A356701 Records values in A081119.
5, 7, 10, 16, 26, 32
Offset: 1
Examples
a(6) = 32 since A356699(6) = 1025, and the equation y^2 = x^3 + 1025 has 32 integral solutions.
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.
a(6) = 32 since A356699(6) = 1025, and the equation y^2 = x^3 + 1025 has 32 integral solutions.
m = 155; f[_List] := ( xm = 2 xm; ym = Ceiling[xm^(3/2)]; Complement[Range[m], Outer[Plus, Range[0, ym]^2, -Range[-xm, xm]^3] //Flatten //Union]); xm=10; FixedPoint[f, {}] (* Jean-François Alcover, Apr 28 2011 *)
a(4)=4 refers to (x,y) = (2,+-2) and (5,+-11).
(* This naive approach gives correct results up to n=1000 *) xmax[] = 10^4; Do[ xmax[n] = 10^5, {n, {366, 775, 999}}]; Do[ xmax[n] = 10^6, {n, {207, 307, 847}}]; f[n] := (x = Floor[n^(1/3)] - 1; s = {}; While[ x <= xmax[n], x++; y2 = x^3 - n; If[y2 >= 0, y = Sqrt[y2]; If[ IntegerQ[y], AppendTo[s, y]]]]; s); a[n_] := (fn = f[n]; If[fn == {}, 0, 2 Length[fn] - If[ First[fn] == 0, 1, 0]]); Table[ an = a[n]; Print["a[", n, "] = ", an]; an, {n, 1, 100}] (* Jean-François Alcover, Mar 06 2012 *)
(* Assuming every term is a cube *) xmax = 2000; r[n_] := Reap[ Do[ rpos = Reduce[y^2 == x^3 + n, y, Integers]; If[rpos =!= False, Sow[rpos]]; rneg = Reduce[y^2 == (-x)^3 + n, y, Integers]; If[rneg =!= False, Sow[rneg]], {x, 1, xmax}]]; ok[n_] := Which[ rn = r[n]; rn[[2]] === {}, False, Length[rn[[2]]] > 1, False, ! FreeQ[rn[[2, 1]], Or], False, True, True]; ok[n_ /; !IntegerQ[n^(1/3)]] = False; ok[1]=False; A179145 = Reap[ Do[ If[ok[n], Print[n]; Sow[n]], {n, 1, 320000}]][[2, 1]] (* Jean-François Alcover, Apr 12 2012 *)
From _Raphie Frank_, Sep 28 2012: (Start) 35*(528 - 15) + 0 = 17955 = a(6), 35*(4095 - 120) + 3 = 139128 = a(7), 35*(17955 - 528) + 15 = 609960 = a(8), 35*(139128 - 4095) + 120 = 4726275 = a(9). (End) From _Raphie Frank_, Feb 02 2013: (Start) a(7) = 139128 and a(9) = 4726275. a(9) = (2*(sqrt(8*a(7) + 1) - 1)/2 + 3*sqrt(a(7) + 1) + 1)^2 - 1 = (2*(sqrt(8*139128 + 1) - 1)/2 + 3*sqrt(139128 + 1) + 1)^2 - 1 = 4726275. a(9) = 1/2*((3*(sqrt(8*a(7) + 1) - 1)/2 + 4*sqrt(a(7) + 1) + 1)^2 + (3*(sqrt(8*a(7) + 1) - 1)/2 + 4*sqrt(a(7) + 1) + 1)) = 1/2*((3*(sqrt(8*139128 + 1) - 1)/2 + 4*sqrt(139128 + 1) + 1)^2 + (3*(sqrt(8*139128 + 1) - 1)/2 + 4*sqrt(139128 + 1) + 1)) = 4726275. (End) From _Vladimir Pletser_, Apr 30 2017: (Start) For n=2, b(n)=5, a(n)=15 For n=5, b(n)=90, a(n)= 4095 For n = 3, A006451(n) = 15. Therefore, A000217(A006451(n)) = A000217(15) = 120. (End)
I:=[0,3,15,120,528,4095]; [n le 6 select I[n] else 35*(Self(n-2) - Self(n-4)) + Self(n-6): n in [1..30]]; // Vincenzo Librandi, Dec 21 2015
A006454:=-3*z*(1+4*z+z**2)/(z-1)/(z**2-6*z+1)/(z**2+6*z+1); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation restart: bm2:=-1: bm1:=0: bp1:=2: bp2:=5: print ('0,0','1,3','2,15'); for n from 3 to 1000 do b:= 8*sqrt((bp1^2+bp1)/2+1)+bm2; a:=b*(b+1)/2; print(n,a); bm2:=bm1; bm1:=bp1; bp1:=bp2; bp2:=b; end do: # Vladimir Pletser, Apr 30 2017
Clear[a]; a[0] = a[1] = 1; a[2] = 2; a[3] = 4; a[n_] := 6a[n - 2] - a[n - 4]; Array[a, 40]^2 - 1 (* Vladimir Joseph Stephan Orlovsky, Mar 03 2011 *) LinearRecurrence[{1,34,-34,-1,1},{0,3,15,120,528},30] (* Harvey P. Dale, Feb 18 2023 *)
concat(0, Vec(3*x*(1 + 4*x + x^2) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)) + O(x^30))) \\ Colin Barker, Apr 30 2017
a(1) = A060951(27) = a(729) = 0. - _Jonathan Sondow_, Sep 10 2013
a(n) = ellanalyticrank(ellinit([0, 0, 0, 0, n]))[1] \\ Jianing Song, Aug 24 2022
apply( {A060950(n)=ellrank(ellinit([0, n]))[1]}, [1..99]) \\ For PARI version < 2.14, use ellanalyticrank(...). - M. F. Hasler, Jul 01 2024
a(12) <= 27*A031508(12) <= 27*6533891544658786928 = 176415071705787247056 (from Quer 1987 and Womack). - _Jonathan Sondow_, Sep 10 2013
{A031507(n)=for(k=1, oo, ellrank(ellinit([0, k]))[1]==n && return(k))} \\ Use ellanalyticrank() for PARI version < 2.14. - M. F. Hasler, Jul 01 2024
3 is a term since the equation y^2 = x^3 + 3^3 has no solution other than (-3,0).
y^2 = x^3 + 1 has solutions (y, x) = (0, -1), (1, 0) and (3, 2), hence a(1) = 3. y^2 = x^3 + 6 has no solutions, hence a(6) = 0. y^2 = x^3 + 17 has 8 solutions (see A029727, A029728), hence a(17) = 8. y^2 = x^3 + 27 has solution (y, x) = (0, -3), hence a(27) = 1.
[ #{ Abs(p[2]) : p in IntegralPoints(EllipticCurve([0, n])) }: n in [1..122] ];
(* This naive approach gives correct results up to n=1000 *) xmax[] = 10^4; Do[ xmax[n] = 10^5, {n, {297, 377, 427, 885, 899}}]; Do[ xmax[n] = 10^6, {n, {225, 353, 618}}]; f[n] := (x = -Ceiling[n^(1/3)] - 1; s = {}; While[x <= xmax[n], x++; y2 = x^3 + n; If[y2 >= 0, y = Sqrt[y2]; If[IntegerQ[y], AppendTo[s, y]]]]; s); a[n_] := a[n] = (fn = f[n]; an = If[fn == {}, 0, 2 Length[fn] - If[First[fn] == 0, 1, 0]]; If[EvenQ[an], an/2, (an + 1)/2]); Table[ Print["a[", n, "] = ", a[n] ]; a[n], {n, 1, 105}] (* Jean-François Alcover, Feb 20 2012 *) A081119 = Cases[Import["https://oeis.org/A081119/b081119.txt", "Table"], {, }][[All, 2]]; a[n_] := With[{an = A081119[[n]]}, If[EvenQ[an], an/2, (an + 1)/2]]; a /@ Range[10000] (* Jean-François Alcover, Nov 24 2019 *)
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