A010701 -id:A010701 - cp's OEIS frontend

A153284 a(n) = n + Sum_{j=1..n-1} (-1)^j * a(j) for n >= 2, a(1) = 1.

1, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1

Offset: 1

Walter Carlini, Dec 23 2008

Row sums of triangle A153860. - Gary W. Adamson, Jan 03 2009

1 followed by interleaving of A000012 and A010701. - Klaus Brockhaus, Jan 04 2009

			a(1)=1, a(2)=2-a(1)=2-1=1, a(3)=3+a(2)-a(1)=3+1-1=3, a(4)=4-a(3)+a(2)-a(1)=4-3+1-1=1, a(5)=5+1-3+1-1=3, a(6)=6-3+1-3+1-1=1, a(7)=7+1-3+1-3+1-1, etc.

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

Equals A010684 with the addition of the leading term of 1
The first sequence of a family that includes A153285 and A153286
Cf. A153860.
Cf. A000012 (all 1's sequence), A010701 (all 3's sequence). - Klaus Brockhaus, Jan 04 2009

S:=[ 1 ]; for n in [2..105] do Append(~S, n + &+[ (-1)^j*S[j]: j in [1..n-1] ]); end for; S; // Klaus Brockhaus, Jan 04 2009

a(n)=1 if n is 1 or even; a(n)=3 if n is odd other than 1.

G.f.: x*(1 + x + 2*x^2)/((1+x)*(1-x)). - Klaus Brockhaus, Jan 04 2009 and Oct 15 2009

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

A134771 A134770 interleaved with threes.

Original entry on oeis.org

1, 3, 5, 3, 21, 3, 77, 3, 277, 3, 1005, 3, 3693, 3, 13725, 3, 51477, 3, 194477, 3, 739021, 3, 2821725, 3, 10816621, 3, 41602397, 3, 160466397, 3, 620470077, 3, 2404321557, 3, 9334424877, 3, 36300541197, 3, 141381055197, 3, 551386115277, 3, 2153031497757, 3

Offset: 0

Gary W. Adamson, Nov 10 2007

Previous name was: A007318^(-2) * A134770.
Second inverse binomial transform of A134770.
A134770 interleaved with threes.

			First few terms of the sequence are (1, 3, 5, 3, 21, 3, 77, ...), since A134770 = (1, 3, 5, 21, 77, ...).

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A000984, A010701, A134770.

A134771:= func< n | (n mod 2) eq 1 select 3 else 2*(n+2)*Catalan(Floor(n/2))-3 >;
[A134771(n): n in [0..50]]; // G. C. Greubel, Oct 13 2023

Table[If[OddQ[n], 3, 4*Binomial[n,n/2] -3], {n,0,50}] (* G. C. Greubel, Oct 13 2023 *)

def A134771(n): return 4*((n+1)%2)*binomial(n, n//2) - 3*(-1)^n
[A134771(n) for n in range(41)] # G. C. Greubel, Oct 13 2023

From G. C. Greubel, Oct 13 2023: (Start)
a(n) = 2*(1 + (-1)^n)*binomial(n, n/2) - 3*(-1)^n.
G.f.: 2/sqrt(1-4*x^2) - 3/(1+x).
E.g.f.: 4*BesselI(0, 2*x) - 3*exp(-x). (End)

Name changed by G. C. Greubel, Oct 13 2023

A176040 Periodic sequence: Repeat 3, 1.

Original entry on oeis.org

3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3

Offset: 0

Klaus Brockhaus, Apr 07 2010

Interleaving of A010701 and A000012.
Also continued fraction expansion of (3+sqrt(21))/2.
Also decimal expansion of 31/99.
Essentially first differences of A014601.
Inverse binomial transform of 3 followed by A020707.
Second inverse binomial transform of A052919 without initial term 2.
Third inverse binomial transform of A007582 without initial term 1.
Exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 2*x^2 + 2*x^3 + 3*x^4 + 3*x^5 + ... is the o.g.f. for A008619. - Peter Bala, Mar 13 2015

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

Cf. A153284, A010701 (all 3's sequence), A000012 (all 1's sequence), A090458 (decimal expansion of (3+sqrt(21))/2), A010684 (repeat 1, 3), A014601 (congruent to 0 or 3 mod 4), A020707 (2^(n+2)), A052919, A007582 (2^(n-1)*(1+2^n)), A008619.

&cat[ [3, 1]: n in [0..52] ];
[ 2+(-1)^n: n in [0..104] ];

PadRight[{},120,{3,1}] (* or *) LinearRecurrence[{0,1},{3,1},120] (* Harvey P. Dale, Mar 11 2015 *)

a(n) = 2+(-1)^n.
a(n) = a(n-2) for n > 1; a(0) = 3, a(1) = 1.
a(n) = -a(n-1)+4 for n > 0; a(0) = 3.
a(n) = 3*((n+1) mod 2)+(n mod 2).
a(n) = A010684(n+1).
G.f.: (3+x)/((1-x)*(1+x)).
From Amiram Eldar, Jan 01 2023: (Start)
Multiplicative with a(2^e) = 3, and a(p^e) = 1 for p >= 3.
Dirichlet g.f.: zeta(s)*(1+2^(1-s)). (End)

A340666 A(n,k) is derived from n by replacing each 0 in its binary representation with a string of k 0's; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 2, 3, 0, 1, 4, 3, 1, 0, 1, 8, 3, 4, 3, 0, 1, 16, 3, 16, 5, 3, 0, 1, 32, 3, 64, 9, 6, 7, 0, 1, 64, 3, 256, 17, 12, 7, 1, 0, 1, 128, 3, 1024, 33, 24, 7, 8, 3, 0, 1, 256, 3, 4096, 65, 48, 7, 64, 9, 3, 0, 1, 512, 3, 16384, 129, 96, 7, 512, 33, 10, 7

Offset: 0

Alois P. Heinz, Jan 15 2021

			Square array A(n,k) begins:
  0, 0,  0,   0,    0,     0,      0,       0,        0, ...
  1, 1,  1,   1,    1,     1,      1,       1,        1, ...
  1, 2,  4,   8,   16,    32,     64,     128,      256, ...
  3, 3,  3,   3,    3,     3,      3,       3,        3, ...
  1, 4, 16,  64,  256,  1024,   4096,   16384,    65536, ...
  3, 5,  9,  17,   33,    65,    129,     257,      513, ...
  3, 6, 12,  24,   48,    96,    192,     384,      768, ...
  7, 7,  7,   7,    7,     7,      7,       7,        7, ...
  1, 8, 64, 512, 4096, 32768, 262144, 2097152, 16777216, ...
  ...

Alois P. Heinz, Antidiagonals n = 0..200, flattened

Columns k=0-2, 4 give: A038573, A001477, A084471, A084473.
Rows n=0..17, 19 give: A000004, A000012, A000079, A010701, A000302, A000051(k+1), A007283, A010727, A001018, A087289, A007582(k+1), A062709(k+2), A164346, A181565(k+1), A005009, A181404(k+3), A001025, A199493, A253208(k+1).
Main diagonal gives A340667.
Cf. A000120, A023416.

A:= (n, k)-> Bits[Join](subs(0=[0$k][], Bits[Split](n))):
seq(seq(A(n, d-n), n=0..d), d=0..12);
# second Maple program:
A:= proc(n, k) option remember; `if`(n<2, n,
     `if`(irem(n, 2, 'r')=1, A(r, k)*2+1, A(r, k)*2^k))
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);

A[n_, k_] := FromDigits[IntegerDigits[n, 2] /. 0 -> Sequence @@ Table[0, {k}], 2];
Table[A[n, d-n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, Feb 02 2021 *)

A000120(A(n,k)) = A000120(n) = log_2(A(n,0)+1).

A023416(A(n,k)) = k * A023416(n) for n >= 1.

A013656 a(n) = n(9n-2).

Original entry on oeis.org

0, 7, 32, 75, 136, 215, 312, 427, 560, 711, 880, 1067, 1272, 1495, 1736, 1995, 2272, 2567, 2880, 3211, 3560, 3927, 4312, 4715, 5136, 5575, 6032, 6507, 7000, 7511, 8040, 8587, 9152, 9735, 10336, 10955, 11592, 12247, 12920, 13611, 14320, 15047, 15792, 16555

Offset: 0

David W. Wilson

For n>0, numbers such that sqrt(a(n)) has the continued fraction {k;[1,1,1,2k]}, where the part in [] is repeated and k is of the form 3m+2 (A016789). - Bruno Berselli, May 30 2013

For n >= 1, the continued fraction expansion of sqrt(4*a(n)) is [6n-1; {3, 3n-1, 3, 12n-2}]. - Magus K. Chu, Sep 18 2022

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A010701, A017257, A185019.

Cf. A016789, A061039, A144454.

[n*(9*n-2): n in [0..60]]; // G. C. Greubel, Mar 11 2022

Table[n(9n-2),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,7,32},50] (* Harvey P. Dale, Jul 07 2012 *)

a(n)=n*(9*n-2) \\ Charles R Greathouse IV, Jun 17 2017

a(n+1) = A144454(9*n+7) = A061039(27*n+21). - Paul Curtz, Nov 05 2008
a(n) = a(n-1) + 18*n - 11 with n>0, a(0)=0. - Vincenzo Librandi, Nov 22 2010
a(0)=0, a(1)=7, a(2)=32, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Jul 07 2012
From G. C. Greubel, Mar 11 2022: (Start)
G.f.: x*(7 - 11*x)/(1-x)^3.
E.g.f.: x*(7 + 9*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = -(psi(7/9)+gamma)/2 = (A354640-A001620)/2 = 0.22000753... - R. J. Mathar, Apr 22 2024

A147296 a(n) = n(9n+2).

Original entry on oeis.org

0, 11, 40, 87, 152, 235, 336, 455, 592, 747, 920, 1111, 1320, 1547, 1792, 2055, 2336, 2635, 2952, 3287, 3640, 4011, 4400, 4807, 5232, 5675, 6136, 6615, 7112, 7627, 8160, 8711, 9280, 9867, 10472, 11095, 11736, 12395, 13072, 13767, 14480, 15211, 15960

Offset: 0

Paul Curtz, Nov 05 2008

For n >= 1, the continued fraction expansion of sqrt(4*a(n)) is [6n; {1, 1, 1, 3n-1, 1, 1, 1, 12n}]. - Magus K. Chu, Sep 17 2022

Reply to V. Librandi, A147296 (SeqFan list). - _M. F. Hasler_, Mar 01 2009
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Equals first 9-fold decimation of A144454.

Cf. A010701, A017173, A147296, A185019.

Table[n(9n+2),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,11,40},50] (* Harvey P. Dale, Dec 19 2014 *)

A147296(n) = n*(9*n + 2) \\ M. F. Hasler, Mar 01 2009

a(n) = n*(9*n + 2), as conjectured by V. Librandi. - M. F. Hasler, Mar 01 2009
G.f.: x*(11+7*x)/(1-x)^3. - Jaume Oliver Lafont, Aug 30 2009
a(n) = floor((3*n + 1/3)^2). - Reinhard Zumkeller, Apr 14 2010
From Elmo R. Oliveira, Dec 15 2024: (Start)
E.g.f.: exp(x)*x*(11 + 9*x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. (End)

More terms from M. F. Hasler, Mar 01 2009

A021025 Decimal expansion of 1/21.

Original entry on oeis.org

0, 4, 7, 6, 1, 9, 0, 4, 7, 6, 1, 9, 0, 4, 7, 6, 1, 9, 0, 4, 7, 6, 1, 9, 0, 4, 7, 6, 1, 9, 0, 4, 7, 6, 1, 9, 0, 4, 7, 6, 1, 9, 0, 4, 7, 6, 1, 9, 0, 4, 7, 6, 1, 9, 0, 4, 7, 6, 1, 9, 0, 4, 7, 6, 1, 9, 0, 4, 7, 6, 1, 9, 0, 4, 7, 6, 1, 9, 0, 4, 7, 6, 1, 9, 0, 4, 7, 6, 1, 9, 0, 4, 7, 6, 1, 9, 0, 4, 7

Offset: 0

N. J. A. Sloane

			0.047619047...

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

Cf. A010701 (1/3), A020806 (1/7).

Join[{0}, RealDigits[10/21, 10, 105][[1]]] (* Alonso del Arte, Mar 07 2019 *)

1/21. \\ Charles R Greathouse IV, Mar 08 2019

A021733 Decimal expansion of 1/729.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

729 = 3^6 = 9^3 = 27^2.
Period is 81 = 9^2 (see example for all 81 digits of the repeating part).
Repeating part in the form of 9 X 9 square table:
  1, 3, 7, 1, 7, 4, 2, 1, 1,
  2, 4, 8, 2, 8, 5, 3, 2, 2,
  3, 5, 9, 3, 9, 6, 4, 3, 3,
  4, 7, 0, 5, 0, 7, 5, 4, 4,
  5, 8, 1, 6, 1, 8, 6, 5, 5,
  6, 9, 2, 7, 2, 9, 7, 6, 6,
  8, 0, 3, 8, 4, 0, 8, 7, 7,
  9, 1, 4, 9, 5, 1, 9, 8, 9,
  0, 2, 6, 0, 6, 3, 1, 0, 0.
Note that each column consists of 9 consecutive (cyclically repeated) digits out of 10. The missing digits in columns from left to right are {7, 6, 5, 4, 3, 2, 0, 9, 8}, which form also a cycle of 9 out of 10 consecutive digits in reverse order, all digits except 1. - Alexander Adamchuk, Dec 28 2013

			1/729 = 0.00137174211248285322359396433470507544581618655692729766\
803840877914951989026063100 (period 81). - _Alexander Adamchuk_, Dec 28 2013

Cf. A068542 (period of the fraction 1/3^n).

Cf. A010701 (1/3), A000012 (1/9), A021031 (1/27), A021085 (1/81).

RealDigits[1/729, 10, 100] (* Alexander Adamchuk, Dec 28 2013 *)

PARI
```
1/729. \\ Michel Marcus, Oct 28 2019
```

Equals Sum_{k>=1} (k*(k+1)/2)/10^(k+2). - Davide Rotondo, Jun 11 2025

A194880 The numerators of the inverse Akiyama-Tanigawa algorithm from A001045(n).

Original entry on oeis.org

0, -1, -1, -4, -5, -2, -7, -8, -3, -10, -11, -4, -13, -14, -5, -16, -17, -6, -19, -20, -7, -22, -23, -8, -25, -26, -9, -28, -29, -10, -31, -32, -11, -34, -35, -12, -37, -38, -13, -40, -41, -14, -43, -44, -15, -46, -47, -16, -49, -50, -17, -52, -53, -18, -55, -56, -19, -58, -59, -20

Offset: 0

Paul Curtz, Sep 07 2011

0,  -1, -1, -4/3, -5/3, -2, -7/3, -8/3, -3, -10/3, -11/3, -4, -13/4, -14/3, -5,    = a(n)/b(n),
1,    0,  1,  4/3,  5/3,  2,  7/3,  8/3,  3,
1,   -2, -1, -4/3, -5/3, -2, -7/3, -8/3, -3,
3,   -2,  1,  4/3,  5/3,  2,  7/3,  8/3,  3,
5,   -6, -1, -4/3, -5/3, -2, -7/3, -8/3, -3,
11, -10,  1,  4/3,  5/3,  2,  7/3,  8/3,  3,
21, -22, -1, -4/3, -5/3, -2, -7/3, -8/3, -3,
Vertical: A001045(n), -A078008(n), (-1)^(n+1)*A000012(n), (-1)^(n+1)*A010709(n)/A010701(n), (-1)^(n+1)*A010716(n+1)/A010701(n), A007395(n), .. .
a(n)=0, 1 before (-A145064(n+1)=-A051176(n+3).
b(n)=1, 1 before A169609(n). b(n)=1, 1, 1 before A144437(n+1).
a(n+5)-a(n+2)=b(n+5)  (=-1,-3,-3,=-A169609(n)).

G. C. Greubel, Table of n, a(n) for n = 0..5000
D. Merlini, R. Sprugnoli, and M. C. Verri, The Akiyama-Tanigawa Transformation, Integers, 5 (1) (2005) #A05.
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

a[0]=0; a[1]=-1; a[n_] := (-n-1)/Max[1, 2*Mod[n, 3]-1]; Table[a[n], {n, 0, 59}] (* Jean-François Alcover, Sep 18 2012 *)

a(3*n)=-3*n-1 except a(0)=0; a(3*n+1)=-3*n-2 except a(1)=-1; a(3*n+2)=-n-1.
From Chai Wah Wu, May 07 2024: (Start)
a(n) = 2*a(n-3) - a(n-6) for n > 7.
G.f.: x*(x^6 + x^5 - 3*x^3 - 4*x^2 - x - 1)/(x^6 - 2*x^3 + 1). (End)

cp's OEIS Frontend

A153284 a(n) = n + Sum_{j=1..n-1} (-1)^j * a(j) for n >= 2, a(1) = 1.

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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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A134771 A134770 interleaved with threes.

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A176040 Periodic sequence: Repeat 3, 1.

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A340666 A(n,k) is derived from n by replacing each 0 in its binary representation with a string of k 0's; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A013656 a(n) = n*(9*n-2).

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A147296 a(n) = n*(9*n+2).

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A013656 a(n) = n(9n-2).

A147296 a(n) = n(9n+2).