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.

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)

A067275 Number of Fibonacci numbers A000045(k), k <= 10^n, which end in 4.

Original entry on oeis.org

0, 1, 7, 67, 667, 6667, 66667, 666667, 6666667, 66666667, 666666667, 6666666667, 66666666667, 666666666667, 6666666666667, 66666666666667, 666666666666667, 6666666666666667, 66666666666666667, 666666666666666667, 6666666666666666667, 66666666666666666667, 666666666666666666667
Offset: 0

Views

Author

Joseph L. Pe, Feb 21 2002

Keywords

Comments

Numbers n such that the digits of A000326(n), the n-th pentagonal number, begin with n. [original definition of this sequence]
The sequence 1,7,67.... has a(n) = 6*10^n/9+3/9. It is the second binomial transform of 6*A001045(3n)/3 + (-1)^n. In general the second binomial transform of k*Jacobsthal(3n)/3+(-1)^n is k*10^n/9 + (1-k/9) = 1, 1+k, 1+11k, 1+111k, ... - Paul Barry, Mar 24 2004
Except for the first two terms, these are the 3-automorphic numbers ending in 7. - Eric M. Schmidt, Aug 28 2012
From Wolfdieter Lang, Feb 08 2017: (Start)
This sequence appears in curious identities based on the Armstrong numbers 370 = A005188(11), 371 = A005188(12) and 407 = A005188(13).
For such identities based on 153 = A005188(10) see a comment in A246057 with the van der Poorten et al. reference.
For 370 these identities are A002277(n)^3 + a(n+1)^3 + 0(n)^3 = A002277(n)*10^(2*n) + a(n+1)*10^n + 0(n) = A281858(n), where 0(n) means n 0's.
For 371 these identities are A002277(n)^3 + a(n+1)^3 + (0(n-1)1)^3 = A002277(n)*10^(2*n) + a(n+1)*10^n + 0(n-1)1 = A281860(n), where 0(n-1)1 means n-1 0's followed by a 1.
For 407 these identities are A093137(n)^3 + 0(n)^3 + a(n+1)^3 = concatenation(A093137(n), 0(n), a(n+1)) = A281859(n). (End)

Examples

			a(2) = 7 because 7 of the first 10^2 Fibonacci numbers end in 4.
From _Wolfdieter Lang_, Feb 08 2017: (Start)
Curious cubic identities:
3^3 + 7^3 + 0^3 = 370, 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, ...
4^3 + 0^3 + 7^3 = 407, 34^3 + (00)^3 + 67^3 = 340067 , 334^3 + (000)^3 + 677^3 = 334000677, ... (End)

Links

Crossrefs

Cf. A072702, A093137, A109344, A246057, A281858, A281859, A281860.
Cf. A000045, A000326, A001045, A002277, A005188, A073553, A199682.

Programs

  • Mathematica
    s = Fibonacci@ Range[10^5]; Table[Count[Take[s, 10^n], m_ /; Mod[m, 10] == 4], {n, 0, Floor@ Log10@ Length@ s}] (* or *) Table[Boole[n > 0] Ceiling[10^n/15], {n, 0, 20}] (* or *) CoefficientList[Series[x (1 - 4 x)/((1 - x) (1 - 10 x)), {x, 0, 20}], x] (* Michael De Vlieger, Feb 08 2017 *)
  • PARI
    a(n)=(10^n+13)\15 \\ Charles R Greathouse IV, Jun 05 2011

Formula

a(n) = ceiling((2/30)*10^n) - Benoit Cloitre, Aug 27 2002
From Paul Barry, Mar 24 2004: (Start)
G.f.: x*(1 - 4*x)/((1 - x)*(1 - 10*x)).
a(n) = 10^n/15 + 1/3 for n>0. (End)
a(n) = 10*a(n-1) - 3 for n>1. - Vincenzo Librandi, Dec 07 2010 [Immediate consequence of the previous formula, R. J. Mathar]
From Eric M. Schmidt, Oct 28 2012: (Start)
For n>0, a(n) = A199682(n-1)/3 = (2*10^(n-1) + 1)/3.
For n>=2, a(n+1) = a(n) + 6*10^n. (End)
From Elmo R. Oliveira, Jul 22 2025: (Start)
E.g.f.: (-6 + 5*exp(x) + exp(10*x))/15.
a(n) = 11*a(n-1) - 10*a(n-2) for n >= 3.
a(n) = A073553(n)/2 for n >= 1. (End)

Extensions

A073552 merged into this sequence by Eric M. Schmidt, Oct 28 2012

A281857 Numbers occurring in a curious cubic identity.

Original entry on oeis.org

153, 165033, 166500333, 166650003333, 166665000033333, 166666500000333333, 166666650000003333333, 166666665000000033333333, 166666666500000000333333333, 166666666650000000003333333333, 166666666665000000000033333333333, 166666666666500000000000333333333333
Offset: 1

Views

Author

Wolfdieter Lang, Feb 07 2017

Keywords

Comments

See A246057 for the van der Poorten et al. reference and a comment.
153 is the Armstrong number A005188(10). [Typo corrected by Jeremy Tan, Feb 25 2023]

Examples

			1^3 + 5^3 + 3^3 = 153, 16^3 + 50^3 + 33^3 = 165033, 166^3 + 500^3 + 333^3 = 166500333, ...

Links

Crossrefs

Cf. A002277, A005188, A093143, A246057, A281858, A281859, A281860.

Programs

  • Mathematica
    Table[FromDigits@ Join[ReplacePart[ConstantArray[6, n], 1 -> 1], ReplacePart[ConstantArray[0, n], 1 -> 5], ConstantArray[3, n]], {n, 12}] (* Michael De Vlieger, Feb 08 2017 *)
  • PARI
    Vec(9*x*(17 - 550*x + 33500*x^2) / ((1 - x)*(1 - 10*x)*(1 - 100*x)*(1 - 1000*x)) + O(x^15)) \\ Colin Barker, Feb 08 2017
  • PARI
    a(n) = (((10^n - 4)/6)^3) + ((10^n/2)^3) + (((10^n - 1)/3)^3) \\ Jean-Jacques Vaudroz, Aug 11 2024

Formula

a(n) = A246057(n-1)^3 + A093143(n)^3 + A002277(n)^3, n >= 1.
From Colin Barker, Feb 08 2017: (Start)
G.f.: 9*x*(17 - 550*x + 33500*x^2) / ((1 - x)*(1 - 10*x)*(1 - 100*x)*(1 - 1000*x)).
a(n) = (-2 + 2^(1+n)*5^n - 100^n + 1000^n) / 6.
a(n) = 1111*a(n-1) - 112110*a(n-2) + 1111000*a(n-3) - 1000000*a(n-4) for n>4. (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.

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