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 12 results. Next

A224269 Consider the spiral of Theodorus (A072895). This sequence gives the number of k successive triangles which is closer to 360 degrees than any previous k triangles.

Original entry on oeis.org

17, 53, 185, 396, 4926, 9086, 20291, 28083, 440835, 579644, 1819320, 3032895, 8305458, 15159436, 29824343, 46104922, 88019569, 89143145, 94929121, 107958869, 227428224, 402409536, 527154160
Offset: 1

Views

Author

Robert G. Wilson v, Apr 02 2013

Keywords

Comments

Any a(i) or a(i)+1 must belong to A072895.
These entries correspond to 1, 2, 4, 6, 22, 30, 45, 53, 211, 242, 429, 554, 917, 1239, 1738, 2161, 2986, 3005, 3101, 3307, 4800, 6385, 7308, ..., turns around the axis. Use the formula in A072895 to check the entries.
Search limit: 10000 turns about the axis.

Examples

			a(1) = 17 because the first 16 right triangles result in 351.15042° (8.84958° before the original axis) and the first 17 right triangles result in 364.78344°. 17 right triangles are within 4.78344° of the original axis.
a(2) = 53 because the first 54 right triangles result in 727.48834° and the first 53 right triangles result in 719.73897°.  This is closer to the original axis than 16 and is within 0.2610252°.
a(3) # 109 nor 110 because the first 109 right triangles result in 1079.12463° and the first 110 right triangles result in 1084.57110°. Neither angle is closer to the original axis (1080°) than 53. Therefore the third turn around the center is not close to the original axis than twice around.
a(3) = 185 because the first 186 right triangles result in 1444.08227° (4.08227° after the original axis) and the first 185 right triangles result in 1439.88864°.  This is closer to the original axis than 53 and is within 0.11136°.
		

Crossrefs

Cf. A072895.

Programs

  • Mathematica
    lmt = Infinity; lst = {}; k = n = 1; s = 0; While[n < 1001, While[s < 2Pi*n, s = N[s + ArcTan[ 1 / Sqrt@ k], 32]; k++]; a = s - 2Pi*n; b = 2Pi*n - (s - ArcTan[1/Sqrt[k - 1]]); If[Min[a, b] < lmt, lmt = Min[a, b]; If[a < b, AppendTo[lst, {n, k - 1}]; Print[{n, k - 1}], AppendTo[lst, {n, k - 2}]; Print[{n, k - 2}]]]; n++]; Last@ Transpose@ lst
    k=minDist=1;lst={};K=-2.1577829966594462209291427868295777235; num[n_]:=Module[{a=-(K/2)+n Pi,b},b=a^2-1/6;If[Floor[b]==Floor[b+1/(144 a^2)],Floor[b],Undefined]]; While[k<40000000,n=num[k];If[!NumberQ[n],Print[k," Stop"];Break[]]; a=2Pi-Mod[K+2 Sqrt[n]+1/(6 Sqrt[n]),2Pi]; b=Mod[K+2 Sqrt[n+1]+1/(6 Sqrt[n+1]),2Pi]; If[aHerbert Kociemba, Jul 18 2013 *)

A295338 Number of points of the outer discrete Theodorus spiral on sheet S_n, n >= 1. First differences of A072895.

Original entry on oeis.org

17, 37, 56, 76, 95, 115, 136, 154, 175, 194, 214, 234, 254, 273, 293, 313, 332, 352, 372, 392, 411, 432, 450, 471, 490, 511, 529, 550, 569, 590, 608, 629, 648, 668, 688, 708, 727, 747, 767, 786, 806, 826, 846, 865, 885
Offset: 1

Views

Author

Wolfdieter Lang, Dec 01 2017

Keywords

Comments

The punctured sheets S_n start here with n=1: S_n := rho*exp(i*phi), rho > 0, 2*Pi*(n-1) <= phi < 2*Pi*n.
For details and references see A072895.

Crossrefs

Cf. A072895.

Formula

a(n) = b(n+1) - b(n), n >= 1, with b(n) = A072895(n), and a(1) = 17.

A227626 Consider the spiral of Theodorus (A072895). This sequence is closely related to A224269 and gives the number of k successive revolutions such that the triangles are closer to 360 degrees than any previous triangles.

Original entry on oeis.org

1, 2, 4, 6, 22, 30, 45, 53, 211, 242, 429, 554, 917, 1239, 1738, 2161, 2986, 3005, 3101, 3307, 4800, 6385, 7308, 15148, 16668, 19287, 28103, 72754, 143406, 457425, 955117, 1129313, 2290339, 7362039, 11374333, 11711400, 11778444, 11896240, 14221855, 31972242
Offset: 1

Views

Author

Herbert Kociemba, Jul 18 2013

Keywords

Crossrefs

Programs

  • Mathematica
    k=minDist=1; lst={}; K=-2.1577829966594462209291427868295777235; num[n_] := Module[{a=-(K/2)+n Pi,b}, b=a^2-1/6; If[Floor[b]==Floor[b+1/(144 a^2)], Floor[b], Undefined]] While[k<40000000, n=num[k]; If[!NumberQ[n], Print[k," Stop"]; Break[]]; a=2Pi-Mod[K+2 Sqrt[n]+1/(6 Sqrt[n]),2Pi]; b=Mod[K+2 Sqrt[n+1]+1/(6 Sqrt[n+1]),2Pi]; If[a
    				

A105459 Decimal expansion of Hlawka's Schneckenkonstante K = -2.157782... (negated).

Original entry on oeis.org

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

Views

Author

David Brink, Jun 13 2011

Keywords

Examples

			-2.157782996659446220929142786829577723504139598607562455...
		

References

  • P. J. Davis, Spirals from Theodorus to Chaos, A K Peters, Wellesley, MA, 1993.

Crossrefs

Cf. A185051 for continued fraction expansion.

Programs

  • Maple
    evalf(Sum((-1)^k*Zeta(k + 1/2)/(2*k+1), k=0..infinity), 120); # Vaclav Kotesovec, Mar 01 2016
  • Mathematica
    RealDigits[ NSum[(-1)^k*Zeta[k + 1/2]/(2 k + 1), {k, 0, Infinity}, Method -> "AlternatingSigns", AccuracyGoal -> 2^6, PrecisionGoal -> 2^6, WorkingPrecision -> 2^7], 10, 2^7][[1]] (* Robert G. Wilson v, Jul 11 2013 *)
  • PARI
    sumalt(k=0,(-1)^k*zeta(k+1/2)/(2*k+1)) \\ M. F. Hasler, Mar 31 2022

Formula

Sum_{x=1..n-1} arctan(1/sqrt(x)) = 2*sqrt(n) + K + o(1). [Corrected by M. F. Hasler, Mar 31 2022]
Equals Sum_{k>=0} (-1)^k*zeta(k+1/2)/(2*k+1). - Robert B Fowler, Oct 23 2022

A226317 Decimal expansion of the constant of Theodorus.

Original entry on oeis.org

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

Views

Author

Walter Gautschi (wxg(AT)cs.purdue.edu), Robert G. Wilson v, and Jean-François Alcover, Apr 15 2013

Keywords

Comments

The decimal expansion of the Sum {k>=1} 1/(k^(3/2) + k^(1/2)).
This constant was first identified by Professor Philip J. Davis.
This constant is not in Steven R. Finch, Mathematical Constants, Cambridge, 2003, nor is it in the Inverse Symbolic Calculator (originally by Simon Plouffe & the Borwein brothers).

Examples

			1.86002507922119030718069591571714332466652412152345149304919950359788...
		

References

  • Philip J. Davis, Spirals: From Theodorus to Chaos, AK Peters, 1993.
  • Julian R. Havil, The Irrationals: A Story of the Numbers You Can't Count On, Princeton University Press, Princeton NJ, 2012, page 277.

Crossrefs

Programs

  • Maple
    Digits := 102: evalf(sum((k^(3/2) + k^(1/2))^(-1), k=1..infinity));
    # Peter Luschny, Feb 28 2022
  • Mathematica
    digits = 100; 2/Sqrt[Pi]*NIntegrate[(-Exp[t^2])*Log[1 - Exp[-t^2]] - 1, {t, 0, Infinity}, WorkingPrecision -> digits] // RealDigits[#, 10, digits]& // First
    (* or *)
    a = NSum[1/(k^(3/2) + k^(1/2)), {k, 1, Infinity}, AccuracyGoal -> 2^8, PrecisionGoal -> 2^8, WorkingPrecision -> 2^8, NSumTerms -> 2^15]; RealDigits[a, 10, 105][[1]]
  • PARI
    sumpos(k=1,1/sqrt(k)/(1+k)) \\ Charles R Greathouse IV, Aug 29 2013
    
  • PARI
    sumalt(k=0,zeta(k+3/2)*(-1)^k) \\ Charles R Greathouse IV, Aug 29 2013

Formula

Sum_{k>=1} 1/(k^(3/2) + k^(1/2)).
Equals -(2/sqrt(Pi)) * Integral_{x>=0} (exp(x^2)*log(1-exp(-x^2))+1) dx (Waldvogel, 2008). - Amiram Eldar, Jul 19 2022

A137515 Maximal number of right triangles in n turns of Pythagoras's snail.

Original entry on oeis.org

16, 53, 109, 185, 280, 395, 531, 685, 860, 1054, 1268, 1502, 1756, 2029, 2322, 2635, 2967, 3319, 3691, 4083, 4494, 4926, 5376, 5847, 6337, 6848, 7377, 7927, 8496, 9086, 9694, 10323, 10971, 11639, 12327, 13035, 13762, 14509, 15276, 16062, 16868, 17694
Offset: 1

Views

Author

Sébastien Dumortier, Apr 23 2008, Apr 25 2008

Keywords

Comments

Pythagoras's snail: begin the snail with an isosceles triangle (side = 1 unit). Then new triangle's right-angle sides are the previous hypotenuse and 1 unit length side.
From one term to the next one, the number added grows by 18, 19, 20 or 21 (tested up to 5000 terms).
To restate the comment immediately above: the second differences of the terms of the sequence consist of 18, 19, 20, or 21. - Harvey P. Dale, May 20 2019

Examples

			17 triangles are needed to close the first turn. So there are 16 triangles in this turn. From the beginning, there are 53 triangles before closing the second turn... etc.
		

Crossrefs

Programs

  • Mathematica
    w[n_] := ArcSin[ 1/Sqrt[n+1] ]//N; s[1] = w[1]; s[n_] := s[n] = s[n-1] + w[n]; a[n_] := (an = 1; While[ s[an] < 2*Pi*n, an++]; an-1); Table[ an = a[n]; Print[an]; an, {n, 1, 42}] (* Jean-François Alcover, Feb 24 2012 *)
  • Python
    from math import asin, sqrt, pi
    hyp=2
    som=0
    n=1
    while n<500:
        if som+asin(1/sqrt(hyp))/pi*180>n*360:
            print(hyp-2, end=', ')
            n=n+1
        som=som+asin(1/sqrt(hyp))/pi*180
        hyp=hyp+1

Formula

a(n) = A072895(n) - 1. - Robert G. Wilson v, Feb 27 2013

A295339 Least k for the inner Theodorus spiral to complete n revolutions.

Original entry on oeis.org

15, 52, 108, 184, 279, 394, 530, 684, 859, 1053, 1267, 1501, 1755, 2028, 2321, 2634, 2966, 3318, 3690, 4082, 4493, 4925, 5375, 5846, 6336, 6847, 7376, 7926, 8495, 9085, 9693, 10322, 10970, 11638, 12326, 13034, 13761, 14508, 15275
Offset: 1

Views

Author

Wolfdieter Lang, Dec 13 2017

Keywords

Comments

Here the points of the inner discrete Theodurus spiral in the complex plane are zhat(k) = rho(k)*exp(i*phihat(k)) with rho(k) = sqrt(k) and phihat(k) starts with phihat(1) = Pi/2 and is not restricted to be <= 2*Pi, it is phihat(k) = Sum_{j=0..k-1} (2*alpha(j+1) - alpha(j)) with alpha(j) = arctan(1/sqrt(j)), for k >= 1. The formula is phihat(k) = phi(k) + alpha(k), with the recurrence for the arguments of the outer spiral phi(k) = phi(k-1) + alpha(k-1), k >= 2, with phi(1) = 0.
If one considers punctured sheets S_n = rho*exp(i*phi_n), with rho > 0 and 2*Pi*(n-1) <= phi_n < 2*Pi*n, for n >= 1, then on sheet S_n there are a(n) - a(n-1) = A296179(n) points zhat, where a(0) = 0.
An analytic continuation of Davis's interpolation of the outer spiral is given in the Waldvogel link (see Figure 2 there). The point zhat(k) (called G_k on Figure 1 there) on the inner spiral is obtained from mirroring the point z(k) (called F_k there) of the outer spiral on the hypotenuse O,z(k+1), for k >= 1. In the present case the arguments phihat(k) of zhat(k) are taken positive.
Conjecture: a(n) = A072895(n) - 2, n >= 1. This follows from the conjecture that the sequences K := {floor(phi(k)/(2*Pi)}{k >= 1} with phi given above, and Khat:= {floor(phihat(k)/(2*Pi)}{k >= 1} with phihat given above satisfy Khat(k-2) = K(k), for k >= 3. Note that phihat(k-2) - phi(k) = alpha(k-2) - alpha(k-1) =: delta(k) = arctan((sqrt(k-1) - sqrt(k-2))/(1 + sqrt((k-1)*(k-2)))) > 0, for k >= 3. Therefore the conjecture is that delta(k) < 2*Pi*(1 - frac(phi(k)/(2*Pi))), for k >= 3, or, equivalently, phihat(k-2) < 2*Pi*(K(k) + 1), for k >= 3.

References

  • P. J. Davis, Spirals from Theodorus to Chaos, A K Peters, Wellesley, MA, 1993.

Crossrefs

Cf. A072895 (outer spiral), A296179.

Formula

a(n) = -1 + first position of n in the sequence
Khat:= {floor(phihat(k)/(2*Pi))}_{ k>= 1}, with phihat given in a comment above in terms of phi.
Conjecture: a(n) = A072895(n) - 2, n >= 1 (see the comment above).

A352741 Numbers on the Spiral of Theodorus that come closer to the x-axis than their neighbors; a(0) = 0 and a(1) = 1.

Original entry on oeis.org

0, 1, 7, 18, 33, 54, 80, 110, 146, 186, 231, 282, 337, 397, 462, 532, 607, 687, 772, 861, 956, 1056, 1160, 1270, 1384, 1504, 1628, 1757, 1891, 2030, 2174, 2323, 2477, 2636, 2800, 2969, 3142, 3321, 3504, 3693, 3886, 4084, 4288, 4496, 4709, 4927, 5150, 5378, 5611, 5849, 6091, 6339
Offset: 0

Views

Author

M. F. Hasler, Mar 31 2022

Keywords

Comments

The Spiral of Theodorus (a.k.a. Pythagorean spiral or Pythagoras's snail) results from constructing the square roots sqrt(2), sqrt(3), sqrt(4), ... as hypotenuse of the right triangle having the previous one as longer leg, and the shorter leg equal to 1, starting with the segment [0, 1] for sqrt(1). In the complex plane, this construction corresponds to a sequence z(n+1) = z(n) + i*z(n)/|z(n)|, starting with z(1) = 1 (and z(0) = 0 by convention).
This sequence lists those n for which the endpoint is closer to the x-axis than the preceding and next one (plus the two initial points 0 and 1 which are the only ones lying directly on the x-axis). In terms of the complex sequence z(n), this means the indices n such that abs(Im(z(n)) <= abs(Im(z(n +- 1))).
The two initial terms correspond to the points (0, 0) and (1, 0), which can be regarded as the start of the Pythagorean spiral, and are the only points exactly on the x-axis.
There is a smooth complex-valued function z: [0,oo) -> C, t |-> z(t), which interpolates the spiral in non-integer values. In terms of this function, a(n) = round(t(n)) where t(n) is the n-th zero of the imaginary part of z (if indexing starts with 0 for t(0) = 0, then t(1) = 1, and t(2) ~ 6.8 where z(6.8) = -1). (This function has a natural extension to the whole of R including also the negative real line, but we don't consider negative arguments here.)

Crossrefs

Cf. A072895, A137515, A105459 (Hlawka's "snail constant").

Programs

  • PARI
    z=!t=y=0; [n | n<-[0..10^4], ![t < t=y, y > y=abs(imag(z))]*z*=1+I/abs(z)]

Formula

a(2n+1) = A072895(n) or A072895(n)+1, for all n > 0.

A172164 Differences between numbers of triangles entirely contained in two consecutive turns of Pythagoras's snail (Theodorus spiral).

Original entry on oeis.org

20, 19, 20, 19, 20, 21, 18, 21, 19, 20, 20, 20, 19, 20, 20, 19, 20, 20, 20, 19, 21, 18, 21, 19, 21, 18, 21, 19, 21, 18, 21, 19, 20, 20, 20, 19, 20, 20, 19, 20, 20, 20, 19, 20, 20, 20, 19, 21, 18, 21, 19, 20, 20, 20, 19, 20, 20, 20, 19, 20, 20, 19, 20, 20, 20, 19, 21, 18, 21
Offset: 2

Views

Author

Sébastien Dumortier, Jan 27 2010

Keywords

Comments

Conjecture : The terms are only 18,19,20,21 (From the first thousand turns, there are 2,3% of 18, 36,5% of 19, 46,2% of 20 and 15% of 21). No period found. Probably due to Pi transcendence.
From the first one hundred thousand turns, there are 1.662% 18s, 36.350% 19s, 48.393% 20s and 13.595% 21s. - Robert G. Wilson v, Mar 31 2013
From the first 10 Million turns, there are 1.69208% 18s, 36.33984% 19s, 48.32320% 20s and 13.64488% 21s. - Herbert Kociemba, Jul 15 2013

Examples

			In the first turn, 16 triangles are complete. In the 2nd turn, there are 36 triangles completely included. The difference is 20.
		

Crossrefs

Programs

  • Mathematica
    (* Obtain the sequence of A072895 and set it equal to lst. *); Differences[lst, 2] (* Robert G. Wilson v, Mar 31 2013 *)
  • Python
    # See A137515 for Python code, and then OooCalc for more.

Formula

The second forward difference of A072895. - Robert G. Wilson v, Mar 31 2013

A296181 First point of the discrete Theodorus spiral in the fourth quadrant for the n-th revolution, for n >= 1.

Original entry on oeis.org

12, 44, 95, 166, 256, 367, 497, 647, 816, 1006, 1215, 1444, 1692, 1961, 2249, 2557, 2884, 3231, 3598, 3985, 4392, 4818, 5264, 5730, 6215, 6720, 7245, 7790
Offset: 1

Views

Author

Wolfdieter Lang, Jan 05 2018

Keywords

Comments

This sequence is used in a conjecture on points z_k of the discrete (outer) Theodorus spiral living on quadrant IV of the complex plane of sheet S_n, where S_n := {r*exp(i*phi), r > 0, 2*Pi*(n-1) <= phi < 2*Pi*n}. This corresponds to the n-th revolution, for n >= 1.
This conjecture is 2*Pi - varphi(A072895(n)) > arctan(a(n)), n >= 1, with varphi(k) = phi(k) - 2*Pi*floor(phi(k)/(2*Pi)) where z_k = sqrt(k)*exp(i*phi(k)).
This conjecture implies a conjecture relating points of the discrete inner spiral to those of the outer ones, namely Khat(k-2) := floor(phihat(k-2)/(2*Pi)) = K(k) =: floor(phi(k)/(2*Pi)) for k >= 3, where zhat_k = sqrt(k)*exp(i*phihat(k)) is a point of the discrete inner Theodorus spiral, given in terms of z_k by zhat(k) = ((k-1 + 2*sqrt(k)*i )/(k+1))*z_k. This implies phihat(k) = phi(k) + arctan((sqrt(k-1) - sqrt(k-2))/(1 + sqrt((k-1)*(k-2)))). The implied conjecture Khat(k-2) = K(k), k >= 3, for the other three quadrants of each sheet S_n can be proved. For the inner spiral see the Waldvogel link.
If the implied conjecture is true then A295339(n) = A072895(n) - 2, for n >= 1, hence A296179(n) = A295338(n), for n >= 2.
For the conjecture and the proof for the first three quadrants for each sheet S_n see the W. Lang link. - Wolfdieter Lang, Jan 24 2018

Examples

			a(1) = 12 because phi(11) - 3*Pi/2 is about -0.1869017440 (Maple 10 digits), that is, KIV(11) = -1 + 1 = 0 (not n = 1) but phi(12) - 3*Pi/2 is about +0.1059410277, that is, KIV(12) = 0 + 1 = 1 (on sheet S_1).
a(2) = 44 because  phi(43) - 3*Pi/2 is about 6.270091849, that is KIV(43) = 0 + 1 = 1 (not n = 2) but varphi(44) - 3*Pi/2 is about 6.421424486, that is KIV(44) = 1 + 1 = 2 (on sheet S_2).
		

Crossrefs

Formula

a(n) is the smallest index k for which KIV(k) = n, with KIV(k):= floor((phi(k) - 3*Pi/2)/(2*Pi)) + 1, for k >= 1, where phi(k) is the polar angle of the point z_k = sqrt(n)*exp(i*phi(k)) of the (outer) discrete Theodorus spiral.
Showing 1-10 of 12 results. Next