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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.

User: Joseph Damico

Joseph Damico's wiki page.

Joseph Damico has authored 3 sequences.

A330138 List of pairs [j,k] ordered by increasing j and k according to the procedure described in the Comments.

Original entry on oeis.org

3, 4, 5, 15, 8, 17, 5, 12, 13, 35, 12, 37, 21, 20, 29, 7, 24, 25, 63, 16, 65, 45, 28, 53, 27, 36, 45, 9, 40, 41, 99, 20, 101, 77, 36, 85, 55, 48, 73, 33, 56, 65, 11, 60, 61, 143, 24, 145, 117, 44, 125, 91, 60, 109, 65, 72, 97, 39, 80, 89, 13, 84, 85
Offset: 1

Views

Author

Joseph Damico, Dec 02 2019

Keywords

Comments

Procedure: The sequence lists two indices, j and k. Let (a, b, and c) represent the two legs and the hypotenuse of a right-angled triangle. The required computations are as follows:
(1) compute 2*j^2 (call this p, it is c-a)
(2) compute (2*k-1)^2 (call this q, it is c-b)
(3) compute 2j*(2k-1) (call this r, it is a+b-c)
So c = p+q+r; a = c-p; b = c-q.
Most, but not all, of the Pythagorean triples generated here are primitive (i.e., they have sides that are relatively prime). In the first 105 Pythagorean triples, there are 14 that are not primitive. The first exception is the 9th in the list: 27,36,45, where j=3 and k=2.

Examples

			As the values of j and k increase, so do the lengths of the sides of the triangles. So it makes sense to order the triangles based on the sum (j+k).  When j=k=1, the sum (j+k)=2, and there is only one triangle (3,4,5).  When the sum of j and k is 3, there are two possibilities: j=1, k=2 (15,8,17) or j=2, k=1 (5,12,13).  When the sum is 4, there are three possibilities. Each time the sum (j+k) increases by one, the number of triangles also increases by one. And the total number of triangles grows according to sequence A000217 (the triangular numbers).
Value of j+k:    2, 3, 4,  5,  6,  7,  8,  9, 10, 11, 12, ...
Total triangles: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, ...
So it is clear that triangle number 61 would be in the range of triangles ranked 56 through 66, where (j+k)=12.  In fact, exactly 6 beyond the triangle associated with the values j=10, k=1.  So j=6, k=6 is associated with the sixty-first triangle. And the computations give us this Pythagorean triple: a=253, b=204, c=325.

Links

Crossrefs

Cf. A103606.

Programs

  • Mathematica
    Table[Map[{#2 + #3, #1 + #3, #1 + #2 + #3} & @@ {2 #1^2, (2 #2 - 1)^2, 2 #1 (2 #2 - 1)} & @@ {#, n - #} &, Range[n - 1]], {n, 7}] // Flatten (* Michael De Vlieger, Dec 14 2019 *)

A307677 a(0) = a(1) = a(2) = a(3) = 1; thereafter a(n) = a(n-1) + a(n-2) + a(n-4).

Original entry on oeis.org

1, 1, 1, 1, 3, 5, 9, 15, 27, 47, 83, 145, 255, 447, 785, 1377, 2417, 4241, 7443, 13061, 22921, 40223, 70587, 123871, 217379, 381473, 669439, 1174783, 2061601, 3617857, 6348897, 11141537, 19552035, 34311429, 60212361, 105665327, 185429723, 325406479, 571048563, 1002120369
Offset: 0

Views

Author

Joseph Damico, Apr 21 2019

Keywords

Comments

A079398, A103609, A003269, A306276, A126116, and A000288 are the other six sequences which have characteristic equations of the form x^4 = ax^3 + bx^2 + cx + 1 in which a, b, and c are equal to either 0 or 1 -- but not all three of them are equal to zero. (Each of those sequences begins with 1,1,1,1.)
A005251 has the same characteristic equation, and each successive term is determined by the same operation, namely, a(n) = a(n-1) + a(n-2) + a(n-4). However, it has different starting values: (0,1,1,1) instead of (1,1,1,1).
The characteristic equation of this sequence is x^4 = x^3 + x^2 + 1. Lim_{n->infinity} a(n+1)/a(n) = 1.754877666...

Links

Crossrefs

Cf. A000288, A003269, A005251, A005314, A079398, A103609, A126116, A306276.

Programs

  • Magma
    [n le 4 select 1 else Self(n-1) +Self(n-2) +Self(n-4): n in [1..51]]; // G. C. Greubel, Oct 23 2024
  • Mathematica
    LinearRecurrence[{1,1,0,1}, {1,1,1,1}, 51] (* G. C. Greubel, Oct 23 2024 *)
  • PARI
    Vec((1 - x^2 - x^3) / ((1 + x)*(1 - 2*x + x^2 - x^3)) + O(x^40)) \\ Colin Barker, Apr 25 2020
  • SageMath
    @CachedFunction # a = A307677
def a(n): return 1 if n<4 else a(n-1) +a(n-2) +a(n-3)
[a(n) for n in range(51)] # G. C. Greubel, Oct 23 2024

Formula

From Colin Barker, Apr 25 2020: (Start)
G.f.: (1 - x^2 - x^3) / ((1 + x)*(1 - 2*x + x^2 - x^3)).
a(n) = a(n-1) + a(n-2) + a(n-4) for n>3. (End)
a(n) = (1/5)*((-1)^n + 2*(2*A005314(n+1) - A005314(n) - 2*A005314(n-1))). - G. C. Greubel, Oct 23 2024

A306276 a(0) = a(1) = a(2) = a(3) = 1; thereafter a(n) = a(n-2) + a(n-3) + a(n-4).

Original entry on oeis.org

1, 1, 1, 1, 3, 3, 5, 7, 11, 15, 23, 33, 49, 71, 105, 153, 225, 329, 483, 707, 1037, 1519, 2227, 3263, 4783, 7009, 10273, 15055, 22065, 32337, 47393, 69457, 101795, 149187, 218645, 320439, 469627, 688271, 1008711, 1478337, 2166609, 3175319, 4653657, 6820265
Offset: 0

Views

Author

Joseph Damico, Feb 02 2019

Keywords

Comments

The characteristic equation of this sequence is x^4 = x^2 + x + 1. The characteristic equation of A000930 is x^3 = x^2 + 1 [1], which can be rewritten as x^4 = x^3 + x [2]. By substituting the value of x^3 from equation [1] in equation [2], we get x^4 = (x^2 + 1) + x, which is the characteristic equation for this sequence. Hence the ratio a(n+1)/a(n) has the same limit as the A000930 sequence does, about 1.465571231.

Links

Crossrefs

Cf. A000930, A092526.

Programs

  • Mathematica
    Nest[Append[#, Total@ #[[-4 ;; -2]] ] &, {1, 1, 1, 1}, 40] (* or *)
CoefficientList[Series[(x^3 - x - 1)/(x^4 + x^3 + x^2 - 1), {x, 0, 43}], x] (* Michael De Vlieger, Feb 09 2019 *)

Formula

G.f.: (x^3 - x - 1)/(x^4 + x^3 + x^2 - 1).
a(n) = a(n-2) + a(n-3) + a(n-4) for n >= 4, a(n) = 1 for n < 4.
Lim_{n->infinity} a(n+1)/a(n) = A092526.

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