33,33,36,36,25,25,2,25,4,3,4,6,6,8,222,8,95,223,99,98,95,88,222,94,
%U 93,94,95,92,226,88,83,92,225,92
%N Least positive integer m such that m + n divides prime(m) + prime(n).
%C Conjecture: a(n) exists for any n > 0. Moreover, a(n) < n*(n-1) for all n > 2. - _Zhi-Wei Sun_, Sep 25 2014
%C A247869(n) = (prime(a(n)) + prime(n)) / (a(n) + n). - _Reinhard Zumkeller_, Sep 27 2014
%C 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
%C 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
%C 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
%H Zhi-Wei Sun, <a href="/A247824/b247824.txt">Table of n, a(n) for n = 1..100000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1409.5685">A new theorem on the prime-counting function</a>, arXiv:1409.5685, 2014.
%H Zhi-Wei Sun, <a href="http://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;b6af193d.1409">m+n divides prime(m)+prime(n) for some n>0</a>, a message to Number Theory List, Sept. 27, 2014.
%H Zhi-Wei Sun, <a href="http://dx.doi.org/doi:10.1007/s11139-015-9702-z">A new theorem on the prime-counting function</a>, Ramanujan J. 42(2017), no.1, 59-67.
%e a(2) = 5 since 5 + 2 = 7 divides prime(5) + prime(2) = 11 + 3 = 14.
%e a(10409) = 69804276 since 69804276 + 10409 = 69814685 divides prime(10409) + prime(69804276) = 109481 + 1396184219 = 1396293700 = 20*69814685.
%e a(35980) = 180302246 since 35980 + 180302246 = 180338226 divides prime(35980) + prime(180302246) = 427727 + 3786675019 = 3787102746 = 21*180338226.
%e a(79276) = 3141281384 since 79276 + 3141281384 = 3141360660 divides prime(79276) + prime(3141281384) = 1010431 + 75391645409 = 75392655840 = 24*3141360660.
%t 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}]
%t 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 *)
%o (PARI) a(n) = {m = 1; while ((prime(m) + prime(n)) % (m + n), m++); m;} \\ _Michel Marcus_, Sep 25 2014
%o (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
%o (Haskell)
%o import Data.List (genericIndex)
%o a247824 n = genericIndex a247824_list (n - 1)
%o a247824_list = f ips where
%o f ((x, p) : xps) = head
%o [y | (y, q) <- ips, (p + q) `mod` (x + y) == 0] : f xps
%o ips = zip [1..] a000040_list
%o -- _Reinhard Zumkeller_, Sep 27 2014
%Y Cf. A000040, A247600, A247793.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Sep 24 2014
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