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 13 results. Next

A180027 Partial sums of A100706.

Original entry on oeis.org

1, 112, 11223, 1122334, 112233445, 11223344556, 1122334455667, 112233445566778, 11223344556677889, 1122334455667789000, 112233445566778900111, 11223344556677890011222, 1122334455667789001122333, 112233445566778900112233444, 11223344556677890011223344555, 1122334455667789001122334455666
Offset: 0

Views

Author

Mark Dols, Aug 07 2010

Keywords

Comments

Up to n=8 the digits of a(n) sum up to n^2.
Similar to this, A014824 (1,12,123,1234,...) is a representation of the triangular numbers; (1,1112,1112223,1112223334,...) of the pentagonal numbers;(1,11112,111122223,...) of the hexagonal numbers, and so on. A nice thing about this sequence(s) is that the (represented) value of the integer matches the partial sums of the number of digits in the sequence.
f(n) = 100*f(n-1) + A100706(n) gives a mirrored version of this sequence, and f(n) = 10*f(n-1) + A100706(n) the symmetrical version (A002477).

Crossrefs

Programs

Formula

a(n) = Sum_{k=0..n} A100706(k). - Michel Marcus, Mar 12 2023

Extensions

More terms and edited by Michel Marcus, Mar 12 2023

A002275 Repunits: (10^n - 1)/9. Often denoted by R_n.

Original entry on oeis.org

0, 1, 11, 111, 1111, 11111, 111111, 1111111, 11111111, 111111111, 1111111111, 11111111111, 111111111111, 1111111111111, 11111111111111, 111111111111111, 1111111111111111, 11111111111111111, 111111111111111111, 1111111111111111111, 11111111111111111111
Offset: 0

Views

Author

Keywords

Comments

R_n is a string of n 1's.
Base-4 representation of Jacobsthal bisection sequence A002450. E.g., a(4)= 1111 because A002450(4)= 85 (in base 10) = 64 + 16 + 4 + 1 = 1*(4^3) + 1*(4^2) + 1*(4^1) + 1. - Paul Barry, Mar 12 2004
Except for the first two terms, these numbers cannot be perfect squares, because x^2 != 11 (mod 100). - Zak Seidov, Dec 05 2008
For n >= 0: a(n) = (A000225(n) written in base 2). - Jaroslav Krizek, Jul 27 2009, edited by M. F. Hasler, Jul 03 2020
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=10, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010
Except 0, 1 and 11, all these integers are Brazilian numbers, A125134. - Bernard Schott, Dec 24 2012
Numbers n such that 11...111 = R_n = (10^n - 1)/9 is prime are in A004023. - Bernard Schott, Dec 24 2012
The terms 0 and 1 are the only squares in this sequence, as a(n) == 3 (mod 4) for n>=2. - Nehul Yadav, Sep 26 2013
For n>=2 the multiplicative order of 10 modulo the a(n) is n. - Robert G. Wilson v, Aug 20 2014
The above is a special case of the statement that the order of z modulo (z^n-1)/(z-1) is n, here for z=10. - Joerg Arndt, Aug 21 2014
From Peter Bala, Sep 20 2015: (Start)
Let d be a divisor of a(n). Let m*d be any multiple of d. Split the decimal expansion of m*d into 2 blocks of contiguous digits a and b, so we have m*d = 10^k*a + b for some k, where 0 <= k < number of decimal digits of m*d. Then d divides a^n - (-b)^n (see McGough). For example, 271 divides a(5) and we find 2^5 + 71^5 = 11*73*271*8291 and 27^5 + 1^5 = 2^2*7*31*61*271 are both divisible by 271. Similarly, 4*271 = 1084 and 10^5 + 84^5 = 2^5*31*47*271*331 while 108^5 + 4^5 = 2^12*7*31*61*271 are again both divisible by 271. (End)
Starting with the second term this sequence is the binary representation of the n-th iteration of the Rule 220 and 252 elementary cellular automaton starting with a single ON (black) cell. - Robert Price, Feb 21 2016
If p > 5 is a prime, then p divides a(p-1). - Thomas Ordowski, Apr 10 2016
0, 1 and 11 are only terms that are of the form x^2 + y^2 + z^2 where x, y, z are integers. In other words, a(n) is a member of A004215 for all n > 2. - Altug Alkan, May 08 2016
Except for the initial terms, the binary representation of the x-axis, from the left edge to the origin, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 737", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. - Robert Price, Mar 17 2017
The term "repunit" was coined by Albert H. Beiler in 1964. - Amiram Eldar, Nov 13 2020
q-integers for q = 10. - John Keith, Apr 12 2021
Binomial transform of A001019 with leading zero. - Jules Beauchamp, Jan 04 2022

References

  • Albert H. Beiler, Recreations in the Theory of Numbers: The Queen of Mathematics Entertains, New York: Dover Publications, 1964, chapter XI, p. 83.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 235-237.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 197-198.
  • Samuel Yates, Peculiar Properties of Repunits, J. Recr. Math. 2, 139-146, 1969.
  • Samuel Yates, Prime Divisors of Repunits, J. Recr. Math. 8, 33-38, 1975.

Crossrefs

Programs

  • Haskell
    a002275 = (`div` 9) . subtract 1 . (10 ^)
    a002275_list = iterate ((+ 1) . (* 10)) 0
    -- Reinhard Zumkeller, Jul 05 2013, Feb 05 2012
    
  • Magma
    [(10^n-1)/9: n in [0..25]]; // Vincenzo Librandi, Nov 06 2014
    
  • Maple
    seq((10^k - 1)/9, k=0..30); # Wesley Ivan Hurt, Sep 28 2013
  • Mathematica
    Table[(10^n - 1)/9, {n, 0, 19}] (* Alonso del Arte, Nov 15 2011 *)
    Join[{0},Table[FromDigits[PadRight[{},n,1]],{n,20}]] (* Harvey P. Dale, Mar 04 2012 *)
  • Maxima
    a[0]:0$
    a[1]:1$
    a[n]:=11*a[n-1]-10*a[n-2]$
    A002275(n):=a[n]$
    makelist(A002275(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
    
  • PARI
    a(n)=(10^n-1)/9; \\ Michael B. Porter, Oct 26 2009
    
  • PARI
    my(x='x+O('x^30)); concat(0, Vec(x/((1-10*x)*(1-x)))) \\ Altug Alkan, Apr 10 2016
    
  • Python
    print([(10**n-1)//9 for n in range(100)]) # Michael S. Branicky, Apr 30 2022
  • Sage
    [lucas_number1(n, 11, 10) for n in range(21)]  # Zerinvary Lajos, Apr 27 2009
    

Formula

a(n) = 10*a(n-1) + 1, a(0)=0.
a(n) = A000042(n) for n >= 1.
Second binomial transform of Jacobsthal trisection A001045(3n)/3 (A015565). - Paul Barry, Mar 24 2004
G.f.: x/((1-10*x)*(1-x)). Regarded as base b numbers, g.f. x/((1-b*x)*(1-x)). - Franklin T. Adams-Watters, Jun 15 2006
a(n) = 11*a(n-1) - 10*a(n-2), a(0)=0, a(1)=1. - Lekraj Beedassy, Jun 07 2006
a(n) = A125118(n,9) for n>8. - Reinhard Zumkeller, Nov 21 2006
a(n) = A075412(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
a(n) = a(n-1) + 10^(n-1) with a(0)=0. - Vincenzo Librandi, Jul 22 2010
a(n) = A242614(n,A242622(n)). - Reinhard Zumkeller, Jul 17 2014
E.g.f.: (exp(9*x) - 1)*exp(x)/9. - Ilya Gutkovskiy, May 11 2016
a(n) = Sum_{k=0..n-1} 10^k. - Torlach Rush, Nov 03 2020
Sum_{n>=1} 1/a(n) = A065444. - Amiram Eldar, Nov 13 2020
From Elmo R. Oliveira, Aug 02 2025: (Start)
a(n) = A002283(n)/9 = A105279(n)/10.
a(n) = A010785(A017173(n-1)) for n >= 1. (End)

A138118 Concatenation of 2n-1 digits 1 and n digits 0.

Original entry on oeis.org

10, 11100, 11111000, 11111110000, 11111111100000, 11111111111000000, 11111111111110000000, 11111111111111100000000, 11111111111111111000000000, 11111111111111111110000000000
Offset: 1

Views

Author

Omar E. Pol, Mar 29 2008

Keywords

Comments

a(n) is also A147537(n) written in base 2. [From Omar E. Pol, Nov 08 2008]

Examples

			n .......... a(n)
1 ........... 10
2 ......... 11100
3 ....... 11111000
4 ..... 11111110000
5 ... 11111111100000
		

Crossrefs

Programs

  • Mathematica
    FromDigits/@Table[Join[PadRight[{},2n-1,1],PadRight[{},n,0]],{n,15}] (* Harvey P. Dale, Dec 09 2011 *)

Formula

O.g.f.: 10*(1+100x)/[(-1+1000x)*(-1+10x)]. a(n)=A100706(n)*10^n = 10*a(n-1)+11*1000^n. - R. J. Mathar, Apr 03 2008

A299960 a(n) = (4^(2*n+1) + 1) / 5.

Original entry on oeis.org

1, 13, 205, 3277, 52429, 838861, 13421773, 214748365, 3435973837, 54975581389, 879609302221, 14073748835533, 225179981368525, 3602879701896397, 57646075230342349, 922337203685477581, 14757395258967641293, 236118324143482260685, 3777893186295716170957
Offset: 0

Views

Author

M. F. Hasler, Feb 22 2018

Keywords

Comments

It is easily seen that 4^(2n+1)+1 is divisible by 5 for all n, since 4 = -1 (mod 5). For even powers this does not hold.
The aerated sequence 1, 0, 13, 0, 205, 0, 3277, ... is a linear divisibility sequence of order 4. It is the case P1 = 0, P2 = -5^2, Q = 4 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. Cf. A007583, A095372 and A100706. - Peter Bala, Aug 28 2019
Let G be a sequence satisfying G(i) = 2*G(i-1) + G(i-2) - 2*G(i-3) for arbitrary integers i and without regard to the initial values of G. Then a(n) = (G(i)*2^(4*n+2) + G(i+8*n+4))/(5*G(i+4*n+2)) as long as G(i+4*n+2) != 0. - Klaus Purath, Feb 02 2021
Ch. Gerbr asks (personal comm.) whether we can prove that 13 is the only prime in this sequence. We can prove divisibility conditions for many residue classes of the index n (cf. formulas), but have not yet found a complete covering set. - M. F. Hasler, Jan 07 2025

Examples

			For n = 0, a(0) = (4^1+1)/5 = 5/5 = 1.
For n = 1, a(1) = (4^3+1)/5 = 65/5 = 13.
		

Crossrefs

Cf. A299959 for the smallest prime factor.

Programs

Formula

a(n) = A052539(2*n+1)/5 = A015521(2*n+1) = A014985(2*n+1) = A007910(4*n+1) = A007909(4*n+1) = A207262(n+1)/5.
O.g.f.: (1 - 4*x)/(1 - 17*x + 16*x^2). - Peter Bala, Aug 28 2019
a(n) = 17*a(n-1) - 16*a(n-2). - Wesley Ivan Hurt, Oct 02 2020
From Klaus Purath, Feb 02 2021: (Start)
a(n) = (2^(4*n+2)+1)/5.
a(n) = (A061654(n) + A001025(n))/2.
a(n) = A091881(n+1) + 7*A131865(n-1) for n > 0.
(End)
E.g.f.: (exp(x) + 4*exp(16*x))/5. - Stefano Spezia, Feb 02 2021
We have d | a(n) for all n in R, for the following pairs (d, R) of divisors d and residue classes R: (13, 1 + 3Z), (5, 2 + 5Z), (29, 3 + 7Z), (397, 5 + 11Z),
(53, 6 + 13Z), (137, 8 + 17Z), (229, 9 + 19Z), (277, 11 + 23Z),
(107367629, 14 + 29Z), (5581, 15 + 31Z), (149, 18 + 27Z), (10169, 20 + 41Z),
(173, 21 + 43Z), (3761, 23 + 47Z), (15358129, 26 + 53Z), (1181, 29 + 59Z),
(733, 30 + 61Z), (269, 33 + 67Z), (569, 35 + 71Z),(293, 36 + 73Z), (317, 39 + 79Z),
(997, 41 + 83Z), (1069, 44 + 89Z), (389, 48 + 97Z), (809, 50 + 101Z),
(41201, 51 + 103Z), (857, 53 + 107Z), (5669, 54 + 109Z), (58309, 56 + 113Z),
(509, 63 + 127Z), (269665073, 65 + 131Z), (189061, 68 + 137Z), (557, 69 + 139Z),
(1789, 74 + 149Z), (653, 81 + 163Z), (9413, 90 + 181Z), (3821, 95 + 191Z),
(773, 96 + 193Z), (4729, 98 + 197Z), (797, 99 + 199Z), ... - M. F. Hasler, Jan 07 2025

A099814 Bisection of A002275.

Original entry on oeis.org

0, 11, 1111, 111111, 11111111, 1111111111, 111111111111, 11111111111111, 1111111111111111, 111111111111111111, 11111111111111111111, 1111111111111111111111, 111111111111111111111111, 11111111111111111111111111, 1111111111111111111111111111, 111111111111111111111111111111, 11111111111111111111111111111111
Offset: 0

Views

Author

N. J. A. Sloane, Nov 19 2004

Keywords

Comments

Except for the first term (replace 0 with 1) this is the binary representation of the n-th iteration of the elementary cellular automaton starting with a single ON (black) cell for Rule 189. - Robert Price, Feb 21 2016

Crossrefs

Cf. A002275, A100706 (other bisection).

Programs

Formula

For n>0, 1 repeated 2n times.
a(n) = a(n-1) + 11*100^(n-1) with a(0)=0. - Vincenzo Librandi, Nov 16 2010
a(n) = (100^n - 1)/9. - Bernard Schott, Apr 09 2023
From Stefano Spezia, Jul 27 2024: (Start)
G.f.: 11*x/((1 - x)*(1 - 100*x)).
E.g.f.: exp(x)*(exp(99*x) - 1)/9. (End)

Extensions

More terms from Stefan Steinerberger, Apr 29 2006

A258372 Smallest nonnegative number k not starting or ending with the digit 1 that forms a prime when it is sandwiched between n ones to the left of k and n ones to the right of k.

Original entry on oeis.org

0, 3, 4, 8, 36, 8, 5, 72, 28, 6, 79, 212, 23, 6, 73, 24, 52, 62, 3, 28, 220, 53, 75, 58, 228, 9, 265, 89, 214, 86, 215, 4, 7, 39, 295, 40, 87, 216, 97, 6, 264, 53, 287, 223, 4, 239, 259, 25, 57, 364, 49, 38, 93, 86, 27, 30, 80, 24, 6, 356, 50, 645, 395, 206
Offset: 1

Views

Author

Felix Fröhlich, May 28 2015

Keywords

Comments

n = 1 is the only case where a(n) = 0, since for any n > 1, A138148(n) is divisible by A002275(n).
No n exists such that a(n) = 2, since any number of the form A100706(n)+A011557(n) is of the form A000533(n)*A002275(n+1) (see comment by Robert Israel in A107123).
a(n) = 3 iff n is in A107123.
a(n) = 4 iff n is in A107124.
If k has an even number of digits and is a multiple of 11, then k is not a term. If k = (10^r+1)(10^m-1)/9 for some m > 0, r >= 0, then k is not a term. If A272232(k) = 0, then k is not a term. - Chai Wah Wu, Nov 08 2019

Examples

			a(1) = 0, because 101 is prime.
a(5) = 36, because the smallest x >= 0 such that 11111_x_11111 (where '_' denotes concatenation) is prime is 36. The decimal expansion of that prime is 111113611111.
		

Crossrefs

Programs

  • Mathematica
    Table[k = 0; s = Table[1, {n}]; While[Or[!PrimeQ[FromDigits[s ~Join~ IntegerDigits[k] ~Join~ s]], Or[First@ IntegerDigits@ k == 1, Last@ IntegerDigits@ k == 1]], k++]; k, {n, 64}] (* Michael De Vlieger, May 28 2015 *)
  • PARI
    a000042(n) = (10^n-1)/9
    a(n) = my(k=0); while(k==10 || k%10==1 || k\(10^(#Str(k)-1))==1 || !ispseudoprime(eval(Str(a000042(n), k, a000042(n)))), k++); k

A267887 Binary representation of the n-th iteration of the "Rule 237" elementary cellular automaton starting with a single ON (black) cell.

Original entry on oeis.org

1, 10, 11111, 1111111, 111111111, 11111111111, 1111111111111, 111111111111111, 11111111111111111, 1111111111111111111, 111111111111111111111, 11111111111111111111111, 1111111111111111111111111, 111111111111111111111111111, 11111111111111111111111111111
Offset: 0

Views

Author

Robert Price, Jan 21 2016

Keywords

References

  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Crossrefs

Programs

  • Mathematica
    rule=237; rows=20; ca=CellularAutomaton[rule,{{1},0},rows-1,{All,All}]; (* Start with single black cell *) catri=Table[Take[ca[[k]],{rows-k+1,rows+k-1}],{k,1,rows}]; (* Truncated list of each row *) Table[FromDigits[catri[[k]]],{k,1,rows}]   (* Binary Representation of Rows *)

Formula

Conjectures from Colin Barker, Jan 22 2016 and Apr 17 2019: (Start)
a(n) = 101*a(n-1)-100*a(n-2) for n>3.
G.f.: (1-91*x+10201*x^2-10100*x^3) / ((1-x)*(1-100*x)).
(End)

A267889 Binary representation of the n-th iteration of the "Rule 239" elementary cellular automaton starting with a single ON (black) cell.

Original entry on oeis.org

1, 110, 11111, 1111111, 111111111, 11111111111, 1111111111111, 111111111111111, 11111111111111111, 1111111111111111111, 111111111111111111111, 11111111111111111111111, 1111111111111111111111111, 111111111111111111111111111, 11111111111111111111111111111
Offset: 0

Views

Author

Robert Price, Jan 21 2016

Keywords

Comments

Seems to differ from A267887 and A100706 only at a(1). - R. J. Mathar, Jun 24 2025

References

  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Crossrefs

Cf. A267871.

Programs

  • Mathematica
    rule=239; rows=20; ca=CellularAutomaton[rule,{{1},0},rows-1,{All,All}]; (* Start with single black cell *) catri=Table[Take[ca[[k]],{rows-k+1,rows+k-1}],{k,1,rows}]; (* Truncated list of each row *) Table[FromDigits[catri[[k]]],{k,1,rows}]   (* Binary Representation of Rows *)

Formula

Conjectures from Colin Barker, Jan 22 2016 and Apr 17 2019: (Start)
a(n) = 101*a(n-1)-100*a(n-2) for n>3.
G.f.: (1+9*x+101*x^2-100*x^3) / ((1-x)*(1-100*x)).
(End)

A274743 Repunits with odd indices multiplied by 99, i.e., 99*(1, 111, 11111, 1111111, ...).

Original entry on oeis.org

99, 10989, 1099989, 109999989, 10999999989, 1099999999989, 109999999999989, 10999999999999989, 1099999999999999989, 109999999999999999989, 10999999999999999999989, 1099999999999999999999989, 109999999999999999999999989, 10999999999999999999999999989
Offset: 1

Views

Author

Rodolfo A. Fiorini, Jul 04 2016

Keywords

Comments

It is apparent that the reciprocals of the terms in the sequence give an increasing sequence of periodic terms similar to A095372, but with the initial term equal to "01". The leading zero is important (see links). Furthermore, the reciprocals of the terms give a sequence of even growing periods, starting from 2, with delta = 4 (i.e., 2, 6, 10, 14, 18, ...).
Adding "11" to each term gives the binary representation of the n-th iteration of "Rule 14" elementary cellular automaton starting with a single ON (black cell) as in A266299.

Examples

			a(2) = 101*10989 - 100*99 = 1099989.
		

Crossrefs

Programs

Formula

a(n) = 101*a(n-1) - 100*a(n-2) for n>2, with a(0)= 99 and a(1)= 10989.
a(n) = 99*A100706(n-1).
G.f.: 99*x*(1 + 10*x)/((1 - x)*(1 - 100*x)). - Ilya Gutkovskiy, Jul 04 2016
a(n) = 11*(10^(2*n-1)-1). - Wesley Ivan Hurt, Jul 04 2016
E.g.f.: 11*(9 - 10*exp(x) + exp(100*x))/10. - Stefano Spezia, Aug 05 2024

A267937 Binary representation of the n-th iteration of the "Rule 251" elementary cellular automaton starting with a single ON (black) cell.

Original entry on oeis.org

1, 101, 11111, 1111111, 111111111, 11111111111, 1111111111111, 111111111111111, 11111111111111111, 1111111111111111111, 111111111111111111111, 11111111111111111111111, 1111111111111111111111111, 111111111111111111111111111, 11111111111111111111111111111
Offset: 0

Views

Author

Robert Price, Jan 22 2016

Keywords

Comments

Seems to differ from A267889, A267887 and A100706 only at a(1). - R. J. Mathar, Jun 24 2025

References

  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Crossrefs

Cf. A267936.

Programs

  • Mathematica
    rule=251; rows=20; ca=CellularAutomaton[rule,{{1},0},rows-1,{All,All}]; (* Start with single black cell *) catri=Table[Take[ca[[k]],{rows-k+1,rows+k-1}],{k,1,rows}]; (* Truncated list of each row *) Table[FromDigits[catri[[k]]],{k,1,rows}]   (* Binary Representation of Rows *)

Formula

Conjectures from Colin Barker, Jan 23 2016 and Apr 16 2019: (Start)
a(n) = 101*a(n-1)-100*a(n-2) for n>3.
G.f.: (1+1010*x^2-1000*x^3) / ((1-x)*(1-100*x)).
(End)
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