A082101
Primes of form 2^k + 3^k.
Original entry on oeis.org
m=0: 1+1, m=1: 2+3, m=2: 4+9, m=4: 16+81.
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a={}; Do[If[PrimeQ[p=2^n+3^n], AppendTo[a, p]], {n, 0, 10^3}]; a (* Vladimir Joseph Stephan Orlovsky, Aug 07 2008 *)
Select[Table[2^k+3^k,{k,0,100}],PrimeQ] (* Harvey P. Dale, May 14 2014 *)
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print1(2);for(n=0,99,if(ispseudoprime(t=2^(2^n)+3^(2^n)),print1(", "t))) \\ Charles R Greathouse IV, Jul 19 2011
A294132
Sorted list of prime factors of numbers of the form 3^(2^m) + 2^(2^m) with m >= 0.
Original entry on oeis.org
5, 13, 17, 97, 257, 401, 769, 1153, 3041, 14177, 65537, 286721, 1810433, 2752513, 4043777, 7340033, 13631489, 23068673, 72222721, 319291393, 348061697, 625483777, 3937533953, 54498164737, 106216554497, 121899667073, 151597350913, 342456532993
Offset: 1
The first 5 such numbers are 5, 13, 97, 6817, 43112257, 1853024483819137. Their prime factorizations are (5), (13), (97), (17) (401), (14177) (3041), (1153) (1607133116929). - _N. J. A. Sloane_, Oct 29 2017
- Arkadiusz Wesolowski, Table of n, a(n) for n = 1..48
- Anders Björn and Hans Riesel, Factors of generalized Fermat numbers, Math. Comp. 67 (1998), no. 221, pp. 441-446.
- Anders Björn and Hans Riesel, Table errata to “Factors of generalized Fermat numbers”, Math. Comp. 74 (2005), no. 252, p. 2099.
- Anders Björn and Hans Riesel, Table errata 2 to "Factors of generalized Fermat numbers", Math. Comp. 80 (2011), pp. 1865-1866.
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print1(5, ", "); forprime(p=13, 342456532993, z=znorder(Mod(3/2, p)); if(2^ispower(z)==z, print1(p, ", ")));
A094316
Primes p for which 2^j+p^j is also prime for j in {0,2,8,512}.
Original entry on oeis.org
13, 4133, 1831343, 2320583, 3828673, 9173893, 23658377, 24037537, 42489677, 56253203, 78222863, 96325093, 99846337, 110453773, 110468653, 117748427, 122173187, 130937467, 138072163, 146981537, 174978913, 184050553, 186927817
Offset: 1
Smallest such prime is 13 and the relevant four primes are
2, 173, 815730977 and a 571-digit prime.
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{ta=Table[0, {100}], u=1}; {exponents, {a, b, c, d}={0, 2, 8, 512}} Do[s0=Prime[j]^a+2^a;s1=Prime[j]^b+2^b;s2=Prime[j]^c+2^c;s3=Prime[j]^d+2^d; If[PrimeQ[s0]&&PrimeQ[s1]&&PrimeQ[s2]&&PrimeQ[s3], Print[{j, Prime[j]}];ta[[u]]=Prime[j];u=u+1], {j, 1, 1000000}] ta
Showing 1-3 of 3 results.
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