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 1-10 of 15 results. Next

A307110 Index of matching grid point in the bijection between two infinite quadratic grids with one grid rotated by Pi/4 around the common point (0,0), using an enumeration of the grid points by A305575 and A305576.

Original entry on oeis.org

0, 1, 6, 3, 8, 2, 11, 4, 9, 5, 15, 7, 19, 14, 10, 16, 17, 18, 12, 20, 13, 26, 27, 28, 25, 21, 22, 23, 24, 38, 31, 40, 33, 42, 35, 44, 29, 30, 51, 32, 53, 34, 55, 36, 49, 57, 58, 59, 60, 62, 39, 64, 41, 66, 43, 68, 37, 46, 47, 48, 45, 50, 63, 52, 65, 54, 67
Offset: 0

Views

Author

Hugo Pfoertner, following a proposal by Rainer Rosenthal, Mar 28 2019

Keywords

Comments

In a discussion in the newsgroup de.sci.mathematik, Klaus Nagel (see links) described a bijection P: G -> H between the grid points of two Cartesian grids G{Z X Z} and H{Z X Z} rotated against each other by Pi/4 around the only common point (0,0). This is a variation of the marriage problem asking for a matching in the infinite bipartite graph of the vertices of G U H with small distance d=|P(g)-g| for all points g in G.
Points within the grids are addressed by (i,j) in grid G and by (k,m) in grid H.
The plane is divided into horizontal strips of width cos(Pi/8) = sqrt(sqrt(2)+2)/2, with the x-axis as centerline of strip 0. Grid G is rotated by Pi/8, grid H by -Pi/8.
Assuming proper boundary conditions, there is exactly one grid point of G per grid line i=const and one grid point of grid H per grid line k=const inside each strip.
The intersections of the grid lines i=const from the rotated grid G and of lines k=const from the rotated grid H with the centerline of the strip are determined. The grid points inside the strip are paired such that the distance of the intersection points of lines i=const of grid G and of lines k=const of grid H with the strip centerline is minimized.
This bijection achieves a maximum of all mutual Euclidean distances of all pairs of cos(Pi/8)=0.9238795... (the strip width).
It is conjectured that the least possible maximum distance within pairs can be reduced to sqrt(5)*sin(Pi/8)=0.855706... (A386241), but not further, and that this can be achieved by "local repairs" of the result of the strip bijection, i.e. by reassigning the connections in groups of 4 pairs, one of which being the pair with d>0.8557... and 3 pairs in the vicinity of the violating pair, but potentially addressing points in neighbor strips. The conjecture is supported by extensive numerical results, but an announced proof by Klaus Nagel remained unpublished.
For the current sequence no repair is applied. The first repairs are required beyond i^2+j^2=40. The affected sequence terms for n>=124 are visible in the b-file of A307731.
The results of the matching are shown by enumerating the grid points of grid G according to the sequence pair A305575(n) for i and A305576(n) for j.
After finding the indices of the bijection partners (k,m) in grid H using Klaus Nagel's method, the position L where A305575(L)=k and A305576(L)=m is determined by table lookups, and the unique result is a(n)=L.
The sequence is a permutation of the natural numbers.

Examples

			The following table shows the first few matched pairs of grid points:
    Grid G     Grid H      Grid H rotated
   n  i  j  a(n) k  m  (k,m) rotated by -Pi/4  distance of matched points
   0  0  0    0  0  0    0.000000  0.000000   0.000000
   1  1  0    1  1  0    0.707107 -0.707107   0.765367
   2  0  1    6 -1  1    0.000000  1.414214   0.414214
   3 -1  0    3 -1  0   -0.707107  0.707107   0.765367
   4  0 -1    8  1 -1    0.000000 -1.414214   0.414214
   5  1  1    2  0  1    0.707107  0.707107   0.414214
   6 -1  1   11 -2  0   -1.414214  1.414214   0.585786
   7 -1 -1    4  0 -1   -0.707107 -0.707107   0.414214
   8  1 -1    9  2  0    1.414214 -1.414214   0.585786
   9  2  0    5  1  1    1.414214  0.000000   0.585786
  10  0  2   15 -1  2    0.707107  2.121320   0.717439
  11 -2  0    7 -1 -1   -1.414214  0.000000   0.585786
  12  0 -2   19  1 -2   -0.707107 -2.121320   0.717439
  13  2  1   14  1  2    2.121320  0.707107   0.317025
		

Crossrefs

Programs

  • PARI
    /* It is assumed that the files a305575 and a305576 contain the second column of the corresponding b-files */
    a305575=readvec(a305575); a305576=readvec(a305576);
    p(i,j)={my(C=cos(Pi/8),S=sin(Pi/8),T=S/C,gx=i*C-j*S,gy=i*S+j*C,k,xm,ym,v=[0,0]);
    k=round(gy/C); ym=C*k; xm=gx+(gy-ym)*T;
      v[1]=round((xm-ym*T)*C);  v[2]=round((ym+v[1]*S)/C);  v}
    findpos(v)={for(k=1,#a305575,if(v[1]==a305575[k]&&v[2]==a305576[k],return(k-1)))}
    for(n=1,67,print1(findpos(p(a305575[n],a305576[n])),", "))

A047621 Numbers that are congruent to {3, 5} mod 8.

Original entry on oeis.org

3, 5, 11, 13, 19, 21, 27, 29, 35, 37, 43, 45, 51, 53, 59, 61, 67, 69, 75, 77, 83, 85, 91, 93, 99, 101, 107, 109, 115, 117, 123, 125, 131, 133, 139, 141, 147, 149, 155, 157, 163, 165, 171, 173, 179, 181, 187, 189, 195, 197, 203, 205, 211, 213, 219, 221, 227, 229
Offset: 1

Views

Author

Keywords

Comments

Numbers k for which Jacobi symbol J(2,k) = -1, so 2 (as well as 2^k) is not a square mod k. - Antti Karttunen, Aug 27 2005, corrected by Jianing Song, Nov 05 2019, see also A329095.
Numbers n whose multiplicative order modulo 2^k is 2^(k - 2) for k >= 4. For k = 3, the numbers whose multiplicative order modulo 8 is 2 are in sequence A047484. - Jianing Song, Apr 29 2018

Crossrefs

Row 1 of A112070. Complement of A047522 relative to A005408. Primes in this sequence: A003629.
Subsequence of A329095.

Programs

  • GAP
    a:=[3];; for n in [2..60] do a[n]:=8*n-a[n-1]-8; od; a; # Muniru A Asiru, Dec 04 2018
  • Haskell
    a047621 n = a047621_list !! (n-1)
    a047621_list = 3 : 5 : map (+ 8) a047621_list
    -- Reinhard Zumkeller, Jul 05 2013
    
  • Mathematica
    LinearRecurrence[{1, 1, -1}, {3, 5, 11}, 100] (* Jean-François Alcover, Jul 31 2018 *)

Formula

a(n) = 8*n - a(n-1) - 8 (with a(1) = 3). - Vincenzo Librandi, Aug 06 2010
G.f.: x*(3 + 2*x + 3*x^2) / ( (1 + x)*(x - 1)^2 ). - R. J. Mathar, Oct 08 2011
A089911(3*a(n)) = 10. - Reinhard Zumkeller, Jul 05 2013
a(n) = 8*floor((n - 1)/2) + 4 + (-1)^n. - Gary Detlefs, Dec 03 2018
From Franck Maminirina Ramaharo, Dec 03 2018: (Start)
a(n) = 4*n - 2 - (-1)^n.
E.g.f.: 3 - (2 - 4*x)*exp(x) - exp(-x). (End)
a(n + 2) = a(n) + 8. - David A. Corneth, Dec 03 2018
Sum_{n>=1} (-1)^(n+1)/a(n) = (sqrt(2)-1)*Pi/8. - Amiram Eldar, Dec 11 2021
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sec(Pi/8) (1/A144981).
Product_{n>=1} (1 + (-1)^n/a(n)) = 2*sin(Pi/8) (A101464). (End)

A179260 Decimal expansion of the connective constant of the honeycomb lattice.

Original entry on oeis.org

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

Views

Author

Jonathan Vos Post, Jul 06 2010

Keywords

Comments

This is the case n=8 of the ratio Gamma(1/n)*Gamma((n-1)/n)/(Gamma(2/n)*Gamma((n-2)/n)). - Bruno Berselli, Dec 13 2012
An algebraic integer of degree 4: largest root of x^4 - 4x^2 + 2. - Charles R Greathouse IV, Nov 05 2014
This number is also the length ratio of the shortest diagonal (not counting the side) of the octagon and the side. This ratio is A121601 for the longest diagonal. - Wolfdieter Lang, May 11 2017 [corrected Oct 28 2020]
From Wolfdieter Lang, Apr 29 2018: (Start)
This constant appears in a historic problem posed by Adriaan van Roomen (Adrianus Romanus) in his Ideae mathematicae from 1593, solved by Viète. See the Havil reference, problem 3, pp. 69-74. See also the comments in A302711 with the Romanus link and his Exemplum tertium.
This problem is equivalent to R(45, 2*sin(Pi/120)) = 2*sin(3*Pi/8) with a special case of monic Chebyshev polynomials of the first kind, named R, given in A127672. For the constant 2*sin(Pi/120) see A302715. (End)

Examples

			1.84775906502257351225636637879357657364483325172728497223019546256107001500...
		

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 5.10, p. 333.
  • Julian Havil, The Irrationals, A Story of the Numbers You Can't Count On, Princeton University Press, Princeton and Oxford, 2012, pp. 69-74.
  • Neal Madras and Gordon Slade, Self-avoiding walks, Probability and its Applications, Birkhäuser Boston, Inc. Boston, MA, 1993.

Crossrefs

Programs

Formula

sqrt(2+sqrt(2)) = (2/1)(6/7)(10/9)(14/15)(18/17)(22/23)... (see Sondow-Yi 2010).
Equals 1/A154739. - R. J. Mathar, Jul 11 2010
Equals 2*A144981. - Paul Muljadi, Aug 23 2010
log (A001668(n)) ~ n log k where k = sqrt(2+sqrt(2)). - Charles R Greathouse IV, Nov 08 2013
2*cos(Pi/8) = sqrt(2+sqrt(2)). See a remark on the smallest diagonal in the octagon above. - Wolfdieter Lang, May 11 2017
Equals also 2*sin(3*Pi/8). See the comment on van Roomen's third problem above. - Wolfdieter Lang, Apr 29 2018
Equals i^(1/4) + i^(-1/4). - Gary W. Adamson, Jul 06 2022
Equals Product_{k>=0} ((8*k + 2)*(8*k + 6))/((8*k + 1)*(8*k + 7)). - Antonio Graciá Llorente, Feb 24 2024
Equals Product_{k>=1} (1 - (-1)^k/A047522(k)). - Amiram Eldar, Nov 22 2024

A182168 Decimal expansion of imaginary part of i^(1/4).

Original entry on oeis.org

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

Views

Author

Stanislav Sykora, May 16 2012

Keywords

Comments

Also sin(Pi/8) or sine of 22.5 degrees.
The real part of i^(1/4) or cos(Pi/8) is A144981.
A quartic number of denominator 2 and minimal polynomial 8*x^4 - 8*x^2 + 1. - Charles R Greathouse IV, Jan 09 2022

Examples

			0.382683432365089771728459984...
		

Crossrefs

Cf. A144981.

Programs

Formula

Equals sqrt(2-sqrt(2)) / 2 = A101464/2. - Bernard Schott, Apr 12 2022
This^2 + A144981^2=1. - R. J. Mathar, Aug 31 2025
Smallest positive root of 8*x^4-8*x^2+1=0. - R. J. Mathar, Aug 31 2025

A188582 Decimal expansion of sqrt(2) - 1.

Original entry on oeis.org

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

Views

Author

Robert G. Wilson v, Apr 04 2011

Keywords

Comments

"In his Book 'The Theory of Poker,' David Sklansky coined the phrase 'Fundamental Theorem of Poker,' a tongue-in-cheek reference to the Fundamental Theorem of Algebra and Fundamental Theorem of Calculus from introductory texts on those two subjects. The constant [sqrt(2) - 1] appears so often in poker analysis that we will in the same vein go so far as to call it 'the golden mean of poker,' and we call it 'r' for short. We will see this value in a number of important results throughout this book." [Chen and Ankenman]
If a triangle has sides whose lengths form a harmonic progression in the ratio 1/(1 - d) : 1 : 1/(1 + d) then the triangle inequality condition requires that d be in the range 1 - sqrt(2) < d < sqrt(2) - 1. - Frank M Jackson, Oct 01 2013
This constant is the 6th smallest radius r < 1 for which a compact packing of the plane exists, with disks of radius 1 and r. - Jean-François Alcover, Sep 02 2014, after Steven Finch
This constant is also the largest argument of the arctangent function in the Viète-like formula for Pi given by Pi/2^(k+1) = arctan(sqrt(2 - a_(k-1))/a_k), where the index k >= 2 and the nested radicals are defined by recurrence using the relations a_k = sqrt(2 + a_(k-1)), a_1 = sqrt(2). When k = 2 the argument of the arctangent function sqrt(2 - a_1)/a_2 = sqrt(2 - sqrt(2))/sqrt(2 + sqrt(2)) = sqrt(2) - 1 is largest. Consequently, at k = 2 the Viète-like formula for Pi can be written as Pi/8 = arctan(sqrt(2 - sqrt(2))/sqrt(2 + sqrt(2))) = arctan(sqrt(2) - 1) (after Abrarov-Quine, see the article). - Sanjar Abrarov, Jan 07 2017
If r and R are respectively the inradius and the circumradius of a triangle, then the ratio r/R <= 1/2 (Euler inequality), and this maximum value 1/2 is obtained when the triangle is equilateral. Now, for a right triangle, the ratio r/R <= this constant = sqrt(2) - 1, and this maximum value sqrt(2) - 1 is obtained when the right triangle is isosceles. This is the answer to the question 1 of the Olympiade Mathématique Belge Maxi in 2008. - Bernard Schott, Sep 07 2022

Examples

			0.414213562373095048801688724209698078569671875376948073...
		

References

  • B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, p. 140, Entry 25.
  • Bill Chen and Jerrod Ankenman, The Mathematics of Poker, Chpt 14 - You Don't Have To Guess: No-Limit Bet Sizing, p. 153, ConJelCo, LLC, Pittsburgh PA 2006.
  • Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, pp. 396 and 486.

Crossrefs

Cf. A002193, A014176, A020807, A120731, A182168 (sin(Pi/8)), A144981 (cos(Pi/8)).

Programs

  • Magma
    Sqrt(2) - 1; // G. C. Greubel, Jan 31 2018
  • Mathematica
    RealDigits[ Sqrt[2] - 1, 10, 111][[1]]
  • PARI
    sqrt(2) - 1 \\ G. C. Greubel, Jan 31 2018
    

Formula

Equals exp(asinh(cos(Pi))) = exp(asinh(-1)). - Geoffrey Caveney, Apr 23 2014
Equals tan(Pi/8) = A182168 / A144981 = 1 / A014176. - Bernard Schott, Apr 12 2022
From Antonio Graciá Llorente, Mar 15 2024: (Start)
Equals Product_{k >= 0} ((8*k - 1)*(8*k + 9))/((8*k - 5)*(8*k + 13)).
Equals Product_{k >= 1} A047554(k)/A047447(k). (End)
From Peter Bala, Mar 24 2024: (Start)
An infinite family of continued fraction expansions for this constant can be obtained from Berndt, Entry 25, by setting n = 1/2 and x = 8*k + 2 for k >= 0.
For example, taking k = 0 and k = 1 yields
Equals 1/(2 + (1*3)/(4 + (5*7)/(4 + (9*11)/(4 + (13*15)/(4 + ... + (4*n + 1)*(4*n + 3)/(4 + ...)))))) and
Equals (21/5) * 1/(10 + (1*3)/(20 + (5*7)/(20 + (9*11)/(20 + (13*15)/(20 + ... + (4*n + 1)*(4*n + 3)/(20 + ...)))))). (End)
Tan(arctan(c) + arctan(c^3)) = 1/2. - Gary W. Adamson, Apr 04 2024

A232735 Decimal expansion of the real part of I^(1/7), or cos(Pi/14).

Original entry on oeis.org

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

Views

Author

Stanislav Sykora, Nov 29 2013

Keywords

Comments

The corresponding imaginary part is in A232736.
Root of the equation -7 + 56*x^2 - 112*x^4 + 64*x^6 = 0. - Vaclav Kotesovec, Apr 04 2021

Examples

			0.974927912181823607018131682993931217232785800619997437648...
		

Crossrefs

Cf. A232736 (imaginary part), A010503 (real(I^(1/2))), A010527 (real(I^(1/3))), A144981 (real(I^(1/4))), A019881 (real(I^(1/5))), A019884 (real(I^(1/6))), A232737 (real(I^(1/8))), A019889 (real(I^(1/9))), A019890 (real(I^(1/10))).

Programs

  • Magma
    R:= RealField(100); Cos(Pi(R)/14); // G. C. Greubel, Sep 19 2022
    
  • Mathematica
    RealDigits[Cos[Pi/14],10,120][[1]] (* Harvey P. Dale, Dec 15 2018 *)
  • SageMath
    numerical_approx(cos(pi/14), digits=120) # G. C. Greubel, Sep 19 2022

Formula

2*this^2 -1 = A073052. - R. J. Mathar, Aug 29 2025
Equals 2F1(-1/14,1/14;1/2;1) . - R. J. Mathar, Aug 31 2025

A307731 Results of strip bijection as described in A307110 with additional application of local repairs to reduce the maximum wobbling distance S from S1=cos(Pi/8) to S2=sqrt(5)*sin(Pi/8).

Original entry on oeis.org

0, 1, 6, 3, 8, 2, 11, 4, 9, 5, 15, 7, 19, 14, 10, 16, 17, 18, 12, 20, 13, 26, 27, 28, 25, 21, 22, 23, 24, 38, 31, 40, 33, 42, 35, 44, 29, 30, 51, 32, 53, 34, 55, 36, 49, 57, 58, 59, 60, 62, 39, 64, 41, 66, 43, 68, 37, 46, 47, 48, 45, 50, 63, 52, 65, 54, 67
Offset: 0

Views

Author

Hugo Pfoertner, Apr 25 2019

Keywords

Comments

The terms visible in the data section are identical with those of A307110. The first difference occurs at a(124)=141, A307110(124)=125.
The wobbling distance S is the mutual Euclidean distance of the pairs matched by a bijection.
.
- - - G - -\- - - - - - / - G - - - - -\- - - - - G -/- - - - - - - - - G
| + \ / | \ +|/ |
| + \ / | \ + | |
| H | H | |
| / \ | / \ | |
| / \ . . . / |\ |
| / . \ | / . | \ /|
|/ \ | / . | \ / |
/| . \| / . | \ / |
/ | |\ / .| \ / |
H + | . | +H # # # # . H + + |
- \ + G - - - - - - - - - - G+ -\- - - - # # # # #G.- - - - - - -\- -+ +G
\ | . #| \ \ +| . / \ |
\ # | D \ +| / \ |
| \ . # | \ \ +| / \
| \ <--------r=S1------C \ + | ./ |
| \ # | B \ + | / |
| . # | B \ + / |
| .H+ | B H |. |
| / \++ | B / #. |
| / . \++ | B / #| \ |
| / . \+++ | B / . # \ |
- - - G / - - - - - . - \ -+G - - - - -B/ - - . +G - - \ - - - - - - - G
+ / . \ |# / B +E+. | . \ + /
+/ | . # /. +M+ | . \ + / |
+/ | | \# / +E+ b | .\ +/ |
H | | H+ . b | .H |
\ | | / .\ b | / .\ |
\ | / . \ b | / . \ |
\ / | . \ b | / \ |
| \ / | . \ b | / . \
| \ / |. \ b | / |
| \ / |. \ c--------r=S1------>. |
- - - G+++++- \ - / - - - - G.- - - - - - - - - HdG - - - - - - - - -.- G
| ++++H +. / \ . |
| / \ |+ / | \ |
| / \ |+. / | \ . |
| / \ | + / | \ . /
|/ \ | + . / | \ . / |
/ \ + . / | \ . / |
/ | | \ + . | . / |
H | | H . | . H |
\ | | / \ . | . / + |
\ | / \ . / + |
- - - G - - - - - - - - - / G - - - - -\- - - - - G - - - / - - - - - - G
.
The ASCII graphics above shows the situation after the application of the strip bijection, as it is described in A307110, for a position in the grids containing a "long" junction exceeding the length S2. The linked graphics file "Construction of repair" shows a similar configuration, but without labels.
All junctions resulting from the strip bijection are marked by plus signs. The long junction is marked by embedded letters "E". There are 6 possible orientations of E-junctions (called E for short), but the method for their elimination is identical for all cases.
The target of the method is to achieve a local reconnecting, which replaces 4 junctions by circularly shifted new junctions. To determine the affected grid points, the following steps are performed:
From the midpoint (marked by M in the figure) of E construct a bisecting line Bb perpendicular to E. Draw two circles, one on each side of E with centers on B and b at distance S1 from M. E is a tangent at M of these circles with radius r = S1. The two circles are marked by dots ".." in the figure.
For the two circle centers C and c determine the distances D and d of the respective closest grid points in lattice G. The position (c) of the circle center, for which this minimum distance is smaller, indicates on which side of E no reconnecting is required. A circle with radius S1 around c contains only one grid point of G and one of H. All other grid points of both lattices lie outside of this circle.
The side of E with the larger distance between circle center and closest grid point is where the circular shift of junctions is to be performed. The circle around C with radius r = S1 contains 3 grid points of lattice G and 3 grid points of lattice H.
After having found c, it is possible to replace the geometric determination of the 3 grid point pairs on the opposite side of E by a lookup in a table of differences between the coordinates of M and c rounded to nearest integers, leading to a unique identification of the 6 occurring cases. The function "repair" in the PARI program implements this selection.
The 4 new junctions are marked by "###" in the figure. They replace the 4 previous "+++" junctions, including the long junction E. The maximum of their lengths does not exceed S2, approaching S2 for length of E approaching S1. The limiting case for the 4 rearranged junctions are two of length S2 and two of length sin(Pi/8) = 0.38268...
The described repair is applied to all occurrences of bijection distances exceeding S2 within the overlay of the two lattices. Numerical experiments with random points on square lattices of huge size show that approximately 0.956 % (roughly 1/105) of the grid points lead to a bijection distance S > S2 after the application of the strip bijection. No counterexample for the validity of the method is known, but a formal proof is missing.
In the ring-wise one-dimensional mapping of the bijection as given by A307110, the first affected position is n = 124. The table in the example section shows the corresponding changes for this earliest repair together with the listing of another repair with different orientation of E.
All affected index positions have to be exchanged in the one-dimensional list. Due to the occurrence frequency of E-junctions the current sequence is expected to differ from A307110 for roughly 4% of the terms.
The PARI program provided as external file is self-contained, including the code for generation of the rings used for 1d-mapping, A305575 and A305576, and the code for the strip bijection of A307110. To generate a b-file of 10000 terms, the corresponding code lines at the end of the program have to be activated.

Examples

			The table shows the first re-matched pairs of grid points together with the result of the unmodified strip bijection:
    Grid G          Grid H             Grid H rotated
   n     i    j    a(n)   k    m   (k,m) rotated by -Pi/4  distance of
                                                           matched points
  124   -6    2    141   -6   -3    -6.363961   2.121320   0.383648
  140   -6    3    125   -6   -2    -5.656854   2.828427   0.383649
  180   -7    3    173   -7   -2    -6.363961   3.535534   0.831470 < S2
  172   -7    2    181   -7   -3    -7.071068   2.828427   0.831470 < S2
  ...
  266   -6    7    256   -9    1    -5.656854   7.071068   0.350428
  309   -6    8    320  -10    1    -6.363961   7.778174   0.426232
  279   -5    8    328  -10    2    -5.656854   8.485281   0.816673 < S2
  235   -5    7    268   -9    2    -4.949747   7.778174   0.779795 < S2
compared to (unmodified):                                        S2=0.855706..
                A307110(n)
  124   -6    2    125   -6   -2    -5.656854   2.828427   0.896683 > S2
  140   -6    3    173   -7   -2    -6.363961   3.535534   0.647506
  180   -7    3    181   -7   -3    -7.071068   2.828427   0.185709
  172   -7    2    141   -6   -3    -6.363961   2.121320   0.647506
  ...
  266   -6    7    320  -10    1    -6.363961   7.778174   0.859083 > S2
  309   -6    8    328  -10    2    -5.656854   8.485281   0.594346
  279   -5    8    268   -9    2    -4.949747   7.778174   0.227446
  235   -5    7    256   -9    1    -5.656854   7.071068   0.660688
		

Crossrefs

Programs

  • PARI
    \\ See Pfoertner link.

A195697 First denominator and then numerator in a fraction expansion of log(2) - Pi/8.

Original entry on oeis.org

2, 1, 3, -1, 12, 1, 30, 1, 35, -1, 56, 1, 90, 1, 99, -1, 132, 1, 182, 1, 195, -1, 240, 1, 306, 1, 323, -1, 380, 1, 462, 1, 483, -1, 552, 1, 650, 1, 675, -1, 756, 1, 870, 1, 899, -1, 992, 1, 1122, 1, 1155, -1, 1260
Offset: 1

Views

Author

Mohammad K. Azarian, Sep 25 2011

Keywords

Comments

The minus sign in front of a fraction is considered the sign of the numerator.

Examples

			1/2 - 1/3 + 1/12 + 1/30 - 1/35 + 1/56 + 1/90 - 1/99 + 1/132 + 1/182 - 1/195 + 1/240 + ... = [(1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + (1/7 - 1/8) + (1/9 - 1/10) + (1/11 - 1/12) + ... ] - (1/2)*[(1 - 1/3) + (1/5 - 1/7) + (1/9 - 1/11) + (1/13 - 1/15) + ... ] = log(2) - Pi/8.
		

References

  • Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.
  • Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968).

Crossrefs

Formula

log(2) - Pi/8 = Sum_{n>=1} (-1)^(n+1)*(1/n) + (-1/2)*Sum_{n>=0} (-1)^n*(1/(2*n+1)).
Empirical g.f.: x*(2+x+x^2-2*x^3+9*x^4+2*x^5+14*x^6-2*x^7+3*x^8+2*x^9+3*x^10-2*x^11+x^13) / ((1-x)^3*(1+x)^3*(1-x+x^2)^2*(1+x+x^2)^2). - Colin Barker, Dec 17 2015

A195909 First numerator and then denominator in a fraction expansion of log(2) - Pi/8.

Original entry on oeis.org

1, 2, -1, 3, 1, 12, 1, 30, -1, 35, 1, 56, 1, 90, -1, 99, 1, 132, 1, 182, -1, 195, 1, 240, 1, 306, -1, 323, 1, 380, 1, 462, -1, 483, 1, 552, 1, 650, -1, 675, 1, 756, 1, 870, -1, 899, 1, 992, 1, 1122, -1, 1155, 1
Offset: 1

Views

Author

Mohammad K. Azarian, Sep 26 2011

Keywords

Examples

			1/2 - 1/3 + 1/12 + 1/30 - 1/35 + 1/56 + 1/90 - 1/99 + 1/132 + 1/182 - 1/195 + 1/240 + ... = [(1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + (1/7 - 1/8) + (1/9 - 1/10) + (1/11 - 1/12) + ... ] - (1/2)*[(1 - 1/3) + (1/5 - 1/7) + (1/9 - 1/11) + (1/13 - 1/15) + ... ] = log(2) - Pi/8.
		

References

  • Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.
  • Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968).

Crossrefs

Formula

log(2) - Pi/8 = Sum_{n>=1} (-1)^(n+1)*(1/n) + (-1/2)*Sum_{n>=0} (-1)^n*(1/(2*n+1)).
Empirical g.f.: x*(1+2*x-2*x^2+x^3+2*x^4+9*x^5-2*x^6+14*x^7+2*x^8+3*x^9-2*x^10+3*x^11+x^12) / ((1-x)^3*(1+x)^3*(1-x+x^2)^2*(1+x+x^2)^2). - Colin Barker, Dec 17 2015

A195913 The denominator in a fraction expansion of log(2)-Pi/8.

Original entry on oeis.org

2, 3, 12, 30, 35, 56, 90, 99, 132, 182, 195, 240, 306, 323, 380, 462, 483, 552, 650, 675, 756, 870, 899, 992, 1122, 1155, 1260, 1406, 1443, 1560, 1722, 1763, 1892, 2070, 2115, 2256, 2450, 2499, 2652, 2862, 2915
Offset: 1

Views

Author

Mohammad K. Azarian, Sep 25 2011

Keywords

Comments

The minus sign in front of a fraction is considered the sign of the numerator and hence the sign of the fraction does not appear in this sequence.

Examples

			1/2 - 1/3 + 1/12 + 1/30 - 1/35 + 1/56 + 1/90 - 1/99 + 1/132 + 1/182 - 1/195 + 1/240 + ... = [(1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + (1/7 - 1/8) + (1/9 - 1/10) + (1/11 - 1/12) + ...] - (1/2)*[(1 - 1/3) + (1/5 - 1/7) + (1/9 - 1/11) + (1/13 - 1/15) + ... ] = log(2) - Pi/8.
		

References

  • Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968).

Crossrefs

Formula

log(2) - Pi/8 = Sum_{n>=1} (-1)^(n+1)*(1/n) + (-1/2)*Sum_{n>=0} (-1)^n*(1/(2*n+1)).
Empirical g.f.: x*(2+x+9*x^2+14*x^3+3*x^4+3*x^5) / ((1-x)^3*(1+x+x^2)^2). - Colin Barker, Dec 17 2015
From Bernard Schott, Aug 11 2019: (Start)
k >= 1, a(3*k) = (4*k-1) * 4*k,
k >= 0, a(3*k+1) = (4*k+1) * (4*k+2),
k >= 0, a(3*k+2) = (4*k+1) * (4*k+3).
The even terms a(3*k) and a(3*k+1) come from log(2) and the odd terms a(3*k+2) come from - Pi/8. (End)
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