A010686 -id:A010686 - cp's OEIS frontend

A141430 a(n) = A000111(n) mod 9.

1, 1, 1, 2, 5, 7, 7, 2, 8, 7, 4, 2, 2, 7, 1, 2, 5, 7, 7, 2, 8, 7, 4, 2, 2, 7, 1, 2, 5, 7, 7, 2, 8, 7, 4, 2, 2, 7, 1, 2, 5, 7, 7, 2, 8, 7, 4, 2, 2, 7

Offset: 0

Paul Curtz, Aug 06 2008

After the initial 1,1, the sequence is periodic with period 12.

This sequence's periodic part is a shuffled version of the two period-6 sequences A070366 and A010697. The sequence contains only the digits 1, 2, 4, 5, 7 and 8 (those of A141425).

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

Cf. A000111, A004099, A010686, A014021.

def A141430(n): return (2, 7, 1, 2, 5, 7, 7, 2, 8, 7, 4, 2)[n%12] if n>1 else 1 # Chai Wah Wu, Apr 17 2023

a(n) = A000111(n) mod 9 = A004099(n) mod 9.
a(n+12) = a(n), n > 1.
a(n) + a(n+6) = 9, n > 1.
a(n+11-p) - a(n+p) = 6 (p=0 or 5), 0 (p=1 or 4), -3 (p=2 or 3), any n > 1.
G.f.: (6x^8-5x^7+x^6+2x^5+3x^4+x^3+1) / ((1-x)(x^2+1)(x^4-x^2+1)). - R. J. Mathar, Dec 05 2008
a(n) = 9/2 +(-1)^floor(n/2)*A010686(n)/2 - 3*A014021(n), n > 1. - R. J. Mathar, Dec 05 2008
a(n) = 9/2 - (3/2)*cos(Pi*n/6) + (1/2)*3^(1/2)*sin(Pi*n/6) - (1/2)*cos(Pi*n/2) - (5/2)*sin(Pi*n/2) - (3/2)*cos(5*Pi*n/6) - (1/2)*3^(1/2)*sin(5*Pi*n/6). - Richard Choulet, Dec 12 2008

Edited by R. J. Mathar, Dec 05 2008

A057566 Number of collinear triples in a 3 X n rectangular grid.

Original entry on oeis.org

0, 1, 2, 8, 20, 43, 78, 130, 200, 293, 410, 556, 732, 943, 1190, 1478, 1808, 2185, 2610, 3088, 3620, 4211, 4862, 5578, 6360, 7213, 8138, 9140, 10220, 11383, 12630, 13966, 15392, 16913, 18530, 20248, 22068, 23995, 26030, 28178, 30440, 32821, 35322

Offset: 0

John W. Layman, Oct 04 2000

Paolo Xausa, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

Second differences give A047264. Third differences are periodic {5, 1, 5, 1, ...} and form A010686. See A000938 for the n X n grid.

LinearRecurrence[{3, -2, -2, 3, -1}, {0, 1, 2, 8, 20}, 50] (* Paolo Xausa, Feb 22 2024 *)

Conjecture: a(n) = 5*floor((2n^3 - 3n^2 - n)/24) + floor((2(n-1)^3 - 3(n-1)^2 - (n-1))/24) + n, which fits all of the listed terms.
From R. J. Mathar, May 23 2010: (Start)
a(n) = 3*a(n-1) - 2*a(n-2) - 2*a(n-3) + 3*a(n-4) - a(n-5) = n^3/2 - n^2 + n + (1-(-1)^n)/4.
G.f.: x*(1 - x + 4*x^2 + 2*x^3)/((1+x)*(x-1)^4). (End)

A171654 Period 10: repeat 0, 1, 6, 7, 2, 3, 8, 9, 4, 5.

Original entry on oeis.org

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

Offset: 0

Paul Curtz, Dec 14 2009

The repeating part contains all ten digits.

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

a(n) = A171507(n+1) mod 10.
a(2n) = A059629(n) mod 10; a(2n+1) = A139788(n).
abs(a(n+1)-a(n)) = A010686(n).
G.f.: -x*(1 + 6*x + 7*x^2 + 2*x^3 + 3*x^4 + 8*x^5 + 9*x^6 + 4*x^7 + 5*x^8) / ( (x-1)*(1+x)*(x^4 + x^3 + x^2 + x + 1)*(x^4 - x^3 + x^2 - x + 1) ). - R. J. Mathar, Jul 13 2011

More terms from Jinyuan Wang, Feb 26 2020

A173078 a(n) = (52^n - 2(-1)^n - 9)/3.

Original entry on oeis.org

1, 3, 11, 23, 51, 103, 211, 423, 851, 1703, 3411, 6823, 13651, 27303, 54611, 109223, 218451, 436903, 873811, 1747623, 3495251, 6990503, 13981011, 27962023, 55924051, 111848103, 223696211, 447392423, 894784851, 1789569703, 3579139411

Offset: 1

Paul Curtz, Feb 09 2010

The sequence and higher-order differences in subsequent rows are
    1,  3,  11,  23, 51, 103, 211, 423, 851, 1703, 3411, 6823, 13651
    2,  8,  12,  28, 52, 108, 212, 428, 852, 1708, 3412, 6828, 13652
    6,  4,  16,  24, 56, 104, 216, 424, 856, 1704, 3416, 6824, 13656
   -2,  12,  8,  32, 48, 112, 208, 432, 848, 1712, 3408, 6832, 13648
   14,  -4, 24,  16, 64,  96, 224, 416, 864, 1696, 3424, 6816, 13664
  -18,  28, -8,  48, 32, 128, 192, 448, 832, 1728, 3392, 6848,  1363
   46, -36, 56, -16, 96,  64, 256, 384, 896, 1664, 3456, 6784,  1369
The main diagonal 1,8,16,... is essentially A000079.
A subdiagonal is 2, 4, 8, 16, ... A155559.
Other diagonals are 3, 12, 24, 48, ... = 3*A151821, 6, 12, 24, ... = A082505 and -2, -4, -8, -16, ..., a negated variant of A171449.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

List([1..40], n-> (5*2^n - 2*(-1)^n - 9)/3); # G. C. Greubel, Dec 01 2019

[5*2^n/3-2*(-1)^n/3-3: n in [1..40]]; // Vincenzo Librandi, Aug 05 2011

seq( (5*2^n -2*(-1)^n -9)/3, n=1..40); # G. C. Greubel, Dec 01 2019

LinearRecurrence[{2,1,-2},{1,3,11},40] (* Harvey P. Dale, Oct 01 2018 *)

vector(40, n, (5*2^n - 2*(-1)^n - 9)/3) \\ G. C. Greubel, Dec 01 2019

[(5*2^n - 2*(-1)^n - 9)/3 for n in (1..40)] # G. C. Greubel, Dec 01 2019

a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3).
a(n+1) - 2*a(n) = A010686(n-1).
a(n) = A084214(n+1) - 3.
G.f.: x*(1 + x + 4*x^2) / ( (1-x)*(1-2*x)*(1+x) ).
a(2n+3) - a(2n+1) = 10*A000302(n).
E.g.f.: (-2*exp(-x) + 6 - 9*exp(x) + 5*exp(2*x))/3. - G. C. Greubel, Dec 01 2019

A176260 Periodic sequence: Repeat 5, 1.

Original entry on oeis.org

5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5

Offset: 0

Klaus Brockhaus, Apr 13 2010

Interleaving of A010716 and A000012.
Also continued fraction expansion of (5+3*sqrt(5))/2.
Also decimal expansion of 17/33.
Essentially first differences of A047264.
Binomial transform of 5 followed by -A122803 without initial terms 1, -2.
Inverse binomial transform of 5 followed by A007283 without initial term 3.
Second inverse binomial transform of A168607 without initial term 3.
Exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 3*x^2 + 3*x^3 + 6*x^4 + 6*x^5 + ... is the o.g.f. for A008805. - Peter Bala, Mar 13 2015

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

Cf. A010716 (all 5's sequence), A000012 (all 1's sequence), A090550 (decimal expansion of (5+3*sqrt(5))/2), A010686 (repeat 1, 5), A047264 (congruent to 0 or 5 mod 6), A122803 (powers of -2), A007283 (3*2^n), A168607 (3^n+2), A008805.

&cat[ [5, 1]: n in [0..52] ];
[ 3+2*(-1)^n: n in [0..104] ];

a(n) = 3+2*(-1)^n.
a(n) = a(n-2) for n > 1; a(0) = 5, a(1) = 1.
a(n) = -a(n-1)+6 for n > 0; a(0) = 5.
a(n) = 5*((n+1) mod 2)+(n mod 2).
a(n) = A010686(n+1).
G.f.: (5+x)/(1-x^2).
From Amiram Eldar, Jan 01 2023: (Start)
Multiplicative with a(2^e) = 5, and a(p^e) = 1 for p >= 3.
Dirichlet g.f.: zeta(s)*(1+2^(2-s)). (End)
E.g.f.: 5*cosh(x) + sinh(x). - Stefano Spezia, Feb 09 2025

A021070 Decimal expansion of 1/66.

Original entry on oeis.org

0, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5

Offset: 0

N. J. A. Sloane

			0.015151515151515151515151515...

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

A010686 shifted right.

Join[{0},RealDigits[1/66,10,120][[1]]] (* or *) PadRight[{0},120,{5,1}] (* Harvey P. Dale, Dec 26 2021 *)

Multiplicative with a(2^e) = 5, a(p^e) = 1 otherwise. - David W. Wilson, Jun 12 2005

Dirichlet g.f.: zeta(s) * (1 + 1/2^(s-2)). - Amiram Eldar, Jun 09 2025

A173261 Array T(n,k) read by antidiagonals: T(n,2k)=1, T(n,2k+1)=n, n>=2, k>=0.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 1, 4, 1, 2, 1, 5, 1, 3, 1, 1, 6, 1, 4, 1, 2, 1, 7, 1, 5, 1, 3, 1, 1, 8, 1, 6, 1, 4, 1, 2, 1, 9, 1, 7, 1, 5, 1, 3, 1, 1, 10, 1, 8, 1, 6, 1, 4, 1, 2, 1, 11, 1, 9, 1, 7, 1, 5, 1, 3, 1, 1, 12, 1, 10, 1, 8, 1, 6, 1, 4, 1, 2, 1, 13, 1, 11, 1, 9, 1, 7, 1, 5, 1, 3, 1, 1, 14, 1, 12, 1, 10, 1, 8, 1, 6, 1, 4, 1, 2

Offset: 2

Paul Curtz, Feb 14 2010

One may define another array B(n,0) = -1, B(n,k) = T(n,k-1) + 2*B(n,k-1), n>=2, which also starts in columns k>=0, as follows:
  -1, -1, 0, 1,  4,  9, 20,  41,  84, 169,  340,  681, 1364 ...: A084639;
  -1, -1, 1, 3,  9, 19, 41,  83, 169, 339,  681, 1363, 2729;
  -1, -1, 2, 5, 14, 29, 62, 125, 254, 509, 1022, 2045, 4094;
  -1, -1, 3, 7, 19, 39, 83, 167, 339, 679, 1363, 2727, 5459 ...: -A173114;
B(n,k) = (n-1)*A001045(k) - T(n,k).
First differences are B(n,k+1) - B(n,k) = (n-1)*A001045(k).

			The array T(n,k) starts in row n=2 with columns k>=0 as:
  1,  2, 1,  2, 1,  2, 1,  2, 1,  2, 1,  2 ... A000034;
  1,  3, 1,  3, 1,  3, 1,  3, 1,  3, 1,  3 ... A010684;
  1,  4, 1,  4, 1,  4, 1,  4, 1,  4, 1,  4 ... A010685;
  1,  5, 1,  5, 1,  5, 1,  5, 1,  5, 1,  5 ... A010686;
  1,  6, 1,  6, 1,  6, 1,  6, 1,  6, 1,  6 ... A010687;
  1,  7, 1,  7, 1,  7, 1,  7, 1,  7, 1,  7 ... A010688;
  1,  8, 1,  8, 1,  8, 1,  8, 1,  8, 1,  8 ... A010689;
  1,  9, 1,  9, 1,  9, 1,  9, 1,  9, 1,  9 ... A010690;
  1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10 ... A010691.
Antidiagonal triangle begins as:
  1;
  1,  2;
  1,  3,  1;
  1,  4,  1,  2;
  1,  5,  1,  3,  1;
  1,  6,  1,  4,  1,  2;
  1,  7,  1,  5,  1,  3,  1;
  1,  8,  1,  6,  1,  4,  1,  2;
  1,  9,  1,  7,  1,  5,  1,  3,  1;
  1, 10,  1,  8,  1,  6,  1,  4,  1,  2;
  1, 11,  1,  9,  1,  7,  1,  5,  1,  3,  1;
  1, 12,  1, 10,  1,  8,  1,  6,  1,  4,  1,  2;
  1, 13,  1, 11,  1,  9,  1,  7,  1,  5,  1,  3,  1;
  1, 14,  1, 12,  1, 10,  1,  8,  1,  6,  1,  4,  1,  2;

G. C. Greubel, Antidiagonals n = 0..50 of the array, flattened

Cf. A000034, A010684, A010685, A010686, A010687, A010688, A010689, A010690, A010691.

Cf. A001045, A024206, A084639, A093178, A173114.

T[n_, k_]:= (1/2)*((n+3) - (n+1)*(-1)^k);
Table[T[n-k, k], {n,2,17}, {k,2,n}]//Flatten (* G. C. Greubel, Dec 03 2021 *)

flatten([[(1/2)*((n-k+3) - (n-k+1)*(-1)^k) for k in (2..n)] for n in (2..17)]) # G. C. Greubel, Dec 03 2021

From G. C. Greubel, Dec 03 2021: (Start)
T(n, k) = (1/2)*((n+3) - (n+1)*(-1)^k).
Sum_{k=0..n} T(n-k, k) = A024206(n).
Sum_{k=0..floor((n+2)/2)} T(n-2*k+2, k) = (1/16)*(2*n^2 4*n -5*(1 +(-1)^n) + 4*sin(n*Pi/2)) (diagonal sums).
T(2*n-2, n) = A093178(n). (End)

A216639 A027642(6*n+6)/(sequence of period 2:repeat 42,210).

Original entry on oeis.org

1, 13, 19, 13, 341, 9139, 43, 221, 19, 270413, 1541, 667147, 79, 16211, 6479, 21437, 103, 996151, 1, 11086933, 103759, 20033, 6533, 11341499, 51491, 8545667, 3097, 16211, 59, 34408161359, 1, 4137341, 5826521, 1339, 219666403, 72719023, 223, 2977, 1501, 45423164501, 83

Offset: 0

Paul Curtz, Sep 12 2012

nonn

Is a(n) always an integer?  Is there an a(n) ending with 5?
It appears (tested for n <= 800) that a(n) mod 9 is always one of {1, 2, 4, 5, 7, 8}.
There is a similar sequence of ratios A027642(10n+1)/(66*A010686(n)) which starts 1, 1, 217, 41, 1, 172081, 71, 697, 4123, 101, 23, 7055321, 131, 2059, 32767, 697, 1, 21896102683,...
a(n) is always an integer: 42 = 2*3*7 and 1, 2, and 6 divide 12n+6; 210 = 2*3*5*7 and 1, 2, 4, and 6 divide 12n+12. a(n) never ends in 5 (or 0) since 12n+6 is not divisible by 4 hence the (12n+6)-th Bernoulli denominator is not divisible by 5, and Bernoulli denominators are squarefree and hence the (12n+12)-th Bernoulli denominator, divided by 210, cannot be divisible by 5. - Charles R Greathouse IV, Sep 12 2012
The previous comments argue that 3 or 5 are never prime divisors of a(n). In addition (tested up to n <=900), 7 apparently is also a non-divisor of a(n). In summary, the prime divisors appear all to be in A140461. - Jean-François Alcover, Sep 17 2012

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
A. Joyal, Les nombres de Bernoulli (in French), 2003.

Cf. A002445, A027762, A165734, A165949.

A010686 := proc(n)
    op((n mod 2)+1,[1,5]) ;
end proc:
A216639 := proc(n)
    A027642(6*n+6)/42/A010686(n) ;
end proc: # R. J. Mathar, Sep 21 2012

a(n)=denominator(bernfrac(6*n+6))/if(n%2,210,42) \\ Charles R Greathouse IV, Sep 12 2012

a(n)=my(t=1);fordiv(3*n+3,d,if(isprime(2*d+1),t*=2*d+1));t/if(n%2,105,21) \\ Charles R Greathouse IV, Sep 12 2012

a(n) = A027642(6*n+6)/(42*A010686(n)).

a(20)-a(40) from Charles R Greathouse IV, Sep 12 2012

A135537 Period 4: repeat [7, 5, 2, 4].

Original entry on oeis.org

7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5, 2, 4, 7, 5

Offset: 0

Paul Curtz, Feb 22 2008

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

Cf. A010686, A010693, A021408, A135536.

&cat [[7, 5, 2, 4]^^30]; // Wesley Ivan Hurt, Jul 08 2016

seq(op([7, 5, 2, 4]), n=0..50); # Wesley Ivan Hurt, Jul 08 2016

Flatten[Table[{7,5,2,4},{20}]] (* Harvey P. Dale, Jun 22 2011 *)

a(n) = A135536(n) mod 9.
O.g.f.: ((x+5)/(x^2+1)+9*(1-x))/2. a(n) = ((-1)^floor(n/2) * A010686(n+1) + 9)/2. - R. J. Mathar, Feb 23 2008
From Wesley Ivan Hurt, Jul 08 2016: (Start)
a(n) = a(n-1) - a(n-2) + a(n-3) for n>2, a(n) = a(n-4) for n>3.
a(n) = A021408(n) for n>1. (End)

cp's OEIS Frontend

A141430 a(n) = A000111(n) mod 9.

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A057566 Number of collinear triples in a 3 X n rectangular grid.

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A171654 Period 10: repeat 0, 1, 6, 7, 2, 3, 8, 9, 4, 5.

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A173078 a(n) = (5*2^n - 2*(-1)^n - 9)/3.

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A176260 Periodic sequence: Repeat 5, 1.

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A021070 Decimal expansion of 1/66.

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A173261 Array T(n,k) read by antidiagonals: T(n,2k)=1, T(n,2k+1)=n, n>=2, k>=0.

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A216639 A027642(6*n+6)/(sequence of period 2:repeat 42,210).

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A173078 a(n) = (52^n - 2(-1)^n - 9)/3.