Morris Neene - cp's OEIS frontend

A258928 a(n) = number of integral points on the elliptic curve y^2 = x^3 - (n^2)*x + 1, considering only nonnegative values of y.

3, 6, 11, 9, 15, 13, 14, 17, 26, 12, 12, 11, 12, 19, 20, 11, 19, 36, 12, 17, 16, 11, 19, 16, 15, 27, 17, 17, 18, 16, 12, 15, 17, 11, 12, 11, 28, 16, 12, 11, 15, 24, 27, 11, 17, 12, 26, 15, 17, 15, 12, 15, 17, 27, 12, 14, 16, 15, 16, 24, 12, 41, 17, 16, 12, 11, 17, 16, 16, 15, 23, 15, 16, 20, 15

Offset: 0

Morris Neene, Jun 14 2015

nonn

For n>3, the number of integral points on y = x^3 - (n^2)*x + 1 is at least 11. These 11 points correspond to the solutions x = {-1, 0, n, -n, n + 2, -n + 2, n^2 - 1, n^2 - 2n + 2, n^2 + 2n + 2, n^4 + 2n, n^4 - 2n}.

			a(0) = 3 because the integer points on y^2 = x^3 + 1 are (-1, 0), (0, 1), and (2, 3).

Robert Israel, Table of n, a(n) for n = 0..209

Cf. A081119, A081120, A259191.

def f(n):
  R. = QQ[]
  E = EllipticCurve(y^2 - x^3 + n^2*x - 1)
  return len(E.integral_points(both_signs=false))
[f(x) for x in range(40)]  # Robert Israel, Apr 23 2021

More terms from Robert Israel, Apr 23 2021

A259048 u(1) = v(1) = 1, u(n) = u(n-1) + v(n-1), v(n) = u(n-1)^2 + v(n-1)^2, a(n) = u(n).

Original entry on oeis.org

1, 2, 4, 12, 92, 6636, 42839036, 1834614576635532, 3365810487617338033584723922844, 11328680238554850474377984661704304183660014108982249765031212

Offset: 1

Morris Neene, Jun 17 2015

nonn

			u(2) = 2, v(2) = 2; u(3) = 4, v(3) = 8; u(4) = 12, v(4) = 80; u(5) = 92, v(5) = 6544.

I:=[1,2]; [n le 2 select I[n] else Self(n-1)^2+Self(n-1)-2*Self(n-1)*Self(n-2)+2*Self(n-2)^2: n in [1..11]]; // Vincenzo Librandi, Jun 18 2015

RecurrenceTable[{x[n+ 2] == x[n+1]^2 + x[n+1] - 2*x[n+1]*x[n] + 2*x[n]^2, x[1] == 1, x[2] == 2 }, x, {n, 10}]

first(m)={my(u=vector(m),v=vector(m));v[1]=1;u[1]=1;for(i=2,m,u[i] = u[i-1] + v[i-1];v[i] = (u[i-1])^2 + (v[i-1])^2);u;} \\ Anders Hellström, Aug 20 2015

def main(size):
    u=[1]; v=[1]; a=[1]
    for i in range(1,size-1):
        u.append(u[i-1]+v[i-1])
        v.append(u[i-1]**2+v[i-1]**2)
        a.append(u[i])
    return a # Anders Hellström, Jul 10 2015

a(n) = a(n-1)^2 + a(n-1) - 2*a(n-1)*a(n-2) + 2*a(n-2)^2, a(1) = 1, a(2) = 2.

A258967 a(1)=1, a(2)=2, a(3)=3, a(n) = ceiling(sqrt(a(n-1)a(n-2)a(n-3))), n>3.

Original entry on oeis.org

1, 2, 3, 3, 5, 7, 11, 20, 40, 94, 275, 1017, 5128, 37871, 444415, 9290130, 395420005, 40404949540, 12183091294648, 13951642918891149, 82872169787001239679, 3753148776564192982863648, 2083123034674803589767277778237, 25454214863632278822694894280883452911

Offset: 1

Morris Neene, Jun 15 2015

nonn

			a(4) = ceiling(sqrt(1*2*3)) = 3;
a(5) = ceiling(sqrt(2*3*3)) = 5;
a(6) = ceiling(sqrt(3*3*5)) = 7.

Harvey P. Dale, Table of n, a(n) for n = 1..39

Cf. A258875, A254400.

I:=[1, 2, 3]; [n le 3 select I[n] else Ceiling(Sqrt(Self(n-1)*Self(n-2)*Self(n-3))): n in [1..23]];

RecurrenceTable[{a[n] == Ceiling[Sqrt[a[n - 1] a[n - 2] a[n - 3]]], a[1] == 1, a[2] == 2, a[3] == 3}, a, {n, 1, 23}] (* Michael De Vlieger, Jul 02 2015 *)
a[1] = 1; a[2] = 2; a[3] = 3; a[n_] := a[n] = Ceiling[ Sqrt[ a[n - 1]*a[n - 2]*a[n - 3]]]; Array[a, 23] (* Robert G. Wilson v, Aug 12 2015 *)
nxt[{a_,b_,c_}]:={b,c,Ceiling[Sqrt[a*b*c]]}; NestList[nxt,{1,2,3},30][[All,1]] (* Harvey P. Dale, Sep 09 2021 *)

first(m)={my(v=vector(m));v[1]=1;v[2]=2;v[3]=3;for(i=4,m,v[i]=ceil(sqrt(v[i-1]*v[i-2]*v[i-3])));v;} \\ Anders Hellström, Aug 20 2015

a(n) is approximately k^(c^n), where c is the real root of x^3 - (x^2 + x + 1)/2 = 0 equal to (1 + (64 - 3*sqrt(417))^(1/3) + (64 + 3*sqrt(417))^(1/3))/6, and k is approximately 1.7450496...

Corrected and extended by Harvey P. Dale, Sep 09 2021

A259191 Number of integral solutions to y^2 = x^3 + n*x^2 + n (with y nonnegative).

Original entry on oeis.org

3, 0, 0, 4, 1, 0, 0, 0, 2, 0, 0, 1, 1, 0, 0, 3, 1, 0, 0, 0, 0, 0, 0, 1, 8, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 6, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 6, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 2, 0

Offset: 1

Morris Neene, Jun 20 2015

nonn

If n is square there are at least two solutions, corresponding to x = 0 and x = -n. If n = 2^(2k) there are at least three solutions, corresponding to x = 0, x = -2^(2k), and x = 2^(6k-2) + 2^(2k). If n = 2k^2 + 2k, there is at least one solution, corresponding to x = 1.

Cf. A081119, A081120.

for i in range(1,31):
    E=EllipticCurve([0,i,0,0,i])
    print(len(E.integral_points()))

A259189 Semiprimes of the form n^3 + 2.

Original entry on oeis.org

10, 218, 514, 731, 1333, 2199, 2746, 3377, 4915, 5834, 6861, 8002, 9263, 12169, 15627, 29793, 35939, 42877, 54874, 59321, 68923, 117651, 125002, 132653, 148879, 185195, 205381, 314434, 405226, 421877, 474554, 531443, 592706, 658505, 704971

Offset: 1

Morris Neene, Jun 20 2015

nonn

Intersection of A001358 and A084380. - Michel Marcus, Jun 20 2015
Since there are no squares of the form n^3 + 2, all semiprimes in this sequence are products of distinct primes.
No term in A040034 divides any term in this sequence.

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000

Cf. A001358 (semiprimes), A084380 (n^3+2), A144953 (primes of same form).

Cf. A237040 (similar sequence with n^3+1).

IsSP:=func;[r:n in [1..1000]|IsSP(r) where r is 2+n^3];

Select[Range[100]^3 + 2, PrimeOmega[#] == 2 &] (* Alonso del Arte, Jun 20 2015 *)

is(n)=bigomega(n^3 + 2)==2 \\ Anders Hellström, Sep 07 2015

use ntheory ":all"; my @sp = grep { scalar(factor($))==2 } map { $**3+2 } 1..100; say "@sp"; # Dana Jacobsen, Sep 07 2015

A258911 a(1)=a(2)=1, a(n) = ceiling(Pi*a(n-1) - a(n-2)), n>2.

Original entry on oeis.org

1, 1, 3, 9, 26, 73, 204, 568, 1581, 4399, 12239, 34051, 94736, 263571, 733297, 2040150, 5676024, 15791606, 43934770, 122233545, 340073237, 946138039, 2632307076, 7323498533, 20375142114, 56686898249, 157712000980

Offset: 1

Morris Neene, Jun 14 2015

nonn

Ratio of consecutive terms approaches (Pi + sqrt(Pi^2 - 4))/2, or approximately 2.782159649779516149316 (A189039).

			a(3) = ceiling(Pi*1 - 1) = 3;
a(4) = ceiling(Pi*3 - 1) = 9;
a(5) = ceiling(Pi*9 - 3) = 26.

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A189039.

I:=[1,1]; [n le 2 select I[n] else Ceiling(pi*Self(n-1)-Self(n-2)): n in [1..200]];

RecurrenceTable[{a[n] == Ceiling[Pi*a[n - 1] - a[n - 2]], a[1] == 1,
  a[2] == 1}, a, {n,1,50}] (* G. C. Greubel, Jun 03 2017 *)
nxt[{a_,b_}]:={b,Ceiling[b*Pi-a]}; NestList[nxt,{1,1},30][[All,1]] (* Harvey P. Dale, Apr 02 2020 *)

main(size)={my(v=vector(size),i);v[1]=1;v[2]=1;for(i=3,size,v[i]=ceil(Pi*v[i-1]-v[i-2]));return(v);} /* Anders Hellström, Jul 14 2015 */

A258948 a(1)=1, a(2)=2; for n>2, a(n) = (1/2)a(n-1)a(n-2) + a(n-1) + a(n-2).

Original entry on oeis.org

1, 2, 4, 10, 34, 214, 3886, 419902, 816293374, 171382426877950, 69949169911638289022974, 5994029248777394614754727872037912574, 209638685189029793998133268981457005889853767752082771673086

Offset: 1

Morris Neene, Jun 15 2015

a(n) + 2 = (1/2)*(a(n-1) + 2)*(a(n-2) + 2), from which the general formula can be proved using the method shown in A063896.

			a(3) = (1/2)*2*1 + 2 + 1 = 4;
a(4) = (1/2)*4*2 + 4 + 2 = 10;
a(5) = (1/2)*10*4 + 10 + 4 = 34;
a(6) = 2*(3^3)(2^2) - 2 = 214.

Cf. A000045, A063896, A100701, A254132.

[n le 2 select n else Self(n-1)*Self(n-2)/2+Self(n-1)+Self(n-2): n in [1..13]];

[2*3^Fibonacci(n-2)*2^Fibonacci(n-3)-2: n in [1..20]]; // Vincenzo Librandi, Jun 17 2015

Table[2 3^Fibonacci[n-2] 2^Fibonacci[n-3] - 2, {n, 1, 20}] (* Vincenzo Librandi, Jun 17 2015 *)

a(n) = 2*(3^fibonacci(n-2))*(2^fibonacci(n-3)) - 2; \\ Michel Marcus, Jun 17 2015

a(n) = 2 * 3^A000045(n-2) * 2^A000045(n-3) - 2, where A000045(n) is the n-th Fibonacci number.

A258875 a(1) = a(2) = a(3) = 1; for n > 3, a(n) = ceiling((a(n-1) + a(n-2) + a(n-3))/2).

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 4, 5, 6, 8, 10, 12, 15, 19, 23, 29, 36, 44, 55, 68, 84, 104, 128, 158, 195, 241, 297, 367, 453, 559, 690, 851, 1050, 1296, 1599, 1973, 2434, 3003, 3705, 4571, 5640, 6958, 8585, 10592, 13068, 16123, 19892, 24542, 30279

Offset: 1

Morris Neene, Jun 13 2015

nonn

First 14 terms are the same as A179241.
Ratio of consecutive terms approaches the real root of x^3 - (x^2 + x + 1)/2 = 0, whose approximate value is 1.2337519285, and whose exact value is (1 + (64 - 3*sqrt(417))^(1/3) + (64 + 3*sqrt(417))^(1/3))/6.
Same as A180235 for n > 5. - Georg Fischer, Oct 09 2018

Vincenzo Librandi, Table of n, a(n) for n = 1..2000

[n le 3 select 1 else Ceiling((Self(n-1)+Self(n-2)+ Self(n-3))/2): n in [1..60]]; // Vincenzo Librandi, Oct 10 2018

a(4) = ceiling((1+1+1)/2) = 2;
a(5) = ceiling((1+1+2)/2) = 2;
a(6) = ceiling((1+2+2)/2) = 3.

RecurrenceTable[{a[n] == Ceiling[(a[n - 1] + a[n - 2] + a[n - 3])/2], a[1] == a[2] == a[3] == 1}, a, {n, 1, 49}] (* Michael De Vlieger, Jun 20 2015 *)
nxt[{a_,b_,c_}]:={b,c,Ceiling[(a+b+c)/2]}; NestList[nxt,{1,1,1},50][[All,1]] (* Harvey P. Dale, Feb 03 2022 *)

lista(nn) = {va = vector(nn, n, if (n<=3, 1)); for (n=4, nn, va[n] = ceil((va[n-1]+va[n-2]+va[n-3])/2);); va;} \\ Michel Marcus, Jun 17 2015

A258898 a(1)=a(2)=1, a(n) = ceiling(e*a(n-1) - a(n-2)) for n>2.

Original entry on oeis.org

1, 1, 2, 5, 12, 28, 65, 149, 341, 778, 1774, 4045, 9222, 21023, 47925, 109251, 249051, 567740, 1294227, 2950334, 6725613, 15331778, 34950481, 79673480, 181624492, 414033077, 943834098, 2151574001, 4904750412, 11180919918

Offset: 1

Morris Neene, Jun 14 2015

nonn

Ratio of consecutive terms approaches A189040, (e + sqrt(e^2 - 4))/2.

			a(2) = ceiling(e*1 - 1) = 2;
a(3) = ceiling(e*2 - 1) = 5;
a(4) = ceiling(e*5 - 2) = 12;
a(5) = ceiling(e*12 - 5) = 28.

Cf. A001113, A189040.

I:=[1,1]; [n le 2 select I[n] else Ceiling(Exp(1)*Self(n-1)-Self(n-2)): n in [1..200]];

a:= proc(n) option remember; `if`(n<3, 1,
      ceil(exp(1)*a(n-1)-a(n-2)))
    end:
seq(a(n), n=1..40);  # Alois P. Heinz, Jun 18 2015

nxt[{a_,b_}]:={b,Ceiling[E*b-a]}; NestList[nxt,{1,1},30][[All,1]] (* Harvey P. Dale, Dec 02 2017 *)

A258692 Integers n such that n(n + 2)(n + 4) + 1 is a perfect square.

Original entry on oeis.org

-4, -3, -2, 0, 1, 2, 8, 10, 18, 112, 1272

Offset: 1

Morris Neene, Jun 12 2015

This sequence is finite as there are finitely many integer solutions to the elliptic curve y^2 = x(x + 2)(x + 4) + 1 = x^3 + 6x^2 + 8x + 1. The x values of the integer solutions are {-4, -3, -2, 0, 1, 2, 8, 10, 18, 112, 1272}. This equation has more integer and natural number solutions than the equation that defines sequence A121234.

			1 * 3 * 5 + 1 = 16 = 4^2, so 4 is in the sequence.
2 * 4 * 6 + 1 = 49 = 7^2, so 2 is in the sequence.
3 * 5 * 7 + 1 = 106 = 2 * 53, so 3 is not in the sequence.

Cf. A121234.

P := PolynomialRing(Integers()); {x: x in Sort([ p[1] : p in IntegralPoints(EllipticCurve(n^3 + 6*n^2 + 8*n + 1)) ])};

Select[Range[-10, 100], IntegerQ[Sqrt[#(# + 2)(# + 4) + 1]] &] (* Alonso del Arte, Jun 12 2015 *)

[i[0] for i in EllipticCurve([0, 6, 0, 8, 1]).integral_points()] # Seiichi Manyama, Aug 26 2019

cp's OEIS Frontend

User: Morris Neene

A258928 a(n) = number of integral points on the elliptic curve y^2 = x^3 - (n^2)*x + 1, considering only nonnegative values of y.

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A259048 u(1) = v(1) = 1, u(n) = u(n-1) + v(n-1), v(n) = u(n-1)^2 + v(n-1)^2, a(n) = u(n).

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A258967 a(1)=1, a(2)=2, a(3)=3, a(n) = ceiling(sqrt(a(n-1)*a(n-2)*a(n-3))), n>3.

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A259191 Number of integral solutions to y^2 = x^3 + n*x^2 + n (with y nonnegative).

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A259189 Semiprimes of the form n^3 + 2.

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A258911 a(1)=a(2)=1, a(n) = ceiling(Pi*a(n-1) - a(n-2)), n>2.

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A258948 a(1)=1, a(2)=2; for n>2, a(n) = (1/2)*a(n-1)*a(n-2) + a(n-1) + a(n-2).

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A258967 a(1)=1, a(2)=2, a(3)=3, a(n) = ceiling(sqrt(a(n-1)a(n-2)a(n-3))), n>3.

A258948 a(1)=1, a(2)=2; for n>2, a(n) = (1/2)a(n-1)a(n-2) + a(n-1) + a(n-2).

A258692 Integers n such that n(n + 2)(n + 4) + 1 is a perfect square.