Displaying 1-10 of 11 results found.
Generalized Fermat numbers: 3^(2^n)+1, n >= 0.
+10
16
4, 10, 82, 6562, 43046722, 1853020188851842, 3433683820292512484657849089282, 11790184577738583171520872861412518665678211592275841109096962
COMMENTS
Generalized Fermat numbers (Ribenboim (1996))
F_n(a) := F_n(a,1) = a^(2^n) + 1, a >= 2, n >= 0, can't be prime if a is odd (as is the case for this sequence). - Daniel Forgues, Jun 19-20 2011
FORMULA
a(0) = 4; a(n) = (a(n-1)-1)^2 + 1, n >= 1.
a(n) = 2*a(n-1)*a(n-2)*...*a(1)*a(0) + 2, n >= 0, where for n = 0, we get 2*(empty product, i.e., 1) + 2 = 4 = a(0).
The above formula implies the GCD of any pair of terms is 2, which means that the terms of (3^(2^n)+1)/2 ( A059917) are pairwise coprime. - Daniel Forgues, Jun 20 & 22 2011
EXAMPLE
a(0) = 3^(2^0)+1 = 3^1+1 = 4 = 2*(1)+2 = 2*(empty product)+2;
a(1) = 3^(2^1)+1 = 3^2+1 = 10 = 2*(4)+2;
a(2) = 3^(2^2)+1 = 3^4+1 = 82 = 2*(4*10)+2;
a(3) = 3^(2^3)+1 = 3^8+1 = 6562 = 2*(4*10*82)+2;
a(4) = 3^(2^4)+1 = 3^16+1 = 43046722 = 2*(4*10*82*6562)+2;
a(5) = 3^(2^5)+1 = 3^32+1 = 1853020188851842 = 2*(4*10*82*6562*43046722)+2;
PROG
(PARI) { for (n=0, 11, write("b059919.txt", n, " ", 3^(2^n) + 1); ) } \\ Harry J. Smith, Jun 30 2009
CROSSREFS
Cf. A000215 (Fermat numbers: 2^(2^n) + 1, n >= 0).
Generalized Fermat numbers: 10^(2^n) + 1, n >= 0.
+10
14
11, 101, 10001, 100000001, 10000000000000001, 100000000000000000000000000000001, 10000000000000000000000000000000000000000000000000000000000000001
COMMENTS
As for standard Fermat numbers 2^(2^n) + 1, a number (2b)^m + 1 (with b > 1) can only be prime if m is a power of 2. On the other hand, out of the first 12 base-10 Fermat numbers, only the first two are primes.
Also, binary representation of Fermat numbers (in decimal, see A000215).
FORMULA
a(0) = 11; a(n) = (a(n - 1) - 1)^2 + 1.
a(n) = 9*a(n-1)*a(n-2)*...*a(1)*a(0) + 2, n >= 0, where for n = 0, we get 9*(empty product, i.e., 1)+ 2 = 11 = a(0). - Daniel Forgues, Jun 20 2011
EXAMPLE
a(0) = 10^1 + 1 = 11 = 9*(1) + 2 = 9*(empty product) + 2.
a(1) = 10^2 + 1 = 101 = 9*(11) + 2.
a(2) = 10^4 + 1 = 10001 = 9*(11*101) + 2.
a(3) = 10^8 + 1 = 100000001 = 9*(11*101*10001) + 2.
a(4) = 10^16 + 1 = 10000000000000001 = 9*(11*101*10001*100000001) + 2.
a(5) = 10^32 + 1 = 100000000000000000000000000000001 = 9*(11*101*10001*100000001*10000000000000001) + 2.
CROSSREFS
Cf. A000215 (Fermat numbers: 2^(2^n) + 1, n >= 0).
Generalized Fermat numbers: 6^(2^n) + 1, n >= 0.
+10
13
7, 37, 1297, 1679617, 2821109907457, 7958661109946400884391937, 63340286662973277706162286946811886609896461828097
COMMENTS
The next term is too large to include.
As for standard Fermat numbers 2^(2^n) + 1, a number (2b)^m + 1 (with b > 1) can only be prime if m is a power of 2. On the other hand, out of the first 13 base-6 Fermat numbers, only the first three are primes.
Either the sequence of (standard) Fermat numbers contains infinitely many composite numbers or the sequence of base-6 Fermat numbers contains infinitely many composite numbers (cf. https://mathoverflow.net/a/404235/1593). - José Hernández, Nov 09 2021 Since all powers of 6 are congruent to 6 (mod 10), all terms of this sequence are congruent to 7 (mod 10). - Daniel Forgues, Jun 22 2011 There are only 5 known Fermat primes of the form 2^(2^n) + 1: {3, 5, 17, 257, 65537}. There are only 2 known base-10 generalized Fermat primes of the form 10^(2^n) + 1: {11, 101}. - Alexander Adamchuk, Mar 17 2007
FORMULA
a(0) = 7, a(n) = (a(n-1)-1)^2 + 1, n >= 1.
a(n) = 5*a(n-1)*a(n-2)*...*a(1)*a(0) + 2, n >= 0, where for n = 0, we get 5*(empty product, i.e., 1)+ 2 = 7 = a(0). This implies that the terms are pairwise coprime. - Daniel Forgues, Jun 20 2011
EXAMPLE
a(0) = 6^1+1 = 7 = 5*(1)+2 = 5*(empty product)+2;
a(1) = 6^2+1 = 37 = 5*(7)+2;
a(2) = 6^4+1 = 1297 = 5*(7*37)+2;
a(3) = 6^8+1 = 1679617 = 5*(7*37*1297)+2;
a(4) = 6^16+1 = 2821109907457 = 5*(7*37*1297*1679617)+2;
a(5) = 6^32+1 = 7958661109946400884391937 = 5*(7*37*1297*1679617*2821109907457)+2;
CROSSREFS
Cf. A000215 (Fermat numbers: 2^(2^n) + 1, n >= 0). Cf. A019434 (Fermat primes of the form 2^(2^n) + 1). Cf. A123669, A123599, A056993, A126032, A178428, A059919, A199591, A078304, A152581, A080176, A199592, A152585.
Generalized Fermat numbers: 7^(2^n)+1, n >= 0.
+10
13
8, 50, 2402, 5764802, 33232930569602, 1104427674243920646305299202, 1219760487635835700138573862562971820755615294131238402
COMMENTS
Generalized Fermat numbers F_n(a) := F_n(a,1) = a^(2^n)+1, a >= 2, n >= 0, can't be prime if a is odd (as is the case for the current sequence) (Ribenboim (1996)).
All factors of generalized Fermat numbers F_n(a,b) := a^(2^n)+b^(2^n), a >= 2, n >= 0, are of the form k*2^m+1, k >= 1, m >=0 (Riesel (1994, 1998)). (This only expresses that the factors are odd, which means that it only applies to odd generalized Fermat numbers.) (End)
FORMULA
a(0) = 8, a(n)=(a(n-1)-1)^2+1, n >= 1.
a(n) = 6*a(n-1)*a(n-2)*...*a(1)*a(0) + 2, n >= 0, where for n = 0, we get 6*(empty product, i.e., 1)+ 2 = 8 = a(0). This means that the GCD of any pair of terms is 2. - Daniel Forgues, Jun 20 2011
EXAMPLE
a(0) = 7^1+1 = 8 = 6*(1)+2 = 6*(empty product)+2.
a(1) = 7^2+1 = 50 = 6*(8)+2.
a(2) = 7^4+1 = 2402 = 6*(8*50)+2.
a(3) = 7^8+1 = 5764802 = 6*(8*50*2402)+2.
a(4) = 7^16+1 = 33232930569602 = 6*(8*50*2402*5764802)+2.
a(5) = 7^32+1 = 1104427674243920646305299202 = 6*(8*50*2402*5764802*33232930569602)+2.
CROSSREFS
Cf. A000215 (Fermat numbers: 2^(2^n)+1, n >= 0).
Generalized Fermat numbers: 12^(2^n) + 1, n >= 0.
+10
13
13, 145, 20737, 429981697, 184884258895036417, 34182189187166852111368841966125057, 1168422057627266461843148138873451659428421700563161428957815831003137
COMMENTS
There appears to be no divisibility rule for this sequence.
13 is the only prime up to 12^(2^15)+1.
FORMULA
a(0) = 13; a(n)=(a(n-1)-1)^2 + 1, n >= 1.
a(n) = 11*a(n-1)*a(n-2)*...*a(1)*a(0) + 2, n >= 0, where for n = 0, we get 11*(empty product, i.e., 1)+ 2 = 13 = a(0). This implies that the terms, all odd, are pairwise coprime. - Daniel Forgues, Jun 20 2011
EXAMPLE
a(0) = 12^1+1 = 13 = 11(1)+2 = 11(empty product)+2.
a(1) = 12^2+1 = 145 = 11(13)+2.
a(2) = 12^4+1 = 20737 = 11(13*145)+2.
a(3) = 12^8+1 = 429981697 = 11(13*145*20737)+2.
a(4) = 12^16+1 = 184884258895036417 = 11(13*145*20737*429981697)+2.
a(5) = 12^32+1 = 34182189187166852111368841966125057 = 11(13*145*20737*429981697*184884258895036417)+2.
PROG
(PARI) g(a, n) = if(a%2, b=2, b=1); for(x=0, n, y=a^(2^x)+b; print1(y", "))
(Python)
CROSSREFS
Cf. A000215 (Fermat numbers: 2^(2^n)+1, n >= 0).
Generalized Fermat numbers: 5^(2^n) + 1, n >= 0.
+10
12
6, 26, 626, 390626, 152587890626, 23283064365386962890626, 542101086242752217003726400434970855712890626
FORMULA
a(0) = 6; a(n) = (a(n-1)-1)^2 + 1, n >= 1.
a(0) = 6, a(1) = 26; a(n) = a(n-1) + 4*5^(2^(n-1))*Product_{i=0..n-2} a(i), n >= 2.
a(0) = 6, a(1) = 26; a(n) = a(n-1)^2 - 2*(a(n-2)-1)^2, n >= 2.
a(0) = 6; a(n) = 4*(Product_{i=0..n-1} a(i)) + 2, n >= 1.
EXAMPLE
a(0) = 5^(2^0) + 1 = 5^1 + 1 = 6 = 4*(2^0) + 2;
a(1) = 5^(2^1) + 1 = 5^2 + 1 = 26 = 4*(2^1*3) + 2;
a(2) = 5^(2^2) + 1 = 5^4 + 1 = 626 = 4*(2^2*3*13) + 2;
a(3) = 5^(2^3) + 1 = 5^8 + 1 = 390626 = 4*(2^3*3*13*313) + 2;
a(4) = 5^(2^4) + 1 = 5^16 + 1 = 152587890626 = 4*(2^4*3*13*313*195313) + 2;
a(5) = 5^(2^5) + 1 = 5^32 + 1 = 23283064365386962890626 = 4*(2^5*3*13*313*195313*76293945313) + 2;
MATHEMATICA
Table[5^2^n + 1, {n, 0, 6}]
PROG
(Magma) [5^2^n+1 : n in [0..6]]
(PARI) for(n=0, 6, print1(5^2^n+1, ", "))
Generalized Fermat numbers: a(n) = 8^(2^n) + 1, n >= 0.
+10
9
9, 65, 4097, 16777217, 281474976710657, 79228162514264337593543950337, 6277101735386680763835789423207666416102355444464034512897
COMMENTS
These numbers are all composite. We rewrite 8^(2^n) + 1 = (2^(2^n))^3 + 1.
Then by the identity a^n + b^n = (a+b)*(a^(n-1) - a^(n-2)*b + ... + b^(n-1)) for odd n, 2^(2^n) + 1 divides 8^(2^n) + 1. All factors of generalized Fermat numbers F_n(a,b) := a^(2^n)+b^(2^n), a >= 2, n >= 0, are of the form k*2^m+1, k >= 1, m >=0 (Riesel (1994)). - Daniel Forgues, Jun 19 2011
FORMULA
a(0)=9, a(n) = (a(n-1) - 1)^2 + 1, n >= 1.
EXAMPLE
For n = 3, 8^(2^3) + 1 = 16777217. Similarly, (2^8)^3 + 1 = 16777217. Then 2^8 + 1 = 257 and 16777217/257 = 65281.
PROG
(PARI) g(a, n) = if(a%2, b=2, b=1); for(x=0, n, y=a^(2^x)+b; print1(y", "))
CROSSREFS
Cf. A000215 (Fermat numbers: 2^(2^n) + 1, n >= 0).
Odd prime factors of generalized Fermat numbers of the form 11^(2^m) + 1 with m >= 0.
+10
8
3, 17, 61, 193, 257, 7321, 15361, 51329, 65537, 163841, 6304673, 15190529, 70254593, 1691123713, 1760464897, 3221225473, 3489660929, 4696846849, 6874464257, 53401878529, 111489577217, 149300051969, 184683593729, 206158430209, 447600088289, 1819992391681
COMMENTS
Odd primes p other than 5 such that the multiplicative order of 11 (mod p) is a power of 2.
REFERENCES
Hans Riesel, Common prime factors of the numbers A_n=a^(2^n)+1, BIT 9 (1969), pp. 264-269.
MATHEMATICA
Delete[Select[Prime@Range[2, 10^5], IntegerQ@Log[2, MultiplicativeOrder[11, #]] &], 2]
Number of odd prime factors (with multiplicity) of generalized Fermat number 11^(2^n) + 1.
+10
2
1, 1, 1, 2, 2, 3, 2, 5, 6
EXAMPLE
b(n) = (11^(2^n) + 1)/2.
Complete Factorizations
b(0) = 2*3
b(1) = 61
b(2) = 7321
b(3) = 17*6304673
b(4) = 51329*447600088289
b(5) = 193*257*21283620033217629539178799361
b(6) = 316955440822738177*P49
b(7) = 15361*111489577217*574341646346402207998363393*
4018529583345312964042058778793458689*P55
b(8) = 15190529*4696846849*19618834249745000485889*
4393553986026616439660661873903822389581313*
290103547098489711747952055517085778590240759297*P138
PROG
(PARI) a001222(n) = bigomega(n)
a199592(n) = 11^(2^n)+1
a(n) = if(n==0, 1, a001222( a199592(n))-1) \\ Felix Fröhlich, Jul 25 2016
EXTENSIONS
a(8) was found in 2006 by Bruce Dodson
Numbers k such that 3*2^k + 1 is a prime factor of a generalized Fermat number 11^(2^m) + 1 for some m.
+10
1
6, 30, 36, 66, 276, 353, 2816, 3189, 34350, 48150, 80190, 1832496, 2291610, 5082306, 10829346
MATHEMATICA
lst = {}; Do[p = 3*2^n + 1; If[PrimeQ[p] && IntegerQ@Log[2, MultiplicativeOrder[11, p]], AppendTo[lst, n]], {n, 3189}]; lst
PROG
(Magma) SetDefaultRealField(RealField(350)); IsInteger := func<k | k eq Floor(k)>; [n: n in [2..353] | IsPrime(k) and IsInteger(Log(2, Modorder(11, k))) where k is 3*2^n+1];
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