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.

A033316 Value of D for incrementally largest values of minimal x satisfying Pell equation x^2-Dy^2=1.

Original entry on oeis.org

1, 2, 5, 10, 13, 29, 46, 53, 61, 109, 181, 277, 397, 409, 421, 541, 661, 1021, 1069, 1381, 1549, 1621, 2389, 3061, 3469, 4621, 4789, 4909, 5581, 6301, 6829, 8269, 8941, 9949, 12541, 13381, 16069, 17341, 24049, 24229, 25309, 29269, 30781, 32341, 36061
Offset: 1

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Author

Keywords

Comments

Equally, value of D for incrementally largest values of minimal y satisfying Pell equation x^2-Dy^2=1.
Values of n where A002349 (or A002350) sets a new record.

Crossrefs

Programs

  • Mathematica
    PellSolve[(m_Integer)?Positive] := Module[{cf, n, s}, cf = ContinuedFraction[ Sqrt[m]]; n = Length[ Last[cf]]; If[ OddQ[n], n = 2*n]; s = FromContinuedFraction[ ContinuedFraction[ Sqrt[m], n]]; {Numerator[s], Denominator[s]}]; f[n_] := If[ !IntegerQ[ Sqrt[n]], PellSolve[n][[1]], 1]; a = b = -1; t = {}; Do[b = f[n]; If[b > a, t = Append[t, n]; a = b], {n, 1, 40500}]; t

Extensions

More terms from Robert G. Wilson v, Apr 15 2003

A033315 Incrementally largest values of minimal x satisfying Pell equation x^2 - D*y^2 = 1.

Original entry on oeis.org

1, 3, 9, 19, 649, 9801, 24335, 66249, 1766319049, 158070671986249, 2469645423824185801, 159150073798980475849, 838721786045180184649, 25052977273092427986049, 3879474045914926879468217167061449
Offset: 1

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Author

Keywords

Crossrefs

Records in A033313 (or A002350).
Cf. A033316 (corresponding values of D).

Programs

  • Mathematica
    PellSolve[(m_Integer)?Positive] := Module[{cf, n, s}, cf = ContinuedFraction[Sqrt[m]]; n = Length[Last[cf]]; If[n == 0, Return[{}]]; If[OddQ[n], n = 2 n]; s = FromContinuedFraction[ ContinuedFraction[ Sqrt[m], n]]; {Numerator[s], Denominator[s]}];
    xx = DeleteCases[PellSolve /@ Range[10^5], {}][[All, 1]];
    Reap[Module[{x, record = 0}, Sow[1]; For[i = 1, i <= Length@xx, i++, x = xx[[i]]; If[x > record, record = x; Sow[x]]]]][[2, 1]] (* Jean-François Alcover, Nov 21 2020, after N. J. A. Sloane in A002349 *)

A068310 n^2 - 1 divided by its largest square divisor.

Original entry on oeis.org

3, 2, 15, 6, 35, 3, 7, 5, 11, 30, 143, 42, 195, 14, 255, 2, 323, 10, 399, 110, 483, 33, 23, 39, 3, 182, 87, 210, 899, 15, 1023, 17, 1155, 34, 1295, 38, 1443, 95, 1599, 105, 1763, 462, 215, 506, 235, 138, 47, 6, 51, 26, 2703, 78, 2915, 21, 3135, 203, 3363, 870, 3599
Offset: 2

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Author

Lekraj Beedassy, Feb 25 2002

Keywords

Comments

In other words, squarefree part of n^2-1.
Least m for which x^2 - m*y^2 = 1 has a solution with x = n.

Examples

			a(6) = 35, as 6^2 - 1 = 35 itself is squarefree.
7^2-1 = 48 = A005563(6), whose largest square divisor is A008833(48) = 16, so a(7) = 48/16 = 3.
		

Crossrefs

Programs

  • Haskell
    a068310 n = f 1 $ a027746_row (n^2 - 1) where
       f y [] = y
       f y [p] = y*p
       f y (p:ps'@(p':ps)) | p == p' = f y ps
                           | otherwise = f (y*p) ps'
    -- Reinhard Zumkeller, Nov 26 2011
  • Mathematica
    a[n_] := Times@@(#[[1]] ^ Mod[ #[[2]], 2]&/@FactorInteger[n^2-1])
    Table[(n^2-1)/Max[Select[Divisors[n^2-1],IntegerQ[Sqrt[#]]&]],{n,2,60}] (* Harvey P. Dale, Dec 08 2019 *)
  • PARI
    a(n) = core(n*n - 1); \\ David Wasserman, Mar 07 2005
    

Formula

a(n) = A007913(n^2-1).
a(n) = A005563(n-1) / A008833(n^2 - 1). - Reinhard Zumkeller, Nov 26 2011; corrected by Georg Fischer, Dec 10 2022

Extensions

Edited by Dean Hickerson, Mar 19 2002
Entry revised by N. J. A. Sloane, Apr 27 2007

A128972 n^3 - 1 divided by its largest cube divisor.

Original entry on oeis.org

7, 26, 63, 124, 215, 342, 511, 91, 37, 1330, 1727, 2196, 2743, 3374, 4095, 614, 17, 254, 7999, 9260, 10647, 12166, 13823, 1953, 17575, 19682, 813, 24388, 26999, 29790, 32767, 4492, 39303, 42874, 46655, 1876, 54871, 59318, 63999, 8615, 74087, 79506
Offset: 2

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Author

Jonathan Vos Post, Apr 28 2007

Keywords

Comments

In other words, cubefree part of n^3-1, or cubefree kernel of n^3-1. Cube analog of A068310.

Examples

			a(9) = (9^3-1)/8 = (2^3 * 7 * 13)/(2^3) = 728/8 = 91.
a(10) = (10^3-1)/27 = (3^3 * 37)/(3^3) = 999/27 = 37.
a(18) = (18^3-1)/343 = (7^3 * 17)/(7^3) = 5831/343 = 17.
		

Crossrefs

Programs

  • Maple
    a:= n -> mul(f[1]^(f[2] mod 3), f = ifactors(n^3-1)[2]):
    seq(a(n),n=2..100); # Robert Israel, Sep 24 2014

Formula

a(n) = A062378(A068601(n)) = A062378(n^3-1).

Extensions

More terms from Carl R. White, Nov 09 2010

A128251 n^4 - 1 divided by its largest fourth power divisor.

Original entry on oeis.org

15, 5, 255, 39, 1295, 150, 4095, 410, 9999, 915, 20735, 1785, 38415, 3164, 65535, 5220, 104975, 8145, 159999, 12155, 234255, 17490, 331775, 24414, 456975, 33215, 614655, 44205, 809999, 57720, 1048575, 74120, 1336335, 93789, 1679615, 117135
Offset: 2

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Author

Jonathan Vos Post, May 03 2007

Keywords

Comments

In other words, biquadratefree part of n^4-1, or biquadratefree kernel of n^4-1. Fourth power analog of what A128972 is to cubes and A068310 is to squares. A046100 Biquadratefree numbers. A008835 Largest 4th power dividing n.

Examples

			a(3) = 5 because (3^4 - 1)/16 = 80/16 = (2^4 * 5)/(2^4) = 5.
a(5) = 39 because (5^4 - 1)/16 = 624/16 = (2^4 * 3 * 13)/(2^4) = 39.
a(7) = 150 because (7^4 - 1)/16 = 2400/16 = (2^5 * 3 * 5^2)/(2^4) = 150.
a(9) = 410 because (9^4 - 1)/16 = 6560/16 = (2^5 * 5 * 41)/(2^4) = 410.
a(63) = 61535 because (63^4 - 1)/256 = 15752960/256 = (2^8 * 5 * 31 * 397)/(2^8) = 61535.
		

Crossrefs

Formula

a(n) = (n^4 - 1)/A008835(n^4 - 1) = (A000583(n)-1)/A008835((A000583(n)-1)).

A354672 Numbers x with property that x is not the smallest possible value in the Pellian equation x^2 - D*y^2 = 1 with D = squarefree part of (x^2 - 1).

Original entry on oeis.org

7, 17, 26, 31, 49, 71, 97, 99, 127, 161, 199, 241, 244, 287, 337, 362, 391, 449, 485, 511, 577, 647, 721, 799, 846, 881, 967, 1057, 1151, 1249, 1351, 1457, 1567, 1681, 1799, 1921, 2024, 2047, 2177, 2311, 2449, 2591, 2737, 2887, 2889, 3041, 3199, 3361, 3363
Offset: 1

Views

Author

Herbert Kociemba, Jun 02 2022

Keywords

Comments

Alternatively numbers k such that A033314(k) <> A068310(k).
Conjecture: this sequence is equivalent to the sorted distinct values of cos(m*arccos(k)), where m and k are integers greater than 1. - Jennifer Buckley, Apr 23 2024

Examples

			a(2)=17. The squarefree part of 17^2 - 1 = 288 is D = 2. But the smallest possible solution to x^2 - 2*y^2 = 1 is not x = 17 but x = 3 (with y = 2).
15 is not a term: the squarefree part of 15^2 - 1 = 224 is D = 14 and x^2 - 14*y^2 = 1 has indeed the minimal solution x = 15 (and y = 4).
		

Programs

  • Mathematica
    squarefreepart[n_] :=
      Times @@ Power @@@ ({#[[1]], Mod[#[[2]], 2]} & /@ FactorInteger[n]);
    a = {}; NMAX = 3400; dict // Clear;
    For[n = 2, n <= NMAX, n++, s = squarefreepart[n^2 - 1];
    If[ ! IntegerQ[dict[s]], dict[s] = 1, AppendTo[a, n]]]; a

A354713 Number of solutions (n, D) for Pell equation n^2 - D*y^2 = 1 with fixed n.

Original entry on oeis.org

1, 2, 1, 2, 1, 3, 2, 3, 2, 2, 1, 2, 1, 3, 1, 6, 1, 4, 1, 2, 1, 3, 2, 3, 4, 2, 2, 2, 1, 4, 1, 4, 1, 4, 1, 4, 1, 3, 1, 3, 1, 2, 2, 2, 2, 3, 2, 6, 2, 4, 1, 4, 1, 6, 1, 3, 1, 2, 1, 2, 2, 4, 2, 4, 1, 2, 1, 2, 1, 6, 1, 6, 2, 2, 2, 2, 1, 3, 3, 3, 3, 2, 1, 2
Offset: 2

Views

Author

Herbert Kociemba, Jun 04 2022

Keywords

Comments

a(n) can be computed as the number of divisors of the square root of the largest square dividing n^2 - 1.
A067874 gives n with a(n) = 1.

Examples

			a(17) = 6 because there are 6 possible solutions to 17^2 - D*y^2 = 1: 17^2 - 2*12^2 = 1, 17^2 - 8*6^2 = 1, 17^2 - 18*4^2 = 1, 17^2 - 32*3^2 = 1, 17^2 - 72*2^2 = 1 and 17^2 - 288*1^2 = 1. D = 18 is the smallest of the 6 D values, where the (17,y) pair is minimal and hence A033314(17) = 18.
		

Crossrefs

Programs

  • Mathematica
    squarefreepart[n_] := Times @@ Power @@@ ({#[[1]], Mod[#[[2]], 2]} & /@ FactorInteger[n]);
    a[n_] :=  Divisors[Sqrt[(n^2 - 1)/squarefreepart[n^2 - 1]]] // Length; Table[a[n], {n, 2, 85}]
  • PARI
    f(n) = sqrtint(n/core(n)) \\ A000188
    a(n) = numdiv(f(n^2-1)); \\ Michel Marcus, Jun 05 2022

Formula

a(n) = A000005(A000188(n^2-1)).
Showing 1-7 of 7 results.