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 1-9 of 9 results.

A024088 a(n) = 8^n - 1.

Original entry on oeis.org

0, 7, 63, 511, 4095, 32767, 262143, 2097151, 16777215, 134217727, 1073741823, 8589934591, 68719476735, 549755813887, 4398046511103, 35184372088831, 281474976710655, 2251799813685247, 18014398509481983
Offset: 0

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Author

Keywords

Comments

Numbers whose base 8 or octal representation is 777777.......7. - Zerinvary Lajos, Feb 03 2007

Crossrefs

Programs

  • GAP
    List([0..30], n-> 8^n -1); # G. C. Greubel, Aug 03 2019
  • Magma
    [8^n -1: n in [0..20]]; // G. C. Greubel, Aug 03 2019
    
  • Mathematica
    8^Range[0,20]-1 (* or *) LinearRecurrence[{9,-8},{0,7},20] (* Harvey P. Dale, Jan 04 2017 *)
  • PARI
    vector(20, n, n--; 8^n -1) \\ G. C. Greubel, Aug 03 2019
    
  • Sage
    [gaussian_binomial(3*n,1,2) for n in range(0,20)] # Zerinvary Lajos, May 28 2009
    
  • Sage
    [stirling_number2(3*n+1,2) for n in range(0,20)] # Zerinvary Lajos, Nov 26 2009
    
  • Sage
    [8^n-1 for n in (0..20)] # Bruno Berselli, Nov 11 2015
    

Formula

From Mohammad K. Azarian, Jan 14 2009: (Start)
G.f.: 1/(1-8*x) - 1/(1-x).
E.g.f.: exp(8*x) - exp(x). (End)
a(n) = A000225(n)*A001576(n). - Reinhard Zumkeller, Feb 15 2009
a(n) = 8*a(n-1) + 7 for n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
a(n) = Sum_{i=1..n} 7^i*binomial(n,n-i) for n>0, a(0)=0. - Bruno Berselli, Nov 11 2015
a(n) = A001018(n) - 1. - Sean A. Irvine, Jun 19 2019
Sum_{n>=1} 1/a(n) = A248725. - Amiram Eldar, Nov 13 2020

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

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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

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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

A248721 Decimal expansion of Sum_{k>=1} 1/(4^k - 1).

Original entry on oeis.org

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

Views

Author

Robert G. Wilson v, Oct 12 2014

Keywords

Examples

			0.4210976860334237772959908879677130489614413363241154046059207967127713704887...
		

Crossrefs

Programs

  • Maple
    evalf(sum(1/(4^k-1), k=1..infinity),120) # Vaclav Kotesovec, Oct 18 2014
    # second program with faster converging series after Joerg Arndt
    evalf( add( (1/4)^(n^2)*(1 + 2/(4^n - 1)), n = 1..13), 105); # Peter Bala, Jan 30 2022
  • Mathematica
    x = 1/4; 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/(4^k-1)) \\ Michel Marcus, Oct 18 2014

Formula

Equals Sum_{k>=1} x^(k^2)*(1+x^k)/(1-x^k) where x = 1/4 (the Lambert series evaluated at 1/4). - Joerg Arndt, Jun 03 2020
Equals Sum_{k>=1} d(k)/4^k, where d(k) is the number of divisors of k (A000005). - Amiram Eldar, Jun 22 2020

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

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

A048333 Numbers that are repdigits in base 8.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 9, 18, 27, 36, 45, 54, 63, 73, 146, 219, 292, 365, 438, 511, 585, 1170, 1755, 2340, 2925, 3510, 4095, 4681, 9362, 14043, 18724, 23405, 28086, 32767, 37449, 74898, 112347, 149796, 187245, 224694, 262143, 299593, 599186, 898779
Offset: 0

Views

Author

Patrick De Geest, Feb 15 1999

Keywords

Comments

For the general case, the sequence of numbers that are repdigits in base b > 1 satisfies the recurrence a(n) = (b+1)*a(n-b+1) - b*a(n-2*(b-1)) for n >= 2(b-1) with g.f.: (sum_{1 <= i < b} i*x^i)/(1 - (b+1)*x^(b-1) + bx^(2(b-1))). - Chai Wah Wu, May 30 2016

Crossrefs

Programs

  • Mathematica
    Union[Flatten[Table[FromDigits[PadRight[{}, n, d], 8], {n, 0, 40}, {d, 7}]]] (* Vincenzo Librandi, Feb 06 2014 *)
    LinearRecurrence[{0,0,0,0,0,0,9,0,0,0,0,0,0,-8},{0,1,2,3,4,5,6,7,9,18,27,36,45,54},50] (* Harvey P. Dale, Dec 09 2018 *)
  • PARI
    is(n)=#Set(digits(n,8))==1 \\ Charles R Greathouse IV, Feb 15 2017

Formula

From Chai Wah Wu, May 30 2016: (Start)
a(n) = 9*a(n-7) - 8*a(n-14) for n > 13.
G.f.: x*(7*x^6 + 6*x^5 + 5*x^4 + 4*x^3 + 3*x^2 + 2*x + 1)/(8*x^14 - 9*x^7 + 1). (End)
Sum_{n>=1} 1/a(n) = (363/20) * A248725 = 2.92153624531838250201... - Amiram Eldar, Jan 21 2022

Extensions

Changed offset from 1 to 0 by Vincenzo Librandi, Feb 06 2014
Showing 1-9 of 9 results.