A003010 -id:A003010 - cp's OEIS frontend

A135928 Digital roots of the Mersenne primes.

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

A217870 Decimal expansion of sqrt(sqrt(2 + sqrt(3))).

Original entry on oeis.org

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

Offset: 1

Arkadiusz Wesolowski, Oct 13 2012

It can be used for a Lucas-Lehmer test of prime numbers.

The value is equal to e^(log(2 + sqrt(3))/4) = e^A182023.

			1.389910663524147717911548811992210102196089903539205052651822014331759...

Arkadiusz Wesolowski, Table of n, a(n) for n = 1..1000
Gottfried Helms, The Lucas-Lehmer-test for Mersenne-numbers and the number lambda ~ 1.389910663524...

Cf. A182023, A003010.

SetDefaultRealField(RealField(100)); Sqrt(Sqrt(2 + Sqrt(3))); // G. C. Greubel, Sep 29 2018

evalf(sqrt(sqrt(2+sqrt(3))),120); # Muniru A Asiru, Sep 30 2018

RealDigits[N[Sqrt@Sqrt[2 + Sqrt[3]], 200]][[1]]

fpprec : 100$ bfloat(sqrt(sqrt(2 + sqrt(3)))); /* Martin Ettl, Oct 15 2012 */

default(realprecision, 200); x=sqrt(sqrt(2+sqrt(3))); for(n=1, 200, d=floor(x); x=(x-d)*10; print1(d, ", "));

Equals (2+sqrt(3))^(1/4). - Vaclav Kotesovec, Oct 18 2014

A182023 Decimal expansion of log(2+sqrt(3))/4.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Apr 06 2012

			0.32923947423120417715626158682699211100674549286687911994211...

Gottfried Helms and others, Postings to Sequence Fans Mailing List, Apr 05 2012

Index entries for transcendental numbers

Cf. A003010.

Equals A065918/4.

SetDefaultRealField(RealField(100)); Log(2 + Sqrt(3)) / 4; // Vincenzo Librandi, May 16 2019

RealDigits[N[Log[2 + Sqrt[3]] / 4, 150]][[1]] (* Vincenzo Librandi, May 16 2019 *)

log(sqrt(3)+2)/4 \\ Charles R Greathouse IV, May 15 2019

A227615 Number of bits necessary to represent u(n) in binary, where u is the Lucas-Lehmer sequence: u(0) = 100 (in binary); for n>0, u(n) = u(n-1)^2 - 2.

Original entry on oeis.org

3, 4, 8, 16, 31, 61, 122, 244, 487, 973, 1946, 3892, 7783, 15565, 31130, 62259, 124517, 249033, 498066, 996131, 1992262, 3984524, 7969047, 15938093, 31876185, 63752369, 127504737, 255009473, 510018945, 1020037890, 2040075780

Offset: 0

Olivier de Mouzon, Jul 17 2013

a(0)=3, a(1)=4 and for n>=1, a(n+1) is 2*a(n) or 2*a(n)-1.
It seems the rule to decide between the 2 is not straightforward. So you actually need to compute u(n) to have its required number of bits.
Yet, for n>=1, we have a lower bound: a(n) >= 2^n and an upper bound: a(n) <= 2^(n+1).

			For n=2, u(2) = 194, log_2(u(2)) is between 7.5 and 7.6, so E(log_2(u(2))) = 7, so a(2) = E(log_2(u(2))) + 1 = 8. And indeed, u(2) = 194 (in base 10) = 11000010 in base 2 requires 8 bits (all bits above are 0).

Cf. A003010, A070939, A227608.

lista(nn) = {a = 4; print1(#binary(a), ", "); for (n=1, nn, a = a^2-2; print1(#binary(a), ", "););} \\ Michel Marcus, Apr 04 2016

a(n) = E(log_2(u(n))) + 1, where E(x) is the integer part of x and u is defined by: u(0) = 4 (or 100 in binary) and for n>0, u(n) = u(n-1)^2 - 2.

a(n) = A070939(A003010(n)). - Michel Marcus, Apr 04 2016

Terms from a(19) on from Michel Marcus, Apr 04 2016

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

A382437 a(n) = a(n-1)^2 + 4 * a(n-1), with a(0) = 2.

Original entry on oeis.org

2, 12, 192, 37632, 1416317952, 2005956546822746112, 4023861667741036022825635656102100992, 16191462721115671781777559070120513664958590125499158514329308740975788032

Offset: 0

V. Barbera, Mar 25 2025

nonn

Paolo Xausa, Table of n, a(n) for n = 0..10

Cf. A002812, A003010.

NestList[#*(4 + #) &, 2, 8] (* Paolo Xausa, Apr 01 2025 *)

a(n)=if(n, a(n-1)^2 + 4*a(n-1), 2);
vector(8, i, a(i-1))

a(n) = A003010(n) - 2.
a(n)/2 = A002812(n) - 1.
For n > 1: a(n) = 3 * 2^(2*n) * Product_{i = 0..n-2} A002812(i)^2.
Conjecture: a(n) = Sum_{k=1..2^n} (2^n * 2^k * binomial(2^n + k - 1, 2*k - 1) / k).

A105850 a(n+1) = a(n)^2 - 2^(2^n+1) with a(1) = 8.

Original entry on oeis.org

8, 56, 3104, 9634304, 92819813433344, 8615517765800787268541087744, 74227146372828989101844394431169040459931374583287906304

Offset: 1

Douglas Stones (dssto1(AT)student.monash.edu.au), Apr 22 2005

nonn

Lehmer used this sequence in his proof of the Lucas-Lehmer test.

D. H. Lehmer, On Lucas's Test for the Primality of Mersenne's Numbers, J. London Math. Soc., 10 (1935) 162-165.

Cf. A003010.

nxt[{n_,a_}]:={n+1,a^2-2^(2^n+1)}; Transpose[NestList[nxt,{1,8},7]][[2]] (* Harvey P. Dale, Jul 24 2013 *)

a(n) = 2^(2^(n-1))*A003010(n).

A261532 Hall's sequence g_n.

Original entry on oeis.org

7, 589, 11064985, 7835767761026353, 7859104819806710982081319640824417, 15811975313589523224392147529414564125936123169432771986649715567359169

Offset: 3

Joerg Arndt, Aug 24 2015

nonn

James S. Hall, A remark on the primeness of Mersenne numbers, J. London Math. Soc. 28, (1953), 285-287.

Cf. A002812, A003010.

[n le 1 select 7 else 2*(Self(n-1)-1)*(1+2^(n+1)*(Self(n-1)-1)) + 1: n in [1..6]]; // Vincenzo Librandi, Aug 24 2015

N=11;  g=vector(N);  g[3] = 7;
for (n=3, N-1, g[n+1] = 2*(g[n]-1)*(1+2^n*(g[n]-1))+1 );
vector(N-3, j, g[j+2])

a(n) = g(n) where g(3) = 7 and g(n+1) = 2*(g(n)-1)*(1+2^n*(g(n)-1)) + 1.

cp's OEIS Frontend

A135928 Digital roots of the Mersenne primes.

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A217870 Decimal expansion of sqrt(sqrt(2 + sqrt(3))).

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A182023 Decimal expansion of log(2+sqrt(3))/4.

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A227615 Number of bits necessary to represent u(n) in binary, where u is the Lucas-Lehmer sequence: u(0) = 100 (in binary); for n>0, u(n) = u(n-1)^2 - 2.

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

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A382437 a(n) = a(n-1)^2 + 4 * a(n-1), with a(0) = 2.

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A105850 a(n+1) = a(n)^2 - 2^(2^n+1) with a(1) = 8.

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