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 1-3 of 3 results.

A242703 Decimal expansion of sqrt(7)/2.

Original entry on oeis.org

1, 3, 2, 2, 8, 7, 5, 6, 5, 5, 5, 3, 2, 2, 9, 5, 2, 9, 5, 2, 5, 0, 8, 0, 7, 8, 7, 6, 8, 1, 9, 6, 3, 0, 2, 1, 2, 8, 5, 5, 1, 2, 9, 5, 9, 1, 5, 4, 1, 2, 2, 5, 0, 9, 0, 1, 8, 4, 1, 6, 7, 2, 2, 9, 6, 0, 0, 5, 3, 4, 4, 1, 1, 6, 1, 5, 1, 4, 1, 8, 1, 3, 8, 8, 0, 1, 9, 6, 4, 4, 3, 2, 3, 7, 2, 7
Offset: 1

Views

Author

Alonso del Arte, May 20 2014

Keywords

Comments

Absolute value of the imaginary part of any of the nontrivial divisors of 2 in O_Q(sqrt(-7)).
The incircle of a triangle with sides of lengths 4, 5, 6 units respectively has a radius of sqrt(7)/2.
With a different offset, decimal expansion of 5 * sqrt(7).
From Wolfdieter Lang, Nov 18 2017: (Start)
In a regular hexagon inscribed in a circle with a radius of 1 unit the three distinct distances between any vertex and the middle of the sides are 1/2, sqrt(7)/2 and sqrt(13)/2.
The continued fraction expansion of sqrt(7)/2 is 1, repeat(3, 10, 3, 2). The convergents are given in A294972/A294973. (End)

Examples

			1.32287565553229529525...

Links

Crossrefs

Cf. A010527, A020837, A040022, A294972/A294973.

Programs

  • Mathematica
    RealDigits[Sqrt[7]/2, 10, 100][[1]]
  • PARI
    { default(realprecision, 20080); x=sqrt(7)/2; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b242703.txt", n, " ", d)); } \\ Iain Fox, Nov 18 2017

Formula

(1/2 - sqrt(-7)/2)(1/2 + sqrt(-7)/2) = 2.
Equals A010465/2. - R. J. Mathar, Feb 23 2021

A294973 Denominators of the continued fraction convergents to sqrt(7)/2.

Original entry on oeis.org

1, 3, 31, 96, 223, 765, 7873, 24384, 56641, 194307, 1999711, 6193440, 14386591, 49353213, 507918721, 1573109376, 3654137473, 12535521795, 129009355423, 399563588064, 928136531551, 3183973182717, 32767868358721, 101487578258880, 235743024876481, 808716652888323, 8322909553759711, 25777445314167456, 59877800182094623, 205410845860451325
Offset: 0

Views

Author

Wolfdieter Lang, Nov 18 2017

Keywords

Comments

The numerators are given in A294972.
The continued fraction of sqrt(7)/2 is [1, repeat(3,10,3,2)].

Links

Crossrefs

Cf. A242703, A294972.

Programs

  • Mathematica
    Denominator[Convergents[Sqrt[7]/2, 30]] (* Vaclav Kotesovec, Nov 19 2017 *)
  • PARI
    Vec((1 + 3*x - x^2)*(1 + 32*x^2 + x^4) / ((1 - 16*x^2 + x^4)*(1 + 16*x^2 + x^4)) + O(x^40)) \\ Colin Barker, Nov 21 2017

Formula

From Colin Barker, Nov 19 2017: (Start)
G.f.: (1 + 3*x - x^2)*(1 + 32*x^2 + x^4) / ((1 - 16*x^2 + x^4)*(1 + 16*x^2 + x^4)).
a(n) = 254*a(n-4) - a(n-8) for n > 7.
(End)
The g.f. is correct: denominator recurrence a(n) = b(n)*a(n-1) + a(n-2), a(-1) = 0, a(0) = 1, (a(-2) = a(0) = 1) with b(n) modulo 4 from the continued fraction given above: b(0) = 1, b(4*(k+1)) = 2, b(4*k+1) = 3 = b(4*k+3) and b(4*k+2) = 10, for k >= 0. The 4-section is G(x) = Sum_{k>=0} a(k)*x^k = G_0(x^4) + x*G_1(x^4) + x^2*G_2(x^4) + x^3*G_3(x^4) with G_j(x) = Sum_{k>=0} a(4*k+j)*x^k, for j=0..3. The recurrence leads to four equations (omit the x here): G_1 = 3*G_0 + x*G_3, G_2 = 10*G_1 + G_4, G_3 = 3*G_2 + G_1, G_0 = 2*x*G_3 +1 + x*G_2 (using a(-2) = 1). This can be solved to obtain for G(x) = (1 + 3*x + 31*x^2 + 96*x^3 - 31*x^4 + 3*x^5 - x^6)/(1 - 254*x^4 + x^8), and the numerator and denominator factorize like given in the above conjecture. - Wolfdieter Lang, Nov 19 2017

A295331 Numerators of continued fraction convergents to sqrt(13)/2 = A295330.

Original entry on oeis.org

1, 2, 9, 128, 521, 649, 1819, 2468, 11691, 166142, 676259, 842401, 2361061, 3203462, 15174909, 215652188, 877783661, 1093435849, 3064655359, 4158091208, 19697020191, 279916373882, 1139362515719, 1419278889601, 3977920294921, 5397199184522, 25566717033009, 363331237646648, 1478891667619601, 1842222905266249, 5163337478152099, 7005560383418348, 33185579011825491
Offset: 0

Views

Author

Wolfdieter Lang, Nov 20 2017

Keywords

Comments

The denominators are given in A295332.
The continued fraction expansion of sqrt(13)/2 is [1, repeat(1, 4, 14, 4, 1, 2)] = [b(0), repeat(b(1), b(2), b(3), b(4), b(5), b(6))].

Examples

			The convergents a(n)/A295332 begin: 1, 2, 9/5, 128/71, 521/289, 649/360, 1819/1009, 2468/1369, 11691/6485, 166142/92159, 676259/375121, 842401/467280, 2361061/1309681, 3203462/1776961, 15174909/8417525, 215652188/119622311, 877783661/486906769, 1093435849/606529080, ...

Links

Crossrefs

Cf. A294972/A294973 (sqrt(7/2)), A295330, A295332.

Programs

  • Mathematica
    Numerator[Convergents[Sqrt[13]/2,40]] (* or *) LinearRecurrence[ {0,0,0,0,0,1298,0,0,0,0,0,-1},{1,2,9,128,521,649,1819,2468,11691,166142,676259,842401},40] (* Harvey P. Dale, Jan 23 2019 *)

Formula

G.f.: (1 + 2*x + 9*x^2 + 128*x^3 + 521*x^4 + 649*x^5 + 521*x^6 - 128*x^7 + 9*x^8 + 2*x^9 + x^10 - x^11) / ((1 - 3*x - x^2)*(1 + 3*x - x^2)*(1 + 3*x + 10*x^2 - 3*x^3 + x^4)*(1 - 3*x + 10*x^2 + 3*x^3 + x^4)). For the derivation see the period 4 example for the denominators of the convergents of sqrt(7)/2 given in A294972. Here the period is 6 and the input for the recurrence a(n) = b(n)*a(n-1) + a(n-2) is a(-1) = 1 = a(0) (a(-2) = 1 - b(0) = 0) with the b(n) modulo 6 given above. Note that a(6*k) = 2*a(6*k-1) + a(6*k-2) is valid only for k >= 1 because a(0) = 1 is not equal to 2*a(-1) + a(-2) = 2. The denominator in the unfactorized form is 1 - 1298*x^6 + x^12.
a(n) = 1298*a(n-6) - a(n-12), n >= 12, with inputs a(0)..a(11).
Showing 1-3 of 3 results.

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