cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 11-17 of 17 results.

A226116 Numbers k such that one of 2^k-1 or 2^k+1 is semiprime, but not both.

Original entry on oeis.org

3, 4, 5, 6, 7, 9, 12, 13, 17, 19, 20, 28, 31, 32, 37, 40, 41, 43, 49, 59, 61, 64, 67, 79, 83, 92, 97, 103, 104, 109, 127, 128, 131, 137, 139, 148, 149, 191, 197, 227, 241, 256, 269, 271, 281, 293, 313, 356, 373, 379, 421, 457, 487, 523, 596, 692, 701, 727, 809, 881, 971, 983, 997, 1004, 1061, 1063
Offset: 1

Views

Author

Irina Gerasimova, May 28 2013

Keywords

Examples

			2^3-1=7 is not a semiprime but 2^3+1 =9 is, so 3 is in the sequence.
2^4-1 =15 is a semiprime but 2^4+1 =17 is not, so 4 is in the sequence.
2^8-1 =255 is a 3-prime (not a 2-prime) and 2^8+1 =257 is a prime (not a 2-prime), so 8 is not in the sequence.

Crossrefs

Cf. A001358, A046051, A054992, A092558, A092559.

Programs

  • PARI
    isok(n) = {nbm = bigomega(2^n-1); nbp = bigomega(2^n+1); return (((nbm == 2) || (nbp == 2)) && ! ((nbm == 2) && (nbp == 2)));} \\ Michel Marcus, Aug 23 2013

Extensions

Original sequence of 4 small numbers replaced by a wider sequence. - R. J. Mathar, Jun 13 2013

A309358 Numbers k such that 10^k + 1 is a semiprime.

Original entry on oeis.org

4, 5, 6, 7, 8, 19, 31, 53, 67, 293, 586, 641, 922, 2137, 3011
Offset: 1

Views

Author

Hugo Pfoertner, Jul 29 2019

Keywords

Comments

a(16) > 12000.
10^k + 1 is composite unless k is a power of 2, and it can be conjectured that it is composite for all k > 2, cf. A038371 and A185121. - M. F. Hasler, Jul 30 2019
Suppose k is odd. Then k is a term if and only if (10^k+1)/11 is prime. - Chai Wah Wu, Jul 31 2019

Examples

			a(1) = 4 because 10^4 + 1 = 10001 = 73*137.

Crossrefs

Cf. A003021, A038371, A046413, A057934, A062397, A092559.
Odd terms in sequence: A001562.

Programs

  • Magma
    IsSemiprime:=func; [n: n in [2..200] | IsSemiprime(s) where s is 10^n+1]; // Vincenzo Librandi, Jul 31 2019
  • Mathematica
    Select[Range[200], Plus@@Last/@FactorInteger[10^# + 1] == 2 &] (* Vincenzo Librandi, Jul 31 2019 *)

A366582 Numbers k such that 6^k + 1 is a semiprime.

Original entry on oeis.org

3, 8, 11, 12, 31, 43, 47, 59, 62, 107, 382, 514, 734, 811
Offset: 1

Views

Author

Sean A. Irvine, Oct 13 2023

Keywords

Examples

			11 is in this sequence because 6^11+1 = 7*51828151 is a semiprime.

Crossrefs

Cf. A062394, A001358, A004062, A092559.

A115398 Numbers k such that both k^2+1 and 2^k + 1 are semiprimes.

Original entry on oeis.org

3, 5, 11, 12, 19, 28, 61, 64, 79, 92, 101, 104, 199, 356, 596, 692, 1709, 3539, 3824
Offset: 1

Views

Author

Zak Seidov, Mar 08 2006

Keywords

Comments

Intersection of A085722 and A092559.

Examples

			11 is a term because 11^2 + 1 = 122 = 2*61 (semiprime) and 2^11 + 1 = 2049 = 3*683 (semiprime).

Crossrefs

Cf. A085722, A092559.

Programs

  • Magma
    IsSemiprime:=func; [n: n in [2..700] | IsSemiprime(n^2+1) and IsSemiprime(2^n+1)]; // Vincenzo Librandi, Oct 10 2013
  • Mathematica
    Select[Range[700],PrimeOmega[#^2+1]==PrimeOmega[2^#+1]==2&] (* Harvey P. Dale, Apr 14 2019 *)
  • PARI
    isok(n) = (bigomega(n^2+1) == 2) && (bigomega(2^n+1) == 2); \\ Michel Marcus, Oct 10 2013

Extensions

a(17)-a(19) from Robert Israel, Nov 27 2023

A242076 Numbers k for which (2^k + 1)/F is prime where F is a Fermat number.

Original entry on oeis.org

3, 5, 6, 7, 11, 12, 13, 17, 19, 20, 23, 28, 31, 40, 43, 61, 79, 92, 96, 101, 104, 127, 148, 167, 191, 199, 313, 347, 356, 596, 692, 701, 1004, 1228, 1268, 1709, 2617, 3539, 3824, 5807, 10501, 10691, 11279, 12391, 14479, 42737
Offset: 1

Views

Author

J. Lowell, May 03 2014

Keywords

Comments

Conjecture: 6 is the only term whose prime factorization contains a single 2.
The largest odd divisor of each term is prime, that is, subsequence of A038550. - J. Lowell, Apr 13 2018
This sequence contains only certain terms from A092559 and certain multiples of 32. - Jon E. Schoenfield, Apr 18 2018 [with thanks to J. Lowell]

Examples

			12 is a term because (2^12 + 1)/17 = 241, a prime number.

Crossrefs

Cf. A000215 (Fermat numbers), A066263.

Programs

  • Sage
    def a(n):
    num = 2^n + 1
    k = 0
    while k < log(n, 2):
        if num % (2^(2^k) + 1) == 0 and is_prime(Integer(num/(2^(2^k)+1))):
            return True
        k = k + 1
    return False          # Ralf Stephan, May 15 2014

Extensions

More terms from Ralf Stephan, May 15 2014
a(40)-a(46) from Jon E. Schoenfield, Apr 14 2018
Wrong property removed by J. Lowell, Apr 14 2018

A348177 a(n) is the number of pair of positive integers (x, y) with 1 <= x <= y such that sum s = x + y and product p = x * y satisfy s + p = 2^n, with n > 0.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 1, 0, 3, 2, 1, 1, 1, 3, 5, 0, 1, 7, 1, 1, 5, 3, 1, 3, 7, 7, 9, 1, 3, 23, 1, 1, 11, 7, 15, 7, 3, 7, 5, 1, 3, 31, 1, 3, 31, 15, 3, 3, 3, 31, 23, 3, 3, 31, 23, 3, 11, 3, 7, 7, 1, 15, 31, 1, 31, 31, 3, 5, 11, 47, 3, 15, 3, 15, 47, 7, 31, 383, 1, 3
Offset: 1

Views

Author

Bernard Schott, Oct 05 2021

Keywords

Comments

That is a generalization of a problem proposed by French site Diophante in link.
Some results:
x and y satisfy (x+1)*(y+1) = 2^n + 1.
x and y are both even, so 2 <= x <= y < 2^n.
There is only one case such that x = y, it is for n = 3 with x = y = 2 (Examples).
a(n) = 0 iff 2^n+1 is Fermat prime (A019434), hence iff n = 1, 2, 4, 8, 16.
a(n) = 1 iff 2^n+1 is semiprime (n is in A092559).

Examples

			For n = 3, only (x=y=2) satisfy s = 2+2 = 4, p = 2*2 = 4 and s+p = 8 = 2^3, hence a(3) = 1.
For n = 6, only (x=4, y=12) satisfy s = 4+12 = 16, p = 4*12 = 48 and s+p = 64 = 2^6 hence a(6) = 1.
For n = 9, (2,170), (8,56), (18,26) are the 3 solutions, with 172+340=512=2^9, 64+448=512, 44+468=512, hence a(9) = 3.
For n = 10, (4, 204) and (24, 40) are the 2 solutions, with 208+816=1024=2^10 and 64+960=1024, hence a(10) = 2.

Links

Crossrefs

Cf. A046798, A092559, A019434.

Programs

  • Maple
    with(numtheory):
M := seq(ceil((tau(2^n+1)-2)/2), n=1..100);
  • Mathematica
    a[3] = 1; a[n_] := DivisorSigma[0, 2^n + 1]/2 - 1; Array[a, 80] (* Amiram Eldar, Oct 05 2021 *)
  • PARI
    a(n) = ceil((numdiv(2^n+1) - 2)/2); \\ Michel Marcus, Oct 11 2021

Formula

For n<>3, the number of positive pairs solution (x,y) is a(n) = (tau(2^n+1) - 2)/2.
For n = 3, there is only one pair solution and a(3) = (tau(2^3+1) - 1)/2 = 1, with (x, y) = (2, 2).
a(n) = ceiling((tau(2^n+1) - 2)/2) = ceiling((A046798(n)-2)/2) is the general formula.

A366648 Numbers k such that 4^k + 1 is a semiprime.

Original entry on oeis.org

3, 6, 10, 14, 16, 20, 32, 46, 52, 64, 74, 128, 178, 298, 346, 502, 614, 634, 1912, 60394, 92116
Offset: 1

Views

Author

Sean A. Irvine, Oct 16 2023

Keywords

Examples

			14 is in this sequence because 4^14+1 = 17*15790321 is a semiprime.

Crossrefs

Cf. A062394, A001358, A004062, A092559.

Formula

The even terms of A092559 divided by 2. - Max Alekseyev, Jan 04 2024

Extensions

a(19)-a(21) from Max Alekseyev, Jan 04 2024
Previous Showing 11-17 of 17 results.

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