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 1-2 of 2 results.

A354337 a(n) is the integer w such that (L(2*n)^2, -L(2*n + 1)^2, w) is a primitive solution to the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 125, where L(n) is the n-th Lucas number (A000032).

Original entry on oeis.org

19, 149, 1039, 7139, 48949, 335519, 2299699, 15762389, 108037039, 740496899, 5075441269, 34787591999, 238437702739, 1634276327189, 11201496587599, 76776199786019, 526231901914549, 3606847113615839, 24721697893396339, 169445038140158549, 1161393569087713519
Offset: 1

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Author

XU Pingya, Jun 20 2022

Keywords

Comments

Subsequence of A017377.

Examples

			2*(L(4)^2)^3 + 2*(-L(5)^2)^3 + (149)^3 = 2*(49)^3 + 2*(-121)^3 + (149)^3 = 125, a(2) = 149.
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{7,-1},{19,149},21]-1 + LucasL[2*Range[21]-3]^2

Formula

a(n) = (125 - 2*A005248(n)^6 + 2*A002878(n)^6)^(1/3).
a(n) = Lucas(4*n+2) + Lucas(4n-1) - 3 = 2*A056914(n)-3 = 15*A092521(n) + A288913(n-1).
a(n) = 2*A081017(n) - 1.
a(n) = 10*A089508(n) + 9.
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: x*(19 - 3*x - x^2)/((1 - x)*(1 - 7*x + x^2)). - Stefano Spezia, Jun 22 2022

A356716 a(n) is the integer w such that (c(n)^2, -d(n)^2, -w) is a primitive solution to the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 11^3, where c(n) = F(n+2) + (-1)^n * F(n-3), d(n) = F(n+1) + (-1)^n * F(n-4) and F(n) is the n-th Fibonacci number (A000045).

Original entry on oeis.org

5, 19, 31, 101, 179, 655, 1189, 4451, 8111, 30469, 55555, 208799, 380741, 1431091, 2609599, 9808805, 17886419, 67230511, 122595301, 460804739, 840280655, 3158402629, 5759369251, 21648013631, 39475304069, 148377692755, 270567759199, 1016995835621, 1854499010291
Offset: 1

Views

Author

XU Pingya, Aug 24 2022

Keywords

Comments

Conjecture:
(i) For all k > 2, 2*x^3 + 2*y^3 + z^3 = A089270(k)^3 have primitive solutions form (c(n)^2, -d(n)^2, -w(n)) with d(n) = 3*d(n-2) - d(n-4), c(n) = d(n+2) - d(n) and w(n) = 8*w(n-2) - 8*w(n-4) + w(n-6).
(ii) This sequence is a subsequence of A089270.
From XU Pingya, Jun 07 2024: (Start)
Several positive examples of conjecture:
When A089270(4,5,6,7) = {19,29,31,41}, d(n) can be taken as:
(1/2) * (F(n+3) + (-1)^n * F(n-6));
((1-(-1)^n)/2) * (F(n+3) + F(n-4)) + ((1+(-1)^n)/2) * (F(n+3) - F(n-4));
((1-(-1)^n)/2) * (2*F(n-1) + 3*F(n-3)) + ((1+(-1)^n)/2) * (3*F(n-2) + 2*F(n-4));
and
((1-(-1)^n)/2) * (2*F(n+1) + F(n-5)) + ((1+(-1)^n)/2) * (F(n+2) + 2*F(n-4)).
When A089270(17) = 121, d(n) can be taken as d(1,2,3,4) = {-3,0,7,11}. (End)
From XU Pingya, Jul 17 2024: (Start)
Furthermore, we observe that if (x, y) (y < x/2) is the solution of the Diophantine equation x^2 + x * y - y^2 = A089270(k). Let
d(2*n-1) = x * F(2*n-2) - y * F(2*n-3), c(2*n-1) = d(2*n+1) - d(2*n-1);
d(2*n) = x * F(2*n-2) + y * F(2*n-1), c(2*n) = d(2*n+2) - d(2*n).
Then such c(n) and d(n) satisfy the conjecture. (End)

Examples

			For n=3, 2 * ((F(5) - F(0))^2)^3 + 2 * (-(F(4) - F(-1))^2)^3 + (-31)^3 = 2 * 25^3 - 2 * 4^3 - 31^3 = 1331, a(3) = 31.
		

Crossrefs

Programs

  • Mathematica
    Table[(-1331+2*((Fibonacci[n+2]+(-1)^n*Fibonacci[n-3]))^6-2*(Fibonacci[n+1]+(-1)^n*Fibonacci[n-4])^6)^(1/3), {n,28}]

Formula

a(n) = (-1331 + 2 * A237132(n)^6 - 2 * A228208(n-1)^6)^(1/3).
a(n) = ((1-(-1)^n)/2) * (-1 + 6 * Sum_{k=0..n-1} Fibonacci(4*k-1) + 14 * Sum_{k=0..n-2} Fibonacci(4*k+1)) + ((1+(-1)^n)/2) * (-1 + 6 * Sum_{k=0..n-1} Fibonacci(4*k-1) + 14 * Sum_{k=0..n-1} Fibonacci(4*k+1)).
a(n) = ((1-(-1)^n)/2) * (-1 + 6*A206351(n) + 14*A081016(n-2)) + ((1+(-1)^n)/2) * (-1 + 6*A206351(n) + 14*A081016(n-1)).
From Stefano Spezia, Aug 25 2022: (Start)
G.f.: x*(5 + 14*x - 23*x^2 - 28*x^3 - x^4)/((1 - x)*(1 - 3*x + x^2)*(1 + 3*x + x^2)).
a(n) = a(n-1) + 7*a(n-2) - 7*a(n-3) - a(n-4) + a(n-5) for n > 5. (End)
From XU Pingya, Jul 17 2024: (Start)
a(2*n-1) = (F(2*n) + F(2*n-2) + F(2*n-5))^2 + (F(2*n) + F(2*n-2) + F(2*n-5)) * (F(2*n-2) + F(2*n-4) + F(2*n-7)) - (F(2*n-2) + F(2*n-4) + F(2*n-7))^2;
a(2*n) = (F(2*n+2) + F(2*n-3))^2 + (F(2*n+2) + F(2*n-3)) * (F(2*n) + F(2*n-5)) - (F(2*n) + F(2*n-5))^2. (End)
Showing 1-2 of 2 results.