A000215 -id:A000215 - cp's OEIS frontend

A372894 A positive integer 2^k*m, where m is odd and k >= 0, belongs to this sequence iff the Jacobi symbol (F_n/m) = -1 for only a finite number of Fermat numbers F_n = A000215(n).

1, 2, 4, 8, 9, 13, 15, 16, 17, 18, 21, 25, 26, 30, 32, 34, 35, 36, 42, 49, 50, 52, 60, 64, 68, 70, 72, 81, 84, 97, 98, 100, 104, 117, 120, 121, 123, 128, 135, 136, 140, 144, 153, 162, 168, 169, 189, 193, 194, 195, 196, 200, 205, 208, 221, 225, 234, 240, 241, 242, 246, 255, 256, 257, 270, 272, 273, 280, 287, 288, 289

Offset: 1

Jianing Song, May 15 2024

nonn

Can be seen as the opposite of A129802.
Let m be an odd number and ord(2,m) = 2^r*d be the multiplicative order of 2 modulo m, where d is odd, then 2^2^n + 1 is congruent to one of 2^2^r + 1, 2^2^(r+1) + 1, ..., 2^2^(r+ord(2,d)-1) + 1 modulo m, so it suffices to check these ord(2,d) numbers.
Note that if m > 1, then m does not divide 2^2^n + 1 for n >= r, otherwise we would have 2^(2^n*d) = (2^ord(2,m))^2^(n-r) == 1 (mod m) and 2^(2^n*d) = (2^2^n)^d == (-1)^d == -1 (mod m). As a result, m is a term if and only if the Jacobi symbol ((2^2^n + 1)/m) is equal to 1 for m = r, r+1, ..., r+ord(2,d)-1.
By definition, a squarefree number that is a product of elite primes (A102742) or anti-elite primes (A128852) is a term if and only if its number of elite factors is even. But a squarefree term can have factors that are neither elite nor anti-elite, the smallest being 341 = 11*31.
Contains divisors of Fermat numbers >= 17 (A307843 \ {3,5}) since they are products of elite primes.

			For n >= 1, we have 2^2^n + 1 == 65, 5, 17, 257 (mod 341) respectively for n == 0, 1, 2, 3 (mod 4). As we have (65/341) = (5/341) = (17/341) = (257/341) = 1, 341 is a term.

Jianing Song, Table of n, a(n) for n = 1..10000

Cf. A000215, A102742, A129802, A307843.

Prime elements of this sequence are given by A128852.

isA372894(n) = n = (n >> valuation(n,2)); my(d = znorder(Mod(2, n)), StartPoint = valuation(d, 2), LengthTest = znorder(Mod(2, d >> StartPoint))); for(i = StartPoint, StartPoint + LengthTest - 1, if(kronecker(lift(Mod(2, n)^2^i + 1), n) == -1, return(0))); 1

A057755 Number of digits in n-th Fermat number (A000215).

Original entry on oeis.org

1, 1, 2, 3, 5, 10, 20, 39, 78, 155, 309, 617, 1234, 2467, 4933, 9865, 19729, 39457, 78914, 157827, 315653, 631306, 1262612, 2525223, 5050446, 10100891, 20201782, 40403563, 80807125, 161614249, 323228497, 646456994, 1292913987, 2585827973

Offset: 0

Robert G. Wilson v, Oct 30 2000

Also number of digits of A001146(n) and A051179(n). - Michel Marcus, Dec 21 2018

			a(6) = 20 because 2^(2^6) + 1 = 18446744073709551617 which is a twenty-digit number.

John H. Conway and R. K. Guy, The Book of Numbers, Copernicus, an imprint of Springer-Verlag, NY, 1995, page 139.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000 (first 100 terms from Jinyuan Wang)
R. Mestrovic, Euclid's theorem on the infinitude of primes: a historical survey of its proofs (300 BC--2012) and another new proof, arXiv preprint arXiv:1202.3670 [math.HO], 2012-2018. - From _N. J. A. Sloane_, Jun 13 2012
Eric Weisstein's World of Mathematics, Fermat Number

Cf. A000215, A001146, A051179.

List([0..18],n->Size(ListOfDigits(2^(2^n)+1))); # Muniru A Asiru, Dec 20 2018

[Floor(2^n*Log(10,2)/Log(10,10))+1: n in [0..40]]; // Vincenzo Librandi, Nov 08 2018

seq(length(2^(2^n)),n=0..20); # Zerinvary Lajos, Apr 20 2008

Table[ Floor[ 2^n * N[ Log[ 10, 2 ], 24 ] + 1 ], {n, 0, 43} ]

for(n=0, 50, print(n, " ", floor(2^n*log(2)/log(10))+1); ) \\ Jinyuan Wang, Nov 07 2018

a(n) = floor(log_10(F_n)+1) (F_n is the n-th Fermat number). - Ivan Panchenko, Sep 06 2009

A162634 Numerators of fractions with denominators A000215(n) approximating the Thue-Morse constant.

Original entry on oeis.org

1, 2, 7, 106, 27031, 1771476586, 7608434000728254871, 140350834813144189858090274002849666666, 47758914269546354982683078068829456704164423862093743397580034411621752859031

Offset: 0

Vladimir Shevelev, Jul 08 2009, Jul 14 2009

nonn

One can prove that if in the sequence of numbers N for which A010060(N+2^n)= A010060(N) you replace the odious (evil) terms by 1's (0's), then we obtain 2^(n+1)-periodic (0,1)-sequence; if you write it in the form .xx...,i.e., as a binary infinite fraction, then the corresponding fraction has the form a(n)/A000215(n). These fractions very fast converge to the Thue-Morse constant .4124540336401...; e.g a(5)/(2^32+1) approximates this constant up to 10^(-9). These approximations differ from A074072-A074073. Conjecture. For n>=1, the fraction a(n)/A000215(n) is a convergent corresponding to the continued fraction for the Thue-Morse constant.

V. Shevelev, Equations of the form t(x+a)=t(x) and t(x+a)=1-t(x) for Thue-Morse sequence,

Cf. A010060, A000215, A085394, A085395, A081706, A161627, A161639, A074072, A074073.

a(n)=if(n<=1, [1,2][n+1], 1+(2^(2^(n-1))-1)*a(n-1)); /* Joerg Arndt, Mar 11 2013 */

a(1)=2, and, for n>=2, a(n) = 1 + (2^(2^(n-1))-1) * a(n-1).

Added more terms, Joerg Arndt, Mar 11 2013

A308695 a(n) is the minimum positive integer m such that m * 2^(n + 2) + 1 is a prime number which does not divide ((F(n + 2) - 1)^m - 1)/(F(n + 2) - 2), where F(n) is the n-th Fermat number (A000215).

Original entry on oeis.org

1, 2, 1, 8, 4, 2, 1, 128, 64, 32, 16, 8, 4, 2, 1, 6300, 3150, 26, 13, 579, 1069378, 534689, 10, 5, 387304, 193652, 96826, 48413, 141015, 298082, 149041, 2958, 1479, 51418638746, 25709319373, 20, 10, 5, 6, 3

Offset: 0

Lorenzo Sauras Altuzarra, Feb 11 2020

nonn

Note that some terms are obtained by dividing the previous one by 2.
From Jinyuan Wang, Feb 18 2020: (Start)
a(n) is the least m such that q = m*2^(n + 2) + 1 is a prime factor of F(n + 2) - 2. Proof: if r(n + 2)/s(n + 2) = ((F(n + 2) - 1)^m - 1)/(F(n + 2) - 2) is not divisible by q, then q divides s(n + 2) because r(n + 2) is always divisible by q (by Fermat's little theorem). Also note that if F(n + 2) - 1 == 1 (mod q), then r(n + 2)/s(n + 2) = Sum_{i = 0..m-1} A001146(n + 2)^i == m (mod q). In conclusion, prime q = m*2^(n + 2) + 1 does not divide r(n + 2)/s(n + 2) if and only if q divides F(n + 2) - 2 = Product_{i = 0..n + 1} F(i).
a(n) always exists because prime factors of F(n) are of the form k*2^(n + 2) + 1. a(n) is not greater than the smallest such k. (End)
a(40) <= 74327396788321657. - Max Alekseyev, Nov 17 2022

			2 is the minimum positive integer m such that m * 2^(1 + 2) + 1 is a prime number (note that 2 * 2^(1 + 2) + 1 = 17) which does not divide ((F(1 + 2) - 1)^m - 1)/(F(1 + 2) - 2) (note that ((F(1 + 2) - 1)^2 - 1)/(F(1 + 2) - 2) = 257, which is a prime number).

Lorenzo Sauras Altuzarra, Some arithmetical problems that are obtained by analyzing proofs and infinite graphs, arXiv:2002.03075 [math.NT], 2020.

Cf. A000215 (Fermat numbers), A001146.

A308695:=proc(n)
   local m:
   m:=1:
   while not isprime(m*2^(n+2)+1) or (2^(2^(n+2))-1) mod (m*2^(n+2)+1) != 0 do
      m:=m+1:
   od:
   return m:
end proc:

Array[Block[{m = 1}, While[Nand[PrimeQ[#4], Mod[((#3 - 1)^#1 - 1)/(#3 - 2), #4] != 0] & @@ {m, #, 2^(2^(# + 2)) + 1, m*2^(# + 2) + 1}, m++]; m] &, 14] (* Michael De Vlieger, Feb 14 2020 *)

F(n) = 2^(2^n) + 1;
a(n) = {my(m=1); while (!isprime(p=(m*2^(n+2)+1)) || !((((F(n+2)-1)^m-1)/ (F(n+2)-2)) % p), m++); m;} \\ Michel Marcus, Feb 14 2020

a(n) = {my(d=4*2^n, q=1); for(m=1, oo, q+=d; if(ispseudoprime(q) && Mod(2, q)^d==1, return(m))); } \\ Jinyuan Wang, Feb 18 2020

a(15)-a(32) from Jinyuan Wang, Feb 18 2020

a(33)-a(39) from Max Alekseyev, Nov 17 2022

A332414 Positive integers r such that A(1,r) = A(2,r - 1) = ... = A(r,1) = 0, where A denotes the function mapping every pair of positive integers (m,n) into 1 if m * 2^(n + 2) + 1 is a prime number dividing F(n + 2) - 2, where F(n) denotes the n-th Fermat number (i.e., F(n) = A000215(n)); and into 0 otherwise.

Original entry on oeis.org

1, 3, 4, 5, 8, 11, 12, 16, 19, 20, 21, 22, 23, 26, 28, 29, 32, 33, 34, 35, 36, 37, 38, 39, 44, 46, 47, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 72, 74, 75, 76, 78, 80, 82, 84, 85, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 99, 100, 101

Offset: 1

Lorenzo Sauras Altuzarra, Feb 12 2020

nonn

Note that this sequence is a subsequence of A332416.

Prime q = m*2^(n + 2) + 1 does not divide ((F(n + 2) - 1)^m - 1)/(F(n + 2) - 2) if and only if q divides F(n + 2) - 2 = Product_{i = 0..n + 1} F(i). Direct implication is Theorem 2.26 of my article (see the links) and reciprocal implication is due to Wang (see A308695).

			3 is a term of this sequence, because A(1,3) = A(2,2) = A(3,1) = 0.

Lorenzo Sauras Altuzarra, Some arithmetical problems that are obtained by analyzing proofs and infinite graphs, arXiv:2002.03075 [math.NT], 2020.

Cf. A000215 (Fermat numbers), A308695, A332416.

A332414:=proc(n)
   local c, i, k, q, r, v:
   c:=0:
   i:=0:
   r:=1:
   while c < n do
      for k from 0 to r-1 do
         q:=(k+1)*2^(r-k+2)+1:
         if not isprime(q) or (2^(2^(r-k+2)) - 1) mod q != 0 then
            i:=i+1:
         fi:
      od:
      if i = r then
         v:=r:
         c:=c+1:
      fi:
      i:=0:
      r:=r+1:
   od:
   return v:
end proc:

Select[Range@ 29, NoneTrue[Transpose@ {#, Reverse@ #} &@ Range@ #, And[PrimeQ[#4], Mod[((#3 - 1)^#1 - 1)/(#3 - 2), #4] != 0] & @@ {#1, #2, 2^(2^(#2 + 2)) + 1, #1*2^(#2 + 2) + 1} & @@ # &] &] (* Michael De Vlieger, Feb 14 2020 *)

isA(m, t) = ispseudoprime(q=4*m*2^t+1) && Mod(2, q)^(4*2^t)==1;
isok(r) = sum(i=1, r, isA(i, r-i+1)) == 0; \\ Jinyuan Wang, Feb 18 2020

a(17)-a(67) from Jinyuan Wang, Feb 18 2020

A332416 Positive integers r such that B(1,r) = B(2,r - 1) = ... = B(r,1) = 0, where B denotes the function mapping every pair of positive integers (m,n) into 1 if m * 2^(n + 2) + 1 is a prime number dividing F(n), where F(n) denotes the n-th Fermat number (i.e., F(n) = A000215(n)); and into 0 otherwise.

Original entry on oeis.org

1, 3, 4, 5, 6, 7, 8, 11, 12, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 24, 25, 26, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 44, 46, 47, 48, 50, 51, 52, 53, 54, 55, 56, 57, 58, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 78, 79

Offset: 1

Lorenzo Sauras Altuzarra, Feb 12 2020

nonn

Note that A332414 is a subsequence of this sequence.

Prime q = m*2^(n + 2) + 1 does not divide ((F(n + 2) - 1)^m - 1)/F(n) if and only if q divides F(n). Direct implication is Theorem 2.24 of my article (see the links). Proof of the reciprocal implication (by Wang): A001146(n) = 2^(2^n) == - 1 (mod q), so ((F(n + 2) - 1)^m - 1)/F(n) = Sum_{i = 0..4*m-1} (-1)^(i+1)*(2^(2^n))^i == -4*m (mod q).

			3 is a term of this sequence, because B(1,3) = B(2,2) = B(3,1) = 0.

Lorenzo Sauras Altuzarra, Some arithmetical problems that are obtained by analyzing proofs and infinite graphs, arXiv:2002.03075 [math.NT], 2020.

Cf. A000215 (Fermat numbers), A001146, A332414.

A332416:=proc(n)
   local c, i, k, q, r, v:
   c:=0:
   i:=0:
   r:=1:
   while c < n do
      for k from 0 to r-1 do
         q:=(k+1)*2^(r-k+2)+1:
         if not isprime(q) or (2^(2^(r-k)) + 1) mod q != 0 then
            i:=i+1:
         fi:
      od:
      if i = r then
         v:=r:
         c:=c+1:
      fi:
      i:=0:
      r:=r+1:
   od:
   return v:
end proc:

Select[Range@ 29, NoneTrue[Transpose@ {#, Reverse@ #} &@ Range@ #, And[PrimeQ[#4], Mod[((#3 - 1)^#1 - 1)/(2^(2^#2) + 1), #4] != 0] & @@ {#1, #2, 2^(2^(#2 + 2)) + 1, #1*2^(#2 + 2) + 1} & @@ # &] &] (* Michael De Vlieger, Feb 14 2020 *)

isB(m, t) = ispseudoprime(q=4*m*2^t+1) && Mod(2, q)^(2^t)==-1;
isok(r) = sum(i=1, r, isB(i, r-i+1)) == 0; \\ Jinyuan Wang, Feb 18 2020

a(25)-a(68) from Jinyuan Wang, Feb 18 2020

A063897 a(n) is the least k such that k - A000215(j), j=0..n, are all primes.

Original entry on oeis.org

5, 8, 22, 274, 65704, 4295145556, 18446744073810262144

Offset: 0

Felice Russo, Aug 29 2001

Is this sequence finite?

The prime k-tuples conjecture implies that the sequence is infinite. - Robert Israel, Jul 11 2016

			For j=0 a(0)=5 because 5-3 is prime.
For j=1 a(1)=8 because 8-5, 8-3 are all primes.
For j=2 a(2)=22 because 22-17, 22-5, 22-3 are all primes.
For j=3 a(3)=274 because 274-257, 274-17, 274-5, 274-3 are all primes.

Cf. A000215.

f:= proc(n) local r, j, good;
    for r from 2^(2^n)+4 by 2 do
       good:= true;
       for j from 0 to n do
          if not isprime(r - 2^(2^j)-1) then good:= false; break fi
       od;
       if good then return(r) fi
    od
end proc:
f(0):= 5:
map(f, [$0..5]); # Robert Israel, Jul 11 2016

okprime(mink, vecf) = {for (i = 1, #vecf, if (! isprime(mink - vecf[i]), return (0));); return (1);}
a(n) = {mink = 2^(2^n) + 2; vecf = vector(n+1, i, 2^(2^(i-1)) + 1); while (! okprime(mink, vecf), mink++); mink;} \\ Michel Marcus, Sep 28 2013

18446744073810262144 from Thomas Baruchel, Oct 21 2003

A133026 Concatenation of next n Fermat numbers A000215.

Original entry on oeis.org

3, 517, 257655374294967297

Offset: 1

Omar E. Pol, Nov 09 2007

a(4) has 292 digits and is too large to include.

Amiram Eldar, Table of n, a(n) for n = 1..4

Cf. A000215, A053067, A133013.

a[n_] := FromDigits[Flatten[IntegerDigits[Table[2^(2^(k - 1)) + 1, {k, n*(n - 1)/2 + 1, n*(n + 1)/2}]]]]; Array[a, 3] (* Amiram Eldar, Jul 09 2025 *)

A133045 Successive digits of Fermat numbers A000215(n).

Original entry on oeis.org

3, 5, 1, 7, 2, 5, 7, 6, 5, 5, 3, 7, 4, 2, 9, 4, 9, 6, 7, 2, 9, 7, 1, 8, 4, 4, 6, 7, 4, 4, 0, 7, 3, 7, 0, 9, 5, 5, 1, 6, 1, 7, 3, 4, 0, 2, 8, 2, 3, 6, 6, 9, 2, 0, 9, 3, 8, 4, 6, 3, 4, 6, 3, 3, 7, 4, 6, 0, 7, 4, 3, 1, 7, 6, 8, 2, 1, 1, 4, 5, 7

Offset: 1

Omar E. Pol, Nov 10 2007

G. C. Greubel, Table of n, a(n) for the first 14 Fermat numbers

Cf. A000215, A033308.

IntegerDigits[Table[2^(2^n) + 1, {n, 0, 10}]] // Flatten (* G. C. Greubel, Oct 03 2017 *)

A273037 Discriminator of the first n Fermat numbers A000215.

Original entry on oeis.org

1, 3, 5, 11, 11, 19, 19, 23, 23, 23, 47, 53, 53, 53, 59, 59, 59, 59, 59, 59, 59, 59, 59, 59, 59, 59, 59, 59, 59, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 163, 163, 167, 167, 167, 167, 167, 167, 167, 167, 167, 167, 167, 167, 167, 167, 167, 167, 167, 167

Offset: 1

Jeffrey Shallit, May 13 2016

nonn

The discriminator of a sequence is the least positive integer k such that the first n terms are pairwise inequivalent, modulo k.

Cf. A000215.

cp's OEIS Frontend

A372894 A positive integer 2^k*m, where m is odd and k >= 0, belongs to this sequence iff the Jacobi symbol (F_n/m) = -1 for only a finite number of Fermat numbers F_n = A000215(n).

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A057755 Number of digits in n-th Fermat number (A000215).

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A162634 Numerators of fractions with denominators A000215(n) approximating the Thue-Morse constant.

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A308695 a(n) is the minimum positive integer m such that m * 2^(n + 2) + 1 is a prime number which does not divide ((F(n + 2) - 1)^m - 1)/(F(n + 2) - 2), where F(n) is the n-th Fermat number (A000215).

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A063897 a(n) is the least k such that k - A000215(j), j=0..n, are all primes.

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