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.

Showing 1-8 of 8 results.

A049094 Numbers m such that 2^m - 1 is divisible by a square > 1.

Original entry on oeis.org

6, 12, 18, 20, 21, 24, 30, 36, 40, 42, 48, 54, 60, 63, 66, 72, 78, 80, 84, 90, 96, 100, 102, 105, 108, 110, 114, 120, 126, 132, 136, 138, 140, 144, 147, 150, 155, 156, 160, 162, 168, 174, 180, 186, 189, 192, 198, 200, 204, 210, 216, 220, 222, 228, 231, 234, 240
Offset: 1

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Author

Keywords

Comments

Conjecture: 2^n-1 is squarefree iff gcd(n,2^n-1)=1. If true, the conjecture would imply that Mersenne numbers (A001348) are squarefree. - Vladeta Jovovic, Apr 12 2002. But the conjecture is not true: counterexamples are n = 364 and n = 1755, i.e., gcd(364,2^364-1) = 1 and (2^364-1) mod 1093^2 = 0 and gcd(1755,2^1755-1) = 1 and (2^1755-1) mod 3511^2 = 0, cf. A001220. - Vladeta Jovovic, Nov 01 2005. The conjecture is true with assumption that n is not a multiple of A002326((q-1)/2), where q is a Wieferich prime A001220. - Thomas Ordowski, Nov 17 2015
If d|n and 2^d-1 is not squarefree, then 2^n-1 cannot be squarefree. For example, if 6 is in the sequence then 6*d is also. - Enrique Pérez Herrero, Feb 28 2009
If p(p-1)|n then p^2|2^n-1, where p is an odd prime. - Thomas Ordowski, Jan 22 2014
The primitive elements of this sequence are A237043. - Charles R Greathouse IV, Feb 05 2014
Dilcher & Ericksen prove that this sequence is exactly the set of numbers divisible by either t(p)p for a Wieferich prime p>2 or t(p) for a non-Wieferich prime p, where t(p) is the order of 2 modulo p (see Proposition 3.1). - Kellen Myers, Jun 09 2015
If d^2 divides 2^n-1 then d divides n, where n is not a multiple of 364, 1755, ...; i.e., A002326((q-1)/2) for Wieferich primes q, A001220. - Thomas Ordowski, Nov 15 2015
(1, 2^n-1, 2^n) is an abc triple iff 2^n-1 is not squarefree. - William Hu, Jul 04 2024

Examples

			a(2)=12 because 2^12 - 1 = 4095 = 5*(3^2)*7*13 is divisible by a square.
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, A3.

Crossrefs

Programs

  • Magma
    [n: n in [1..250] | not IsSquarefree(2^n-1)]; // Vincenzo Librandi, Jul 14 2015
  • Maple
    N:= 250:
    B:= Vector(N):
    for n from 1 to N do
      if B[n] <> 1 then
        F:= ifactors(2^n-1,easy)[2];
        if max(seq(t[2],t=F)) > 1 or (hastype(F,symbol)
                and not numtheory:-issqrfree(2^n-1)) then
           B[[seq(n*k,k=1..floor(N/n))]]:= 1;
        fi
      fi;
    od:
    select(t -> B[t]=1, [$1..N]); # Robert Israel, Nov 17 2015
  • Mathematica
    Select[Range[240], !SquareFreeQ[2^#-1]&] (* Vladimir Joseph Stephan Orlovsky, Mar 18 2011 *)
  • PARI
    default(factor_add_primes,1);
    is(n)=my(f=factor(n>>valuation(n,2))[,1],N,o); for(i=1,#f,if(n%(f[i]-1) == 0, return(1))); N=2^n-1; fordiv(n,d,f=factor(2^d-1)[,1]; for(i=1,#f, if(d==n,return(!issquarefree(N))); o=valuation(N,f[i]); if(o>1, return(1)); N/=f[i]^o)) \\ Charles R Greathouse IV, Feb 02 2014
    
  • PARI
    is(n)=!issquarefree(2^n-1) \\ Charles R Greathouse IV, Feb 04 2014
    

Extensions

More terms from Vladeta Jovovic, Apr 12 2002
Definition corrected by Jonathan Sondow, Jun 29 2010

A069226 a(n) = gcd(n, 2^n + 1).

Original entry on oeis.org

2, 1, 1, 3, 1, 1, 1, 1, 1, 9, 5, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 27, 1, 1, 5, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 25, 3, 1, 1, 1, 11, 1, 3, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 17, 3, 5, 1, 1, 1, 1, 3, 1, 1, 13, 1, 1, 81, 1, 1, 1, 1, 1, 3, 1, 1, 5, 1, 1, 3, 1, 1, 1, 1, 1, 9, 1
Offset: 0

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Author

Vladeta Jovovic, Apr 12 2002

Keywords

Comments

First occurrence of n: a(1)=1, a(3)=3, a(10)=5, a(9)=9, a(55)=11, a(78)=13, a(68)=17, a(50)=25, a(27)=27, a(406)=29, a(165)=33, a(666)=37, a(301)=43, a(1378)=53, a(1711)=59, a(390)=65, a(81)=81, a(3403)=83, a(2328)=97, a(495)=99, ... - R. J. Mathar, Dec 14 2016

Crossrefs

Cf. A006521 (fixed points), A014491, A049095, A049096, A060444.

Programs

  • Mathematica
    Table[GCD[n,2^n+1],{n,100}] (* Harvey P. Dale, Dec 12 2012 *)
  • PARI
    A069226(n) = gcd(n, 1+(1<Antti Karttunen, Jan 15 2025

Extensions

Term a(0) = 2 prepended by Antti Karttunen, Jan 15 2025

A069171 Numbers k such that gcd(k, 2^k-1) = 3.

Original entry on oeis.org

6, 12, 24, 30, 48, 66, 78, 96, 102, 114, 132, 138, 150, 174, 186, 192, 204, 222, 228, 246, 258, 264, 276, 282, 318, 348, 354, 366, 372, 384, 390, 402, 426, 438, 444, 456, 474, 492, 498, 510, 516, 528, 534, 552, 564, 570, 582, 606, 618, 636, 642, 654, 678
Offset: 1

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Author

Benoit Cloitre, Apr 09 2002

Keywords

Comments

The number of terms not exceeding 10^m, for m = 1, 2, ..., are 1, 8, 79, 793, 7935, 79349, 793524, 7935094, 79350930, 793509394, ... . Apparently, the asymptotic density of this sequence exists and equals 0.0793509... . - Amiram Eldar, Jun 14 2022

Crossrefs

Cf. A014491.

Programs

  • Mathematica
    Select[Range[700],GCD[#,2^#-1]==3&] (* Harvey P. Dale, Nov 22 2011 *)
    Select[Range[700], GCD[#, PowerMod[2, #, #] - 1] == 3 &] (* Amiram Eldar, Jun 14 2022 *)

A250208 Ratio of the primitive part of 2^n-1 to the product of primitive prime factors of 2^n-1.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 5, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Eric Chen, Mar 02 2015

Keywords

Comments

As with A178764, it can be shown that all terms are either 1 or prime.
a(2*3^n) = 3 (n>=1).
a(4*5^n) = 5 (n>=1).
a(3*7^n) = 7 (n>=1).
a(10*11^n) = 11 (n>=1).
a(12*13^n) = 13 (n>=1).
a(8*17^n) = 17 (n>=1).
a(18*19^n) = 19 (n>=1).
...
a(A014664(k)*prime(k)^n) = prime(k).
For other n (while Phi_n(2) is squarefree), a(n) = 1.
a(n) != 1 for n = {6, 18, 20, 21, 54, 100, 110, 136, 147, 155, 156, 162, ...}.
At least, a(A049093(n)) = 1. (In fact, since Phi_n(2) is not completely factored for n = 991, 1207, 1213, 1217, 1219, 1229, 1231, 1237, 1243, 1249, ..., so it is unknown whether they are squarefree or not, but it is likely that Phi_n(2) is squarefree for all n except 364 and 1755 (because it is likely 1093 and 3511 are the only two Wieferich primes), so a(991), a(1207), a(1213), ..., are likely to be 1.)

Examples

			a(11) = 1 since Phi_11(2) = (2^11-1)/(2-1) = 2047, and the primitive prime factors of 2^11-1 are 23 and 89, so a(11) = 2047/(23*89) = 1.
a(18) = 3 since Phi_18(2) = 2^6 - 2^3 + 1 = 57, and the only primitive prime factor of 2^18-1 is 19, so a(18) = 57/19 = 3.
		

Crossrefs

Programs

  • Mathematica
    a250208[n_] = If[n == 364, 1093, If[n == 1755, 3511, GCD[Cyclotomic[n, 2], n]]]; Table[a250208[n], {n, 0, 200}]
  • PARI
    a(n) = if (n==364, 1093, if (n==1755, 3511, gcd(polcyclo(n, 2), n)));
    
  • PARI
    isprimitive(p, n) = {for (r=1, n-1, if (((2^r-1) % p) == 0, return (0)); ); return (1); }
    ppf(n) = {my(pf = factor(2^n-1)[,1]); prod(k=1,#pf, if (isprimitive(pf[k], n), pf[k], 1));}
    a(n) = if (issquarefree(m=polcyclo(n,2)), gcd(m, n), m/ppf(n)); \\ Michel Marcus, Mar 06 2015

Formula

a(n) = A019320(n) / A064078(n) while Phi_n(2) is squarefree.
a(n) = GCD(Phi_n(2), n) while Phi_n(2) is squarefree.
Notice: a(364) = 1093, a(1755) = 3511. (See A001220.)

A260626 a(n) = gcd(m, 2^m-1) where m is the n-th nonprime positive integer.

Original entry on oeis.org

1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 9, 5, 7, 1, 3, 1, 1, 1, 1, 3, 1, 1, 1, 1, 9, 1, 1, 5, 21, 1, 1, 1, 3, 1, 1, 1, 1, 27, 1, 1, 1, 1, 15, 1, 7, 1, 1, 3, 1, 1, 1, 9, 1, 1, 1, 1, 3, 5, 1, 1, 21, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 3, 1, 1, 25, 3, 1, 7, 1, 27, 11, 1, 1, 3, 1
Offset: 1

Views

Author

Michel Lagneau, Oct 31 2015

Keywords

Comments

2^m - 1 is a nonprime number if m is a nonprime number.

Crossrefs

Programs

  • Maple
    seq(`if`(isprime(m), NULL, igcd(m, 2^m-1)), m=1..150);
  • Mathematica
    GCDnonPrime[n_Integer]:=GCD[2^FixedPoint[n+PrimePi@#&,n+PrimePi@n]-1,FixedPoint[n+PrimePi@#&,n+PrimePi@n]];Array[GCDnonPrime,120]
    GCD[#,2^#-1]&/@Select[Range[200],!PrimeQ[#]&] (* Harvey P. Dale, Aug 25 2019 *)
  • PARI
    for(n=1, 1e3, if(!isprime(n), print1(gcd(n,2^n-1)", "))) \\ Altug Alkan, Nov 01 2015

Formula

a(n) = gcd(A018252(n), 2^A018252(n)-1).
a(n) = A014491(A018252(n)). - Michel Marcus, Nov 01 2015

A172522 Partial sums of A049094.

Original entry on oeis.org

6, 18, 36, 56, 77, 101, 131, 167, 207, 249, 297, 351, 411, 474, 540, 612, 690, 770, 854, 944, 1040, 1140, 1242, 1347, 1455, 1565, 1679, 1799, 1925, 2057, 2193, 2331, 2471, 2615, 2762, 2912, 3067, 3223, 3383, 3545, 3713, 3887, 4067, 4253, 4442, 4634, 4832
Offset: 1

Views

Author

Jonathan Vos Post, Feb 06 2010

Keywords

Comments

The subsequence of primes in this sequence begins: 101, 131, 167, 3067.

Examples

			a(20) = 6 + 12 + 18 + 20 + 21 + 24 + 30 + 36 + 40 + 42 + 48 + 54 + 60 + 63 + 66 + 72 + 78 + 80 + 84 + 90.
		

Crossrefs

Formula

a(n) = SUM[i=1..n] {i such that 2^i - 1 is divisible by a square}.

Extensions

More terms from R. J. Mathar, Feb 16 2010

A298306 The Frobenius number of the set of binary n-th powers, divided out by its GCD.

Original entry on oeis.org

17, 723, 52753, 49790415, 126629
Offset: 2

Views

Author

Jeffrey Shallit, Jan 16 2018

Keywords

Comments

The binary n-th powers are those positive integers whose base-2 representation consists of n consecutive identical blocks. For example, the binary squares 3, 10, 15, ... form sequence A020330. The GCD of the binary n-th powers form sequence A014491. The Frobenius number of a set S with GCD 1 is the largest number not representable as an N-linear combination of members of S.

Examples

			For n = 2 the first few binary squares are 3, 10, 15, 36, ... with GCD 1 and the Frobenius number of (3, 10, 15, 36) is 17.
		

Crossrefs

A375117 Irregular triangle of positive integers, read by rows, the elements of the n-th row being the nonzero remainders, in increasing order, when the Euclidean algorithm is applied to 2^n-1 and n.

Original entry on oeis.org

1, 1, 1, 3, 1, 3, 1, 1, 7, 1, 2, 7, 1, 3, 1, 3, 1, 1, 2, 3, 1, 7, 1, 15, 1, 9, 1, 5, 15, 7, 1, 3, 1, 3, 6, 9, 15, 1, 6, 1, 2, 3, 1, 2, 25, 1, 2, 13, 15, 1, 3, 1, 1, 31, 1, 2, 5, 7, 1, 3, 1, 17, 9, 27, 1, 1, 2, 3, 1, 3, 4, 7, 5, 10, 15, 1, 21, 1, 1, 14, 15, 1, 3, 13, 16
Offset: 2

Views

Author

Mike Jones, Jul 30 2024

Keywords

Examples

			The triangle begins:
    1;
    1;
    1, 3;
    1;
    3;
    1;
    1, 7;
    1, 2, 7;
    1, 3;
    1;
    3;
    1;
    3;
    ...
Row(2) is {1}, because 2^2-1 = 4-1 = 3, and 3 divided by 2 leaves a remainder of 1.
Row(4) is {1, 3}, because 2^4-1 = 16-1 = 15, and 15 divided by 4 leaves a remainder of 3, and 4 divided by 3 leaves a remainder of 1.
		

Crossrefs

Programs

  • PARI
    row(n) = my(x=2^n-1, y=n, ok=1, list=List()); while (ok, my(z=divrem(x, y)); x = y; y = z[2]; if (y==0, ok=0, listput(list, y));); listsort(list); Vec(list); \\ Michel Marcus, Jul 31 2024

Extensions

More terms from Michel Marcus, Jul 31 2024
Showing 1-8 of 8 results.