A001566 -id:A001566 - cp's OEIS frontend

A338304 Decimal expansion of Sum_{k>=0} 1/L(2^k), where L(k) is the k-th Lucas number (A000032).

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

Offset: 1

Amiram Eldar, Oct 21 2020

Erdős and Graham (1980) asked whether this constant is irrational or transcendental.
Badea (1987) proved that it is irrational, and André-Jeannin (1991) proved that it is not a quadratic irrational in Q(sqrt(5)), in contrast to the corresponding sum with Fibonacci numbers, Sum_{k>=0} 1/F(2^k) = (7-sqrt(5))/2 (A079585).
Bundschuh and Pethö (1987) proved that it is transcendental.

			1.49792038099906271987068555399285960807207719857085...

Richard André-Jeannin, A note on the irrationality of certain Lucas infinite series, The Fibonacci Quarterly, Vol. 29, No. 2 (1991), pp. 132-136.
Catalin Badea, The irrationality of certain infinite series, Glasgow Mathematical Journal, Vol. 29, No. 2 (1987), pp. 221-228.
Peter Bundschuh and Attila Pethö, Zur transzendenz gewisser Reihen, Monatshefte für Mathematik, Vol. 104, No. 3 (1987), pp. 199-223, alternative link.
Paul Erdős and Ronald L. Graham, Old and new problems and results in combinatorial number theory, L'enseignement Mathématique, Université de Genève, 1980, pp. 64-65.
Index entries for transcendental numbers

Cf. A000045, A000032, A001566, A079585, A338305.

RealDigits[Sum[1/LucasL[2^n], {n, 0, 10}], 10, 100][[1]]

Equals 1 + Sum_{k>=0} 1/A001566(k).

A088334 Expansion of 1/phi (phi being the golden ratio) as an infinite product: 1/phi = Product_{k=0..n} (1-1/a(k)).

Original entry on oeis.org

3, 14, 611, 1346270, 6557470319843, 155576970220531065681649694, 87571595343018854458033386304178158174356588264390371

Offset: 0

Thomas Baruchel, Nov 07 2003

nonn

The next term is too large to include.

J. Shallit, Problem B-423, The Fibonacci Quarterly 18,1 Feb.(1980)85. Solution 19,1 Feb. (1981) 92. [_Wolfdieter Lang_, Nov 04 2010]

Cf. A001566, A001622, A094214, A101342.

PARI
```
a(n)=if(n<0,0,fibonacci(2^(n+2)-1)+1)
```

a(0) = 3, a(n+1) = (a(n)-1)*A001566(n+1)
a(n) = 1+ceiling(1/2*(1-1/sqrt(5))*phi^(2^(n+2))) where phi=(1+sqrt(5))/2. a(n)==2 (mod 3) for n>0. - Benoit Cloitre, Nov 09 2003
a(n) = b(n+2)+1, n>=0, with b(n):= A101342(n) = F(2^n-1). See the reciprocal of the infinite product of this entry. For a proof see the J. Shallit reference. - Wolfdieter Lang, Nov 04 2010

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

A181393 Numbers of the form Fibonacci(2^c)/Fibonacci(2^b), 1 <= b < c.

Original entry on oeis.org

3, 7, 21, 47, 329, 987, 2207, 103729, 726103, 2178309, 4870847, 10749959329, 505248088463, 3536736619241, 10610209857723, 23725150497407, 115561578124843393729, 255044402921529369959903, 11987086937311880388115441, 83909608561183162716808087, 251728825683549488150424261

Offset: 1

Vladimir Shevelev, Oct 17 2010

nonn

Using an Eratosthenes-like sieve, we find "primes" of the form P_k = Fibonacci(2^(k+1)) / Fibonacci(2^k) = A001566(k-1), k=1,2,..., such that every term has a unique "prime" factorization.

			If k=3, m=1, by the latter formula, we have a(8) = A001566(2)*A001566(3) = 47*2207 = 103729.

Cf. A000045, A001566.

For n >= 1, a((n^2-n+2)/2) = P_n = A001566(n-1); for 1 <= m < k, a((k^2+3*k)/2-m) = Product_{i=m+1..k} A001566(i).

a(12)-a(16) corrected and more terms from Jason Yuen, Feb 10 2025

A293390 Least m such that the exponents in expression for n as a sum of distinct powers of 2 are pairwise distinct mod m; a(0) = 0 by convention.

Original entry on oeis.org

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

Offset: 0

Rémy Sigrist, Oct 08 2017

The set of exponents in expression for n as a sum of distinct powers of 2 corresponds to the n-th row of A133457.
The sum of digits of n in base 2^a(n), say s, can be computed without carry in base 2; the Hamming weight of s equals the Hamming weight of n.
a(n) >= A000120(n) for any n > 0.
Apparently, a(n) = A000120(n) iff n = 0 or n belongs to A100290.
a(n) <= A070939(n) for any n >= 0.
For any sequence s of distinct nonnegative integers (s(n) being defined for n >= 0):
- let D_s be defined for any n > 0 by D_s(n) = a(Sum_{k=0..n-1} 2^s(k)),
- then D_s is the discriminator of s as introduced by Arnold, Benkoski, and McCabe in 1985,
- D_s(1) = 1,
- D_s(n) >= n for any n >= 1,
- D_s(n+1) >= D_s(n) for any n >= 1.

			For n=42:
- 42 = 2^5 + 2^3 + 2^1,
- 5 mod 1 = 3 mod 1,
- 5 mod 2 = 3 mod 2,
- 5 mod 3, 3 mod 3 and 1 mod 3 are all distinct,
- hence a(42) = 3.

Robert Israel, Table of n, a(n) for n = 0..10000
Sajed Haque, Jeffrey Shallit, Discriminators and k-Regular Sequences, arXiv:1605.00092 [cs.DM], 2016.

Cf. A000041, A000045, A000058, A000069, A000108, A000110, A000120, A000215, A001147, A001566, A001969, A002808, A003095, A005823, A016726, A062383, A070939, A076793, A100290, A133457, A173195, A270097, A270151, A270176, A272633, A272881, A272882, A273037, A273041, A273043, A273044, A273056, A273062, A273064, A273068, A273237, A273377

f:= proc(n) local L,D,k;
  L:= convert(n,base,2);
  L:= select(t -> L[t+1]=1, [$0..nops(L)-1]);
  if nops(L) = 1 then return 1 fi;
  D:= {seq(seq(L[j]-L[i],i=1..j-1),j=2..nops(L))};
  D:= `union`(seq(numtheory:-divisors(i),i=D));
  min({$2..max(D)+1} minus D)
end proc:
0, seq(f(i),i=1..100); # Robert Israel, Oct 08 2017

{0}~Join~Table[Function[r, SelectFirst[Range@ 10, Length@ Union@ Mod[r, #] == Length@ r &]][Join @@ Position[#, 1] - 1 &@ Reverse@ IntegerDigits[n, 2]], {n, 86}] (* Michael De Vlieger, Oct 08 2017 *)

a(n) = if (n, my (d=Vecrev(binary(n)), x = []); for (i=1, #d, if (d[i], x = concat(x, i-1))); for (m=1, oo, if (#Set(vector(#x, i, x[i]%m))==#x, return (m))), return (0))

a(2*n) = a(n) for any n >= 0.
a(2^k-1) = k for any k >= 0.
a(n) = 1 iff n = 2^k for some k >= 0.
a(n) = 2 iff n belongs to A173195.
a(Sum_{k=1..n} 2^(k^2)) = A016726(n) for any n >= 1.
a(Sum_{k=1..n} 2^A000069(k)) = A062383(n) for any n >= 1.
a(Sum_{k=0..n} 2^(2^k)) = A270097(n) for any n >= 0.
a(Sum_{k=1..n} 2^A000045(k+1)) = A270151(n) for any n >= 1.
a(Sum_{k=1..n} 2^A000041(k)) = A270176(n) for any n >= 1.
a(A076793(n)) = A272633(n) for any n >= 0.
a(Sum_{k=1..n} 2^A001969(k)) = A272881(n) for any n >= 1.
a(Sum_{k=1..n} 2^A005823(k)) = A272882(n) for any n >= 1.
a(Sum_{k=1..n} 2^A000215(k-1)) = A273037(n) for any n >= 1.
a(Sum_{k=1..n} 2^A000108(k)) = A273041(n) for any n >= 1.
a(Sum_{k=1..n} 2^A001566(k)) = A273043(n) for any n >= 1.
a(Sum_{k=1..n} 2^A003095(k)) = A273044(n) for any n >= 1.
a(Sum_{k=1..n} 2^A000058(k-1)) = A273056(n) for any n >= 1.
a(Sum_{k=1..n} 2^A002808(k)) = A273062(n) for any n >= 1.
a(Sum_{k=1..n} 2^(k!)) = A273064(n) for any n >= 1.
a(Sum_{k=1..n} 2^(k^k)) = A273068(n) for any n >= 1.
a(Sum_{k=1..n} 2^A000110(k)) = A273237(n) for any n >= 1.
a(Sum_{k=1..n} 2^A001147(k)) = A273377(n) for any n >= 1.

A186751 a(0) = 3; thereafter, a(n) = a(n-1)^2 - 4.

Original entry on oeis.org

3, 5, 21, 437, 190965, 36467631221, 1329888126870853950837, 1768602429992068534155014726612412013000565

Offset: 0

Jonathan Vos Post, Feb 26 2011

This is to A001566 as 4 is to 2 (subtrahend). This is the k=4 row of the array A[k,0] = 3, A[k,n] = A[k,n-1]^2 - k; A186750 is the k=3 row; and A001566 is the k=2 row. A003096(n+1) is the k=1 row.

			a(1) = a(0)^2 - 4 = 3^2 - 4 = 5, which is, like a(0), a prime.

Cf. A001566, A003096, A186750.

NestList[#^2-4&,3,10]  (* Harvey P. Dale, Mar 14 2011 *)

A264745 Rectangular array A read by upward antidiagonals in which the entry in row n and column k is defined by A(n,k) = Fibonacci(2^(n-1)(2k-1) + 1), n,k >= 1.

Original entry on oeis.org

1, 2, 3, 5, 13, 8, 34, 233, 89, 21, 1597, 75025, 10946, 610, 55, 3524578, 7778742049, 165580141, 514229, 4181, 144, 17167680177565, 83621143489848422977, 37889062373143906, 365435296162, 24157817, 28657, 377, 407305795904080553832073954

Offset: 1

L. Edson Jeffery, Nov 23 2015

The array exhausts, without duplication, the subsequence of A000045 obtained by removing the first two terms {0,1}.

			The array begins:
.     1           3                  8                        21
.     2          13                 89                       610
.     5         233              10946                    514229
.    34       75025          165580141              365435296162
.  1597  7778742049  37889062373143906  184551825793033096366333

G. C. Greubel, Table of n, a(n) for n = 1..78

Cf. A001906, A033891 (rows 1--2).
Cf. A192222 (column 1).
Cf. A000045, A001511, A001566, A003602, A054582

(* Array: *)
Grid[Table[Fibonacci[2^(n - 1)*(2 k - 1) + 1], {n, 5}, {k, 4}]]
(* Array antidiagonal flattened: *)
Flatten[Table[Fibonacci[2^(n - k)*(2 k - 1) + 1], {n, 7}, {k, n}]]

A(n,k) = A000045(A054582(n-1,k-1) + 1).
A(A001511(m),A003602(m)) = A000045(m+1), m >= 1.
Conjectured g.f. for row n: x*(A000045(2^(n-1)+1) - A000045(2^(n-1)-1)*x)/(1 - A001566(n)*x + x^2), n >= 1.

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

A331038 Residues of the Lucas-Lehmer primality test for M(127) = 2^127 - 1.

Original entry on oeis.org

3, 7, 47, 2207, 4870847, 23725150497407, 562882766124611619513723647, 9932388036497706472820043948129789713, 102423269049837077051675109560558766898, 7949236499829405891753012242872011683, 119093374737774941856311333667076322210

Offset: 0

Sergio Pimentel, Jan 08 2020

Since a(125) = 0, 2^127 - 1 = 170141183460469231731687303715884105727 is prime. This calculation was carried out by hand by Edouard Lucas. It took him 19 years from 1857 to 1876. The method works with a(0) = 3 since M(127) == 3 (mod 4). It also works with a(0) = 4 or a(0) = 10.

Sergio Pimentel, Table of n,a(n) for n = 0..125
Eric Weisstein's World of Mathematics, Lucas Lehmer Test.
Wikipedia, Lucas Lehmer Primality Test.

Cf. A000043, A000668, A001566, A095847.

Cf. also A129219, A129220, A129221, A129222, A129223, A129224, A129225.

NestList[Mod[#^2-2,2^127-1]&, 3,10] (* Stefano Spezia, Mar 28 2025 *)

a(n) = (a(n-1)^2 - 2) mod (2^127-1) with a(0) = 3; a(125) is the final term.

A174652 Partial sums of A002814.

Original entry on oeis.org

1, 3, 20, 5797, 192900159414, 7177905237579946589743785824843591, 369822356418414944143680173221426891716916679027557977938929258031497305419444723776968084111223998808

Offset: 0

Jonathan Vos Post, Mar 25 2010

nonn

Partial sums of an infinite coprime sequence defined by recursion. 3 is the only prime through a(6). Is it computationally feasible to find another?

			a(3) = 1 + 2 + 17 + 5777 = 5797 = 11 * 17 * 31.

Cf. A002814, A000045, A001566, A045529.

a(n) = SUM[i=0..n] A002814(i) = SUM[i=0..n] {a(n) = a(n-1)^3 + 3a(n-1)^2 - 3}.

cp's OEIS Frontend

A338304 Decimal expansion of Sum_{k>=0} 1/L(2^k), where L(k) is the k-th Lucas number (A000032).

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A088334 Expansion of 1/phi (phi being the golden ratio) as an infinite product: 1/phi = Product_{k=0..n} (1-1/a(k)).

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A135928 Digital roots of the Mersenne primes.

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A181393 Numbers of the form Fibonacci(2^c)/Fibonacci(2^b), 1 <= b < c.

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A293390 Least m such that the exponents in expression for n as a sum of distinct powers of 2 are pairwise distinct mod m; a(0) = 0 by convention.

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A186751 a(0) = 3; thereafter, a(n) = a(n-1)^2 - 4.

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A264745 Rectangular array A read by upward antidiagonals in which the entry in row n and column k is defined by A(n,k) = Fibonacci(2^(n-1)*(2*k-1) + 1), n,k >= 1.

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

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A264745 Rectangular array A read by upward antidiagonals in which the entry in row n and column k is defined by A(n,k) = Fibonacci(2^(n-1)(2k-1) + 1), n,k >= 1.