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.

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A072068 Number of integer solutions to the equation 2x^2+y^2+8z^2=m for an odd number m=2n-1.

Original entry on oeis.org

2, 4, 0, 0, 10, 12, 0, 0, 16, 12, 0, 0, 10, 16, 0, 0, 16, 24, 0, 0, 32, 12, 0, 0, 18, 24, 0, 0, 16, 36, 0, 0, 32, 12, 0, 0, 16, 28, 0, 0, 34, 36, 0, 0, 48, 24, 0, 0, 16, 36, 0, 0, 32, 36, 0, 0, 32, 24, 0, 0, 26, 24, 0, 0
Offset: 1

Views

Author

T. D. Noe, Jun 13 2002

Keywords

Comments

Related to primitive congruent numbers A006991.
Assuming the Birch and Swinnerton-Dyer conjecture, the odd number 2n-1 is a congruent number if it is squarefree and a(n) = 2*A072069(n).
Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

Examples

			a(2) = 4 because (1,1,0), (-1,1,0), (1,-1,0) and (-1,-1,0) are solutions when m=3.
G.f. = 2*x + 4*x^2 + 10*x^5 + 12*x^6 + 16*x^9 + 12*x^10 + 10*x^13 + 16*x^14 + 16*x^17 + ...
G.f. = 2*q + 4*q^3 + 10*q^9 + 12*q^11 + 16*q^17 + 12*q^19 + 10*q^25 + 16*q^27 + ...
		

Crossrefs

Programs

  • Mathematica
    maxN=128; soln1=Table[0, {maxN/2}]; xMax=Ceiling[Sqrt[maxN/2]]; yMax=Ceiling[Sqrt[maxN]]; zMax=Ceiling[Sqrt[maxN/8]]; Do[n=2x^2+y^2+8z^2; If[OddQ[n]&&nA072068 = CoefficientList[s, x] // Rest (* Jean-François Alcover, Feb 16 2015, after Michael Somos *)
  • PARI
    {a(n) = my(A); n--; if( n<0, 0, A = x * O(x^n); polcoeff( 2 * eta(x^2 + A)^5 * eta(x^8 + A)^7 / (eta(x + A)^2 * eta(x^4 + A)^5 * eta(x^16 + A)^2), n))}; /* Michael Somos, Dec 26 2019 */

Formula

Expansion of 2 * x * phi(x) * psi(x^4) * phi(x^4) in powers of x where phi(), psi() are Ramanujan theta functions. - Michael Somos, Jun 08 2012
Expansion of 2 * q^(1/2) * eta(q^2)^5 * eta(q^8)^7 / (eta(q)^2 * eta(q^4)^5 * eta(q^16)^2) in powers of q. - Michael Somos, Feb 19 2015

A072069 Number of integer solutions to the equation 2x^2+y^2+32z^2=m for an odd number m=2n-1.

Original entry on oeis.org

2, 4, 0, 0, 6, 4, 0, 0, 4, 4, 0, 0, 2, 8, 0, 0, 12, 8, 0, 0, 16, 12, 0, 0, 10, 16, 0, 0, 12, 20, 0, 0, 16, 4, 0, 0, 12, 12, 0, 0, 14, 20, 0, 0, 20, 8, 0, 0, 4, 20, 0, 0, 8, 12, 0, 0, 24, 8, 0, 0, 14, 8, 0, 0
Offset: 1

Views

Author

T. D. Noe, Jun 13 2002

Keywords

Comments

Related to primitive congruent numbers A006991.
Assuming the Birch and Swinnerton-Dyer conjecture, the odd number 2n-1 is a congruent number if it is squarefree and 2 a(n) = A072068(n).
Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

Examples

			a(2) = 4 because (1,1,0), (-1,1,0), (1,-1,0) and (-1,-1,0) are solutions when m=3.
G.f. = 2*x + 4*x^2 + 6*x^5 + 4*x^6 + 4*x^9 + 4*x^10 + 2*x^13 + 8*x^14 + ... - _Michael Somos_, Dec 26 2019
G.f. = 2*q + 4*q^3 + 6*q^9 + 4*q^11 + 4*q^17 + 4*q^19 + 2*q^25 + 8*q^27 + 12*q^33
+ ...
		

References

  • J. B. Tunnell, A classical Diophantine problem and modular forms of weight 3/2, Invent. Math., 72 (1983), 323-334.

Crossrefs

Programs

  • Mathematica
    maxN=128; soln2=Table[0, {maxN/2}]; xMax=Ceiling[Sqrt[maxN/2]]; yMax=Ceiling[Sqrt[maxN]]; zMax=Ceiling[Sqrt[maxN/32]]; Do[n=2x^2+y^2+32z^2; If[OddQ[n]&&n
    				
  • PARI
    {a(n) = my(A); n--; if( n<0, 0, A = x * O(x^n); polcoeff( 2 * eta(x^2 + A)^5 * eta(x^8 + A)^2 * eta(x^32 + A)^5 / (eta(x + A)^2 * eta(x^4 + A)^3 * eta(x^16 + A)^2 * eta(x^64 + A)^2), n))}; /* Michael Somos, Dec 26 2019 */

Formula

Expansion of 2 * x * phi(x) * psi(x^4) * phi(x^16) in powers of x where phi(), psi() are Ramanujan theta functions. - Michael Somos, Jun 08 2012
Expansion of 2 * q^(1/2) * eta(q^2)^5 * eta(q^8)^2 * eta(q^32)^5 / (eta(q)^2 * eta(q^4)^3 * eta(q^16)^2 * eta(q^64)^2) in powers of q. - Michael Somos, Dec 26 2019

A072070 Number of integer solutions to the equation 4*x^2 + y^2 + 8*z^2 = n.

Original entry on oeis.org

1, 2, 0, 0, 4, 4, 0, 0, 6, 6, 0, 0, 8, 12, 0, 0, 12, 8, 0, 0, 8, 8, 0, 0, 8, 14, 0, 0, 16, 4, 0, 0, 6, 16, 0, 0, 12, 20, 0, 0, 24, 8, 0, 0, 8, 20, 0, 0, 24, 18, 0, 0, 24, 12, 0, 0, 0, 16, 0, 0, 16, 20, 0, 0, 12, 8, 0, 0, 16, 16, 0, 0, 30, 32, 0, 0, 24, 16, 0, 0, 24, 18, 0, 0, 16, 24, 0, 0, 24, 16
Offset: 0

Views

Author

T. D. Noe, Jun 13 2002

Keywords

Comments

Related to primitive congruent numbers A006991.
Assuming the Birch and Swinnerton-Dyer conjecture, the even number 2n is a congruent number if it is squarefree and a(n) = 2 A072071(n).
Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

Examples

			a(4) = 4 because (1, 0, 0), (-1, 0, 0), (0, 2, 0) and (0, -2, 0) are solutions.
G.f. = 1 + 2*q + 4*q^4 + 4*q^5 + 6*q^8 + 6*q^9 + 8*q^12 + 12*q^13 + 12*q^16 + 8*q^17 + ...
		

References

  • J. B. Tunnell, A classical Diophantine problem and modular forms of weight 3/2, Invent. Math., 72 (1983), 323-334.

Crossrefs

Programs

  • Mathematica
    maxN=128; soln3=Table[0, {maxN/2}]; xMax=Ceiling[Sqrt[maxN/8]]; yMax=Ceiling[Sqrt[maxN/2]]; zMax=Ceiling[Sqrt[maxN/16]]; Do[n=4x^2+y^2+8z^2; If[n>0&&n<=maxN/2, s=8; If[x==0, s=s/2]; If[y==0, s=s/2]; If[z==0, s=s/2]; soln3[[n]]+=s], {x, 0, xMax}, {y, 0, yMax}, {z, 0, zMax}]
    a[ n_] := SeriesCoefficient[ EllipticTheta[ 3, 0, q] EllipticTheta[ 3, 0, q^4] EllipticTheta[ 3, 0, q^8], {q, 0, n}]; (* Michael Somos, Jul 23 2018 *)
  • PARI
    {a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x + A)^-2 * eta(x^2 + A)^5 * eta(x^4 + A)^-4 * eta(x^8 + A)^3 * eta(x^16 + A)^3 * eta(x^32 + A)^-2, n))}; /* Michael Somos, Feb 11 2003 */

Formula

Expansion of phi(q) * phi(q^4) * phi(q^8) in powers of q where phi() is a Ramanujan theta function. - Michael Somos, Jun 09 2012
Euler transform of period 32 sequence [2, -3, 2, 1, 2, -3, 2, -2, 2, -3, 2, 1, 2, -3, 2, -5, 2, -3, 2, 1, 2, -3, 2, -2, 2, -3, 2, 1, 2, -3, 2, -3, ...]. - Michael Somos, Feb 11 2003
a(4*n + 2) = a(4*n + 3) = 0. a(4*n) = A014455(n). - Michael Somos, Jun 08 2012
G.f. is a period 1 Fourier series which satisfies f(-1 / (32 t)) = 2^(7/2) (t/i)^(3/2) g(t) where q = exp(2 Pi i t) and g() is the g.f. for A080917. - Michael Somos, Jul 23 2018

A072071 Number of integer solutions to the equation 4x^2+y^2+32z^2=n.

Original entry on oeis.org

1, 2, 0, 0, 4, 4, 0, 0, 4, 2, 0, 0, 0, 4, 0, 0, 4, 4, 0, 0, 8, 0, 0, 0, 0, 6, 0, 0, 0, 4, 0, 0, 6, 4, 0, 0, 12, 12, 0, 0, 16, 8, 0, 0, 0, 12, 0, 0, 8, 10, 0, 0, 24, 4, 0, 0, 0, 12, 0, 0, 0, 12, 0, 0, 12, 8, 0, 0, 16, 8, 0, 0, 20, 12, 0, 0, 0, 8, 0, 0, 8, 6, 0, 0, 16, 16, 0, 0, 0, 4, 0, 0, 0, 8, 0, 0, 8
Offset: 0

Views

Author

T. D. Noe, Jun 13 2002

Keywords

Comments

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).
Related to primitive congruent numbers A006991.
Assuming the Birch and Swinnerton-Dyer conjecture, the even number 2n is a congruent number if it is squarefree and 2 a(n) = A072070(n).

Examples

			a(4) = 4 because (1,0,0), (-1,0,0), (0,2,0) and (0,-2,0) are solutions.
1 + 2*x + 4*x^4 + 4*x^5 + 4*x^8 + 2*x^9 + 4*x^13 + 4*x^16 + 4*x^17 + 8*x^20 + ...
		

References

  • J. B. Tunnell, A classical Diophantine problem and modular forms of weight 3/2, Invent. Math., 72 (1983), 323-334.

Crossrefs

Programs

  • Mathematica
    J12[q_] := Sum[q^n^2, {n, -10, 10}]; CoefficientList[Series[J12[q]J12[q^4]J12[q^32], {q, 0, 100}], q]
  • PARI
    {a(n) = local(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x + A)^-2 * eta(x^2 + A)^5 * eta(x^4 + A)^-4 * eta(x^8 + A)^5 * eta(x^16 + A)^-2 * eta(x^32 + A)^-2 * eta(x^64 + A)^5 * eta(x^128 + A)^-2, n))}

Formula

Expansion of phi(x) * phi(x^4) * phi(x^32) in powers of x where phi() is a Ramanujan theta function.
a(4*n + 2) = a(4*n + 3) = 0. - Michael Somos, Jun 08 2012

Extensions

More terms from Vladeta Jovovic, Jun 16 2002

A273929 Numbers that are congruent to {5, 6, 7} mod 8 and are squarefree.

Original entry on oeis.org

5, 6, 7, 13, 14, 15, 21, 22, 23, 29, 30, 31, 37, 38, 39, 46, 47, 53, 55, 61, 62, 69, 70, 71, 77, 78, 79, 85, 86, 87, 93, 94, 95, 101, 102, 103, 109, 110, 111, 118, 119, 127, 133, 134, 141, 142, 143, 149, 151, 157, 158, 159, 165, 166, 167, 173, 174
Offset: 1

Views

Author

Frank M Jackson, Jun 04 2016

Keywords

Comments

It has been shown, conditional on the Birch Swinnerton-Dyer conjecture, that this sequence is a subset of the primitive congruent numbers (A006991). The union of this sequence with A062695 gives A006991. Also this sequence is the intersection of A047574 and A005117.
The asymptotic density of this sequence is 3/Pi^2 (A104141). - Amiram Eldar, Mar 09 2021

Crossrefs

Programs

  • Magma
    [n: n in [1..250] | n mod 8 in [5, 6, 7] and IsSquarefree(n)]; // Vincenzo Librandi, Jun 06 2016
  • Mathematica
    Select[Range[1000], MemberQ[{5, 6, 7}, Mod[#, 8]] && SquareFreeQ[#] &]
  • PARI
    is(n) = n % 8 > 4 && issquarefree(n) \\ Felix Fröhlich, Jun 04 2016
    

A274264 Number of squarefree integers congruent to {5, 6, 7} mod 8 <= 10^n.

Original entry on oeis.org

3, 33, 308, 3050, 30405, 303979, 3039648, 30396356, 303963597, 3039635407, 30396354916, 303963551200, 3039635509025, 30396355093247, 303963550927371, 3039635509273730, 30396355092701463, 303963550927001730
Offset: 1

Views

Author

Frank M Jackson, Jun 16 2016

Keywords

Comments

Empirically, the limit of a(n)/10^n tends to 3/Pi^2 (A104141) and implies that the asymptotic density of squarefree numbers congruent to {5, 6, 7} mod 8 is half that of the asymptotic density of all squarefree integers (A071172). There is a slight bias towards more squarefree numbers congruent to {5, 6, 7} mod 8 that can be argued heuristically as {1, 2, 3} mod 8 contains a square residue and its equivalence class should contain less squarefree numbers.
Also it has been shown, conditional on the Birch Swinnerton-Dyer conjecture, that all squarefree integers congruent to {5, 6, 7} mod 8 (A273929) are primitive (squarefree) congruent numbers (A006991). However, this property applies only sparsely to squarefree integers congruent to {1, 2, 3} mod 8 (A062695).

Crossrefs

Programs

  • Mathematica
    Table[Length@Select[Range[10^n], MemberQ[{5, 6, 7}, Mod[#, 8]]&& SquareFreeQ[#] &], {n, 1, 8}]

Extensions

a(10)-a(11) from Giovanni Resta, Jun 17 2016
a(7) corrected and a(12)-a(18) added by Hiroaki Yamanouchi, Dec 25 2016

A147779 Squarefree positive integers of the form u*v*(u^2-v^2) for some integer u,v.

Original entry on oeis.org

6, 30, 210, 330, 546, 2310, 2730, 3570, 4290, 5610, 6090, 6630, 7854, 8970, 9690, 10374, 10626, 13566, 18354, 19866, 22134, 25806, 26970, 39270, 43890, 51330, 51414, 52026, 54834, 56730, 59334, 66990, 68034, 71610, 72930, 74046, 75174
Offset: 1

Views

Author

Max Alekseyev, Nov 12 2008

Keywords

Crossrefs

Formula

Squarefree terms of A147778. Squarefree terms of A073120. Squarefree terms of A009112.
Terms of A006991 (primitive congruent numbers) corresponding to right triangles with integer sides.

A375017 Squarefree numbers k such that k is the area of a rational isosceles triangle.

Original entry on oeis.org

3, 7, 10, 11, 14, 15, 17, 19, 23, 26, 30, 31, 35, 39, 42, 43, 46, 47, 51, 55, 58, 59, 62, 67, 69, 71, 74, 77, 78, 79, 82, 83, 87, 91, 94, 95, 97, 103, 105, 106, 107, 110, 111, 113, 115, 119, 122, 123, 127, 130, 131, 138, 139, 142, 143, 151, 154, 155, 158, 159, 163, 165, 167, 170
Offset: 1

Views

Author

Frank M Jackson, Aug 08 2024

Keywords

Comments

All rational isosceles triangles are the join of two identical rational right triangles along one of their common legs. Therefore, for any g, a squarefree congruent number, it can be the area of the right triangle creating a rational isosceles triangle with area 2g. If g is odd then the rational isosceles triangle will have squarefree k = 2g. If g is even then the rational isosceles triangle can be reduced by a factor 4 to give a squarefree value for k = g/2.

Examples

			The congruent number 5 can create a rational right triangle with sides (9/6, 40/6, 41/6) and squarefree area 5. This can create a rational isosceles triangle with sides (3, 41/6, 41/6) or (80/6, 41/6, 41/6) with squarefree area 10.
However the congruent number 6 can create a rational right triangle with sides (3, 4, 5) and squarefree area 6. This can create a rational isosceles triangle with sides (5/2, 5/2, 3) or (4, 5/2, 5/2) with squarefree area 3.
		

Crossrefs

Programs

  • Mathematica
    lst = Last /@ReadList["https://oeis.org/A006991/b006991.txt", {Number, Number}]; lst1={}; Do[If[EvenQ[lst[[n]]], AppendTo[lst1, lst[[n]]/2], AppendTo[lst1, 2 lst[[n]]]], {n, 1, Length@lst}]; (Sort@lst1)[[1 ;; 75]]

A274043 Number of squarefree integers congruent to {1, 2, 3} mod 8 <= 10^n.

Original entry on oeis.org

4, 28, 300, 3033, 30389, 303947, 3039643, 30396338, 303963527, 3039635535, 30396355364, 303963551074, 3039635509269, 30396355092700, 303963550926732, 3039635509266675, 30396355092702331, 303963550927021020
Offset: 1

Views

Author

Frank M Jackson, Jun 18 2016

Keywords

Comments

Empirically, the limit of a(n)/10^n tends to 3/Pi^2 (A104141) and implies that the asymptotic density of squarefree numbers congruent to {1, 2, 3} mod 8 is half that of the asymptotic density of all squarefree integers (A071172). When this sequence is compared with squarefree numbers congruent to {5, 6, 7} mod 8 (A274264) it contains slightly fewer squarefree integers at each of the sampling points, 10^n for n > 1. It can be argued heuristically that, as {1, 2, 3} mod 8 contains a square residue, its equivalence class should contain fewer squarefree numbers.
Also it has been shown, conditional on the Birch Swinnerton-Dyer conjecture, that all squarefree integers congruent to {5, 6, 7} mod 8 (A273929) are primitive congruent numbers (A006991). However, this property applies only sparsely to squarefree integers congruent to {1, 2, 3} mod 8 (A062695).

Crossrefs

Programs

  • Mathematica
    Table[Length@Select[Range[10^n], MemberQ[{1, 2, 3}, Mod[#, 8]]&&SquareFreeQ[#] &], {n, 1, 8}]

Extensions

a(10)-a(11) from Giovanni Resta, Jun 19 2016
a(12)-a(18) from Hiroaki Yamanouchi, Dec 25 2016

A290349 Least multiplier of n such that n*a(n) becomes a congruent number A003273.

Original entry on oeis.org

5, 3, 2, 5, 1, 1, 1, 3, 5, 2, 2, 2, 1, 1, 1, 5, 2, 3, 2, 1, 1, 1, 1, 1, 5, 2, 2, 1, 1, 1, 1, 3, 5, 1, 2, 5, 1, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 5, 3, 2, 1, 1, 1, 1, 1, 5, 2, 2, 1, 1, 1, 1, 5, 1, 3, 2, 2, 1, 1, 1, 3, 3, 2, 2, 2, 1, 1, 1, 1, 5, 2, 2, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Frank M Jackson, Jul 28 2017

Keywords

Comments

Conditional on the Birch and Swinnerton-Dyer conjecture, it can be shown that the only members of this sequence are the Fibonacci numbers {1,2,3,5}. The underlying pattern of three consecutive 1's per octet shows that numbers congruent to {5,6,7} mod 8 are congruent numbers. Also if n is a square then a(n)=5. This is because all congruent numbers can be obtained by multiplying a primitive congruent number A006991 by a positive square number A000290 and 5 is the least congruent number.

Examples

			a(10)=2 as 10*2=20 and 10*3=30 are congruent numbers but 2 is the least multiplier.
		

Crossrefs

Programs

  • Mathematica
    Sfcore[n_] := Module[{m, fac=Select[FactorInteger[n], OddQ[#[[2]]] &]}, If[!SquareFreeQ[n], Times@@Table[fac[[m]][[1]], {m, Length[fac]}], n]]; CongruentQ[n_] := Module[{x, y, z, ok=False}, (Which[! SquareFreeQ[n], Null[], MemberQ[{5, 6, 7}, Mod[n, 8]], ok=True, OddQ[n]&&Length@Solve[x^2+2y^2+8z^2==n, {x, y, z}, Integers]==2Length@Solve[x^2+2y^2+32z^2==n, {x, y, z}, Integers], ok=True, EvenQ[n]&&Length@Solve[x^2+4y^2+8z^2==n/2, {x, y, z}, Integers]==2Length@Solve[x^2+4y^2+32z^2==n/2, {x, y, z}, Integers], ok=True]; ok)]; lst = {}; Do[AppendTo[lst, (Min[Select[n {1, 2, 3, 5}, CongruentQ[Sfcore[#]] &]])/n], {n, 1, 200}]; lst
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