A050614 -id:A050614 - cp's OEIS frontend

A062878 a(n) is the position of A050614(n) in A062877.

1, 3, 6, 15, 24, 60, 102, 255, 384, 960, 1632, 4080, 6168, 15420, 26214, 65535, 98304, 245760, 417792, 1044480, 1579008, 3947520, 6710784, 16776960, 25166208, 62915520, 106956384, 267390960, 404232216, 1010580540, 1717986918, 4294967295, 6442450944

Offset: 0

Antti Karttunen, Jun 26 2001

nonn

In binary this sequence looks like 1, 11, 110, 1111, 11000, 111100, 1100110, 11111111, 110000000, 1111000000, 11001100000, 111111110000, 1100000011000, 11110000111100, 110011001100110, ...
Sequence A282387 may be the same, but I cannot prove nor disprove this beyond a(22). - Robert Price, Feb 13 2017
Agrees with A282387 for at least 1000 terms. - Sean A. Irvine, Apr 14 2023

A050614 = Table[k = Floor[Log[2, n + 1]]; Product[j = 2^(i + 1); l = Fibonacci[j + 1] + Fibonacci[j - 1]; If[BitAnd[2^i, n] == 0, b = 0, b = 1]; l^b, {i, 0, k}], {n, 0, 200}]; A062877 = Union[Total /@ Subsets[Fibonacci[Range[1, 46, 2]]]]; Flatten[Table[Position[ A062877, A050614[[i]] ] - 1, {i, 1, 25}]] (* Robert Price, Feb 13 2017 *)

a(2^n-1) = 2^(2^n) - 1. - Philippe Deléham, Apr 05 2007

a(n) = Sum_{k=0..n} A127872(n,k)*2^k. - Philippe Deléham, Oct 09 2007

a(15)-a(22) from Robert Price, Feb 13 2017

More terms from Sean A. Irvine, Apr 14 2023

A001566 a(0) = 3; thereafter, a(n) = a(n-1)^2 - 2.

Original entry on oeis.org

3, 7, 47, 2207, 4870847, 23725150497407, 562882766124611619513723647, 316837008400094222150776738483768236006420971486980607

Offset: 0

N. J. A. Sloane

Expansion of 1/phi: 1/phi = (1-1/3)*(1-1/((3-1)*7))*(1-1/(((3-1)*7-1)*47))*(1-1/((((3-1)*7-1)*47-1)*2207))... (phi being the golden ration (1+sqrt(5))/2). - Thomas Baruchel, Nov 06 2003
An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
Starting with 7, the terms end with 7,47,07,47,07,..., of the form 8a+7 where a = 0,1,55,121771,... Conjecture: Every a is squarefree, every other a is divisible by 55, the a's are a subset of A046194, the heptagonal triangular numbers (the first, 2nd, 3rd, 6th, 11th, ?, ... terms). - Gerald McGarvey, Aug 08 2004
Also the reduced numerator of the convergents to sqrt(5) using Newton's recursion x = (5/x+x)/2. - Cino Hilliard, Sep 28 2008
The subsequence of primes begins a(n) for n = 0, 1, 2, 3. - Jonathan Vos Post, Feb 26 2011
We have Sum_{n=0..N} a(n)^2 = 2*(N+1) + Sum_{n=1..N+1} a(n), Sum_{n=0..N} a(n)^4 = 5*(Sum_{n=1..N+1} a(n)) + a(N+1)^2 + 6*N -3, etc. which is very interesting with respect to the fact that a(n) = Lucas(2^(n+1)); see W. Webb's problem in Witula-Slota's paper. - Roman Witula, Nov 02 2012
From Peter Bala, Nov 11 2012: (Start)
The present sequence corresponds to the case x = 3 of the following general remarks.
The recurrence a(n+1) = a(n)^2 - 2 with initial condition a(0) = x > 2 has the solution a(n) = ((x + sqrt(x^2 - 4))/2)^(2^n) + ((x - sqrt(x^2 - 4))/2)^(2^n).
We have the product expansion sqrt(x + 2)/sqrt(x - 2) = Product_{n>=0} (1 + 2/a(n)) (essentially due to Euler - see Mendes-France and van der Poorten). Another expansion is sqrt(x^2 - 4)/(x + 1) = Product_{n>=0} (1 - 1/a(n)), which follows by iterating the identity sqrt(x^2 - 4)/(x + 1) = (1 - 1/x)*sqrt(y^2 - 4)/(y + 1), where y = x^2 - 2.
The sequence b(n) := a(n) - 1 satisfies b(n+1) = b(n)^2 + 2*b(n) - 2. Cases currently in the database are A145502 through A145510. The sequence c(n) := a(n)/2 satisfies c(n+1) = 2*c(n)^2 - 1. Cases currently in the database are A002812, A001601, A005828, A084764 and A084765.
(End)
E. Lucas in Section XIX of "The Theory of Simply Periodic Numerical Functions" (page 56 of English translation) equation "(127) (1-sqrt(5))/2 = -1/1 + 1/3 + 1/(3*7) + 1/(3*7*47) + 1/(3*7*47*2207) + ..." - Michael Somos, Oct 11 2022
Let b(n) = a(n) - 3. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. - Peter Bala, Dec 08 2022
The number of digits of a(n) is given by A094057(n+1). - Hans J. H. Tuenter, Jul 29 2025

			From _Cino Hilliard_, Sep 28 2008: (Start)
Init x=1;
x = (5/1 + 1)/2 = 3/1;
x = (5/3 + 3)/2 = 7/3;
x = ((5/7)/3 + 7/3)/2 = 47/21;
x = ((5/47)/21 + 47/21)/2 = 2207/987;
(2207/987)^2 = 5.000004106... (End)

L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 397.
E.-B. Escott, Note #1741, L'Intermédiaire des Mathématiciens, 8 (1901), page 13. - N. J. A. Sloane, Mar 02 2022
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 223.
Édouard Lucas, Nouveaux théorèmes d'arithmétique supérieure, Comptes Rend., 83 (1876), 1286-1288.
Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 7.
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).

Vincenzo Librandi, Table of n, a(n) for n = 0..12
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
Pierre Liardet and Pierre Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
Édouard Lucas, Nouveaux théorèmes d'arithmétique supérieure (annotated scanned copy)
Édouard Lucas, The Theory of Simply Periodic Numerical Functions, Fibonacci Association, 1969. English translation of article "Théorie des Fonctions Numériques Simplement Périodiques, I", Amer. J. Math., 1 (1878), 184-240.
M. Mendes France and A. J. van der Poorten, From geometry to Euler identities, Theoret. Comput. Sci., 65 (1989), 213-220.
E. L. Roettger and H. C. Williams, Some Remarks Concerning the Lucas-Lehmer Primality Test, Journal of Integer Sequences, Vol. 28 (2025), Article 25.2.5. See p. 2.
Chance Sanford, Infinite Series Involving Fibonacci Numbers Via Apéry-Like Formulae, arXiv:1603.03765 [math.NT], 2016.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
Wikipedia, Engel Expansion.
Roman Wituła and Damian Słota, delta-Fibonacci numbers, Applicable Analysis and Discrete Mathematics, Vol. 3, No. 2 (2009), pp. 310-329.
Index entries for sequences of form a(n+1)=a(n)^2 + ....

Lucas numbers (A000032) with subscripts that are powers of 2 greater than 1 (Herbert S. Wilf). Cf. A000045.
Cf. A003010 (starting with 4), A003423 (starting with 6), A003487 (starting with 5).
Cf. A058635. - Artur Jasinski, Oct 05 2008
Cf. A002812, A094874, A145502, A145274, A050614, A088334, A181393, A181419, A186750, A186751, A338304.

a:= n-> simplify(2*ChebyshevT(2^n, 3/2), 'ChebyshevT'):
seq(a(n), n=0..8);

NestList[#^2-2&,3,10] (* Harvey P. Dale, Dec 17 2014 *)
Table[LucasL[2^n], {n, 1, 8}] (* Amiram Eldar, Oct 22 2020 *)

a[0]:3$
a[n]:=a[n-1]^2-2$
A001566(n):=a[n]$
makelist(A001566(n),n,0,7); /* Martin Ettl, Nov 12 2012 */

{a(n) = if( n<1, 3*(n==0), a(n-1)^2 - 2)}; /* Michael Somos, Mar 14 2004 */

g(n,p) = x=1;for(j=1,p,x=(n/x+x)/2;print1(numerator(x)","));
g(5,8) \\ Cino Hilliard, Sep 28 2008

{a(n) = my(w = quadgen(5)); if( n<0, 0, n++; imag( (2*w - 1) * w^2^n ))}; /* Michael Somos, Nov 30 2014 */

{a(n) = my(y = x^2-x-1); if( n<0, 0, n++; for(i=1, n, y = polgraeffe(y)); -polcoeff(y, 1))}; /* Michael Somos, Nov 30 2014 */

a(n) = Fibonacci(2^(n+2))/Fibonacci(2^(n+1)) = A058635(n+2)/A058635(n+1). - Len Smiley, May 08 2000, and Artur Jasinski, Oct 05 2008
a(n) = ceiling(c^(2^n)) where c = (3+sqrt(5))/2 = tau^2 is the largest root of x^2-3*x+1=0. - Benoit Cloitre, Dec 03 2002
a(n) = round(G^(2^n)) where G is the golden ratio (A001622). - Artur Jasinski, Sep 22 2008
a(n) = (G^(2^(n+1))-(1-G)^(2^(n+1)))/((G^(2^n))-(1-G)^(2^n)) = G^(2^n)+(1-G)^(2^n) = G^(2^n)+(-G)^(-2^n) where G is the golden ratio. - Artur Jasinski, Oct 05 2008
a(n) = 2*cosh(2^(n+1)*arccosh(sqrt(5)/2)). - Artur Jasinski, Oct 09 2008
a(n) = Fibonacci(2^(n+1)-1) + Fibonacci(2^(n+1)+1). (3-sqrt(5))/2 = 1/3 + 1/(3*7) + 1/(3*7*47) + 1/(3*7*47*2207) + ... (E. Lucas). - Philippe Deléham, Apr 21 2009
a(n)*(a(n+1)-1)/2 = A023039(2^n). - M. F. Hasler, Sep 27 2009
For n >= 1, a(n) = 2 + Product_{i=0..n-1} (a(i) + 2). - Vladimir Shevelev, Nov 28 2010
a(n) = 2*T(2^n,3/2) where T(n,x) is the Chebyshev polynomial of the first kind. - Leonid Bedratyuk, Mar 17 2011
From Peter Bala, Oct 31 2012: (Start)
Engel expansion of 1/2*(3 - sqrt(5)). Thus 1/2*(3 - sqrt(5)) = 1/3 + 1/(3*7) + 1/(3*7*47) + ... as noted above by Deleham. See Liardet and Stambul.
sqrt(5)/4 = Product_{n>=0} (1 - 1/a(n)).
sqrt(5) = Product_{n>=0} (1 + 2/a(n)). (End)
a(n) - 1 = A145502(n+1). - Peter Bala, Nov 11 2012
a(n) == 2 (mod 9), for n > 1. - Ivan N. Ianakiev, Dec 25 2013
From Amiram Eldar, Oct 22 2020: (Start)
a(n) = A000032(2^(n+1)).
Sum_{k>=0} 1/a(k) = -1 + A338304. (End)
a(n) = (A000045(m+2^(n+2))+A000045(m))/A000045(m+2^(n+1)) for any m>=0. - Alexander Burstein, Apr 10 2021
a(n) = 2*cos(2^n*arccos(3/2)). - Peter Luschny, Oct 12 2022
a(n) == -1 ( mod 2^(n+2) ). - Peter Bala, Nov 07 2022
a(n) = 5*Fibonacci(2^n)^2+2 = 5*A058635(n)^2+2, for n>0. - Jianglin Luo, Sep 21 2023
Sum_{n>=0} a(n)/Fibonacci(2^(n+2)) = A094874 (Sanford, 2016). - Amiram Eldar, Mar 01 2024

A127872 Triangle formed by reading A039599 mod 2.

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1

Offset: 0

Philippe Deléham, Apr 05 2007

Also triangle formed by reading triangles A061554, A106180, A110519, A124574, A124576, A126953, A127543 modulo 2.

			Triangle begins:
1;
1, 1;
0, 1, 1;
1, 1, 1, 1;
0, 0, 0, 1, 1;
0, 0, 1, 1, 1, 1;
0, 1, 1, 0, 0, 1, 1;
1, 1, 1, 1, 1, 1, 1, 1;
0, 0, 0, 0, 0, 0, 0, 1, 1;
0, 0, 0, 0, 0, 0, 1, 1, 1, 1;
0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1;
0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1;
0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1;
0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1;
0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1;
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1; ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A061554, A106180, A110519, A124574, A124576, A126953, A127543.

Cf. A000007, A036987, A001316, A062878, A050614, A000045, A001519

T[0, 0] := 1; T[n_, k_] := Binomial[2*n - 1, n - k] - Binomial[2*n - 1, n - k - 2]; Table[Mod[T[n, k], 2], {n,0,10}, {k,0,n}] // Flatten (* G. C. Greubel, Apr 18 2017 *)

Sum_{k=0..n} T(n,k)*x^k = A000007(n), A036987(n), A001316(n), A062878(n) for x=-1,0,1,2 respectively.

Sum_{k=0..n} T(n,k)*Fibonacci(2*k+1) = A050614(n), see A000045 and A001519. - Philippe Deléham, Aug 30 2007

A062877 0 and numbers representable as a sum of distinct odd-indexed Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 7, 8, 13, 14, 15, 16, 18, 19, 20, 21, 34, 35, 36, 37, 39, 40, 41, 42, 47, 48, 49, 50, 52, 53, 54, 55, 89, 90, 91, 92, 94, 95, 96, 97, 102, 103, 104, 105, 107, 108, 109, 110, 123, 124, 125, 126, 128, 129, 130, 131, 136, 137, 138, 139, 141, 142, 143, 144

Offset: 0

Antti Karttunen, Jun 26 2001

nonn

			F_1 = 1,
F_3 = 2,
F_1 + F_3 = 3,
F_5 = 5,
F_5 + F_1 = 6,
F_5 + F_3 = 7,
F_5 + F_3 + F_1 = 8,
F_7 = 13, ...

R. J. Mathar, Table of n, a(n) for n = 0..3071
A. Karttunen, On Pascal's Triangle Modulo 2 in Fibonacci Representation, The Fibonacci Quarterly, Vol. 42, #1 (2004) pp. 38-46.

A062878 gives the positions of A050614(n) in this sequence. A062879 is bisection.
A036796(n) - 1.
Cf. A022290 (even-indexed Fibonaccis), A054204.

with(combinat); [seq(A062877(j),j=0..265)]; A062877 := n -> add((floor(n/(2^i)) mod 2)*fibonacci((2*i)+1),i=0..floor_log_2(n+1));
floor_log_2 := proc(n) local nn,i; nn := n; for i from -1 to n do if(0 = nn) then RETURN(i); fi; nn := floor(nn/2); od; end;
# alternative
isA062877 := proc(n)
    local fset,fidx,ps ;
    if n = 0 then
        return true;
    end if;
    fset := {} ;
    for fidx from 1 by 2 do
        if combinat[fibonacci](fidx) >n then
            break;
        end if;
        fset := fset union {combinat[fibonacci](fidx)} ;
    end do:
    for ps in combinat[powerset](fset) do
        if n = add(fidx,fidx=ps) then
            return true;
        end if;
    end do:
    return false;
end proc: # R. J. Mathar, Aug 22 2016

Take[Union[Total/@Subsets[Fibonacci[Range[1,20,2]]]],70](* Harvey P. Dale, Dec 21 2013 *)

my(m=Mod('x,'x^2-3*'x+1)); a(n) = subst(lift(subst(Pol(binary(n)), 'x,m)), 'x,2); \\ Kevin Ryde, Nov 25 2020

A050613 Products of distinct terms of 1 and rest from A001566: a(n) = Product_{i=0..floor(log_2(n+1))} L(2^i)^bit(n,i).

Original entry on oeis.org

1, 1, 3, 3, 7, 7, 21, 21, 47, 47, 141, 141, 329, 329, 987, 987, 2207, 2207, 6621, 6621, 15449, 15449, 46347, 46347, 103729, 103729, 311187, 311187, 726103, 726103, 2178309, 2178309, 4870847, 4870847, 14612541, 14612541, 34095929, 34095929

Offset: 0

Antti Karttunen, Dec 02 1999

nonn

Used to produce the rows of A050609.

Also Sum(((C(2((n+((n+1) mod 2)) mod (2^floor(log_2(n)))),i) mod 2)*F(n+((n+1) mod 2)-i)),i=0..2((n+((n+1) mod 2)) mod (2^floor(log_2(n))))) or Sum(((C(2((n-(n mod 2)) mod (2^floor(log_2(n)))),i) mod 2)*L(n-(n mod 2)-i)),i=0..2((n-(n mod 2)) mod (2^floor(log_2(n))))) for all n > 1. Here F(n) and L(n) are n-th Fibonacci (A000045) and Lucas (A000032) numbers respectively.

A. Karttunen, On Pascal's Triangle Modulo 2 in Fibonacci Representation, Fibonacci Quarterly, 42 (2004), 38-46.

Bisection: A050614.

with(combinat); A050613 := n -> product('luc(2^i)^bit_i(n,i)','i'=0..floor_log_2(n+1));
luc := n -> (fibonacci(n-1)+fibonacci(n+1));
bit_i := (n,i) -> `mod`(floor(n/(2^i)),2);
floor_log_2 := proc(n) local nn,i; nn := n; for i from -1 to n do if(0 = nn) then RETURN(i); fi; nn := floor(nn/2); od; end;

A075149 Sum_{i=0..2A053645(n)} (C(2A053645(n),i) mod 2)*A000045(n-i) [where C(r,c) is the binomial coefficient (A007318) and A000045(n) is the n-th Fibonacci number].

Original entry on oeis.org

0, 1, 1, 3, 3, 7, 9, 21, 21, 47, 63, 141, 147, 329, 441, 987, 987, 2207, 2961, 6621, 6909, 15449, 20727, 46347, 46389, 103729, 139167, 311187, 324723, 726103, 974169, 2178309, 2178309, 4870847, 6534927, 14612541, 15248163, 34095929, 45744489

Offset: 0

Antti Karttunen, Sep 05 2002

nonn

A. Karttunen, On Pascal's Triangle Modulo 2 in Fibonacci Representation, Fibonacci Quarterly, 42 (2004), 38-46.

Bisection gives A050614.

with(combinat); [seq(A075149(n,n=0..50)]; A075149 := n -> add((binomial(2*r(n),i) mod 2)*fibonacci(n-i),i=0..2*r(n));
r := n -> n - 2^floor_log_2(n);
floor_log_2 := proc(n) local nn,i; nn := n; for i from -1 to n do if(0 = nn) then RETURN(i); fi; nn := floor(nn/2); od; end;

A050615 Products of distinct terms of Fibonacci(2^(i+2)): a(n) = Product_{i=0..floor(log_2(n+1))} F(2^(i+2))^bit(n,i).

Original entry on oeis.org

1, 3, 21, 63, 987, 2961, 20727, 62181, 2178309, 6534927, 45744489, 137233467, 2149990983, 6449972949, 45149810643, 135449431929, 10610209857723, 31830629573169, 222814407012183, 668443221036549, 10472277129572601

Offset: 0

Antti Karttunen, Dec 02 1999

nonn

Cf. A050613, A050614.

distinct_fib_products := n -> product('fib(2^(i+2))^bit_i(n,i)','i'=0..floor_log_2(n+1));

cp's OEIS Frontend

A062878 a(n) is the position of A050614(n) in A062877.

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A001566 a(0) = 3; thereafter, a(n) = a(n-1)^2 - 2.

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A127872 Triangle formed by reading A039599 mod 2.

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A062877 0 and numbers representable as a sum of distinct odd-indexed Fibonacci numbers.

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Maple

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PARI

A050613 Products of distinct terms of 1 and rest from A001566: a(n) = Product_{i=0..floor(log_2(n+1))} L(2^i)^bit(n,i).

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A075149 Sum_{i=0..2*A053645(n)} (C(2*A053645(n),i) mod 2)*A000045(n-i) [where C(r,c) is the binomial coefficient (A007318) and A000045(n) is the n-th Fibonacci number].

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A050615 Products of distinct terms of Fibonacci(2^(i+2)): a(n) = Product_{i=0..floor(log_2(n+1))} F(2^(i+2))^bit(n,i).

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A075149 Sum_{i=0..2A053645(n)} (C(2A053645(n),i) mod 2)*A000045(n-i) [where C(r,c) is the binomial coefficient (A007318) and A000045(n) is the n-th Fibonacci number].