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 1-9 of 9 results.

A001622 Decimal expansion of golden ratio phi (or tau) = (1 + sqrt(5))/2.

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane

Keywords

Comments

Also decimal expansion of the positive root of (x+1)^n - x^(2n). (x+1)^n - x^(2n) = 0 has only two real roots x1 = -(sqrt(5)-1)/2 and x2 = (sqrt(5)+1)/2 for all n > 0. - Cino Hilliard, May 27 2004
The golden ratio phi is the most irrational among irrational numbers; its successive continued fraction convergents F(n+1)/F(n) are the slowest to approximate to its actual value (I. Stewart, in "Nature's Numbers", Basic Books, 1997). - Lekraj Beedassy, Jan 21 2005
Let t=golden ratio. The lesser sqrt(5)-contraction rectangle has shape t-1, and the greater sqrt(5)-contraction rectangle has shape t. For definitions of shape and contraction rectangles, see A188739. - Clark Kimberling, Apr 16 2011
The golden ratio (often denoted by phi or tau) is the shape (i.e., length/width) of the golden rectangle, which has the special property that removal of a square from one end leaves a rectangle of the same shape as the original rectangle. Analogously, removals of certain isosceles triangles characterize side-golden and angle-golden triangles. Repeated removals in these configurations result in infinite partitions of golden rectangles and triangles into squares or isosceles triangles so as to match the continued fraction, [1,1,1,1,1,...] of tau. For the special shape of rectangle which partitions into golden rectangles so as to match the continued fraction [tau, tau, tau, ...], see A188635. For other rectangular shapes which depend on tau, see A189970, A190177, A190179, A180182. For triangular shapes which depend on tau, see A152149 and A188594; for tetrahedral, see A178988. - Clark Kimberling, May 06 2011
Given a pentagon ABCDE, 1/(phi)^2 <= (A*C^2 + C*E^2 + E*B^2 + B*D^2 + D*A^2) / (A*B^2 + B*C^2 + C*D^2 + D*E^2 + E*A^2) <= (phi)^2. - Seiichi Kirikami, Aug 18 2011
If a triangle has sides whose lengths form a geometric progression in the ratio of 1:r:r^2 then the triangle inequality condition requires that r be in the range 1/phi < r < phi. - Frank M Jackson, Oct 12 2011
The graphs of x-y=1 and x*y=1 meet at (tau,1/tau). - Clark Kimberling, Oct 19 2011
Also decimal expansion of the first root of x^sqrt(x+1) = sqrt(x+1)^x. - Michel Lagneau, Dec 02 2011
Also decimal expansion of the root of (1/x)^(1/sqrt(x+1)) = (1/sqrt(x+1))^(1/x). - Michel Lagneau, Apr 17 2012
This is the case n=5 of (Gamma(1/n)/Gamma(3/n))*(Gamma((n-1)/n)/Gamma((n-3)/n)): (1+sqrt(5))/2 = (Gamma(1/5)/Gamma(3/5))*(Gamma(4/5)/Gamma(2/5)). - Bruno Berselli, Dec 14 2012
Also decimal expansion of the only number x>1 such that (x^x)^(x^x) = (x^(x^x))^x = x^((x^x)^x). - Jaroslav Krizek, Feb 01 2014
For n >= 1, round(phi^prime(n)) == 1 (mod prime(n)) and, for n >= 3, round(phi^prime(n)) == 1 (mod 2*prime(n)). - Vladimir Shevelev, Mar 21 2014
The continuous radical sqrt(1+sqrt(1+sqrt(1+...))) tends to phi. - Giovanni Zedda, Jun 22 2019
Equals sqrt(2+sqrt(2-sqrt(2+sqrt(2-...)))). - Diego Rattaggi, Apr 17 2021
Given any complex p such that real(p) > -1, phi is the only real solution of the equation z^p+z^(p+1)=z^(p+2), and the only attractor of the complex mapping z->M(z,p), where M(z,p)=(z^p+z^(p+1))^(1/(p+2)), convergent from any complex plane point. - Stanislav Sykora, Oct 14 2021
The only positive number such that its decimal part, its integral part and the number itself (x-[x], [x] and x) form a geometric progression is phi, with respectively (phi -1, 1, phi) and a ratio = phi. This is the answer to the 4th problem of the 7th Canadian Mathematical Olympiad in 1975 (see IMO link and Doob reference). - Bernard Schott, Dec 08 2021
The golden ratio is the unique number x such that f(n*x)*c(n/x) - f(n/x)*c(n*x) = n for all n >= 1, where f = floor and c = ceiling. - Clark Kimberling, Jan 04 2022
In The Second Scientific American Book Of Mathematical Puzzles and Diversions, Martin Gardner wrote that, by 1910, Mark Barr (1871-1950) gave phi as a symbol for the golden ratio. - Bernard Schott, May 01 2022
Phi is the length of the equal legs of an isosceles triangle with side c = phi^2, and internal angles (A,B) = 36 degrees, C = 108 degrees. - Gary W. Adamson, Jun 20 2022
The positive solution to x^2 - x - 1 = 0. - Michal Paulovic, Jan 16 2023
The minimal polynomial of phi^n, for nonvanishing integer n, is P(n, x) = x^2 - L(n)*x + (-1)^n, with the Lucas numbers L = A000032, extended to negative arguments with L(n) = (-1)^n*L(n). P(0, x) = (x - 1)^2 is not minimal. - Wolfdieter Lang, Feb 20 2025
This is the largest real zero x of (x^4 + x^2 + 1)^2 = 2*(x^8 + x^4 + 1). - Thomas Ordowski, May 14 2025

Examples

			1.6180339887498948482045868343656381177203091798057628621...

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 24, 112, 123, 184, 190, 203.
  • Michael Doob, The Canadian Mathematical Olympiad & L'Olympiade Mathématique du Canada 1969-1993 - Canadian Mathematical Society & Société Mathématique du Canada, Problem 4, 1975, pages 76-77, 1993.
  • Richard A. Dunlap, The Golden Ratio and Fibonacci Numbers, World Scientific, River Edge, NJ, 1997.
  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, Vol. 94, Cambridge University Press, 2003, Section 1.2.
  • Martin Gardner, The Second Scientific American Book Of Mathematical Puzzles and Diversions, "Phi: The Golden Ratio", Chapter 8, Simon & Schuster, NY, 1961.
  • Martin Gardner, Weird Water and Fuzzy Logic: More Notes of a Fringe Watcher, "The Cult of the Golden Ratio", Chapter 9, Prometheus Books, 1996, pages 90-97.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.5 The Fibonacci and Related Sequences, p. 287.
  • H. E. Huntley, The Divine Proportion, Dover, NY, 1970.
  • Mario Livio, The Golden Ratio, Broadway Books, NY, 2002. [see the review by G. Markowsky in the links field]
  • Gary B. Meisner, The Golden Ratio: The Divine Beauty of Mathematics, Race Point Publishing (The Quarto Group), 2018. German translation: Der Goldene Schnitt, Librero, 2023.
  • Scott Olsen, The Golden Section, Walker & Co., NY, 2006.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 137-139.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Hans Walser, The Golden Section, Math. Assoc. of Amer. Washington DC 2001.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See pp. 36-40.
  • Claude-Jacques Willard, Le nombre d'or, Magnard, Paris, 1987.

Links

Crossrefs

Cf. A000012 (continued fraction coefficients), A000032, A000045, A006497, A080039, A104457, A188635, A192222, A192223, A145996, A139339, A197762, A002163, A094874, A134973.
Cf. A102208, A102769, A131595.
Cf. A302973, A303069, A304022.

Programs

  • Maple
    Digits:=1000; evalf((1+sqrt(5))/2); # Wesley Ivan Hurt, Nov 01 2013
  • Mathematica
    RealDigits[(1 + Sqrt[5])/2, 10, 130] (* Stefan Steinerberger, Apr 02 2006 *)
RealDigits[ Exp[ ArcSinh[1/2]], 10, 111][[1]] (* Robert G. Wilson v, Mar 01 2008 *)
RealDigits[GoldenRatio,10,120][[1]] (* Harvey P. Dale, Oct 28 2015 *)
  • PARI
    default(realprecision, 20080); x=(1+sqrt(5))/2; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b001622.txt", n, " ", d));  \\ Harry J. Smith, Apr 19 2009
  • PARI
    /* Digit-by-digit method: write it as 0.5+sqrt(1.25) and start at hundredths digit */
r=11; x=400; print(1); print(6);
for(dig=1, 110, {d=0; while((20*r+d)*d <= x, d++);
d--; /* while loop overshoots correct digit */
print(d); x=100*(x-(20*r+d)*d); r=10*r+d})
\\ Michael B. Porter, Oct 24 2009
  • PARI
    a(n) = floor(10^(n-1)*(quadgen(5))%10);
alist(len) = digits(floor(quadgen(5)*10^(len-1))); \\ Chittaranjan Pardeshi, Jun 22 2022
  • Python
    from sympy import S
def alst(n): # truncate extra last digit to avoid rounding
  return list(map(int, str(S.GoldenRatio.n(n+1)).replace(".", "")))[:-1]
print(alst(105)) # Michael S. Branicky, Jan 06 2021

Formula

Equals Sum_{n>=2} 1/A064170(n) = 1/1 + 1/2 + 1/(2*5) + 1/(5*13) + 1/(13*34) + ... - Gary W. Adamson, Dec 15 2007
Equals Hypergeometric2F1([1/5, 4/5], [1/2], 3/4) = 2*cos((3/5)*arcsin(sqrt(3/4))). - Artur Jasinski, Oct 26 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
The fractional part of phi^n equals phi^(-n), if n is odd. For even n, the fractional part of phi^n is equal to 1-phi^(-n).
General formula: Provided x>1 satisfies x-x^(-1)=floor(x), where x=phi for this sequence, then:
for odd n: x^n - x^(-n) = floor(x^n), hence fract(x^n) = x^(-n),
for even n: x^n + x^(-n) = ceiling(x^n), hence fract(x^n) = 1 - x^(-n),
for all n>0: x^n + (-x)^(-n) = round(x^n).
x=phi is the minimal solution to x - x^(-1) = floor(x) (where floor(x)=1 in this case).
Other examples of constants x satisfying the relation x - x^(-1) = floor(x) include A014176 (the silver ratio: where floor(x)=2) and A098316 (the "bronze" ratio: where floor(x)=3). (End)
Equals 2*cos(Pi/5) = e^(i*Pi/5) + e^(-i*Pi/5). - Eric Desbiaux, Mar 19 2010
The solutions to x-x^(-1)=floor(x) are determined by x=(1/2)*(m+sqrt(m^2+4)), m>=1; x=phi for m=1. In terms of continued fractions the solutions can be described by x=[m;m,m,m,...], where m=1 for x=phi, and m=2 for the silver ratio A014176, and m=3 for the bronze ratio A098316. - Hieronymus Fischer, Oct 20 2010
Sum_{n>=1} x^n/n^2 = Pi^2/10 - (log(2)*sin(Pi/10))^2 where x = 2*sin(Pi/10) = this constant here. [Jolley, eq 360d]
phi = 1 + Sum_{k>=1} (-1)^(k-1)/(F(k)*F(k+1)), where F(n) is the n-th Fibonacci number (A000045). Proof. By Catalan's identity, F^2(n) - F(n-1)*F(n+1) = (-1)^(n-1). Therefore,(-1)^(n-1)/(F(n)*F(n+1)) = F(n)/F(n+1) - F(n-1)/F(n). Thus Sum_{k=1..n} (-1)^(k-1)/(F(k)*F(k+1)) = F(n)/F(n+1). If n goes to infinity, this tends to 1/phi = phi - 1. - Vladimir Shevelev, Feb 22 2013
phi^n = (A000032(n) + A000045(n)*sqrt(5)) / 2. - Thomas Ordowski, Jun 09 2013
Let P(q) = Product_{k>=1} (1 + q^(2*k-1)) (the g.f. of A000700), then A001622 = exp(Pi/6) * P(exp(-5*Pi)) / P(exp(-Pi)). - Stephen Beathard, Oct 06 2013
phi = i^(2/5) + i^(-2/5) = ((i^(4/5))+1) / (i^(2/5)) = 2*(i^(2/5) - (sin(Pi/5))i) = 2*(i^(-2/5) + (sin(Pi/5))i). - Jaroslav Krizek, Feb 03 2014
phi = sqrt(2/(3 - sqrt(5))) = sqrt(2)/A094883. This follows from the fact that ((1 + sqrt(5))^2)*(3 - sqrt(5)) = 8, so that ((1 + sqrt(5))/2)^2 = 2/(3 - sqrt(5)). - Geoffrey Caveney, Apr 19 2014
exp(arcsinh(cos(Pi/2-log(phi)*i))) = exp(arcsinh(sin(log(phi)*i))) = (sqrt(3) + i) / 2. - Geoffrey Caveney, Apr 23 2014
exp(arcsinh(cos(Pi/3))) = phi. - Geoffrey Caveney, Apr 23 2014
cos(Pi/3) + sqrt(1 + cos(Pi/3)^2). - Geoffrey Caveney, Apr 23 2014
2*phi = z^0 + z^1 - z^2 - z^3 + z^4, where z = exp(2*Pi*i/5). See the Wikipedia Kronecker-Weber theorem link. - Jonathan Sondow, Apr 24 2014
phi = 1/2 + sqrt(1 + (1/2)^2). - Geoffrey Caveney, Apr 25 2014
Phi is the limiting value of the iteration of x -> sqrt(1+x) on initial value a >= -1. - Chayim Lowen, Aug 30 2015
From Isaac Saffold, Feb 28 2018: (Start)
1 = Sum_{k=0..n} binomial(n, k) / phi^(n+k) for all nonnegative integers n.
1 = Sum_{n>=1} 1 / phi^(2n-1).
1 = Sum_{n>=2} 1 / phi^n.
phi = Sum_{n>=1} 1/phi^n. (End)
From Christian Katzmann, Mar 19 2018: (Start)
phi = Sum_{n>=0} (15*(2*n)! + 8*n!^2)/(2*n!^2*3^(2*n+2)).
phi = 1/2 + Sum_{n>=0} 5*(2*n)!/(2*n!^2*3^(2*n+1)). (End)
phi = Product_{k>=1} (1 + 2/(-1 + 2^k*(sqrt(4+(1-2/2^k)^2) + sqrt(4+(1-1/2^k)^2)))). - Gleb Koloskov, Jul 14 2021
Equals Product_{k>=1} (Fibonacci(3*k)^2 + (-1)^(k+1))/(Fibonacci(3*k)^2 + (-1)^k) (Melham and Shannon, 1995). - Amiram Eldar, Jan 15 2022
From Michal Paulovic, Jan 16 2023: (Start)
Equals the real part of 2 * e^(i * Pi / 5).
Equals 2 * sin(3 * Pi / 10) = 2*A019863.
Equals -2 * sin(37 * Pi / 10).
Equals 1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / ...)))).
Equals (2 + 3 * (2 + 3 * (2 + 3 * ...)^(1/4))^(1/4))^(1/4).
Equals (1 + 2 * (1 + 2 * (1 + 2 * ...)^(1/3))^(1/3))^(1/3).
Equals (1 + phi + (1 + phi + (1 + phi + ...)^(1/3))^(1/3))^(1/3).
Equals 13/8 + Sum_{k=0..oo} (-1)^(k+1)*(2*k+1)!/((k+2)!*k!*4^(2*k+3)).
(End)
phi^n = phi * A000045(n) + A000045(n-1). - Gary W. Adamson, Sep 09 2023
The previous formula holds for integer n, with F(-n) = (-1)^(n+1)*F(n), for n >= 0, with F(n) = A000045(n), for n >= 0. phi^n are integers in the quadratic number field Q(sqrt(5)). - Wolfdieter Lang, Sep 16 2023
Equals Product_{k>=0} ((5*k + 2)*(5*k + 3))/((5*k + 1)*(5*k + 4)). - Antonio Graciá Llorente, Feb 24 2024
From Antonio Graciá Llorente, Apr 21 2024: (Start)
Equals Product_{k>=1} phi^(-2^k) + 1, with phi = A001622.
Equals Product_{k>=0} ((5^(k+1) + 1)*(5^(k-1/2) + 1))/((5^k + 1)*(5^(k+1/2) + 1)).
Equals Product_{k>=1} 1 - (4*(-1)^k)/(10*k - 5 + (-1)^k) = Product_{k>=1} A047221(k)/A047209(k).
Equals Product_{k>=0} ((5*k + 7)*(5*k + 1 + (-1)^k))/((5*k + 1)*(5*k + 7 + (-1)^k)).
Equals Product_{k>=0} ((10*k + 3)*(10*k + 5)*(10*k + 8)^2)/((10*k + 2)*(10*k + 4)*(10*k + 9)^2).
Equals Product_{k>=5} 1 + 1/(Fibonacci(k) - (-1)^k).
Equals Product_{k>=2} 1 + 1/Fibonacci(2*k).
Equals Product_{k>=2} (Lucas(k)^2 + (-1)^k)/(Lucas(k)^2 - 4*(-1)^k). (End)

Extensions

Additional links contributed by Lekraj Beedassy, Dec 23 2003
More terms from Gabriel Cunningham (gcasey(AT)mit.edu), Oct 24 2004
More terms from Stefan Steinerberger, Apr 02 2006
Broken URL to Project Gutenberg replaced by Georg Fischer, Jan 03 2009
Edited by M. F. Hasler, Feb 24 2014

A335866 Number of classes of simple difference sets of the Singer type (m^2 + m + 1, m + 1, 1) with m = m(n) = A000961(n), for n >= 1.

Original entry on oeis.org

1, 2, 4, 2, 10, 12, 8, 12, 36, 40, 12, 102, 84, 156, 60, 84, 264, 220, 60, 264, 574, 420, 720, 252, 816, 1180, 768, 144, 840, 1704, 1200, 1176, 432, 2196, 2670, 2112, 3434, 2380, 3024, 2280, 3960, 1296, 1656, 3612, 672, 5764, 5184, 3984, 6120, 4368, 5512, 4752, 9352, 3120, 10034, 9204, 7176, 9360, 7128
Offset: 1

Views

Author

Wolfdieter Lang, Jul 26 2020

Keywords

Comments

For details on these simple difference sets see A333852, with references, and a W. Lang link.
The formula given below was conjectured by Singer for n >= 2 on p. 383. See also the table on p. 384.
This conjecture was later proved by Berman.

Examples

			n = 2, m(2) = 2 = 2^1, a(2) = phi(7)/(3*1) = 6/3 = 2. There are two classes of type (7,3,1) (Fano plane), with representatives {0, 1, 3} and {0, 1, 5}. The two equivalence classes (by elementwise addition of 1, 2, ..., 6 modulo 7) are Dev({0, 1, 3}) = {{0, 1, 3}, {0, 2, 6}, {0, 4, 5}, {1, 2, 4}, {1, 5, 6}, {2, 3, 5}, {3, 4, 6}, and Dev({0, 1, 5}) = {{0, 1, 5}, {0, 2, 3}, {0, 4, 6}, {1, 2, 6}, {1, 3, 4}, {2, 4, 5}, {3, 5, 6}}.

Links

Crossrefs

Cf. A000010, A025474, A000961, A333852, A335865.

Programs

  • PARI
    print1(1); for(q=2, 193 , if(n=isprimepower(q), print1(", ", eulerphi(q^2+q+1)/(3*n)))) \\ Martin Becker, Jun 11 2024

Formula

a(1) = 1, and a(n) = phi(v(n))/(3*e(n)), with phi = A000010 (Euler's totient), v(n) = A335865(n) = m(n)^2 + m(n) + 1, with m(n) = A000961(n), and e(n) = A025474(n), the exponent of the prime power dividing m(n), for n >= 2.

A335865 Moduli a(n) = v(n) for the simple difference sets of Singer type of order m(n) (v(n), m(n)+1, 1) in the additive group modulo v(n) = m(n)^2 + m(n) + 1, with m(n) = A000961(n).

Original entry on oeis.org

3, 7, 13, 21, 31, 57, 73, 91, 133, 183, 273, 307, 381, 553, 651, 757, 871, 993, 1057, 1407, 1723, 1893, 2257, 2451, 2863, 3541, 3783, 4161, 4557, 5113, 5403, 6321, 6643, 6973, 8011, 9507, 10303, 10713, 11557, 11991, 12883, 14763, 15751
Offset: 1

Views

Author

Wolfdieter Lang, Jul 26 2020

Keywords

Comments

For details on these difference sets see A333852, with references, and a W. Lang link.
Because these simple difference sets of Singer type of order m = m(n) in the addive group (Z_{v(n)}, +) = RS(v(n)) = {0, 1, ..., v(n)-1} are also simple symmetric balanced incomplete block designs (BIBD), the number of blocks b(n) is also v(n) = a(n). This is the number of simple difference sets of each of the A335865(n) classes.
From Ed Pegg Jr, May 16 2019, edited by Hugo Pfoertner, May 13 2024: (Start)
(n^2+n+1,n+1) difference sets exist when n is a prime power.
(7,3), (1,2,4)
(13,4), (0,1,3,9)
(21,5), (3,6,7,12,14) (A095029)
(31,6), (1,5,11,24,25,27) (A095030)
(57,8), (0,1,6,15,22,26,45,55) (A095032)
(73,9), (0,1,12,20,26,30,33,35,57) (A095035)
(91,10), (0,2,6,7,18,21,31,54,63,71) (A095036)
(133,12), (1,10,11,13,27,31,68,75,83,110,115,121) (A095038)
(183,14), (1,13,20,21,23,44,61,72,77,86,90,116,122,169) (A095040) (End)
Is a(n) = A138077(n-1)? - R. J. Mathar, Sep 11 2020

Examples

			n = 2, m(2) = 2, a(2) = 2^2 + 2 + 1 = 7. The simple Singer difference set of order 2 is denoted by (7, 3, 1) (Fano plane). There are two classes (A335866(2) = 2) obtained from the representative difference sets {0, 1, 3} and {0, 1, 5} by element-wise addition of 1, 2, ..., 6 taken modulo 7. Each class consists of 7 simple difference sets.

Links

Crossrefs

Cf. A000961, A333852, A335866, A138077.

Formula

a(n) = m(n)^2 + m(n) + 1 , with m(n) = A000961(n), for n >= 1.

Extensions

Comments about difference sets moved from A138077 to here by Max Alekseyev, Apr 05 2022

A353077 Triangle read by rows, where the n-th row consists of the lexicographically earliest solution for n integers in 0..p-1 whose n*(n-1) differences are congruent to 1..p-1 (mod p), where p=n*(n-1)+1. If no solution exists, the n-th row consists of n -1's.

Original entry on oeis.org

0, 0, 1, 0, 1, 3, 0, 1, 3, 9, 0, 1, 4, 14, 16, 0, 1, 3, 8, 12, 18, -1, -1, -1, -1, -1, -1, -1, 0, 1, 3, 13, 32, 36, 43, 52, 0, 1, 3, 7, 15, 31, 36, 54, 63, 0, 1, 3, 9, 27, 49, 56, 61, 77, 81, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 0, 1, 3, 12, 20, 34, 38, 81, 88, 94, 104, 109
Offset: 1

Views

Author

Michel Marcus, Apr 22 2022

Keywords

Comments

Comment from Martin Becker, May 18 2025: (Start)
Rows k with k-1 not a prime power are precisely the rows with -1 values for k <= 2*10^10. Cf. the Gordon (2020) link.
In the b-file, values a(n) > 3 from row 17 onwards depend on the conjecture that all perfect difference sets are Singer type, and were obtained through computer enumeration of Singer type sets.
(End)

Examples

			   n   row
   1 [0];
   2 [0,1];
   3 [0,1,3];
   4 [0,1,3,9];
   5 [0,1,4,14,16];
   6 [0,1,3,8,12,18];
   7 no solution exists;
   8 [0,1,3,13,32,36,43,52];
   9 [0,1,3,7,15,31,36,54,63];
  10 [0,1,3,9,27,49,56,61,77,81];
  11 no solution exists;
  12 [0,1,3,12,20,34,38,81,88,94,104,109];
  13 no solution exists;
  14 [0,1,3,16,23,28,42,76,82,86,119,137,154,175];
  15 no solution exists;
  16 no solution exists.

Links

Crossrefs

Cf. A002061, A058241, A333852, A351690, A383574.

Programs

  • PARI
    isok(n, v) = my(p=n*(n-1)+1); setbinop((x,y)->lift(Mod(x-y, p)), v, v) == [0..p-1];
row(n) = forsubset([n^2-n+1, n], s, my(ds = apply(x->x-1, Vec(s))); if (isok(n, ds), return(ds)););

Extensions

Name and data corrected for "lexicographically earliest solution" by Michel Marcus, May 09 2022
Adjusted to a regular triangle, and rows 1, 2, 7, and 10-12 inserted by Pontus von Brömssen, May 09 2022

A373514 Number of simple difference sets of the Singer type (m^2 + m + 1, m + 1, 1) that are a superset of {0, 1, 3} with m = m(n) = A000961(n), for n >= 1.

Original entry on oeis.org

0, 1, 1, 0, 2, 1, 1, 1, 4, 3, 1, 6, 3, 8, 2, 3, 9, 6, 2, 8, 14, 10, 14, 4, 14, 20, 10, 2, 14, 24, 15, 18, 6, 27, 30, 19, 34, 21, 26, 22, 33, 10, 13, 30, 5, 44, 38, 30, 41, 26, 36, 25, 56, 17, 58, 52, 38, 51, 40, 63, 45, 41, 46, 76, 47, 70, 72, 55, 15, 80, 6
Offset: 1

Views

Author

Martin Becker, Jun 07 2024

Keywords

Examples

			For n=5, m=5, there are 2 Singer type planar difference sets of order 5 containing 0, 1, and 3: {0,1,3,8,12,18} and {0,1,3,10,14,26}. Thus a(5) = 2.
For n=11, m=16, there is only 1 such set: {0,1,3,7,15,31,63,90,116,127,136,181,194,204,233,238,255}. Thus a(11) = 1.

Links

Crossrefs

Cf. A335866, A000961, A373946. Counts sets in A333852 with the property that 3 is also in the set.

A335862 Decimal expansion of the zero x1 of the cubic polynomial x^3 - 2*x^2 - 10*x - 6.

Original entry on oeis.org

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

Views

Author

Wolfdieter Lang, Jun 29 2020

Keywords

Comments

This cubic polynomial P3(x) = x^3 - 2*x^2 - 10*x - 6 is a factor of the characteristic polynomial F(x) of degree 7 of the 7 X 7 adjacency matrix F7 of the Fano graph with nodes (vertices) of degree 6, 5, 5, 5, 3, 3, 3. See the links for the Fano plane. The graph is in fact planar.
The adjacency matrix is F7 = Matrix([[0, 1, 1, 1, 1, 1, 1], [1, 0, 1, 1, 1, 1, 0], [1, 1, 0, 1, 0, 1, 1], [1, 1, 1, 0, 1, 0, 1], [1, 1, 0, 1, 0, 0, 0], [1, 1, 1, 0, 0, 0, 0], [1, 0, 1, 1, 0, 0, 0]]).
The determinant of F7 is 6. The characteristic polynomial is F(x) = x^7 - 15*x^5 - 26*x^4 + 3*x^3 + 24*x^2 + 2*x - 6 = P3(x)*(x^2 + x - 1)^2. The zeros of F(x) (the eigenvalues or spectrum of F7) are: x1, x2 = -A335863 = -1.752517821..., x3 = -A335864 = -0.7588868422..., twice -1 + phi = 0.618033988..., and twice -phi, where phi = A001622.
For the bipartite incidence graph see the links for the Heawood graph.

Examples

			x1 = 4.5114046642267581233392214968131695740218436164...

Links

Crossrefs

Cf. A001622, A335863 (-x2), A335864 (-x3).

Programs

  • Mathematica
    With[{k = 3 Sqrt[3] Sqrt[269] I}, First@ RealDigits[Re[(1/3) (2 + (179 + k)^(1/3) + (179 - k)^(1/3))], 10, 100]] (* Michael De Vlieger, Nov 17 2020 *)

Formula

x1 = (1/3)*(2 + (179 + 3*sqrt(3)*sqrt(269)*i)^(1/3) + ( 179 - 3*sqrt(3)*sqrt(269)*i)^(1/3)), where i is the imaginary unit.

A335863 Decimal expansion of the negative of the zero x2 of the cubic polynomial x^3 - 2*x^2 - 10*x - 6.

Original entry on oeis.org

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

Views

Author

Wolfdieter Lang, Jun 29 2020

Keywords

Comments

For details and links see A335862.

Examples

			-x2 = 1.7525178219298168184898392124373100279...

Links

Crossrefs

Cf. A335862 (x1), A335864 (-x3).

Programs

  • Mathematica
    With[{j = Sqrt[3] I, k = 3 Sqrt[3] Sqrt[269] I}, First@ RealDigits[Re[(1/3) (2 - (1/2) (1 - j) (179 + k)^(1/3) - (1/2) (1 + j) (179 - k)^(1/3))], 10, 99]] (* Michael De Vlieger, Nov 17 2020 *)

Formula

-x2 = (1/3)*(2 - (1/2)*(1 - sqrt(3)*i)*(179 + 3*sqrt(3)*sqrt(269)*i)^(1/3) - (1/2)*(1 + sqrt(3)*i)*(179 - 3*sqrt(3)*sqrt(269)*i)^(1/3)), where i is the imaginary unit.

A335864 Decimal expansion of the negative of the zero x3 of the cubic polynomial x^3 - 2*x^2 - 10*x - 6.

Original entry on oeis.org

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

Views

Author

Wolfdieter Lang, Jun 29 2020

Keywords

Comments

For details and links see A335862.

Examples

			-x3 = 0.758886842296941304849382284375859546...

Links

Crossrefs

Cf. A335862 (x1), A335863 (-x2).

Programs

  • Maple
    evalf((f-> (sqrt(34)*(cos(f)-sin(f)*sqrt(3))-2)/3)(arctan(sqrt(807)*3/179)/3), 120);  # Alois P. Heinz, Nov 17 2020
  • Mathematica
    With[{j = Sqrt[3] I, k = 3 Sqrt[3] Sqrt[269] I}, First@ RealDigits[Re[(1/3) (2 - (1/2) (1 + j) (179 + k)^(1/3) - (1/2) (1 - j) (179 - k)^(1/3))], 10, 97]] (* Michael De Vlieger, Nov 17 2020 *)

Formula

-x3 = (1/3)*(2 - (1/2)*(1 + sqrt(3)*i)*(179 + 3*sqrt(3)*sqrt(269)*i)^(1/3) - (1/2)*(1 - sqrt(3)*i)*(179 - 3*sqrt(3)*sqrt(269)*i)^(1/3)), where i is the imaginary unit.

A383574 Fourth column of A353077.

Original entry on oeis.org

9, 14, 8, -1, 13, 7, 9, -1, 12, -1, 16, -1, -1, 7, 21, -1, 12, -1, -1, -1, 13, -1, 33, -1, 9, -1, 12, -1, 13, 7, -1, -1, -1, -1, 19, -1, -1, -1, 8, -1, 10, -1, -1, -1, 10, -1, 25, -1, -1, -1, 15, -1, -1, -1, -1, -1, 8, -1, 16, -1, -1, 7, -1, -1, 12, -1, -1
Offset: 4

Views

Author

Martin Becker, May 03 2025

Keywords

Comments

Integers 0 to 6 are not in the sequence: For n > 5, the first three columns of A353077 are necessarily -1, -1, -1 or 0, 1, 3, and the fourth column is -1 or > 3, respectively. It is actually > 6 in the second case, as 4 - 3 = 1 - 0, 5 - 3 = 3 - 1, 6 - 3 = 3 - 0, respectively, would violate the distinctness of differences in perfect difference sets.
For n = 2^m + 1, m > 2, a(n) = 7, because 2 is a multiplier of such sets, therefore perfect difference sets containing 1, 2, 4, and 8 with translated sets containing 0, 1, 3, and 7 exist.
If n-1 is a prime power, a(n) != -1, as then there exist Singer type perfect difference sets.
If 4 <= n < 2*10^10 and n-1 is not a prime power, a(n) = -1. Cf. Gordon (2020).
Empirical observations further suggest that:
For n = 3^m + 1, m >= 1, a(n) = 9.
The most frequent positive value is 10.
11 is not in the sequence.

Examples

			For n = 4, there are 4 perfect difference sets containing 0 and 1: {0, 1, 3, 9}, {0, 1, 4, 6}, {0, 1, 5, 11}, and {0, 1, 8, 10}. The lexically earliest is {0, 1, 3, 9}. Its fourth element is 9, thus a(4) = 9.
There are no perfect difference sets with 7 elements. Thus a(7) = -1.

Links

Crossrefs

Fourth column of A353077.
Cf. A000961, A333852, A373514.
Showing 1-9 of 9 results.

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