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A248550
Numbers n such that the smallest prime divisor of n^2+1 is 73.
1
100, 246, 484, 630, 776, 830, 976, 1506, 1706, 1944, 2144, 2236, 2290, 2874, 3020, 3604, 3696, 3750, 3896, 4134, 4426, 4626, 4864, 5064, 5210, 5356, 5594, 5740, 5794, 5940, 6086, 6324, 6470, 6616, 6670, 6816, 7200, 7254, 7346, 7400, 7546, 7930, 7984, 8076
OFFSET
1,1
COMMENTS
Or numbers n such that the smallest prime divisor of n^2+1 is A002313(10).
a(n) == 46 or 100 (mod 146).
No need to completely factorize n^2+1. - David A. Corneth, Apr 29 2017
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)
EXAMPLE
100 is in the sequence because 100^2+1= 73*137.
246 is in the sequence because 246^2+1 isn't divisible by any prime less than 73 and is divisible by 73. - David A. Corneth, Apr 29 2017
MATHEMATICA
lst={}; Do[If[FactorInteger[n^2+1][[1, 1]]==73, AppendTo[lst, n]], {n, 2, 10000}]; lst
Select[Range[8100], FactorInteger[#^2+1][[1, 1]]==73&] (* Harvey P. Dale, Apr 29 2017 *)
p = 73; ps = Select[Range[p - 1], Mod[#, 4] != 3 && PrimeQ[#] &]; Select[Range[8000], Divisible[(nn = #^2 + 1), p] && ! Or @@ Divisible[nn, ps] &] (* Amiram Eldar, Aug 16 2019 *)
PROG
(PARI) is(n) = {my(m=n%146, p=2, n21 = n^2+1, v=[5, 13, 17, 29, 37, 41, 53, 61]);
return(abs(73-m)==27&&sum(i=1, #v, p=nextprime(p+1); valuation(n21, v[i]))==0)}
upto(n) = {my(l=List(), i=54, m=46); while(m<n, if(is(m), listput(l, m)); m+=i; i=146-i); l} \\ David A. Corneth, Apr 29 2017
KEYWORD
nonn,easy
AUTHOR
Michel Lagneau, Oct 08 2014
STATUS
approved