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 1-4 of 4 results.

A002277 a(n) = 3*(10^n - 1)/9.

Original entry on oeis.org

0, 3, 33, 333, 3333, 33333, 333333, 3333333, 33333333, 333333333, 3333333333, 33333333333, 333333333333, 3333333333333, 33333333333333, 333333333333333, 3333333333333333, 33333333333333333, 333333333333333333, 3333333333333333333, 33333333333333333333, 333333333333333333333
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

From Wolfdieter Lang, Feb 08 2017: (Start)
This sequence (for n >= 1) appears in n-families satisfying so-called curious cubic identities based on the Armstrong numbers 153, 370 and 371, A005188(10) - A005188(12).
153 also involves A246057(n-1) and A093143(n). See a comment in A246057 with the van Poorten et al. reference, and A281857.
370 and 371 also involve A067275(n+1). See the comment there, and A281858 and A281860. (End)

Examples

			From _Wolfdieter Lang_, Feb 08 2017: (Start)
Curious cubic identities (see a comment above):
1^3 + 5^3 + 3^3 = 153, 16^3 + 50^3 + 33^3 = 165033, 166^3 + 500^3 + 333^3 = 166500333, ...
3^3 + 7^3 + 0^3 = 370; 336700 = 33^3 + 67^3 + (00)^3 = 336700,  333^3 + 667^3 + (000)^3 = 333667000, ...
3^3 + 7^3 + 1^3 = 371, 33^3 + 67^3 + (01)^3 = 336701, 333^3 + 667^3 + (001)^3 = 333667001, ... (End)

Links

Crossrefs

Cf. A002275, A002276, A002278, A002279, A002280, A002281, A002282, A002283, A010785, A017197.
Cf. A005188, A067275, A075412, A093143, A135702, A178631, A178633, A246057, A281857, A281858, A281860.

Programs

Formula

a(n) = 3*A002275(n).
a(n) = A178631(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
From Vincenzo Librandi, Jul 22 2010: (Start)
a(n) = a(n-1) + 3*10^(n-1) with a(0)=0;
a(n) = 11*a(n-1) - 10*a(n-2) with a(0)=0, a(1)=3. (End)
G.f.: 3*x/((1 - x)*(1 - 10*x)). - Ilya Gutkovskiy, Feb 24 2017
Sum_{n>=1} 1/a(n) = A135702. - Amiram Eldar, Nov 13 2020
E.g.f.: exp(x)*(exp(9*x) - 1)/3. - Stefano Spezia, Sep 13 2023
From Elmo R. Oliveira, Jul 20 2025: (Start)
a(n) = (A246057(n) - 1)/5.
a(n) = A010785(A017197(n-1)) for n >= 1. (End)

A093137 Expansion of (1-7*x)/((1-x)*(1-10*x)).

Original entry on oeis.org

1, 4, 34, 334, 3334, 33334, 333334, 3333334, 33333334, 333333334, 3333333334, 33333333334, 333333333334, 3333333333334, 33333333333334, 333333333333334, 3333333333333334, 33333333333333334, 333333333333333334, 3333333333333333334, 33333333333333333334
Offset: 0

Views

Author

Paul Barry, Mar 24 2004

Keywords

Comments

Second binomial transform of 3*A001045(3n)/3+(-1)^n. Partial sums of A093138. A convex combination of 10^n and 1. In general the second binomial transform of k*Jacobsthal(3n)/3+(-1)^n is 1,1+k,1+11k,1+111k,... This is the case for k=3.
a(n) is the number of n-length sequences of decimal digits whose sum is divisible by 3. - Geoffrey Critzer, Jan 18 2014
This sequence appears in a family of curious cubic identities based on the Armstrong number 407 = A005188(13). See the formula section. For the analog identities based on 153 = A005188(10) see a comment on A246057 with the van der Poorten et al. reference and A281857. For those based on 370 = A005188(11) see A067275, A002277 and A281858. - Wolfdieter Lang, Feb 08 2017

Examples

			a(1)^2 = 16
a(2)^2 = 1156
a(3)^2 = 111556
a(4)^2 = 11115556
a(5)^2 = 1111155556
a(6)^2 = 111111555556
a(7)^2 = 11111115555556
a(8)^2 = 1111111155555556
a(9)^2 = 111111111555555556, etc... (see A102807). - _Philippe Deléham_, Oct 03 2011
Curious cubic identities: 407 = 4^3 + 0^3 + 7^3, 340067 = 34^3 + (00)^3 + 67^3, 334000677 = 334^3 + (000)^3 + 677^3, ... - _Wolfdieter Lang_, Feb 08 2017

References

  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See entry 3334 at p. 168.

Links

Crossrefs

Cf. A005188, A067275, A246057, A281857, A281859.
Cf. A001045, A002277, A093138, A281858.
Cf. A102807 (squared).

Programs

  • Mathematica
    nn=20; r=Solve[{s==4x s+3 x a+3x b+1,a==4x a+3x s+3x b,b==4x b+3x s+3x a},{s,a,b}]; CoefficientList[Series[s/.r,{x,0,nn}],x] (* Geoffrey Critzer, Jan 18 2014 *)
Table[3*10^n/9 + 6/9, {n, 0, 20}] (* or *) NestList[10 # - 6 &, 1, 20] (* Michael De Vlieger, Feb 08 2017 *)
LinearRecurrence[{11,-10},{1,4},20] (* Harvey P. Dale, Oct 07 2017 *)
  • PARI
    Vec((1-7*x)/((1-x)*(1-10*x)) + O (x^30)) \\ Michel Marcus, Feb 09 2017

Formula

a(n) = 3*10^n/9 + 6/9.
a(n) = 10*a(n-1)-6 with a(0)=1. - Vincenzo Librandi, Aug 02 2010
a(n)^3 + 0(n)^3 + A067275(n+1)^3 = concatenation(a(n), 0(n), A067275(n+1)) = A281859(n), where 0(n) denotes n 0's, n >= 1. - Wolfdieter Lang, Feb 08 2017
From Elmo R. Oliveira, Aug 17 2024: (Start)
E.g.f.: exp(x)*(exp(9*x) + 2)/3.
a(n) = 11*a(n-1) - 10*a(n-2) for n > 1. (End)

A281858 Curious cubic identities based on the Armstrong number 370.

Original entry on oeis.org

370, 336700, 333667000, 333366670000, 333336666700000, 333333666667000000, 333333366666670000000, 333333336666666700000000, 333333333666666667000000000, 333333333366666666670000000000, 333333333336666666666700000000000, 333333333333666666666667000000000000
Offset: 1

Views

Author

Wolfdieter Lang, Feb 08 2017

Keywords

Comments

See a comment in A067275, and the analog to the Armstrong number 153 = A005188(10) treated in A281857, 370 = A005188(11).

Examples

			n=1: 370 =  3^3 + 7^3 + 0^3; n=2: 336700 = 33^3 + 67^3 + (00)^3; n=3: 333667000 = 333^3 + 667^3 + (000)^3.

Links

Crossrefs

Cf. A002277, A005188, A067275, A246057, A281857.

Programs

  • Mathematica
    Table[FromDigits@ Join[ConstantArray[3, n], ReplacePart[ConstantArray[6, n], -1 -> 7], ConstantArray[0, n]], {n, 12}] (* Michael De Vlieger, Feb 08 2017 *)
  • PARI
    Vec(10*x*(37 - 7400*x + 100000*x^2) / ((1 - 10*x)*(1 - 100*x)*(1 - 1000*x)) + O(x^30)) \\ Colin Barker, Feb 08 2017

Formula

a(n) = A002277(n)^3 + A067275(n+1)^3 + 0(n)^3, n >= 1, with 0(n) standing for n 0's.
From Colin Barker, Feb 08 2017: (Start)
G.f.: 10*x*(37 - 7400*x + 100000*x^2) / ((1 - 10*x)*(1 - 100*x)*(1 - 1000*x)).
a(n) = 10^n*(1 + 10^n + 100^n) / 3.
a(n) = 1110*a(n-1) - 111000*a(n-2) + 1000000*a(n-3) for n>3. (End)

A281860 Curious identities based on the Armstrong number 371 = A005188(12).

Original entry on oeis.org

371, 336701, 333667001, 333366670001, 333336666700001, 333333666667000001, 333333366666670000001, 333333336666666700000001, 333333333666666667000000001, 333333333366666666670000000001, 333333333336666666666700000000001, 333333333333666666666667000000000001
Offset: 1

Views

Author

Wolfdieter Lang, Feb 08 2017

Keywords

Comments

See a comment in A067275.

Examples

			n=1: 371 = 3^3 + 7^3 + 1^3;
n=2: 336701 = 33^3 + 67^3 + (01)^3;
n=3: 333667001 = 333^3 + 667^3 + (001)^3.

Links

Crossrefs

Cf. A002277, A067275, A281857, A281858, A281859.

Programs

  • Mathematica
    LinearRecurrence[{1111,-112110,1111000,-1000000},{371,336701,333667001,333366670001},20] (* Harvey P. Dale, May 28 2024 *)
  • PARI
    Vec(x*(371 - 75480*x + 1185000*x^2 - 2000000*x^3) / ((1 - x)*(1 - 10*x)*(1 - 100*x)*(1 - 1000*x)) + O(x^30)) \\ Colin Barker, Feb 09 2017

Formula

a(n) = A002277(n) * 10^(2*n) + A067275(n+1) * 10^n + 0(n-1)1, where 0(n-1)1 stands for n-1 0's followed by a 1, for n >= 1.
a(n) = A002277(n)^3 + A067275(n+1)^3 + (0(n-1)1)^3.
From Colin Barker, Feb 09 2017: (Start)
G.f.: x*(371 - 75480*x + 1185000*x^2 - 2000000*x^3)/((1 - x)*(1 - 10*x)*(1 - 100*x)*(1 - 1000*x)).
a(n) = 1111*a(n-1) - 112110*a(n-2) + 1111000*a(n-3) - 1000000*a(n-4) for n>4.
a(n) = (3 + 10^n + 100^n + 1000^n)/3. (End)
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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