A002277 -id:A002277 - cp's OEIS frontend

A246057 a(n) = (5*10^n - 2)/3.

1, 16, 166, 1666, 16666, 166666, 1666666, 16666666, 166666666, 1666666666, 16666666666, 166666666666, 1666666666666, 16666666666666, 166666666666666, 1666666666666666, 16666666666666666, 166666666666666666, 1666666666666666666, 16666666666666666666, 166666666666666666666

Offset: 0

Vincenzo Librandi, Aug 13 2014

a(k-1) = (10^k - 4)/6, together with b(k) = 3*a(k-1) + 2 = A093143(k) and c(k) = 2*a(k-1) + 1 = A002277(k) are k-digit numbers for k >= 1 satisfying the so-called curious cubic identity a(k-1)^3 + b(k)^3 + c(k)^3 = a(k)*10^(2*k) + b(k)*10^k + c(k) (concatenated a(k)b(k)c(k)). This k-family and the proof of the identity has been given in the introduction of the van der Poorten reference. Thanks go to S. Heinemeyer for bringing these identities to my attention. - Wolfdieter Lang, Feb 07 2017

			Curious cubic identities (see a comment and reference above): 1^3 + 5^3 + 3^3 = 153, 16^3 + 50^3 + 33^3 = 165033, 166^3 + 500^3 + 333^3 = 166500333, ... - _Wolfdieter Lang_, Feb 07 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Alf van der Poorten, Kurt Thomsen, and Mark Wiebe, A curious cubic identity and self-similar sums of squares, The Mathematical Intelligencer, Vol. 29(2), pp. 39-41, March 2007.
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. sequences with terms of the form 1k..k where the digit k is repeated n times: A000042 (k=1), A090843 (k=2), A097166 (k=3), A099914 (k=4), A099915 (k=5), this sequence (k=6), A246058 (k=7), A246059 (k=8), A067272 (k=9).

Cf. A002277, A086948, A093143, A323639.

Magma
```
[(5*10^n-2)/3: n in [0..20]];
```
Mathematica
```
Table[(5 10^n - 2)/3, {n, 0, 20}]
```

vector(50, n, (5*10^(n-1)-2)/3) \\ Derek Orr, Aug 13 2014

G.f.: (1 + 5*x)/((1 - x)*(1 - 10*x)).
a(n) = 11*a(n-1) - 10*a(n-2).
E.g.f.: exp(x)*(5*exp(9*x) - 2)/3. - Stefano Spezia, May 02 2025
a(n) = A323639(n+1)/2 = A086948(n+1)/12. - Elmo R. Oliveira, May 07 2025

A004722 Delete all digits 3 from the terms of the sequence of nonnegative integers.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 14, 15, 16, 17, 18, 19, 20, 21, 22, 2, 24, 25, 26, 27, 28, 29, 0, 1, 2, 4, 5, 6, 7, 8, 9, 40, 41, 42, 4, 44, 45, 46, 47, 48, 49, 50, 51, 52, 5, 54, 55, 56, 57, 58, 59, 60, 61, 62, 6, 64, 65, 66, 67, 68, 69, 70, 71, 72, 7, 74, 75, 76

Offset: 0

N. J. A. Sloane

Very similar to A004178, except that 3-repdigits (A002277) are completely removed from the sequence, whereas A004178 has 0's in their place. It is thus guaranteed that a(n) = n only when n < 3. - Alonso del Arte, Oct 18 2012

Alonso del Arte, Table of n, a(n) for n = 0..9999

Cf. A004719, A004720, A004721, A004723, A004724, A004725, A004726, A004727, A004728.

m=1;
for u=0:1000
    v=dec2base(u,10)-'0'; v = v(v~=3);
    if length(v)>0;sol(m)=(str2num(strrep(num2str(v), ' ', ''))); m=m+1; end;
end
sol % Marius A. Burtea, May 07 2019

endAt = 103; Delete[Table[FromDigits[DeleteCases[IntegerDigits[n], 3]], {n, 0, endAt}], Table[{(10^expo - 1)/3 + 1}, {expo, Floor[Log[10, endAt]]}]] (* Alonso del Arte, Apr 29 2019 *)

def A004722(n):
    l = len(str(n))
    m = (10**l-1)//3
    k = n + l - int(n+l < m)
    return 2 if k == m else int(str(k).replace('3','')) # Chai Wah Wu, Apr 20 2021

a(n) = n for -1 < n < 3;
a(n) = A004178(n + 1) for 2 < n < 32,
a(n) = A004178(n + 2) for 31 < n < 331,
a(n) = A004178(n + 3) for 330 < n < 3330,
a(n) = A004178(n + 4) for 3329 < n < 33329, etc. - Alonso del Arte, Oct 21 2012

Sean A. Irvine pointed out erroneous terms in b-file and confirmed correction, Apr 28 2019

Name edited by Felix Fröhlich, Apr 29 2019

A086066 a(n) = Sum_{d in D(n)} 2^d, where D(n) = set of digits of n in decimal representation.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 3, 2, 6, 10, 18, 34, 66, 130, 258, 514, 5, 6, 4, 12, 20, 36, 68, 132, 260, 516, 9, 10, 12, 8, 24, 40, 72, 136, 264, 520, 17, 18, 20, 24, 16, 48, 80, 144, 272, 528, 33, 34, 36, 40, 48, 32, 96, 160, 288, 544, 65, 66, 68, 72, 80

Offset: 0

Reinhard Zumkeller, Jul 08 2003

For bitwise logical operations AND and OR:
a(m) = (a(m) AND a(n)) iff D(m) is a subset of D(n),
(a(m) AND a(n)) = 0 iff D(m) and D(n) are disjoint,
a(m) = (a(m) OR a(n)) iff D(n) is a subset of D(m),
a(m) = a(n) iff D(m) = D(n);
A086067(n) = A007088(a(n)).
From Reinhard Zumkeller, Sep 18 2009: (Start)
a(A052382(n)) mod 2 = 0; a(A011540(n)) mod 2 = 1;
for n > 0: a(A000004(n))=1, a(A000042(n))=2, a(A011557(n))=3, a(A002276(n))=4, a(A111066(n))=6, a(A002277(n))=8, a(A002278(n))=16, a(A002279(n))=32, a(A002280(n))=64, a(A002281(n))=128, a(A002282(n))=256, a(A002283(n))=512;
a(n) <= 1023. (End)

			n=242, D(242) = {2,4}: a(242) = 2^2 + 2^4 = 20.

Nathaniel Johnston, Table of n, a(n) for n = 0..10000

A086066 := proc(n) local d: if(n=0)then return 1: fi: d:=convert(convert(n,base,10),set): return add(2^d[j],j=1..nops(d)): end: seq(A086066(n),n=0..64); # Nathaniel Johnston, May 31 2011

A332130 a(n) = (10^(2n+1)-1)/3 - 3*10^n.

Original entry on oeis.org

0, 303, 33033, 3330333, 333303333, 33333033333, 3333330333333, 333333303333333, 33333333033333333, 3333333330333333333, 333333333303333333333, 33333333333033333333333, 3333333333330333333333333, 333333333333303333333333333, 33333333333333033333333333333, 3333333333333330333333333333333

Offset: 0

M. F. Hasler, Feb 09 2020

Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A002275 (repunits R_n = (10^n-1)/9), A002277 (3*R_n), A011557 (10^n).
Cf. A138148 (cyclops numbers with binary digits), A002113 (palindromes).
Cf. A332120 .. A332190 (variants with different repeated digit 2, ..., 9).
Cf. A332131 .. A332139 (variants with different middle digit 1, ..., 9).

A332130 := n -> (10^(2*n+1)-1)/3-3*10^n;

Array[ ((10^(2 # + 1)-1)/3 - 3*10^#) &, 15, 0]

apply( {A332130(n)=10^(n*2+1)\3-3*10^n}, [0..15])

def A332130(n): return 10**(n*2+1)//3-3*10**n

a(n) = 3*A138148(n) = A002277(2n+1) - 3*10^n.
G.f.: 3*x*(101 - 200*x)/((1 - x)(1 - 10*x)(1 - 100*x)).
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 2.
E.g.f.: exp(x)*(10*exp(99*x) - 9*exp(9*x) - 1)/3. - Stefano Spezia, Jul 13 2024

A332139 a(n) = (10^(2n+1)-1)/3 + 610^n.

Original entry on oeis.org

9, 393, 33933, 3339333, 333393333, 33333933333, 3333339333333, 333333393333333, 33333333933333333, 3333333339333333333, 333333333393333333333, 33333333333933333333333, 3333333333339333333333333, 333333333333393333333333333, 33333333333333933333333333333, 3333333333333339333333333333333

Offset: 0

M. F. Hasler, Feb 09 2020

Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A002275 (repunits R_n = (10^n-1)/9), A002277 (3*R_n), A011557 (10^n).
Cf. A138148 (cyclops numbers with binary digits), A002113 (palindromes).
Cf. A332129 .. A332189 (variants with different repeated digit 2, ..., 8).
Cf. A332130 .. A332138 (variants with different middle digit 0, ..., 8).

A332139 := n -> (10^(2*n+1)-1)/3+6*10^n;

Array[ (10^(2 # + 1)-1)/3 + 6*10^# &, 15, 0]
LinearRecurrence[{111,-1110,1000},{9,393,33933},20] (* Harvey P. Dale, Sep 17 2020 *)

apply( {A332139(n)=10^(n*2+1)\3+6*10^n}, [0..15])

def A332139(n): return 10**(n*2+1)//3+6*10**n

a(n) = 3*A138148(n) + 9*10^n = A002277(2n+1) + 6*10^n = 3*A332113(n).
G.f.: (9 - 606*x + 300*x^2)/((1 - x)(1 - 10*x)(1 - 100*x)).
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 2.

A350814 Numbers m such that the largest digit in the decimal expansion of 1/m is 3.

Original entry on oeis.org

3, 30, 33, 75, 300, 303, 330, 333, 429, 750, 813, 3000, 3003, 3030, 3125, 3300, 3330, 3333, 4290, 4329, 7500, 7575, 8130, 30000, 30003, 30030, 30300, 30303, 31250, 33000, 33300, 33330, 33333, 42900, 43290, 46875, 75000, 75075, 75750, 76923, 81103, 81300, 300000

Offset: 1

Bernard Schott, Jan 30 2022

If m is a term, 10*m is also a term.
3 is the only prime up to 2.6*10^8 (see comments in A333237).
Some subsequences:
{3, 30, 300, ...} = A093138 \ {1}.
{3, 33, 333, ...} = A002277 \ {0}.
{3, 33, 303, 3003, ...} = 3 * A000533.
{3, 303, 30303, 3030303, ...} = 3 * A094028.

			As 1/33 = 0.0303030303..., 33 is a term.
As 1/75 = 0.0133333333..., 75 is a term.
As 1/429 = 0.002331002331002331..., 429 is a term.

Index entries for sequences related to decimal expansion of 1/n

Cf. A000533, A094028, A266385, A333236.
Similar with largest digit k: A333402 (k=1), A341383 (k=2), A333237 (k=9).
Subsequences: A002277 \ {0}, A093138 \ {1}.
Decimal expansion: A010701 (1/3), A010674 (1/33).

Select[Range[10^5], Max[RealDigits[1/#][[1]]] == 3 &] (* Amiram Eldar, Jan 30 2022 *)

from fractions import Fraction
from itertools import count, islice
from sympy import n_order, multiplicity
def repeating_decimals_expr(f, digits_only=False):
    """ returns repeating decimals of Fraction f as the string aaa.bbb[ccc].
        returns only digits if digits_only=True.
    """
    a, b = f.as_integer_ratio()
    m2, m5 = multiplicity(2,b), multiplicity(5,b)
    r = max(m2,m5)
    k, m = 10**r, 10**n_order(10,b//2**m2//5**m5)-1
    c = k*a//b
    s = str(c).zfill(r)
    if digits_only:
        return s+str(m*k*a//b-c*m)
    else:
        w = len(s)-r
        return s[:w]+'.'+s[w:]+'['+str(m*k*a//b-c*m)+']'
def A350814_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda m:max(repeating_decimals_expr(Fraction(1,m),digits_only=True)) == '3',count(max(startvalue,1)))
A350814_list = list(islice(A350814_gen(),10)) # Chai Wah Wu, Feb 07 2022

More terms from Amiram Eldar, Jan 30 2022

A180160 (sum of digits) mod (number of digits) of n in decimal representation.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 0, 1, 2

Offset: 0

Reinhard Zumkeller, Aug 15 2010

a(n) = A007953(n) mod A055642(n);
a(A061383(n)) = 0; a(A180157(n)) > 0;
a(repdigits)=0: a(A010785(n))=0: a(A002275(n))=0: a(A002276(n))=0: a(A002277(n))=0: a(A002278(n))=0: a(4(n))=0: a(A002279(n))=0: a(A002280(n))=0: a(A002281(n))=0: a(A002282(n))=0: a(A002283(n))=0;
A123522 gives smallest m such that a(m) = n.

Reinhard Zumkeller, Table of n, a(n) for n = 0..25000

Cf. A002275-A002283, A007953, A010785, A055642, A061383, A123522, A180157, A374097.

A180160[n_] := If[n == 0, 0, Mod[Total[#], Length[#]] & [IntegerDigits[n]]];
Array[A180160, 100, 0] (* Paolo Xausa, Jun 30 2024 *)
Join[{0},Table[Mod[Total[IntegerDigits[n]],IntegerLength[n]],{n,110}]] (* Harvey P. Dale, Jul 30 2025 *)

A180593 Digital root of 3n.

Original entry on oeis.org

0, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6

Offset: 0

Odimar Fabeny, Sep 10 2010

Decimal expansion of 41/1110. - Enrique Pérez Herrero, Nov 13 2021

Index entries for linear recurrences with constant coefficients, signature (0,0,1).

Cf. A008585 (3*n), A010888 (digital root), A002277.

Cf. A180592, A180594, A180595, A180596, A180597, A180598, A180599.

digitalRoot[n_Integer?Positive] := FixedPoint[Plus@@IntegerDigits[#]&,n]; Table[If[n==0,0,digitalRoot[3*n]], {n,0,200}] (* Vladimir Joseph Stephan Orlovsky, May 02 2011 *)
LinearRecurrence[{0,0,1},{0,3,6,9},120] (* Harvey P. Dale, Sep 03 2020 *)

a(n+1) = 3*A010882(n). - Reinhard Zumkeller, Oct 25 2010
G.f.: (-3*(1 + 2*x + 3*x^2))/(-1 + x^3) for n>0. - Alexander R. Povolotsky, Jun 13 2012
a(n) = A010888(A002277(n)). - Enrique Pérez Herrero, Nov 24 2022
a(n) = A010888(A008585(n)). - Michel Marcus, Nov 24 2022

Edited by N. J. A. Sloane, Sep 23 2010

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

Wolfdieter Lang, Feb 08 2017

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

			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.

Colin Barker, Table of n, a(n) for n = 1..333
Index entries for linear recurrences with constant coefficients, signature (1110,-111000,1000000).

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

Table[FromDigits@ Join[ConstantArray[3, n], ReplacePart[ConstantArray[6, n], -1 -> 7], ConstantArray[0, n]], {n, 12}] (* Michael De Vlieger, Feb 08 2017 *)

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

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)

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

Wolfdieter Lang, Feb 07 2017

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]

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

Colin Barker, Table of n, a(n) for n = 1..333
Index entries for linear recurrences with constant coefficients, signature (1111,-112110,1111000,-1000000).

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

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 *)

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

a(n) = (((10^n - 4)/6)^3) + ((10^n/2)^3) + (((10^n - 1)/3)^3) \\ Jean-Jacques Vaudroz, Aug 11 2024

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)

cp's OEIS Frontend

A246057 a(n) = (5*10^n - 2)/3.

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A004722 Delete all digits 3 from the terms of the sequence of nonnegative integers.

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A086066 a(n) = Sum_{d in D(n)} 2^d, where D(n) = set of digits of n in decimal representation.

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Maple

A332130 a(n) = (10^(2n+1)-1)/3 - 3*10^n.

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A332139 a(n) = (10^(2*n+1)-1)/3 + 6*10^n.

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A350814 Numbers m such that the largest digit in the decimal expansion of 1/m is 3.

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A180160 (sum of digits) mod (number of digits) of n in decimal representation.

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A332139 a(n) = (10^(2n+1)-1)/3 + 610^n.