A135927 -id:A135927 - cp's OEIS frontend

A246723 Decimal expansion of r_1, the smallest radius for which a compact packing of the plane exists, with disks of radius 1 and r_1.

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

Offset: 0

Jean-François Alcover, Sep 02 2014

			0.101020514433643803605431850588217216068105038686659743...

Steven R. Finch, Errata and Addenda to Mathematical Constants, arXiv:2001.00578 [math.HO], 2020-2021, p. 62.
Gabriel Klambauer, Summation of Series, Amer. Math. Monthly, Vol. 87, No. 2 (Feb., 1980), pp. 128-130.
Wikipedia, Engel expansion
Index entries for algebraic numbers, degree 2

Cf. A246724 (r_2), A246725 (r_3), A246726 (r_4), A246727 (r_5), A002193 (r_6 = sqrt(2)-1), A246728 (r_7), A246729 (r_8), A246730 (r_9).

RealDigits[5 - 2*Sqrt[6], 10, 104] // First

Equals 5 - 2*sqrt(6).
Equals Sum_{k>=1} binomial(2*k,k)/((k+1) * 12^k). - Amiram Eldar, Oct 04 2021
Engel expansion of 5 - 2*sqrt(6) = 1/10 + 1/(10*98) + 1/(10*98*9602) + ..., where [10, 98, 9602, ...] = A135927. See Klambauer, p. 130. - Peter Bala, Feb 01 2022
Equals exp(-arccosh(5)). - Amiram Eldar, Jul 06 2023

A084765 a(n) = 2*a(n-1)^2 - 1, a(0)=1, a(1)=5.

Original entry on oeis.org

1, 5, 49, 4801, 46099201, 4250272665676801, 36129635465198759610694779187201, 2610701117696295981568349760414651575095962187244375364404428801

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Jun 04 2003

Product_{k>=1} (1 + 1/a(k)) = sqrt(3/2) (see A010527).
A subsequence of A001079 (cf. formula), which must contain any prime occurring in A001079. The initial term a(0)=1 seems rather unnatural; using the recurrence relation it would yield the constant sequence 1,1,1,... Note that this sequence corresponds to sequence b(n) in Shallit's paper, which starts only at offset n=1. - M. F. Hasler, Sep 27 2009
Since if x is even (x^2-2)/2 = 2*y^2-1 and 10 is even from a(1) onward this is a reduced version of the LL sequence starting with 10 (A135927) as it is reduced by dividing by 2 it is also the difference between two possible LL sequences. - Roderick MacPhee, May 31 2015
For n >= 3, a(n) == 201 (mod 1000) if n is even, a(n) == 801 (mod 1000) if n is odd. - Robert Israel, Jun 01 2015
The next term -- a(8) -- has 128 digits. - Harvey P. Dale, Mar 28 2020

G. C. Greubel, Table of n, a(n) for n = 0..10
Jeffrey Shallit, Rational numbers with non-terminating, non-periodic modified Engel-type expansions, Fib. Quart., 31 (1993), 37-40.
H. S. Wilf, Limit of a sequence, Elementary Problem E 1093, Amer. Math. Monthly 61 (1954), 424-425.

Cf. A001079, A002812, A010527, A084764, A135927.

[n le 2 select 5^(n-1) else 2*Self(n-1)^2-1: n in [1..10]]; // Vincenzo Librandi, Jun 02 2015

1,seq(expand((5+2*sqrt(6))^(2^n)+(5-2*sqrt(6))^(2^n))/2, n=0..10); # Robert Israel, Jun 01 2015

a[n_]:= a[n]= If[n<2, 5^n, 2 a[n-1]^2 -1]; Table[a[n], {n,0,10}]
Join[{1}, NestList[2 #^2 - 1 &, 5, 10]] (* Harvey P. Dale, Mar 28 2020 *)

first(m)={my(v=[1,5]);for(i=3,m,v=concat(v, 2*v[i-1]^2 - 1));v;} \\ Anders Hellström, Aug 22 2015

def A084765(n): return 1 if n==0 else chebyshev_T(2^(n-1), 5)
[A084765(n) for n in range(11)] # G. C. Greubel, May 17 2023

a(n+1) = (x^(2^n) + y^(2^n))/2, with x = 5 + 2*sqrt(6), y = 5 - 2*sqrt(6).
a(n) = A001079(2^(n-1)) with a(0) = 1. - M. F. Hasler, Sep 27 2009
4*sqrt(6)/11 = Product_{n >= 1} (1 - 1/(2*a(n))). See A002812 for some general properties of the recurrence a(n+1) = 2*a(n)^2 - 1. - Peter Bala, Nov 11 2012
a(n) = cos(2^(n-1)*arccos(5)) for n >= 1. - Peter Luschny, Oct 12 2022

A087799 a(n) = 10*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 10.

Original entry on oeis.org

2, 10, 98, 970, 9602, 95050, 940898, 9313930, 92198402, 912670090, 9034502498, 89432354890, 885289046402, 8763458109130, 86749292044898, 858729462339850, 8500545331353602, 84146723851196170, 832966693180608098, 8245520207954884810

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 11 2003

a(n+1)/a(n) converges to (5+sqrt(24)) = 9.8989794... a(0)/a(1)=2/10; a(1)/a(2)=10/98; a(2)/a(3)=98/970; a(3)/a(4)=970/9602; ... etc. Lim a(n)/a(n+1) as n approaches infinity = 0.10102051... = 1/(5+sqrt(24)) = (5-sqrt(24)).
Except for the first term, positive values of x (or y) satisfying x^2 - 10xy + y^2 + 96 = 0. - Colin Barker, Feb 25 2014
A triangle whose sides are a(n) - 1, a(n) and a(n) + 1 is nearly Fleenor-Heronian since its area is the product of an integer and the square root of 2. See A003500. - Charlie Marion, Dec 18 2020

			a(4) = 9602 = 10*a(3) - a(2) = 10*970 - 98 = (5+sqrt(24))^4 + (5-sqrt(24))^4.

T. D. Noe, Table of n, a(n) for n = 0..200
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (10,-1).

Cf. A005248, A006242, A036336, A086927, A135927, A174503.

I:=[2,10]; [n le 2 select I[n] else 10*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Nov 07 2018

a[0] = 2; a[1] = 10; a[n_] := 10a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 17}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{10,-1}, {2,10}, 30] (* G. C. Greubel, Nov 07 2018 *)

polsym(x^2 - 10*x + 1,20) \\ Charles R Greathouse IV, Jun 11 2011

{a(n) = 2 * real( (5 + 2 * quadgen(24))^n )}; /* Michael Somos, Feb 25 2014 */

[lucas_number2(n,10,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

a(n) = (5+sqrt(24))^n + (5-sqrt(24))^n.
G.f.: (2-10*x)/(1-10*x+x^2). - Philippe Deléham, Nov 02 2008
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 5 - sqrt(24). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.09989 80642 72052 68138 ... = 2 + 1/(10 + 1/(98 + 1/(970 + ...))).
Also F(-alpha) = 0.89989 78538 78393 34715 ... has the continued fraction representation 1 - 1/(10 - 1/(98 - 1/(970 - ...))) and the simple continued fraction expansion 1/(1 + 1/((10-2) + 1/(1 + 1/((98-2) + 1/(1 + 1/((970-2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((10^2-4) + 1/(1 + 1/((98^2-4) + 1/(1 + 1/((970^2-4) + 1/(1 + ...))))))). Cf. A174503 and A005248. (End)
a(-n) = a(n). - Michael Somos, Feb 25 2014
From Peter Bala, Oct 16 2019: (Start)
8*Sum_{n >= 1} 1/(a(n) - 12/a(n)) = 1.
12*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 8/a(n)) = 1.
Series acceleration formulas for sums of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/8 - 12*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 12))  and
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/12 + 8*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 8)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(5-sqrt(24)))^2 - 1 )/4 and
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - (theta_3(sqrt(24)-5))^2 )/4, where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499. (End)
E.g.f.: 2*exp(5*x)*cosh(2*sqrt(6)*x). - Stefano Spezia, Oct 18 2019
From Peter Bala, Mar 29 2022: (Start)
a(n) = 2*T(n,5), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
a(2^n) = A135927(n+1) and a(3^n) = A006242(n+1), both for n >= 0. (End)

More terms from Colin Barker, Feb 25 2014

A135928 Digital roots of the Mersenne primes.

Original entry on oeis.org

3, 7, 4, 1, 1, 4, 1, 1, 1, 4, 4, 1, 4, 1, 1, 1, 1, 1, 4, 1, 4, 4, 4, 4, 4, 1, 1, 4, 1, 1, 1, 4, 4, 1, 4, 4, 4, 4, 1, 1, 1, 4, 1, 4, 4, 1, 4, 4

Offset: 1

Ant King, Dec 07 2007

As a consequence of the fact that all prime numbers are of the form 6n-1 or 6n+1 for p>3, all the elements of this sequence after the second will be either 1 or 4, although there is no obvious pattern to their distribution. We can use this result to show that all Mersenne primes after the first are congruent to 1, modulo 6.

			The fourth Mersenne prime is 127, which has a digital root of 1. Hence a(4)=1.

Syed Asadulla, Digital Roots of Mersenne Primes and Even Perfect Numbers, The College Mathematics Journal, Vol. 15, No. 1. (1984), pp. 53-54.
Eric Weisstein's World of Mathematics, Digital Root.

Cf. A000668, A000043, A003010, A001566, A010888, A135927.

DigitalRoot[n_]:=FixedPoint[Plus@@IntegerDigits[ # ]&,n];data1=Select[Range[4500],PrimeQ[2^#-1] &];data2=2^#-1 &/@data1;DigitalRoot/@data2

a(n) = A010888(A000668(n)).

For n > 2, a(n) = (A000043(n) mod 3)^2. - Jens Kruse Andersen, Jul 29 2014

a(40)-a(43) (using A000043) from Jens Kruse Andersen, Jul 29 2014

a(44)-a(48) from mersenne.org added by M Sayer, Jan 05 2023

A319749 a(n) is the numerator of the Heron sequence with h(0)=3.

Original entry on oeis.org

3, 11, 119, 14159, 200477279, 40191139395243839, 1615327685887921300502934267457919, 2609283532796026943395592527806764363779539144932833602430435810559

Offset: 0

Paul Weisenhorn, Sep 27 2018

The denominator of the Heron sequence is in A319750.
The following relationship holds between the numerator of the Heron sequence and the numerator of the continued fraction A041018(n)/A041019(n) convergent to sqrt(13).
n even: a(n)=A041018((5*2^n-5)/3).
n  odd: a(n)=A041018((5*2^n-1)/3).
More generally, all numbers c(n)=A078370(n)=(2n+1)^2+4 have the same relationship between the numerator of the Heron sequence and the numerator of the continued fraction convergent to 2n+1.
sqrt(c(n)) has the continued fraction 2n+1; n,1,1,n,4n+2.
hn(n)^2-c(n)*hd(n)^2=4 for n>1.
From Peter Bala, Mar 29 2022: (Start)
Applying Heron's method (sometimes called the Babylonian method) to approximate the square root of the function x^2 + 4, starting with a guess equal to x, produces the sequence of rational functions [x, 2*T(1,(x^2+2)/2)/x, 2*T(2,(x^2+2)/2)/( 2*x*T(1,(x^2+2)/2) ), 2*T(4,(x^2+2)/2)/( 4*x*T(1,(x^2+2)/2)*T(2,(x^2+2)/2) ), 2*T(8,(x^2+2)/2)/( 8*x*T(1,(x^2+2)/2)*T(2,(x^2+2)/2)*T(4,(x^2+2)/2) ), ...], where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. The present sequence is the case x = 3. Cf. A001566 and A058635 (case x = 1), A081459 and A081460 (essentially the case x = 4). (End)

			A078370(2)=29.
hn(0)=A041046(0)=5; hn(1)=A041046(3)=27; hn(2)=A041046(5)=727;
hn(3)=A041046(13)=528527.

P. Liardet and P. Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
Wikipedia, Engel Expansion
Wikipedia, Methods of computing square roots
Index entries for sequences of form a(n+1)=a(n)^2 + ...

2*T(2^n,x/2) modulo differences of offset: A001566 (x = 3 and x = 7), A003010 (x = 4), A003487 (x = 5), A003423 (x = 6), A346625 (x = 8), A135927 (x = 10), A228933 (x = 18).

Cf. A041018, A041019, A057076, A078370, A081459, A010122, A319750, A041046.

hn[0]:=3:  hd[0]:=1:
for n from 1 to 6 do
hn[n]:=(hn[n-1]^2+13*hd[n-1]^2)/2:
hd[n]:=hn[n-1]*hd[n-1]:
   printf("%5d%40d%40d\n", n, hn[n], hd[n]):
end do:
#alternative program
a := n -> if n = 0 then 3 else simplify( 2*ChebyshevT(2^(n-1), 11/2) ) end if:
seq(a(n), n = 0..7); # Peter Bala, Mar 16 2022

def aupton(nn):
    hn, hd, alst = 3, 1, [3]
    for n in range(nn):
        hn, hd = (hn**2 + 13*hd**2)//2, hn*hd
        alst.append(hn)
    return alst
print(aupton(7)) # Michael S. Branicky, Mar 16 2022

h(n) = hn(n)/hd(n); hn(0)=3; hd(0)=1.
hn(n+1) = (hn(n)^2+13*hd(n)^2)/2.
hd(n+1) = hn(n)*hd(n).
A041018(n) = A010122(n)*A041018(n-1) + A041018(n-2).
A041019(n) = A010122(n)*A041019(n-1) + A041019(n-2).
From Peter Bala, Mar 16 2022: (Start)
a(n) = 2*T(2^(n-1),11/2) for n >= 1, where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
a(n) = 2*T(2^n, 3*sqrt(-1)/2) for n >= 2.
a(n) = ((11 + 3*sqrt(13))/2)^(2^(n-1)) + ((11 - 3*sqrt(13))/2)^(2^(n-1)) for n >= 1.
a(n+1) = a(n)^2 - 2 for n >= 1.
a(n) = A057076(2^(n-1)) for n >= 1.
Engel expansion of (1/6)*(13 - 3*sqrt(13)); that is, (1/6)*(13 - 3*sqrt(13)) = 1/3 + 1/(3*11) + 1/(3*11*119) + .... (Define L(n) = (1/2)*(n - sqrt(n^2 - 4)) for n >= 2 and show L(n) = 1/n + L(n^2-2)/n. Iterate this relation with n = 11. See also Liardet and Stambul, Section 4.)
sqrt(13) = 6*Product_{n >= 0} (1 - 1/a(n)).
sqrt(13) = (9/5)*Product_{n >= 0} (1 + 2/a(n)). See A001566. (End)

a(6) and a(7) added by Peter Bala, Mar 16 2022

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A246723 Decimal expansion of r_1, the smallest radius for which a compact packing of the plane exists, with disks of radius 1 and r_1.

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A084765 a(n) = 2*a(n-1)^2 - 1, a(0)=1, a(1)=5.

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A087799 a(n) = 10*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 10.

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A135928 Digital roots of the Mersenne primes.

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A319749 a(n) is the numerator of the Heron sequence with h(0)=3.

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