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 1-9 of 9 results.

A078357 Minimal positive solution x of Pell equation y^2 - A077426(n)*x^2 = -4.

Original entry on oeis.org

1, 1, 2, 1, 2, 10, 1, 5, 2, 250, 1, 106, 1138, 2, 25, 146, 1, 298, 2, 5, 17, 1, 97, 10, 253970, 2, 1, 3034, 9148450, 2, 746, 10, 157, 126890, 1, 14341370, 5, 2, 110671282, 986, 7586, 1, 530, 130, 173, 2, 11068353370, 21685, 26, 694966754, 1, 17883410, 5528222698, 17, 87922, 2, 5, 41
Offset: 1

Views

Author

Wolfdieter Lang, Nov 29 2002

Keywords

Comments

The corresponding values y are given in A078356.
For the general solution of Pell equation y^2 - A077426(n)*x^2 = -4 see a comment in A078356.
For the conversion of the values given in Perron's table to sequences A078356 and A078357, see comments in A078356.

References

  • O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 30, Satz 3.35, p. 109 and table p. 108).

Programs

  • Mathematica
    $MaxExtraPrecision = 100; A077426 = Select[Range[ 600], ! IntegerQ[Sqrt[#]] && OddQ[ Length[ ContinuedFraction[(Sqrt[#] + 1)/2] // Last]] &]; a[n_] := {y, x} /. {ToRules[ Reduce[y > 0 && x > 0 && y^2 - A077426[[n]]*x^2 == -4, {y, x}, Integers] /. C[1] -> 0]} // Sort // First // Last; Table[a[n], {n, 1, 60}] (* Jean-François Alcover, Jun 21 2013 *)

Extensions

Edited and extended by Max Alekseyev, Mar 03 2010

A078356 Minimal positive solution z of Pell equation z^2 - A077426(n)*t^2 = -4.

Original entry on oeis.org

1, 3, 8, 5, 12, 64, 7, 39, 16, 2136, 9, 1000, 11208, 20, 261, 1552, 11, 3488, 24, 61, 213, 13, 1305, 136, 3528264, 28, 15, 46312, 142022136, 32, 12144, 164, 2613, 2127064, 17, 253724736, 89, 36, 2031654672, 18420, 142528, 19, 10236, 2564, 3447, 40, 223843593936
Offset: 1

Views

Author

Wolfdieter Lang, Nov 29 2002

Keywords

Comments

The corresponding values of t are given in A078357.
Computed from Perron's table (see reference p. 108) which gives the minimal x,y values for the Diophantine equation x^2 - x*y - ((D(m)-1)/4)*y^2 = +1 and -1 for respectively D(m)=A077425(m) and D(m)=A077426(m) (this second case excludes in Perron's table the D values with a 'Teilnenner' in brackets).
The conversion from the x,y values of Perron's table to the minimal a=a(n) and b=b(n) solutions of a^2 - D(n)*b^2 =-4 see a comment in A077428. Here only D values with no 'Teilnenner' in brackets are of interest and a(n)=2*x(n)-y(n) and b(n)=y(n). E.g. D=41, with 'Teilnenner von (sqrt(D)+1)/2' in the notation, explained in an example of A077427, 3,1,2 (period length k=5) and (x,y)=(37,10) which translates to the minimal solution (a,b)=(64,10).
Generic D(n) values are those from A078370(k)=(4*k(k+1)+5), k>=0, which are 5 (mod 8). For such D values the minimal solution is (a,b)=(2*k+1,1) (e.g. D(7)= A077426(7) = 53 = A078370(3) with a(7)= 2*3+1=7 and b(7)=A078357(7)=1).
The general solution of Pell a^2-D(n)*b^2 = -4 with generic D(n)=A078370(k), k>=0, is a(n,m)= (2*k+1)*S(2*m,sqrt(D(n))) and b(n,m)= T(2*m+1,sqrt(D(n))/2)/(sqrt(D(n))/2), m>=0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second, kind. See A053120 resp. A049310.
For non-generic D(n) (not from A078370) the general solution of a^2-D(n)*b^2 = -4 is a(n,m)=a(n)*S(2*m,sqrt(a(n)^2+4)) and b(n,m)= b(n)*T(2*m+1,sqrt(a(n)^2+4)/2)/(sqrt(a(n)^2+4)/2), m>=0, with Chebyshev's polynomials and in this case b(n)>1.

Examples

			41=D(6)=A077426(6) (also A077425(8)), hence a(6)=64 and b(6)=A078357(6)=10 satisfies 64^2 - 41*10^2 = -4.
		

References

  • O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 30, Satz 3.35, p. 109 and table p. 108).

Programs

  • Mathematica
    $MaxExtraPrecision = 100; A077426 = Select[Range[ 500], ! IntegerQ[Sqrt[#]] && OddQ[ Length[ ContinuedFraction[(Sqrt[#] + 1)/2] // Last]] &]; a[n_] := {z, t} /. {ToRules[ Reduce[z > 0 && t > 0 && z^2 - A077426[[n]]*t^2 == -4, {z, t}, Integers] /. C[1] -> 0]} // Sort // First // First; Table[a[n], {n, 1, 50}] (* Jean-François Alcover, Jun 21 2013 *)

Extensions

More terms from R. J. Mathar, Sep 24 2009
Edited by Max Alekseyev, Mar 03 2010

A078370 a(n) = 4*(n+1)*n + 5.

Original entry on oeis.org

5, 13, 29, 53, 85, 125, 173, 229, 293, 365, 445, 533, 629, 733, 845, 965, 1093, 1229, 1373, 1525, 1685, 1853, 2029, 2213, 2405, 2605, 2813, 3029, 3253, 3485, 3725, 3973, 4229, 4493, 4765, 5045, 5333, 5629, 5933, 6245, 6565, 6893, 7229, 7573, 7925, 8285, 8653, 9029
Offset: 0

Views

Author

Wolfdieter Lang, Nov 29 2002

Keywords

Comments

This is the generic form of D in the (nontrivially) solvable Pell equation x^2 - D*y^2 = -4. See A078356, A078357.
1/5 + 1/13 + 1/29 + ... = (Pi/8)*tanh Pi [Jolley]. - Gary W. Adamson, Dec 21 2006
Appears in A054413 and A086902 in relation to sequences related to the numerators and denominators of continued fractions convergents to sqrt((2*n+1)^2 + 4), n = 1, 2, 3, ... . - Johannes W. Meijer, Jun 12 2010
(2*n + 1 + sqrt(a(n)))/2 = [2*n + 1; 2*n + 1, 2*n + 1, ...], n >= 0, with the regular continued fraction with period length 1. This is the odd case. See A087475 for the general case with the Schroeder reference and comments. For the even case see A002522.
Primes in the sequence are in A005473. - Russ Cox, Aug 26 2019
The continued fraction expansion of sqrt(a(n)) is [2n+1; {n, 1, 1, n, 4n+2}]. For n=0, this collapses to [2; {4}]. - Magus K. Chu, Aug 27 2022
Discriminant of the binary quadratic forms y^2 - x*y - A002061(n+1)*x^2. - Klaus Purath, Nov 10 2022
From Klaus Purath, Apr 08 2025: (Start)
There are no squares in this sequence. The prime factors of these terms are always of the form 4*k + 1.
All a(n) = D satisfy the Pell equation (k*x)^2 - D*(m*y)^2 = -1 for any integer n where m = (D - 3)/2. The values for k and the solutions x, y can be calculated using the following algorithm: k = sqrt(D*m^2 - 1), x(0) = 1, x(1) = 4*D*m^2 - 1, y(0) = 1, y(1) = 4*D*m^2 - 3. The two recurrences are of the form (4*D*m^2 - 2, -1).
It follows from the above that this sequence belongs to A031396. (End)

References

  • L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 176.

Crossrefs

Subsequence of A077426 (D values (not a square) for which Pell x^2 - D*y^2 = -4 is solvable in positive integers).
Subsequence of A031396.

Programs

  • Magma
    [4*n^2+4*n+5 : n in [0..80]]; // Wesley Ivan Hurt, Aug 29 2022
  • Mathematica
    Table[4 n (n + 1) + 5, {n, 0, 45}] (* or *)
    Table[8 Binomial[n + 1, 2] + 5, {n, 0, 45}] (* or *)
    CoefficientList[Series[(5 - 2 x + 5 x^2)/(1 - x)^3, {x, 0, 45}], x] (* Michael De Vlieger, Jan 04 2017 *)
  • PARI
    a(n)=4*n^2+4*n+5 \\ Charles R Greathouse IV, Sep 24 2015
    
  • Python
    a= lambda n: 4*n**2+4*n+5 # Indranil Ghosh, Jan 04 2017
    
  • Scala
    (1 to 99 by 2).map(n => n * n + 4) // Alonso del Arte, May 29 2019
    

Formula

a(n) = (2*n + 1)^2 + 4.
a(n) = 4*(n+1)*n + 5 = 8*binomial(n+1, 2) + 5, hence subsequence of A004770 (5 (mod 8) numbers). [Typo fixed by Zak Seidov, Feb 26 2012]
G.f.: (5 - 2*x + 5*x^2)/(1 - x)^3.
a(n) = 8*n + a(n-1), with a(0) = 5. - Vincenzo Librandi, Aug 08 2010
a(n) = A016754(n) + 4. - Leo Tavares, Feb 22 2023
From Elmo R. Oliveira, Oct 31 2024: (Start)
E.g.f.: (5 + 8*x + 4*x^2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

Extensions

More terms from Max Alekseyev, Mar 03 2010

A077428 Minimal (positive) solution a(n) of Pell equation a(n)^2 - D(n)*b(n)^2 = +4 with D(n)= A077425(n). The companion sequence is b(n)=A078355(n).

Original entry on oeis.org

3, 11, 66, 5, 27, 46, 146, 4098, 7, 51, 302, 1523, 258, 25, 4562498, 9, 83, 1000002, 29, 125619266, 402, 82, 68123, 2408706, 11, 123, 33710, 173, 12166146, 190, 578, 3723, 4354, 45371, 23550, 13, 171, 124846, 1703027, 18498, 110, 12448646853698, 786
Offset: 1

Views

Author

Wolfdieter Lang, Nov 29 2002

Keywords

Comments

Computed from Perron's table (see reference p. 108, for n = 1..28) which gives the minimal x,y values for the Diophantine eq. x^2 - x*y - ((D(n)-1)/4)*y^2= +1, resp., -1 if D(n)=A077425(n), resp, D(n)=A077425(n) and D(n) also in A077426.
The conversion from the x,y values of Perron's table to the minimal a=a(n) and b=b(n) solutions of a^2 - D(n)*b^2 =+4 is as follows. If D(n)=A077425(n) but not from A077426 (period length of continued fraction of (sqrt(D(n))+1)/2 is even) then a(n)=2*x(n)-y(n) and b(n)=y(n). E.g. D(4)=21 with Perron's (x,y)=(3,1) and (a,b)=(5,1). 1=b(4)=A078355(4). If D(n)=A077425(n) appears also in A077426 (odd period length of continued fraction of (sqrt(D(n))+1)/2) then a(n)=(2*x-y)^2+2 and b(n)=(2*x-y)*y. E.g. D(7)=37 with Perron's (x,y)=(7,2) leading to (a,b)=(146,24) with 24=b(7)=A078355(7).
The generic D(n) values are those from A078371(k-1) := (2*k+3)*(2*k-1), for k>=1, which are 5 (mod 8). For such D values the minimal solution is (a(n),b(n))=(2*k+1,1) (e.g. D(16)=77= A078371(3) with a(16)=2*4+1=9 and b(16)=A078355(16)=1).
The general solution of Pell a^2-D(n)*b^2 = +4 with generic D(n)=A077425(n)=A078371(k-1), k>=1, is a(n,m)= 2*T(m+1,(2*k+1)/2) and b(n,m)= S(m,2*k+1), m>=0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second, kind. See A053120 resp. A049310.
For non-generic D(n) (not from A078371) the general solution of a^2-D(n)*b^2 = +4 is a(n,m)= 2*T(m+1,a(n)/2) and b(n,m)= b(n)*S(m,a(n)), m>=0, with Chebyshev's polynomials and in this case b(n)>1.

References

  • O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 30, Satz 3.35, p. 109 and table p. 108).

Programs

  • Mathematica
    d = Select[Range[5, 300, 4], !IntegerQ[Sqrt[#]]&]; a[n_] := Module[{a, b, r}, a /. {r = Reduce[a > 0 && b > 0 && a^2 - d[[n]]*b^2 == 4, {a, b}, Integers]; (r /. C[1] -> 0) || (r /. C[1] -> 1) // ToRules} // Select[#, IntegerQ, 1] &] // First; Table[a[n], {n, 1, 43}] (* Jean-François Alcover, Jul 30 2013 *)

Extensions

More terms from Max Alekseyev, Mar 03 2010

A077425 a(n) == 1 (mod 4) (see A016813), but not a square (i.e., not in A000290).

Original entry on oeis.org

5, 13, 17, 21, 29, 33, 37, 41, 45, 53, 57, 61, 65, 69, 73, 77, 85, 89, 93, 97, 101, 105, 109, 113, 117, 125, 129, 133, 137, 141, 145, 149, 153, 157, 161, 165, 173, 177, 181, 185, 189, 193, 197, 201, 205, 209, 213, 217, 221, 229, 233, 237, 241, 245, 249, 253, 257
Offset: 1

Views

Author

Wolfdieter Lang, Nov 29 2002

Keywords

Comments

The Pell equation x^2 - a(n)*y^2 = +4 has infinitely many (integer) solutions (see A077428 and A078355).
These are the odd numbers in A079896. The even ones are 4*A000037. - Wolfdieter Lang, Sep 15 2015
First differences: 8, 4, 4, 8, 4, 4, 4, 4, 8, 4, 4, 4, 4, 4, 4, 8, ... , only 4's and 8's?. - Paul Curtz, Apr 11 2019
Yes. There are only 4's and 8's. Proof: Only multiples of 4 may appear. The 4's correspond to successive composite in A016813, whereas an 8 corresponds to a square. A greater multiple of 4 would imply to have at least 2 consecutive squares in A016813, which is not possible since 2 consecutive squares cannot have a difference of 4. That sequence of 4's and 8's can be obtained with A010052 (without the 1st term) where the 0's are replaced with 4's and 1's replaced with 8's. - Michel Marcus, Apr 16 2019

Crossrefs

Intersection of A016813 and A000037.

Programs

  • Maple
    A077425 := proc(n::integer) local resul,i ; resul := 5 ; i := 1 ; while i < n do resul := resul+4 ; while issqr(resul) do resul := resul+4 ; od ; i:= i+1 ; od ; RETURN(resul) ; end proc:
    seq(A077425(n),n=1..31) ; # R. J. Mathar, Apr 25 2006
  • Mathematica
    Select[Range[5,300,4],!IntegerQ[Sqrt[#]]&] (* Harvey P. Dale, Dec 05 2012 *)
  • PARI
    [n | n <- vector(100,n,4*n+1), !issquare(n)] \\ Charles R Greathouse IV, Mar 11 2014
    
  • PARI
    list(lim)=my(v=List()); for(s=2,sqrtint((lim\=1)+1), forstep(n=s^2 + if(s%2,4,1), min((s+1)^2-1,lim), 4, listput(v,n))); Vec(v) \\ Charles R Greathouse IV, Nov 04 2021
    
  • Python
    from operator import sub
    from sympy import integer_nthroot
    def A077425(n): return n+sub(*integer_nthroot(n,2))<<2|1 # Chai Wah Wu, Oct 01 2024

Extensions

More terms from Max Alekseyev, Mar 03 2010

A077427 Primitive period length of (regular) continued fraction of (sqrt(D(n))+1)/2 for D(n)=A077425(n).

Original entry on oeis.org

1, 1, 3, 2, 1, 4, 3, 5, 2, 1, 6, 3, 3, 4, 9, 2, 1, 7, 2, 9, 3, 6, 7, 7, 2, 1, 10, 4, 7, 4, 3, 5, 8, 5, 10, 2, 1, 12, 5, 3, 4, 15, 3, 14, 4, 12, 4, 16, 2, 1, 9, 2, 19, 2, 16, 6, 3, 8, 11, 5, 6, 9, 15, 2, 1, 10, 10, 4, 6, 19, 3, 4, 3, 16
Offset: 1

Views

Author

Wolfdieter Lang, Nov 29 2002

Keywords

Comments

The Pell equation x^2 - D(n)*y^2 = -4 has (infinitely many integer) solutions if and only if a(n) is odd.

Examples

			a(6)=4 because the (periodic) continued fraction for (sqrt(D(6))+1)/2 = (sqrt(33)+1)/2 = 3.372281324... is [3, periodic(2, 1, 2, 5,)] with period length 4. Because these continued fractions are always of the form [b(0),periodic(b(1),b(2),...,b(2),b(1),2*b(0)-1,)] with the symmetric piece b(1),b(2),..., b(2),b(1), Perron op. cit. writes for this b(0),b(1),b(2),...,(b(k/2)) if the period length k is even and b(0),b(1),b(2),...,b((k-1)/2) if the period length is odd. In this example: k=4 and Perron writes 3,2,(1). Another example: D(8)= A077425(8)=41 leads to Perron's 3,1,2 standing for [3,periodic(1,2,2,1,5,)], the continued fraction for (sqrt(41)+1)/2 which has odd period length a(8)=5.
a(4)=2 is even and D(4)=A077425(4)=21, hence x^2 - 21*y^2 = -4 has no nontrivial integer solution.
a(8)=5 is odd and D(8)=A077425(8)=41, hence x^2 - 41*y^2 = -4 is solvable (with nontrivial integers) as well as x^2 - 41*y^2 = +4.
		

References

  • O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 30, Satz 3.35, p. 109).

Crossrefs

Cf. A077426.

A078361 Minimal positive solution a(n) of Pell equation a(n)^2 - D(n)*b(n)^2 = +4 or -4 with D(n)=A077425(n). The companion sequence is b(n)=A077058(n).

Original entry on oeis.org

1, 3, 8, 5, 5, 46, 12, 64, 7, 7, 302, 39, 16, 25, 2136, 9, 9, 1000, 29, 11208, 20, 82, 261, 1552, 11, 11, 33710, 173, 3488, 190, 24, 61, 4354, 213, 23550, 13, 13, 124846, 1305, 136, 110, 3528264, 28, 1030190, 43, 93102, 73, 7688126, 15, 15, 46312, 77
Offset: 1

Views

Author

Wolfdieter Lang, Nov 29 2002

Keywords

Comments

Computed from Perron's table (see reference p. 108, for n = 1..28) which gives the minimal x,y values for the Diophantine eq. x^2 - x*y - ((D(n)-1)/4)*y^2= +1, resp., -1 if D(n)=A077425(n), resp, D(n)=A077425(n) and D(n) also in A077426 (this second case excludes in Perron's table the D values with a 'Teilnenner' in brackets).
The conversion from the x,y values of Perron's table to the minimal a=a(n) and b=b(n) solutions is a(n)=2*x(n)-y(n) and b(n)=y(n). If D(n)=A077425(n) is not in A077426 then the equation with -4 has no solution and a(n) and b(n) are the minimal solutions of the a(n)^2 - D(n)*b(n)^2 = +4 equation. If D(n)=A077425(n) is in A077426 then the a(n) and b(n) values are the minimal solution of the a(n)^2 - D(n)*b(n)^2 = -4 equation. In this case a(+,n)= a(n)^2+2 and b(+,n)=a(n)*b(n) are the minimal solution of a^2 - D(n)*b^2 = +4.
For Pell equation a^2 - D*b^2 = +4, see A077428 and A078355. For Pell equation a^2 - D*b^2 = -4, see A078356 and A078357.

Examples

			29=D(5)=A077425(5) is A077426(4), hence a(5)=5 and b(5)=A077058(5)=1 solve a^2 - 29*b^2=-4 minimally and a(+,5)=a(5)^2+2=27 with b(+,5)=a(5)*b(5)=5*1=5 solve a^2 - 29*b^2=+4 minimally. See also A077428 with companion A078355.
21=D(4)=A077425(4) is not in A077426, hence a(4)=5 and b(4)=A077058(4)=1 give the solution with minimal positive b of a^2 - 21*b^2=+4.
		

References

  • O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 30, Satz 3.35, p. 109 and table p. 108).

Extensions

More terms from Matthew Conroy, Apr 20 2003

A175155 Numbers m satisfying m^2 + 1 = x^2 * y^3 for positive integers x and y.

Original entry on oeis.org

0, 682, 1268860318, 1459639851109444, 2360712083917682, 86149711981264908618, 4392100110703410665318, 8171493471761113423918890682, 15203047261220215902863544865414318, 5484296027914919579181500526692857773246, 28285239023397517753374058381589688919682, 12439333951782387734360136352377558500557329868
Offset: 1

Views

Author

Michel Lagneau, Feb 27 2010

Keywords

Comments

This sequence is infinite. The fundamental solution of m^2 + 1 = x^2 y^3 is (m,x,y) = (682,61,5), which means the Pellian equation m^2 - 125x^2 = -1 has the solution (m,x) = (682,61) = (m(1),x(1)). This Pellian equation admits an infinity of solutions (m(2k+1),x(2k+1)), k=1,2,..., given by the following recursive relation, starting with m(1)=682, x(1)= 61: m(2k+1) + x(2k+1)*sqrt(125) = (m(1) + x(1)*sqrt(125))^(2k+1).
Squares of these terms are in A060355, since both a(n)^2 and a(n)^2 + 1 are powerful (A001694). - Charles R Greathouse IV, Nov 16 2012
It appears that y = A077426. - Robert G. Wilson v, Nov 16 2012
Also m^2 + 1 is powerful. Other solutions arise from solutions x to x^2 - k^3*y^2 = -1. - Georgi Guninski, Nov 17 2012
Although it is believed that the b-file is complete for all terms m < 10^100, the search only looked for y < 100000. - Robert G. Wilson v, Nov 17 2012

Examples

			For m=682, m^2 + 1 = 465125 = 61^2 * 5^3.
		

References

  • Albert H. Beiler, "The Pellian" (Chap. 22), Recreations in the Theory of Numbers, 2nd ed. NY: Dover, 1966.
  • A. Cayley, Report of a committee appointed for the purpose of carrying on the tables connected with the Pellian equation ..., Collected Mathematical Papers. Vols. 1-13, Cambridge Univ. Press, London, 1889-1897, Vol. 13, pp. 430-443.
  • J. M. De Koninck, Ces nombres qui nous fascinent, Ellipses, 2008, p. 108.

Crossrefs

Programs

  • Maple
    C:=array(0..20,0..20):C[1,1]=1: C[2,1]=1: n1:=682:x1:=61:for nn from 1 by 2 to 15 do:s:=0:for i from 2 to 15 do:for j from 1 to i do:C[i,j]:= C[i-1,j] + C[i-1,j-1]: od:od:for n from 1 by 2 to nn+1 do:s:=s + C[nn+1,n] * n1^(nn-n+1)*x1^(n-1)*125^((n-1)/2):od:print (s):od: # Michel Lagneau
    # 2nd program R. J. Mathar, Mar 16 2016:
    # print (nonsorted!) all solutions of A175155 up to search limit
    with(numtheory):
    # upper limit for solutions n
    nsearchlim := 10^40 :
    A175155y := proc(y::integer)
        local disc;
        disc := y^3 ;
        cfrac(sqrt(disc),periodic,quotients) ;
    end proc:
    for y from 2 do
        if issqrfree(y) then
            # find continued fraction for x^2-(y^3=disc)*y^2=-1, sqrt(disc)
            cf := A175155y(y) ;
            nlen :=  nops(op(2,cf)) ;
            if type(nlen,odd) then
                # fundamental solution
                fuso := numtheory[nthconver](cf,nlen-1) ;
                fusolx := numer(fuso) ;
                fusoly := denom(fuso) ;
                solx := fusolx ;
                soly := fusoly ;
                while solx <= nsearchlim do
                    rhhs := solx^2-y^3*soly^2 ;
                    if rhhs = -1 then
                        # print("n=",solx,"x=",soly,"y=",y^3) ;
                        print(solx) ;
                    end if;
                    # solutions from fundamental solution
                    tempx := fusolx*solx+y^3*fusoly*soly ;
                    tempy := fusolx*soly+fusoly*solx ;
                    solx := tempx ;
                    soly := tempy ;
                end do;
            end if;
        fi;
    end do:
  • Mathematica
    nmax = 10^50; ymax = 100; instances = 10; fi[y_] := n /. FindInstance[0 <= n <= nmax && x > 0 && n^2 + 1 == x^2*y^3, {n, x}, Integers, instances]; yy = Select[Range[1, ymax, 2], !IntegerQ[Sqrt[#]] && OddQ[ Length[ ContinuedFraction[Sqrt[#]][[2]]]]&]; Join[{0}, fi /@ yy // Flatten // Union // Most] (* Jean-François Alcover, Jul 12 2017 *)
  • PARI
    is(n)=ispowerful(n^2+1) \\ Charles R Greathouse IV, Nov 16 2012

Formula

m(1)=682, x(1) = 61 and m(2k+1) + x(2k+1)*sqrt(125) = (m(1) + x(1)*sqrt(125))^(2k+1) m(2k+1) = C(2k+1,0) * m(1)^(2k+1) + C(2k+1,2)*m(1)^(2k-1)*x(1)^2 + ...

Extensions

Added condition that x and y must be positive. Added missing initial term 0. Added warning that b-file has not been proved to be correct - there could be missing entries. - N. J. A. Sloane, Nov 17 2012

A225432 Twice the coefficient of sqrt(q) in e^h, where e is the fundamental unit and h is the class number of Q(sqrt(q)), q prime and congruent to 1 mod 4. (The coefficient lies in (1/2)Z, so twice it is an integer.)

Original entry on oeis.org

1, 1, 2, 1, 2, 10, 1, 5, 250, 106, 1138, 2, 25, 146, 298, 5, 17, 1, 97, 253970, 2, 226, 3034, 9148450, 2050, 10, 157, 126890, 1, 14341370, 5, 110671282, 986, 7586, 530, 130, 173, 5129602, 11068353370, 21685, 694966754, 17883410, 5528222698, 17, 41, 11248618, 60037, 10, 242718010, 24514292738
Offset: 1

Views

Author

Robert R. Bruner, May 07 2013

Keywords

Comments

This also arises in the relation satisfied by Euler classes in the connective K-theory of the classifying space of the group of order pq, where p=(q-1)/2. See p. 39 in Bruner and Greenlees, cited below. Take an irreducible representation of the cyclic group of order q which generates the representations as a ring, induce it up to the group of order pq, and let z be its Euler class in ku^{2p}(BG_{pq}). Then z satisfies the relation z^3 -2bq z^2 + qz = 0. This follows from the arithmetic fact that in Q(sort(q)) we have the relation e^h = a + b sqrt(q), as shown on pp. 39-42 of Bruner and Greenlees.
This is closely related to the subsequence of A078357 containing those entries such that the corresponding entry in A077426 is prime. However, a(22) = 226 (corresponding to e^3 = 1710 + 113*sqrt(229)) does not occur in A078357, and more such terms appear after this.
For the n-th Pythagorean prime q=A002144(n), a(n) is also -1/q of the coefficient of term x in the minimal polynomial of A=Product_{a} 2*sin(a*Pi/q) (where the index runs through all quadratic residues in {1,2,...,q-1}) and B=Product_{b} 2*sin(b*Pi/q) (where the index runs through all quadratic nonresidues in {1,2,...,q-1}). It is easy to show that A*B = p. By the class number formula of real quadratic number fields, one obtains B/A = e^(+-2h), so A+B = sqrt(q)*(e^h+e^(-h)) is exactly q*a(n). - Zichang Wang, Dec 15 2022

References

  • R. R. Bruner and J. P. C. Greenlees, The Connective K-theory of Finite Groups, Memoirs AMS, Vol. 165, No. 785, 2003.
  • T. Mitsuhiro, T. Nakahara and T. Uehara, The Class Number Formula of a Real Quadratic Field and an Estimate of the Value of a Unit, Canadian Mathematical Bulletin, 38(1)(1995), 98-103.

Crossrefs

Programs

  • Magma
    // Magma code to generate all terms for which the prime q is less than or equal to 4N+1 (an initial segment of the sequence). (Note that the brute force computation of the fundamental unit is very inefficient, and will have trouble producing the 39th term.)
    N := 40;
    pr := [4*n+1 : n in [1..N] | IsPrime(4*n+1)];
    for i in [1..#pr] do
       q := pr[i];
       Q := QuadraticField(q);
       h := ClassNumber(Q);
       x := 1;
       while not IsSquare(x*x*q-4) do
          x := x+1;
       end while;
       x := x/2;
       tr,y := IsSquare(x*x*q-1);
       e := y + x*s;
       eh := e^h;
       b := (eh-Trace(eh)/2)/s;
       print i,2*b;
    end for;
  • Mathematica
    (* e.g., first 270 terms *)
    Lq = Select[4*Range[1000] + 1, PrimeQ[#] &];
    Lh = NumberFieldClassNumber[Sqrt[Lq]];
    Le = NumberFieldFundamentalUnits[Sqrt[Lq]];
    Transpose[RootReduce[(Le^(2 Lh) + 1)/(Sqrt[Lq] Le^Lh)]][[1]]
    (* Zichang Wang, Dec 15 2022 *)

Extensions

a(39) onward from Zichang Wang, Dec 15 2022
Showing 1-9 of 9 results.