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 1-6 of 6 results.

A000051 a(n) = 2^n + 1.

Original entry on oeis.org

2, 3, 5, 9, 17, 33, 65, 129, 257, 513, 1025, 2049, 4097, 8193, 16385, 32769, 65537, 131073, 262145, 524289, 1048577, 2097153, 4194305, 8388609, 16777217, 33554433, 67108865, 134217729, 268435457, 536870913, 1073741825, 2147483649, 4294967297, 8589934593
Offset: 0

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Author

Keywords

Comments

Same as Pisot sequence L(2,3).
Length of the continued fraction for Sum_{k=0..n} 1/3^(2^k). - Benoit Cloitre, Nov 12 2003
See also A004119 for a(n) = 2a(n-1)-1 with first term = 1. - Philippe Deléham, Feb 20 2004
From the second term on (n>=1), in base 2, these numbers present the pattern 1000...0001 (with n-1 zeros), which is the "opposite" of the binary 2^n-2: (0)111...1110 (cf. A000918). - Alexandre Wajnberg, May 31 2005
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=5, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^(n-1)* charpoly(A,3). - Milan Janjic, Jan 27 2010
First differences of A006127. - Reinhard Zumkeller, Apr 14 2011
The odd prime numbers in this sequence form A019434, the Fermat primes. - David W. Wilson, Nov 16 2011
Pisano period lengths: 1, 1, 2, 1, 4, 2, 3, 1, 6, 4, 10, 2, 12, 3, 4, 1, 8, 6, 18, 4, ... . - R. J. Mathar, Aug 10 2012
Is the mentioned Pisano period lengths (see above) the same as A007733? - Omar E. Pol, Aug 10 2012
Only positive integers that are not 1 mod (2k+1) for any k>1. - Jon Perry, Oct 16 2012
For n >= 1, a(n) is the total length of the segments of the Hilbert curve after n iterations. - Kival Ngaokrajang, Mar 30 2014
Frénicle de Bessy (1657) proved that a(3) = 9 is the only square in this sequence. - Charles R Greathouse IV, May 13 2014
a(n) is the number of distinct possible sums made with at most two elements in {1,...,a(n-1)} for n > 0. - Derek Orr, Dec 13 2014
For n > 0, given any set of a(n) lattice points in R^n, there exist 2 distinct members in this set whose midpoint is also a lattice point. - Melvin Peralta, Jan 28 2017
Also the number of independent vertex sets, irredundant sets, and vertex covers in the (n+1)-star graph. - Eric W. Weisstein, Aug 04 and Sep 21 2017
Also the number of maximum matchings in the 2(n-1)-crossed prism graph. - Eric W. Weisstein, Dec 31 2017
Conjecture: For any integer n >= 0, a(n) is the permanent of the (n+1) X (n+1) matrix with M(j, k) = -floor((j - k - 1)/(n + 1)). This conjecture is inspired by the conjecture of Zhi-Wei Sun in A036968. - Peter Luschny, Sep 07 2021

References

  • Paul Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 75.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 46, 60, 244.
  • 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).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 141.

Crossrefs

Apart from the initial 1, identical to A094373.
See A008776 for definitions of Pisot sequences.
Column 2 of array A103438.
Cf. A007583 (a((n-1)/2)/3 for odd n).

Programs

  • Haskell
    a000051 = (+ 1) . a000079
    a000051_list = iterate ((subtract 1) . (* 2)) 2
    -- Reinhard Zumkeller, May 03 2012
    
  • Magma
    [2^n+1: n in [0..40]]; // G. C. Greubel, Jan 18 2025
  • Maple
    A000051:=-(-2+3*z)/(2*z-1)/(z-1); # Simon Plouffe in his 1992 dissertation
    a := n -> add(binomial(n,k)*bernoulli(n-k,1)*2^(k+1)/(k+1),k=0..n); # Peter Luschny, Apr 20 2009
  • Mathematica
    Table[2^n + 1, {n,0,40}]
    2^Range[0,40] + 1 (* Eric W. Weisstein, Jul 17 2017 *)
    LinearRecurrence[{3, -2}, {2, 3}, 40] (* Eric W. Weisstein, Sep 21 2017 *)
  • PARI
    a(n)=2^n+1
    
  • PARI
    first(n) = Vec((2 - 3*x)/((1 - x)*(1 - 2*x)) + O(x^n)) \\ Iain Fox, Dec 31 2017
    
  • Python
    def A000051(n): return (1<Chai Wah Wu, Dec 21 2022
    

Formula

a(n) = 2*a(n-1) - 1 = 3*a(n-1) - 2*a(n-2).
G.f.: (2-3*x)/((1-x)*(1-2*x)).
First differences of A052944. - Emeric Deutsch, Mar 04 2004
a(0) = 1, then a(n) = (Sum_{i=0..n-1} a(i)) - (n-2). - Gerald McGarvey, Jul 10 2004
Inverse binomial transform of A007689. Also, V sequence in Lucas sequence L(3, 2). - Ross La Haye, Feb 07 2005
a(n) = A127904(n+1) for n>0. - Reinhard Zumkeller, Feb 05 2007
Equals binomial transform of [2, 1, 1, 1, ...]. - Gary W. Adamson, Apr 23 2008
a(n) = A000079(n)+1. - Omar E. Pol, May 18 2008
E.g.f.: exp(x) + exp(2*x). - Mohammad K. Azarian, Jan 02 2009
a(n) = A024036(n)/A000225(n). - Reinhard Zumkeller, Feb 14 2009
From Peter Luschny, Apr 20 2009: (Start)
A weighted binomial sum of the Bernoulli numbers A027641/A027642 with A027641(1)=1 (which amounts to the definition B_{n} = B_{n}(1)).
a(n) = Sum_{k=0..n} C(n,k)*B_{n-k}*2^(k+1)/(k+1). (See also A052584.) (End)
a(n) is the a(n-1)-th odd number for n >= 1. - Jaroslav Krizek, Apr 25 2009
From Reinhard Zumkeller, Feb 28 2010: (Start)
a(n)*A000225(n) = A000225(2*n).
a(n) = A173786(n,0). (End)
If p[i]=Fibonacci(i-4) and if A is the Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise, then, for n>=1, a(n-1)= det A. - Milan Janjic, May 08 2010
a(n+2) = a(n) + a(n+1) + A000225(n). - Ivan N. Ianakiev, Jun 24 2012
a(A006521(n)) mod A006521(n) = 0. - Reinhard Zumkeller, Jul 17 2014
a(n) = 3*A007583((n-1)/2) for n odd. - Eric W. Weisstein, Jul 17 2017
Sum_{n>=0} 1/a(n) = A323482. - Amiram Eldar, Nov 11 2020

A065442 Decimal expansion of Erdős-Borwein constant Sum_{k>=1} 1/(2^k - 1).

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Nov 18 2001

Keywords

Comments

Also the decimal expansion of the (finite) value of Sum_{ k >= 1, k has no digit equal to 0 in base 2 } 1/k. - Robert G. Wilson v, Aug 03 2010
This constant is irrational (Erdős, 1948; Borwein, 1992). - Amiram Eldar, Aug 01 2020

Examples

			1.60669515241529176378330152319092458048057967150575643577807955369...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 354-361.
  • Paul Halmos, "Problems for Mathematicians, Young and Old", Dolciani Mathematical Expositions, 1991, p. 258.

Crossrefs

See A038631 for continued fraction.

Programs

  • Maple
    # Uses Lambert series, cf. formula by Arndt:
    evalf( add( (1/2)^(n^2)*(1 + 2/(2^n - 1)), n = 1..20 ), 105);
    # Peter Bala, Jan 22 2021
  • Mathematica
    RealDigits[ Sum[1/(2^k - 1), {k, 350}], 10, 111][[1]] (* Robert G. Wilson v, Nov 05 2006 *)
    (* first install irwinSums.m, see reference, then *) First@ RealDigits@ iSum[0, 0, 111, 2] (* Robert G. Wilson v, Aug 03 2010 *)
    RealDigits[(Log[2] - 2 QPolyGamma[0, 1, 2])/Log[4], 10, 100][[1]] (* Fred Daniel Kline, May 23 2011 *)
    x = 1/2; RealDigits[ Sum[ DivisorSigma[0, k] x^k, {k, 1000}], 10, 105][[1]] (* Robert G. Wilson v, Oct 12 2014 after an observation and formula of Amarnath Murthy, see A073668 *)
  • PARI
    a(n)= s=0; for(x=1,n,s=s+1.0/(2^x-1)); s
    
  • PARI
    default(realprecision, 2080); x=suminf(k=1, 1/(2^k - 1)); for (n=1, 2000, d=floor(x); x=(x-d)*10; write("b065442.txt", n, " ", d)) \\ Harry J. Smith, Oct 19 2009
    
  • PARI
    k=1.; suminf(n=1, k>>=1; k^n*(1+k)/(1-k)) \\ Charles R Greathouse IV, Jun 03 2015

Formula

Note: Sum_{k>=1} d(k)/2^k = Sum_{k>=1} 1/(2^k - 1).
Fast computation via Lambert series: 1.60669515... = Sum_{n>=1} x^(n^2)*(1+x^n)/(1-x^n) where x=1/2. - Joerg Arndt, May 24 2011
Equals (1/2) * A211705. - Amiram Eldar, Aug 01 2020
Equals 1/4 + Sum_{k >= 2} (1 + 8^k)/((2^k - 1)*2^(k^2+k)). See Mathematics Stack Exchange link. - Peter Bala, Jan 28 2022
Equals A066766 - A065443. - Amiram Eldar, Oct 16 2022

Extensions

More terms from Randall L Rathbun, Jan 16 2002

A086341 a(n) = 2*2^floor(n/2) - (-1)^n.

Original entry on oeis.org

1, 3, 3, 5, 7, 9, 15, 17, 31, 33, 63, 65, 127, 129, 255, 257, 511, 513, 1023, 1025, 2047, 2049, 4095, 4097, 8191, 8193, 16383, 16385, 32767, 32769, 65535, 65537, 131071, 131073, 262143, 262145, 524287, 524289, 1048575, 1048577, 2097151, 2097153
Offset: 0

Views

Author

Paul Barry, Jul 16 2003

Keywords

Crossrefs

Cf. A016116 (2^floor(n/2)).

Programs

  • Magma
    [2*2^Floor(n/2)-(-1)^n: n in [0..40]]; // Vincenzo Librandi, Aug 16 2011
    
  • Mathematica
    CoefficientList[Series[(1+2x)^2/((1+x)(1-2x^2)),{x,0,50}],x] (* or *) LinearRecurrence[ {-1,2,2},{1,3,3},50] (* Harvey P. Dale, Mar 10 2013 *)
  • PARI
    vector(40, n, n--; 2^(floor(n/2)+1) - (-1)^n) \\ G. C. Greubel, Nov 08 2018

Formula

E.g.f.: 2*cosh(sqrt(2)*x) + 2*sinh(sqrt(2)*x)/sqrt(2) - sinh(x) + cosh(x).
a(n) = (1 + 1/sqrt(2))*sqrt(2)^n + (1 - 1/sqrt(2))*(-sqrt(2))^n - (-1)^n.
G.f.: (1+2*x)^2/((1+x)*(1-2*x^2)). - Colin Barker, Aug 17 2012
a(n) = a(n-1) + 2*a(n-2) + 2*a(n-3); a(0)=1, a(1)=3, a(2)=3. - Harvey P. Dale, Mar 10 2013
From Amiram Eldar, Sep 14 2022: (Start)
Sum_{n>=0} 1/a(n) = A065442 + A323482 - 1/2.
Sum_{n>=0} (-1)^n/a(n) = 2 * A248721. (End)

A179951 Decimal expansion of Sum_{k has exactly two bits equal to 1 in base 2} 1/k.

Original entry on oeis.org

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

Views

Author

Robert G. Wilson v, Aug 03 2010

Keywords

Comments

Obviously for k > 0 in base 2 having no bit equal to 1 the sum is 0 and for 1 bit equal to 1 the sum is 2.

Examples

			Sum_{k>0} 1/A018900(k) = 1.52899956069688841838263949451...
		

Crossrefs

Programs

  • Maple
     evalf( 2*add( (-1)^(n+1)*((4^n + 1)/(4^n - 1))*(1/2)^(n^2), n = 1..18), 100); # Peter Bala, Jan 28 2022
  • Mathematica
    (* first install irwinSums.m, see either reference, then *) First@ RealDigits@ iSum[1, 2, 2^7, 2]

Formula

Equals Sum_{j>=1} Sum_{i=0..j-1} 1/(2^i + 2^j).
From Amiram Eldar, Jun 30 2020: (Start)
Equals Sum_{k>=0} 1/(2^k + 1/2).
Equals 2 * A323482 - 1. (End)
Equals 2*Sum_{n >= 1} (-1)^(n+1)*((4^n + 1)/(4^n - 1))*(1/2)^(n^2). The first 18 terms of the series gives the constant correct to more than 100 decimal places. - Peter Bala, Jan 28 2022

A305436 Number of divisors of n of the form 2^k + 1 for k >= 0.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 0, 1, 2, 2, 0, 2, 0, 1, 2, 1, 1, 3, 0, 2, 1, 1, 0, 2, 1, 1, 2, 1, 0, 3, 0, 1, 2, 2, 1, 3, 0, 1, 1, 2, 0, 2, 0, 1, 3, 1, 0, 2, 0, 2, 2, 1, 0, 3, 1, 1, 1, 1, 0, 3, 0, 1, 2, 1, 2, 3, 0, 2, 1, 2, 0, 3, 0, 1, 2, 1, 0, 2, 0, 2, 2, 1, 0, 2, 2, 1, 1, 1, 0, 4, 0, 1, 1, 1, 1, 2, 0, 1, 3, 2, 0, 3, 0, 1, 2
Offset: 1

Views

Author

Antti Karttunen, Jun 11 2018

Keywords

Comments

a(n) is the number of terms of A000051 that divide n.

Crossrefs

Cf. A000051, A209229, A292315 (positions of zeros), A305435, A323482.
Cf. also A154402.

Programs

Formula

a(n) = Sum_{d|n} A209229(d-1).
a(n) = A305435(n) + A209229(n-1).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = A323482 = 1.264499... . - Amiram Eldar, Dec 31 2023

A379515 Numerators of the partial alternating sums of the reciprocals of the sum of unitary divisors function (A034448).

Original entry on oeis.org

1, 2, 11, 43, 53, 4, 37, 293, 329, 103, 113, 107, 809, 129, 809, 12913, 41119, 39691, 41833, 8081, 33395, 32443, 33871, 10973, 148361, 48275, 7149, 34861, 108119, 319937, 164941, 1761311, 112361, 662011, 5405483, 26502319, 516671461, 508357441, 3620857237, 3556192637
Offset: 1

Views

Author

Amiram Eldar, Dec 23 2024

Keywords

Examples

			Fractions begin with 1, 2/3, 11/12, 43/60, 53/60, 4/5, 37/40, 293/360, 329/360, 103/120, 113/120, 107/120, ...
		

Crossrefs

Programs

  • Mathematica
    usigma[n_] := Times @@ (1 + Power @@@ FactorInteger[n]); usigma[1] = 1; Numerator[Accumulate[Table[(-1)^(n+1)/usigma[n], {n, 1, 50}]]]
  • PARI
    usigma(n) = {my(f = factor(n)); prod(i = 1, #f~, 1 + f[i, 1]^f[i, 2]);}
    list(nmax) = {my(s = 0); for(k = 1, nmax, s += (-1)^(k+1) / usigma(k); print1(numerator(s), ", "))};

Formula

a(n) = numerator(Sum_{k=1..n} (-1)^(k+1)/A034448(k)).
a(n)/A379516(n) = E * log(n) + F + O(log(n)^(5/3) * log(log(n))^(4/3) / n^u), where u > 0, E = A308041 * (2/(A323482 + 1/2) - 1) = 0.10259754363391420806..., and F is a constant.
Showing 1-6 of 6 results.