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 1-10 of 11 results. Next

A024036 a(n) = 4^n - 1.

Original entry on oeis.org

0, 3, 15, 63, 255, 1023, 4095, 16383, 65535, 262143, 1048575, 4194303, 16777215, 67108863, 268435455, 1073741823, 4294967295, 17179869183, 68719476735, 274877906943, 1099511627775, 4398046511103, 17592186044415, 70368744177663, 281474976710655
Offset: 0

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Author

Keywords

Comments

This sequence is the normalized length per iteration of the space-filling Peano-Hilbert curve. The curve remains in a square, but its length increases without bound. The length of the curve, after n iterations in a unit square, is a(n)*2^(-n) where a(n) = 4*a(n-1)+3. This is the sequence of a(n) values. a(n)*(2^(-n)*2^(-n)) tends to 1, the area of the square where the curve is generated, as n increases. The ratio between the number of segments of the curve at the n-th iteration (A015521) and a(n) tends to 4/5 as n increases. - Giorgio Balzarotti, Mar 16 2006
Numbers whose base-4 representation is 333....3. - Zerinvary Lajos, Feb 03 2007
From Eric Desbiaux, Jun 28 2009: (Start)
It appears that for a given area, a square n^2 can be divided into n^2+1 other squares.
It's a rotation and zoom out of a Cartesian plan, which creates squares with side
= sqrt( (n^2) / (n^2+1) ) --> A010503|A010532|A010541... --> limit 1,
and diagonal sqrt(2*sqrt((n^2)/(n^2+1))) --> A010767|... --> limit A002193.
(End)
Also the total number of line segments after the n-th stage in the H tree, if 4^(n-1) H's are added at the n-th stage to the structure in which every "H" is formed by 3 line segments. A164346 (the first differences of this sequence) gives the number of line segments added at the n-th stage. - Omar E. Pol, Feb 16 2013
a(n) is the cumulative number of segment deletions in a Koch snowflake after (n+1) iterations. - Ivan N. Ianakiev, Nov 22 2013
Inverse binomial transform of A005057. - Wesley Ivan Hurt, Apr 04 2014
For n > 0, a(n) is one-third the partial sums of A002063(n-1). - J. M. Bergot, May 23 2014
Also the cyclomatic number of the n-Sierpinski tetrahedron graph. - Eric W. Weisstein, Sep 18 2017

Examples

			G.f. = 3*x + 15*x^2 + 63*x^3 + 255*x^4 + 1023*x^5 + 4095*x^6 + ...
		

References

  • Graham Everest, Alf van der Poorten, Igor Shparlinski, and Thomas Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.

Crossrefs

Programs

  • Haskell
    a024036 = (subtract 1) . a000302
    a024036_list = iterate ((+ 3) . (* 4)) 0
    -- Reinhard Zumkeller, Oct 03 2012
    
  • Maple
    A024036:=n->4^n-1; seq(A024036(n), n=0..30); # Wesley Ivan Hurt, Apr 04 2014
  • Mathematica
    Array[4^# - 1 &, 50, 0] (* Vladimir Joseph Stephan Orlovsky, Nov 03 2009 *)
    (* Start from Eric W. Weisstein, Sep 19 2017 *)
    Table[4^n - 1, {n, 0, 20}]
    4^Range[0, 20] - 1
    LinearRecurrence[{5, -4}, {0, 3}, 20]
    CoefficientList[Series[3 x/(1 - 5 x + 4 x^2), {x, 0, 20}], x]
    (* End *)
  • PARI
    for(n=0, 100, print1(4^n-1, ", ")) \\ Felix Fröhlich, Jul 04 2014
  • Sage
    [gaussian_binomial(2*n,1, 2) for n in range(21)] # Zerinvary Lajos, May 28 2009
    
  • Sage
    [stirling_number2(2*n+1, 2) for n in range(21)] # Zerinvary Lajos, Nov 26 2009
    

Formula

a(n) = 3*A002450(n). - N. J. A. Sloane, Feb 19 2004
G.f.: 3*x/((-1+x)*(-1+4*x)) = 1/(-1+x) - 1/(-1+4*x). - R. J. Mathar, Nov 23 2007
E.g.f.: exp(4*x) - exp(x). - Mohammad K. Azarian, Jan 14 2009
a(n) = A000051(n)*A000225(n). - Reinhard Zumkeller, Feb 14 2009
A079978(a(n)) = 1. - Reinhard Zumkeller, Nov 22 2009
a(n) = A179857(A000225(n)), for n > 0; a(n) > A179857(m), for m < A000225(n). - Reinhard Zumkeller, Jul 31 2010
a(n) = 4*a(n-1) + 3, with a(0) = 0. - Vincenzo Librandi, Aug 01 2010
A000120(a(n)) = 2*n. - Reinhard Zumkeller, Feb 07 2011
a(n) = (3/2)*A020988(n). - Omar E. Pol, Mar 15 2012
a(n) = (Sum_{i=0..n} A002001(i)) - 1 = A178789(n+1) - 3. - Ivan N. Ianakiev, Nov 22 2013
a(n) = n*E(2*n-1,1)/B(2*n,1), for n > 0, where E(n,x) denotes the Euler polynomials and B(n,x) the Bernoulli polynomials. - Peter Luschny, Apr 04 2014
a(n) = A000302(n) - 1. - Sean A. Irvine, Jun 18 2019
Sum_{n>=1} 1/a(n) = A248721. - Amiram Eldar, Nov 13 2020
a(n) = A080674(n) - A002450(n). - Elmo R. Oliveira, Dec 02 2023

Extensions

More terms Wesley Ivan Hurt, Apr 04 2014

A073668 Decimal expansion of Sum_{k>=1} 1/(10^k - 1).

Original entry on oeis.org

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

Views

Author

Robert G. Wilson v, Aug 29 2002

Keywords

Comments

Parallels A000005 up to a(46).
Sum_{k>=1} x^k/(1-x^k) = Sum_{k>=1} tau(k)*x^k. Choosing x = 1/10 gives the result. - Amarnath Murthy, Oct 21 2002

Examples

			0.122324243426244526264428344628264449244... = A065444/9.
		

References

  • Amarnath Murthy, Some interesting results on d(N), the number of divisors of a natural number, page 463, Octogon Mathematical Magazine, Vol. 8 No. 2, October 2000.

Crossrefs

Programs

  • Maple
    evalf(Sum(1/(10^k - 1), k = 1..infinity), 200) # Vaclav Kotesovec, Jul 16 2019
    # second program with faster converging series after Joerg Arndt
    evalf( add( (1/10)^(n^2)*(1 + 2/(10^n - 1)), n = 1..8), 105); # Peter Bala, Jan 30 2022
  • Mathematica
    RealDigits[ N[ Sum[1/(10^k - 1), {k, 1, Infinity}], 120]] [[1]]
  • PARI
    suminf(k=1,1/(10^k-1)) \\ Charles R Greathouse IV, Oct 05 2014

Formula

From Eric Desbiaux, Mar 11 2009: (Start)
Equals Sum_{k >= 1} 1/((2^k*5^k)-1).
Equals Sum_{k >= 1} (1/2^k)*(1/5^k)/(1-((1/2^k)*(1/5^k))).
Sum_{k >= 1} 1/(5^k) = 1/4.
Sum_{k >= 1} 1/(2^k) = 1.
Sum_{k >= 1} (1/5^k)/(1-((1/2^k)*(1/5^k))) = 0.2726344339156...
Sum_{k >= 1} (1/2^k)/(1-((1/2^k)*(1/5^k))) = 1.0582125127815...
Sum_{k >= 1} 1/(1-((1/2^k)*(1/5^k))) - 1 = A073668.
(End)
Fast computation via Lambert series: 0.122324243426... = Sum_{n>=1} x^(n^2)*(1+x^n)/(1-x^n) where x=1/10. - Joerg Arndt, Oct 18 2020

A214369 Decimal expansion of Sum_{n>=1} 1/(3^n-1).

Original entry on oeis.org

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

Views

Author

R. J. Mathar, Jul 14 2012

Keywords

Examples

			Equals 0.6821535026052380667...
		

Crossrefs

Programs

  • Maple
    evalf(sum(1/(3^k-1), k=1..infinity), 120); # Vaclav Kotesovec, Oct 18 2014
    # second program with faster converging series
    evalf( add( (1/3)^(n^2)*(1 + 2/(3^n - 1)), n = 1..14 ), 105); # Peter Bala, Jan 30 2022
  • Mathematica
    RealDigits[ NSum[1/(3^n - 1), {n, 1, Infinity}, WorkingPrecision -> 110, NSumTerms -> 100], 10, 105] // First (* or *) 1 - (Log[2] + QPolyGamma[0, 1, 1/3])/Log[3] // RealDigits[#, 10, 105]& // First (* Jean-François Alcover, Jun 05 2013 *)
    x = 1/3; RealDigits[ Sum[ DivisorSigma[0, k] x^k, {k, 1000}], 10, 105][[1]] (* Robert G. Wilson v, Oct 12 2014 after an observation and the formula of Amarnath Murthy, see A073668 *)
  • PARI
    suminf(n=1, 1/(3^n-1)) \\ Michel Marcus, Mar 11 2017

Formula

Equals Sum_{n>=1} 1/A024023(n).
Equals Sum_{k>=1} d(k)/3^k, where d(k) is the number of divisors of k (A000005). - Amiram Eldar, May 17 2020

Extensions

More terms from Jean-François Alcover, Feb 12 2013

A248722 Decimal expansion of Sum_{k>=1} 1/(5^k - 1).

Original entry on oeis.org

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

Views

Author

Robert G. Wilson v, Oct 12 2014

Keywords

Examples

			0.301733853597972457948162159393991192623009431517157720395791923318379825892...
		

Crossrefs

Programs

  • Maple
    evalf( add( (1/5)^(n^2)*(1 + 2/(5^n - 1)), n = 1..12), 105); # Peter Bala, Jan 30 2022
  • Mathematica
    x = 1/5; RealDigits[ Sum[ DivisorSigma[0, k] x^k, {k, 1000}], 10, 105][[1]] (* after an observation and the formula of Amarnath Murthy, see A073668 *)
  • PARI
    sumpos(k=1,1/(5^k-1)) \\ M. F. Hasler, Oct 15 2014

Formula

Equals Sum_{k>=1} d(k)/5^k, where d(k) is the number of divisors of k (A000005). - Amiram Eldar, Jun 22 2020

A248723 Decimal expansion of the Sum_{k>=1} 1/(6^k - 1).

Original entry on oeis.org

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

Views

Author

Robert G. Wilson v, Oct 12 2014

Keywords

Examples

			0.2341491301348092064851116728138729185463610347865138985224213867102381986628...
		

Crossrefs

Programs

  • Maple
    evalf(sum(1/(6^k-1), k=1..infinity),120); # Vaclav Kotesovec, Oct 18 2014
    # second program with faster converging series
    evalf( add( (1/6)^(n^2)*(1 + 2/(6^n - 1)), n = 1..11), 105); # Peter Bala, Jan 30 2022
  • Mathematica
    x = 1/6; RealDigits[ Sum[ DivisorSigma[0, k] x^k, {k, 1000}], 10, 105][[1]] (* after an observation and the formula of Amarnath Murthy, see A073668 *)
  • PARI
    suminf(k=1, 1/(6^k-1)) \\ Michel Marcus, Oct 18 2014

Formula

Equals Sum_{k>=1} d(k)/6^k, where d(k) is the number of divisors of k (A000005). - Amiram Eldar, Jun 22 2020

A248724 Decimal expansion of Sum_{k>=1} 1/(7^k - 1).

Original entry on oeis.org

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

Views

Author

Robert G. Wilson v, Oct 12 2014

Keywords

Examples

			0.1909100624102615782021996444176911687692684760082664083347711086409996755846...
		

Crossrefs

Programs

  • Maple
    evalf(sum(1/(7^k-1), k=1..infinity),120) # Vaclav Kotesovec, Oct 18 2014
    # second program with faster converging series
    evalf( add( (1/7)^(n^2)*(1 + 2/(7^n - 1)), n = 1..11), 105); # Peter Bala, Jan 30 2022
  • Mathematica
    x = 1/7; RealDigits[ Sum[ DivisorSigma[0, k] x^k, {k, 1000}], 10, 105][[1]] (* after an observation and the formula of Amarnath Murthy, see A073668 *)
  • PARI
    suminf(k=1, 1/(7^k-1)) \\ Michel Marcus, Oct 18 2014

Formula

Equals Sum_{k>=1} d(k)/7^k, where d(k) is the number of divisors of k (A000005). - Amiram Eldar, Jun 22 2020

A248725 Decimal expansion of Sum_{k>=1} 1/(8^k - 1).

Original entry on oeis.org

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

Views

Author

Robert G. Wilson v, Oct 12 2014

Keywords

Examples

			0.16096618431506239680530256414364288555074385602532834636083591864782394085800...
		

Crossrefs

Programs

  • Maple
    evalf(sum(1/(8^k-1), k=1..infinity),120) # Vaclav Kotesovec, Oct 18 2014
    # second program with faster converging series
    evalf( add( (1/8)^(n^2)*(1 + 2/(8^n - 1)), n = 1..10), 105); # Peter Bala, Jan 30 2022
  • Mathematica
    x = 1/8; RealDigits[ Sum[ DivisorSigma[0, k] x^k, {k, 1000}], 10, 105][[1]] (* after an observation and the formula of Amarnath Murthy, see A073668 *)
  • PARI
    suminf(k=1, 1/(8^k-1)) \\ Michel Marcus, Oct 18 2014

Formula

Equals Sum_{k>=1} d(k)/8^k, where d(k) is the number of divisors of k (A000005). - Amiram Eldar, Jun 22 2020

A248726 Decimal expansion of Sum_{k>=1} 1/(9^k - 1).

Original entry on oeis.org

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

Views

Author

Robert G. Wilson v, Oct 12 2014

Keywords

Examples

			0.13904511766218812935872847436908905213936264706781960955103549347967020145366...
		

Crossrefs

Programs

  • Maple
    evalf(sum(1/(9^k-1), k=1..infinity),120) # Vaclav Kotesovec, Oct 18 2014
    # second program with faster converging series
    evalf( add( (1/9)^(n^2)*(1 + 2/(9^n - 1)), n = 1..10), 105); # Peter Bala, Jan 30 2022
  • Mathematica
    x = 1/9; RealDigits[ Sum[ DivisorSigma[0, k] x^k, {k, 1000}], 10, 105][[1]] (* after an observation and the formula of Amarnath Murthy, see A073668 *)
  • PARI
    suminf(k=1, 1/(9^k-1)) \\ Michel Marcus, Oct 18 2014

Formula

Equals Sum_{k>=1} d(k)/9^k, where d(k) is the number of divisors of k (A000005). - Amiram Eldar, Jun 22 2020

A048329 Numbers that are repdigits in base 4.

Original entry on oeis.org

0, 1, 2, 3, 5, 10, 15, 21, 42, 63, 85, 170, 255, 341, 682, 1023, 1365, 2730, 4095, 5461, 10922, 16383, 21845, 43690, 65535, 87381, 174762, 262143, 349525, 699050, 1048575, 1398101, 2796202, 4194303, 5592405, 11184810, 16777215, 22369621, 44739242, 67108863
Offset: 0

Views

Author

Patrick De Geest, Feb 15 1999

Keywords

Examples

			10_10 = 22_4, 15_10 = 33_4, 5461_10 = 1111111_4.
		

Crossrefs

Base 4 repdigits 1,2,3 (trisections): A002450, A020988, A024036.

Programs

  • Magma
    [0] cat  [k:k in [1..10^7]| #Set(Intseq(k,4)) eq 1]; // Marius A. Burtea, Oct 11 2019
  • Maple
    a:= n-> (1+irem(n+2, 3))*(4^iquo(n+2,3)-1)/3:
    seq(a(n), n = 0..45);
  • Mathematica
    Union[Flatten[Table[FromDigits[PadRight[{}, n, d], 4], {n, 0, 40}, {d, 3}]]](* Vincenzo Librandi, Feb 06 2014 *)
    LinearRecurrence[{0,0,5,0,0,-4},{0,1,2,3,5,10},40] (* Harvey P. Dale, Jul 11 2023 *)

Formula

G.f.: x*(1+2*x+3*x^2) / ( (x-1)*(4*x^3-1)*(1+x+x^2) ) with a(n) = 5*a(n-3) - 4*a(n-6). - R. J. Mathar, Mar 15 2015
Sum_{n>=1} 1/a(n) = (11/2) * A248721 = 2.31603727318383077512... - Amiram Eldar, Jan 21 2022

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)
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