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User: Jose Aranda

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Jose Aranda has authored 5 sequences.

A372543 Least k such that the rank of the elliptic curve y^2 = x^3 - k^2*x + 1 is n, or -1 if no such k exists.

Original entry on oeis.org

0, 1, 2, 4, 8, 17, 61, 347, 3778, 11416

Offset: 0

Jose Aranda, Jul 04 2024

This family of curves quickly reaches a moderate value of rank with a relatively small parameter k.

By heuristic search (see links), a(10) <= 216493 and a(11) <= 1448203.

Anna Antoniewicz, On a family of elliptic curves, (2005) Iagellonicae Acta Mathematica, XLIII.
Jose Aranda, Non sequential search for upper bounds (PARI-GP Script)
Cecylia Bocovich, Elliptic Curves of High Rank, (2012) Macalester College, Science Honors Projects.

Cf. A194687, A309060, A309068, A309069, A309146, A309147, A309164, A309165.

a(n,startAt=0)=for(k=startAt, oo, my(t=ellrank(ellinit([-k^2, +1]))); if(t[2]n, warning("k=",k," has rank in ",t[1..2]); next); if(t[1]n, error("Cannot determine if a(",n,") is ",k," or larger; rank is in ",t[1..2])); return(k)) \\ Charles R Greathouse IV, Jul 08 2024

PARI
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\\ See Aranda link.
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A374926 Least k such that the rank of the elliptic curve y^2 = x^3 - x + k^2 is n, or -1 if no such k exists.

Original entry on oeis.org

1, 2, 5, 24, 113, 337, 6310, 78560, 423515, 765617

Offset: 1

Jose Aranda, Jul 24 2024

This family of curves quickly reaches a moderate value of rank with a relatively low "k" parameter. And is fully analyzed in Tadik's work (see link). Tadik finds 11 terms, a rank lower bound and shows the torsion group is always trivial. The evolution of the rank is shown in detail, finding that a(11) <= 1118245045.

I have sequentially checked the first 10 terms, thus proving that they are the least k for each rank.

			The curve C[1] = [-1,1^2] has rank one, with generator [1,-1].The rank of C[2] = [-1,2^2] is 2 because it has two generators:PARI> e=ellinit([-1,2^2] );ellgenerators(e) = [[-1, 2], [0, 2]].If k>1, the curve C[k] always has at least two generators: [0,k], [-1,k], then its minimum rank is two.

Ezra Brown and Bruce T. Myers, Elliptic Curves from Mordell to Diophantus and Back, Amer. Math. Monthly 109 (2002), 639-649.
Edward Vincent Eikenberg, Rational points on some families of Elliptic Curves, PhD thesis, University of Maryland, 2004.
Petra Tadik, The rank of certain subfamilies of the elliptic curve y^2 = x^3 -x +t^2, Ann. Math. Inform. 40 (2012), 145-153.

Cf. A194687, A309060, A309068, A309069.

A364171 a(n) = m is the least m = bc > a(n-1) such that (b+c)n = m-1 where 1 < b <= c < m.

Original entry on oeis.org

6, 21, 40, 105, 126, 301, 456, 657, 910, 1221, 1596, 2041, 2562, 3165, 3856, 4641, 5526, 6517, 7620, 8841, 10186, 11661, 13272, 15025, 16926, 18981, 21196, 23577, 26130, 28861, 31776, 34881, 38182, 41685, 45396, 49321, 53466, 57837, 62440, 67281, 72366, 77701

Offset: 1

Jose Aranda, Jul 12 2023

nonn

Each term is a representative of the class of numbers with quotient n.
A364169 is the smallest m = b*c without requiring an increasing sequence. Sometimes the present sequence is still that minimum, a(n) = A364169(n).
Also subsequence of A364202.
Is a(n) = A062158(n+1) + 1 for n >= 6? - Hugo Pfoertner, Jul 23 2023

			For n = 7, a(7) = 456 because it is the smallest number m > 301 = a(6) that has a pair of distinct proper divisors b = 8 and c = 57 that give b*c = 8*57 = 456 and (b+c)*n = (8 + 57)*7 = 456 - 1.

Cf. A062158, A364169, A364202.

f[kmin_, n_] := f[kmin, n] = Module[{k = kmin + 1}, While[PrimeQ[k] || !AnyTrue[Rest@ Divisors[k], #^2 <= k && k - 1 == (# + k/#)*n &], k++]; k]; Rest@ FoldList[f][Join[{5}, Range[50]]] (* Amiram Eldar, Jul 12 2023 *)

More terms from Amiram Eldar, Jul 12 2023

A364169 Smallest integer m = bc which satisfies (b + c)n = m - 1.

Original entry on oeis.org

6, 21, 40, 105, 126, 301, 204, 273, 550, 1221, 936, 697, 690, 3165, 2176, 4641, 1242, 1333, 4200, 8841, 1786, 3213, 2508, 15025, 9126, 18981, 3700, 6105, 13950, 3901, 3876, 4161, 6106, 5781, 23976, 49321, 8178, 6765, 32800, 67281, 6930, 18565, 7440, 11001, 49726, 8925, 9072, 26977

Offset: 1

Jose Aranda, Jul 12 2023

All terms have b,c > 1.
a(n) is the smallest of a certain n-class. The 1-class would be related to the numbers denoted as "pqrs" in A009112.
From David A. Corneth, Jul 19 2023: (Start)
n < min(b, c) <= 2*n.
Proof of n < min(b, c):
As m = b*c and (b + c)*n = m - 1 we have m - (m - 1) = 1 = b*c - (b + c)*n.
Solving 1 = b*c = (b + c)*n for c gives c = (n*b + 1) / (b - n) > 0.
As 0 < b, c, n we have b - n > 0 so b > n.
Similarily as b = (n*c + 1)/(c - n) we have c > n.
Proof of min(b, c) <= 2*n by contradiction.
Suppose 2*n < min(b, c). Then 2*n + 1 <= min(b, c) as both b and c are integers.
Let b = 2n + b' and c = 2n + c' where 1 <= b', c', n. Then
1 = b*c - (b + c)*n = (2n + c') * (2n + b') - (2n + c' + 2n + b') * n = (b' + c')*n + c'*b' > 1. A contradiction. Q.e.d. (End)

			a(1) = 6 as 6 = 2*3, with (2 + 3)*1 = 6 - 1.
a(2) = 21 as 21 = 3*7, with (3 + 7)*2 = 21 - 1.
a(6) = 301 as 301 = 7*43, with (7 + 43)*6 = 301 - 1.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A009112 ("1-pqrs" numbers), A364171 (increasing m).

f:= proc(n) local t,d,b,c,m;
   d:= max(select(`<=`,numtheory:-divisors(n^2+1),n));
   b:= d+n;
   c:= (n^2+1)/d + n;
   b*c
end proc:
map(f, [$1..100]); # Robert Israel, Jul 19 2023

seq[max_] := Module[{len = Floor[Sqrt[max]/2], s, r}, s = Table[max + 1, {len}]; Do[r = (b*c - 1)/(b + c); If[IntegerQ[r] && r <= len && b*c < s[[r]], s[[r]] = b*c], {b, 2, max}, {c, 2, max/b}]; TakeWhile[s, # <= max &]]; seq[70000] (* Amiram Eldar, Jul 12 2023 *)

a(n) = for (x=1, oo, my(d=divisors(x)); for (i=1, #d\2, b = d[i]; c = x/d[i]; if ((b+c)*n == (x-1), return(x)););); \\ Michel Marcus, Jul 12 2023

a(n) = {forstep(i = 1, oo, n, if(iscan(i, n), return(i)))}
iscan(c, n) = {D = (1 - c)^2 - 4*n^2*c; if(!issquare(D), return(0)); b = ((c - 1) + sqrtint((1-c)^2 - 4*n^2*c)) / (2*n); if(denominator(b) == 1, return(1))} \\ David A. Corneth, Jul 12 2023

a(n) = {res = oo; for(b = n+1, 2*n, c = (n*b + 1)/(b - n); if(denominator(c) == 1, res = min(res, b*c))); res} \\ David A. Corneth, Jul 19 2023

a(n) = my(d = divisors(n^2 + 1), t = d[#d \ 2], b = t+n, c = (n^2 + 1)/t + n); return(b*c) \\ David A. Corneth, Jul 20 2023, adapted from Robert Israel, Jul 19 2023

More terms from Michel Marcus, Jul 12 2023

A364202 Integers m which can be written as m = pq = rs, with 1 <= r < p < q < s <= m and satisfying (p+q) | (s-r).

Original entry on oeis.org

6, 21, 24, 30, 40, 52, 54, 60, 72, 84, 96, 105, 120, 126, 150, 154, 160, 165, 180, 186, 189, 204, 208, 210, 216, 240, 270, 273, 288, 294, 300, 301, 312, 322, 330, 336, 342, 357, 360, 378, 384, 414, 420, 456, 468, 480, 486, 504, 525, 540, 546, 550, 594, 600

Offset: 1

Jose Aranda, Jul 13 2023

nonn

Terms may have multiple solutions p,q,r,s, and each has a least quotient k = (s-r) / (p+q).
Those with k=1 are the congruent numbers (A003273) and others are a more general case.
They all share a simple inter-square characterization. The 4 squares are A = (q-p)^2, B = (p+q)^2, C = ((p+q)*k)^2 and D = (r+s)^2. We have B = A + 4m, C = B*(k^2) and D = C + 4m, where 4m is added exclusively to avoid the use of fractions.

			21 is a term since 21 = 3*7 = 1*21 which has 3+7 = 10 divides 21-1 = 20 (k=2).
So there are 4 squares, in this case, 16, 100, 400 and 484, which are related by this number. In effect, 4*21=+84 jumps from the first to the second, which, multiplied by k^2, gives the third, where +84 gives the fourth.

Jose Aranda, Table of n, a(n) for n = 1..9999

Cf. A003273 (congruent numbers).

isok(k) = my(d=divisors(k)); if (#d >= 4, for (i=1, #d-1, my(r = d[i], s = k/r); if (rMichel Marcus, Jul 17 2023

More terms from Alois P. Heinz, Jul 13 2023

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User: Jose Aranda

A372543 Least k such that the rank of the elliptic curve y^2 = x^3 - k^2*x + 1 is n, or -1 if no such k exists.

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A374926 Least k such that the rank of the elliptic curve y^2 = x^3 - x + k^2 is n, or -1 if no such k exists.

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A364171 a(n) = m is the least m = b*c > a(n-1) such that (b+c)*n = m-1 where 1 < b <= c < m.

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A364169 Smallest integer m = b*c which satisfies (b + c)*n = m - 1.

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A364202 Integers m which can be written as m = p*q = r*s, with 1 <= r < p < q < s <= m and satisfying (p+q) | (s-r).

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A364171 a(n) = m is the least m = bc > a(n-1) such that (b+c)n = m-1 where 1 < b <= c < m.

A364169 Smallest integer m = bc which satisfies (b + c)n = m - 1.

A364202 Integers m which can be written as m = pq = rs, with 1 <= r < p < q < s <= m and satisfying (p+q) | (s-r).