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 1-7 of 7 results.

A083204 Bases for the cubes arising in A083203.

Original entry on oeis.org

2, 4, 7, 11, 16, 22, 9, 17, 289, 1873, 11553, 63553, 4573699, 14073021495, 65713427547321843, 57325015297059875497553701478
Offset: 1

Views

Author

Amarnath Murthy, Apr 28 2003

Keywords

Examples

			a(1)=2 since A083203(1)*A083203(2)+1=1*7+1=8=2^3.
		

Crossrefs

Cf. A083203.

Formula

a(n) = [A083203(n)*A083203(n+1)+1]^(1/3). - R. J. Mathar, Feb 05 2007

Extensions

Corrected and extended by Vladeta Jovovic, May 01 2003
More terms from Michel ten Voorde Jun 23 2003
More terms from R. J. Mathar, Feb 05 2007

A091569 a(1) = 1; for n > 1, a(n) is the smallest positive integer not already used such that a(n)*a(n-1) + 1 is a perfect square.

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 15, 8, 6, 4, 2, 12, 10, 36, 34, 32, 30, 28, 26, 24, 22, 20, 18, 16, 14, 52, 50, 48, 35, 33, 31, 29, 27, 25, 23, 21, 19, 17, 64, 62, 60, 40, 38, 148, 146, 144, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83
Offset: 1

Views

Author

David Wasserman, Mar 04 2004

Keywords

Comments

Does this sequence contain every positive integer? We could get an equally interesting sequence by choosing a(1) to be any other positive integer.
A sequence with the same condition but without the requirement for a(n) to be distinct would end up repeating (1,3) or (2,4), depending on the initial term. - Ivan Neretin, May 26 2015

Examples

			10 is followed by 36 because 10*36+1 = 19^2 and 8 and 12 were already used.
		

Crossrefs

Cf. A083203.

Programs

  • MATLAB
    A = zeros(1, 100); A(1) = 1; used = zeros(1, 1000); used(1) = 1; for i = 2:100; found = 0; k = 0; while found == 0; k = k + 1; if used(k) == 0; s = sqrt(k*A(i - 1) + 1); if s == floor(s); A(i) = k; used(k) = 1; found = 1; end; end; end; end; A
  • Maple
    N:= 10^4: Used:= Vector(N,datatype=integer[4]):
    a[1]:= 1: blocked:= false: Used[1]:= 1:
    for n from 2 to 100 while not(blocked) do
        ndone:= false;
        if n = 2 then T:= [0]
          else T:= select(t -> t^2 mod a[n-1] = 1, [$0..a[n-1]-1])
          fi;
        for s from 0 while not (ndone) do
           for t in T while not (ndone) do
             x:= s * a[n-1] + t;
             if x <= 1 then next fi;
             y:= (x^2-1)/a[n-1];
             if y > N then blocked:= true; ndone:= true
             elif Used[y] = 0 then
                a[n]:= y;
                Used[y]:= 1;
                ndone:= true;
                print(n,y);
             fi
           od
        od
    od:
    seq(a[n],n=1..100); # Robert Israel, May 26 2015
  • Mathematica
    a = {1}; Do[a = Join[a, Select[Complement[Range[(Max[a] + 1)*n], a], IntegerQ[Sqrt[#*a[[-1]] + 1]] &, 1]], {n, 2, 71}]; a (* Ivan Neretin, May 26 2015 *)

A082609 Begin with 1 and then smallest number not included earlier such that a(n)*a(n+1) -1 is a cube.

Original entry on oeis.org

1, 2, 14, 9, 57, 2193, 3059049, 646785, 14861988225, 4007948400041, 227329891050248477628633, 253359999446098770547643849168025163162993, 294907355674806113726323566850834771401836431259814899322458665187277652242833
Offset: 0

Views

Author

Amarnath Murthy, Apr 28 2003

Keywords

Comments

Sequence is infinite. a(n+1) <= {(a(n) - 1)^3 + 1}/a(n).

Crossrefs

Extensions

Corrected and extended by David Wasserman, Oct 18 2004

A082536 a(1) = 1; for n > 1, a(n) = smallest number greater than a(n-1) such that a(n-1)*a(n)+1 is a cube.

Original entry on oeis.org

1, 7, 9, 38, 1561, 3926, 4958, 5284, 9316636, 84705277923, 774752630747122741, 8816203161790126559791912099597266, 333977591776082560884277873242324869870383615439405188
Offset: 1

Views

Author

Amarnath Murthy, Apr 28 2003

Keywords

Examples

			7*9 + 1 = 64 = 8^3, 9*38 + 1 = 343 = 7^3.
		

Crossrefs

Extensions

Corrected and extended by Vladeta Jovovic, May 01 2003
More terms from Michel ten Voorde Jun 23 2003
More terms from Vladeta Jovovic, Jun 29 2003
a(11)-a(13) from David Wasserman, Mar 26 2004

A083205 a(1) = 1, then smallest number not included earlier such that a(n)*a(n+1) + 1 is an n-th power.

Original entry on oeis.org

1, 2, 4, 31, 26129, 466202136816810031, 10584868011442581563053546190350068005080430521324963528734180259722815036145
Offset: 1

Views

Author

Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), Apr 28 2003

Keywords

Comments

The sequence is infinite and a(n+1) <= ([a(n)+1]^n - 1)/a(n) when n is even, or a(n+1) <= ([a(n)-1]^n - 1)/a(n) when n is odd.
To find a(6), we need an x such that x^5 = 1 (mod a(5)); then a(6) = (x^5 - 1)/a(5). The multiplicative group mod a(5) has order phi(a(5)) = 23296, which is not divisible by 5. So the only 5th root of 1 in this group is 1. x = 1 would give a(6) = 0, this is not allowed, so we take x to be the next representative of 1 mod a(5), i.e. a(5)+1. So a(6) = [(a(5)+1)^5 - 1]/a(5). - David Wasserman, Mar 02 2004
Next term is approximately 8.1*10^451. - David Wasserman, May 26 2004

Examples

			a(4) = 31, a(5) = 26129, 31*26129 + 1 = 810000 = 30^4.
		

Crossrefs

Extensions

More terms from David Wasserman, Mar 02 2004

A082606 n-th powers arising in A082605.

Original entry on oeis.org

3, 9, 125, 810000, 12181395632886429300000
Offset: 1

Views

Author

Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), Apr 28 2003

Keywords

Comments

The sequence is infinite.

Crossrefs

Formula

a(n) = A083205(n)*A083205(n+1)+1. - David Wasserman, Sep 21 2004

Extensions

More terms from David Wasserman, Sep 21 2004

A093812 a(1) = 1; for n > 1, a(n) is the smallest positive integer not already used such that a(n)*a(n-1) + 1 is a fourth power.

Original entry on oeis.org

1, 15, 17, 1680, 421, 56086995, 9940915081637, 31271643639794562523698024915, 3090319883395626207270178075850946683940891639487741952, 9721376888636052447184767296534734059409700304429485542107645657085305
Offset: 1

Views

Author

David Wasserman, May 20 2004

Keywords

Comments

The next terms are (to two significant digits) 9.6*10^95, 3.0*10^111, 3.0*10^137, 9.4*10^152, 9.3*10^178.

Examples

			The 4th roots of unity mod 17 are 1, 4, 13 and 16. (4^4 - 1)/17 = 15, which has already been used, so 17 is followed by (13^4 - 1)/17 = 1680.
		

Crossrefs

Showing 1-7 of 7 results.