A307843 -id:A307843 - 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

A119564 Define F(n) = 2^(2^n)+1 = n-th Fermat number, M(n) = 2^n-1 = the n-th Mersenne number. Then a(n) = F(n)-M(n)-1 = 2^(2^n) - 2^n + 1.

Original entry on oeis.org

2, 3, 13, 249, 65521, 4294967265, 18446744073709551553, 340282366920938463463374607431768211329, 115792089237316195423570985008687907853269984665640564039457584007913129639681

Offset: 0

Cino Hilliard, May 31 2006

nonn

The numbers n that divide a(n) are A373580. - Thomas Ordowski, Jun 11 2024

			F(2) = 2^(2^2)+1 = 17, M(2) = 2^2-1 = 3, F(2)-M(2)-1 = 13.

Cf. A119550, A119563, A307843, A373580.

fm4(n) = for(x=0,n,y=2^(2^x)+1-(2^x-1)-1;print1(y","))

a(n) = (2^(2^n) - 1) - (2^n - 2). - Thomas Ordowski, Jun 11 2024

Edited by N. J. A. Sloane, Jun 03 2006

Definition corrected by R. J. Mathar, May 15 2007

A373580 Numbers k that divide 2^(2^k) - 2^k + 1.

Original entry on oeis.org

1, 3, 5, 17, 257, 641, 1605, 4369, 32113, 65537, 94945, 114689, 159441, 164737, 274177, 319489, 974849, 2424833, 3862465, 6700417, 13631489, 13906833, 16843009, 26017793, 42009217, 45592577, 63766529, 70463489, 167772161, 175747457, 825753601, 1214251009, 1227890731, 1251711641

Offset: 1

Thomas Ordowski, Jun 10 2024

nonn

Numbers k that divide A119564(k).
A term is prime if and only if it is in A023394.
If k is in A307843, then it is a term of this sequence.
The terms that are not divisors of Fermat numbers are 1605, 4369, 32113, 94945, ... (they are all composite). Are there infinitely many of them?
Note that 2^(2^k) - 2^k + 1 = (2^(2^k) - 1) - (2^k - 2).

Cf. A023394 (primes in this sequence), A119564, A307843 (subsequence).

q[k_] := Mod[PowerMod[2, 2^k, k] - PowerMod[2, k, k] + 1, k] == 0; Select[Range[1, 10^5, 2], q] (* Amiram Eldar, Jun 10 2024 *)

isok(k) = Mod(Mod(2, k)^(2^k) - Mod(2,k)^k + 1, k) == 0; \\ Michel Marcus, Jun 12 2024

More terms from Amiram Eldar, Jun 10 2024

A360469 Only k >= 0 such that, for every odd r > 0, A093179(n) divides the generalized Fermat number (A007117(n)^r)^(2^k) + 1.

Original entry on oeis.org

3, 3, 5, 3, 7, 7, 9, 8, 11, 11, 13, 10, 15, 15, 17, 16, 19, 19, 21, 19, 23, 23, 25, 24, 27, 27, 29, 25, 31, 31, 33, 32, 35, 35, 37, 35, 39, 39, 41, 40, 43, 43, 45, 42, 47, 47, 49, 48, 51, 51, 53, 51, 55, 55, 57, 56, 59, 59, 61, 56, 63, 63, 65, 64, 67, 67, 69, 67, 71, 71, 73, 72, 75, 75, 77, 74, 79

Offset: 3

Lorenzo Sauras Altuzarra, Feb 27 2023

			A093179(5) = 641, A007117(5) = 5 and the only k >= 0 such that, for every odd r > 0, 641 divides the generalized Fermat number (5^r)^(2^k) + 1 is 5; so a(5) = 5.

Lorenzo Sauras-Altuzarra, Some properties of the factors of Fermat numbers, Art Discrete Appl. Math. (2022).

Cf. A000215 (Fermat numbers), A007117, A007814 (dyadic valuation), A093179, A307843 (divisors of Fermat numbers).

a:=n->n-padic:-ordp(n+2,2):
seq(a(n), n=3..79);

a(n) = n - valuation(n+2, 2);
vector(77,n,a(n+2)) \\ Joerg Arndt, Mar 03 2023

a(n) = n - A007814(n + 2) (due to Jinyuan Wang).

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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A119564 Define F(n) = 2^(2^n)+1 = n-th Fermat number, M(n) = 2^n-1 = the n-th Mersenne number. Then a(n) = F(n)-M(n)-1 = 2^(2^n) - 2^n + 1.

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A373580 Numbers k that divide 2^(2^k) - 2^k + 1.

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A360469 Only k >= 0 such that, for every odd r > 0, A093179(n) divides the generalized Fermat number (A007117(n)^r)^(2^k) + 1.

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