A033115 -id:A033115 - cp's OEIS frontend

A128174 Transform, (1,0,1,...) in every column.

1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1

Offset: 1

Gary W. Adamson, Feb 17 2007

Inverse of the triangle = a tridiagonal matrix with (1,1,1,...) in the superdiagonal, (0,0,0,...) in the main diagonal and (-1,-1,-1,...) in the subdiagonal.
Riordan array (1/(1-x^2), x) with inverse (1-x^2,x). - Paul Barry, Sep 10 2008
The position of 1's in this sequence is equivalent to A246705, and the position of 0's is equivalent to A246706. - Bernard Schott, Jun 05 2019

			First few rows of the triangle are:
  1;
  0, 1;
  1, 0, 1;
  0, 1, 0, 1;
  1, 0, 1, 0, 1; ...

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened

Cf. A004526 (row sums).

a128174 n k = a128174_tabl !! (n-1) !! (k-1)
a128174_row n = a128174_tabl !! (n-1)
a128174_tabl = iterate (\xs@(x:_) -> (1 - x) : xs) [1]
-- Reinhard Zumkeller, Aug 01 2014

[[(1+(-1)^(n-k))/2: k in [1..n]]: n in [1..12]]; // G. C. Greubel, Jun 05 2019

A128174 := proc(n,k)
    if k > n or k < 1 then
        0;
    else
        modp(k+n+1,2) ;
    end if;
end proc: # R. J. Mathar, Aug 06 2016

a128174[r_] := Table[If[EvenQ[n+k], 1, 0], {n, 1, r}, {k, 1, n}]
TableForm[a128174[5]] (* triangle *)
Flatten[a128174[10]] (* data *) (* Hartmut F. W. Hoft, Mar 15 2017 *)
Table[(1+(-1)^(n-k))/2, {n,1,12}, {k,1,n}]//Flatten (* G. C. Greubel, Sep 26 2017 *)

for(n=1,12, for(k=1,n, print1((1+(-1)^(n-k))/2, ", "))) \\ G. C. Greubel, Sep 26 2017

[[(1+(-1)^(n-k))/2 for k in (1..n)] for n in (1..12)] # G. C. Greubel, Jun 05 2019

A lower triangular matrix transform, (1, 0, 1, ...) in every column; n terms of (1, 0, 1, ...) in odd rows; n terms of (0, 1, 0, ...) in even rows.
T(n,k) = [k<=n]*(1+(-1)^(n-k))/2. - Paul Barry, Sep 10 2008
With offset n=1, k=0: Sum_{k=0..n} {T(n,k)*x^k} = A000035(n), A004526(n+1), A000975(n), A033113(n), A033114(n), A033115(n), A033116(n), A033117(n), A033118(n), A033119(n), A056830(n+1) for x=0,1,2,3,4,5,6,7,8,9,10 respectively. - Philippe Deléham, Oct 17 2011
T(n+1,1) = 1 - T(n,1); T(n+1,k) = T(n,k-1), 1 < k <= n+1. - Reinhard Zumkeller, Aug 01 2014

A015531 Linear 2nd order recurrence: a(n) = 4a(n-1) + 5a(n-2).

Original entry on oeis.org

0, 1, 4, 21, 104, 521, 2604, 13021, 65104, 325521, 1627604, 8138021, 40690104, 203450521, 1017252604, 5086263021, 25431315104, 127156575521, 635782877604, 3178914388021, 15894571940104, 79472859700521, 397364298502604, 1986821492513021, 9934107462565104

Offset: 0

Olivier Gérard

Number of walks of length n between any two distinct vertices of the complete graph K_6. Example: a(2)=4 because the walks of length 2 between the vertices A and B of the complete graph ABCDEF are: ACB, ADB, AEB and AFB. - Emeric Deutsch, Apr 01 2004
General form: k=5^n-k. Also: A001045, A078008, A097073, A115341, A015518, A054878, A015521, A109499. - Vladimir Joseph Stephan Orlovsky, Dec 11 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-4, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=charpoly(A,1). - Milan Janjic, Jan 27 2010
Pisano period lengths: 1, 2, 6, 2, 2, 6, 6, 4, 18, 2, 10, 6, 4, 6, 6, 8, 16, 18, 18, 2,... - R. J. Mathar, Aug 10 2012
The ratio a(n+1)/a(n) converges to 5 as n approaches infinity. - Felix P. Muga II, Mar 09 2014
For odd n, a(n) is congruent to 1 (mod 10). For even n > 0, a(n) is congruent to 4 (mod 10). - Iain Fox, Dec 30 2017

Iain Fox, Table of n, a(n) for n = 0..1431 (terms 0..1000 from Vincenzo Librandi)
Jean-Paul Allouche, Jeffrey Shallit, Zhixiong Wen, Wen Wu, Jiemeng Zhang, Sum-free sets generated by the period-k-folding sequences and some Sturmian sequences, arXiv:1911.01687 [math.CO], 2019.
F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014.
Index entries for linear recurrences with constant coefficients, signature (4,5).

A083425 shifted right.

Cf. A033115 (partial sums), A213128.

[Round(5^n/6): n in [0..30]]; // Vincenzo Librandi, Jun 24 2011

seq(round(5^n/6), n=0..25); # Mircea Merca, Dec 28 2010

LinearRecurrence[{4,5},{0,1},30] (* Harvey P. Dale, Jul 09 2017 *)

a(n)=5^n\/6 ; \\ Charles R Greathouse IV, Apr 14 2014

first(n) = Vec(x/((1 - 5*x)*(1 + x)) + O(x^n), -n) \\ Iain Fox, Dec 30 2017

[lucas_number1(n,4,-5) for n in range(0, 22)] # Zerinvary Lajos, Apr 23 2009

From Paul Barry, Apr 20 2003: (Start)
a(n) = (5^n -(-1)^n)/6.
G.f.: x/((1-5*x)*(1+x)).
E.g.f.(exp(5*x)-exp(-x))/6. (End) (corrected by M. F. Hasler, Jan 29 2012)
a(n) = Sum_{k=1..n} binomial(n, k)*(-1)^(n+k)*6^(k-1). - Paul Barry, May 13 2003
a(n) = 5^(n-1) - a(n-1). - Emeric Deutsch, Apr 01 2004
a(n) = ((2+sqrt(9))^n - (2-sqrt(9))^n)/6. - Al Hakanson (hawkuu(AT)gmail.com), Jan 07 2009
a(n) = round(5^n/6). - Mircea Merca, Dec 28 2010
The logarithmic generating function 1/6*log((1+x)/(1-5*x)) = x + 4*x^2/2 + 21*x^3/3 + 104*x^4/4 + ... has compositional inverse 6/(5+exp(-6*x)) - 1, the e.g.f. for a signed version of A213128. - Peter Bala, Jun 24 2012
a(n) = (-1)^(n-1)*Sum_{k=0..(n-1)} A135278(n-1,k)*(-6)^k = (5^n - (-1)^n)/6 = (-1)^(n-1)*Sum_{k=0..(n-1)} (-5)^k. Equals (-1)^(n-1)*Phi(n,-5) when n is an odd prime, where Phi is the cyclotomic polynomial. - Tom Copeland, Apr 14 2014

A056830 Alternate digits 1 and 0.

Original entry on oeis.org

0, 1, 10, 101, 1010, 10101, 101010, 1010101, 10101010, 101010101, 1010101010, 10101010101, 101010101010, 1010101010101, 10101010101010, 101010101010101, 1010101010101010, 10101010101010101, 101010101010101010

Offset: 0

Henry Bottomley, Aug 30 2000

Fibonacci bit-representations of numbers for which there is only one possible representation and for which the maximal and minimal bit-representations (A104326 and A014417) are equal. The numbers represented are equal to the numbers in A000071 (subtract the first term of that sequence). For example, 10101 = 12 because 8+5+1. - Casey Mongoven, Mar 19 2006
Sequence A000975 written in base 2. - Jaroslav Krizek, Aug 05 2009
The absolute value of alternating sum of the first n repunits: a(n) = abs(Sum_{k=0..n} (-1)^k*A002275(n)). - Ilya Gutkovskiy, Dec 02 2015
Binary representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 131", based on the 5-celled von Neumann neighborhood. See A279053 for references and links. - Robert Price, Dec 05 2016

			n  a(n)             A000975(n)   (If a(n) is interpreted in base 2.)
------------------------------
0  0 ....................... 0
1  1 ....................... 1
2  10 ...................... 2 = 2^1
3  101 ..................... 5
4  1010 ................... 10 = 2^1 + 2^3
5  10101 .................. 21
6  101010 ................. 42 = 2^1 + 2^3 + 2^5
7  1010101 ................ 85
8  10101010 .............. 170 = 2^1 + 2^3 + 2^5 + 2^7
9  101010101 ............. 341
10 1010101010 ............ 682 = 2^1 + 2^3 + 2^5 + 2^7 + 2^9
11 10101010101 .......... 1365
12 101010101010 ......... 2730 = 2^1 + 2^3 + 2^5 + 2^7 + 2^9 + 2^11, etc.
- _Bruno Berselli_, Dec 02 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..800
Index entries for linear recurrences with constant coefficients, signature (10,1,-10).
Index entries for 2-automatic sequences.

Partial sums of A015585.

Cf. A000975, A033113, A033114, A033115, A033116, A033117, A033118, A033119, A059848, A062864.

List([0..30], n-> Int(10^(n+1)/99) ); # G. C. Greubel, Jul 14 2019

[Round((20*10^n-11)/198) : n in [0..30]]; // Vincenzo Librandi, Jun 25 2011

A056830 := proc(n) floor(10^(n+1)/99) ; end proc:

CoefficientList[Series[x/((1-x^2)*(1-10*x)), {x,0,30}], x] (* G. C. Greubel, Sep 26 2017 *)

Vec(x/((1-x)*(1+x)*(1-10*x))+O(x^30)) \\ Charles R Greathouse IV, Feb 13 2017

[floor(10^(n+1)/99) for n in (0..30)] # G. C. Greubel, Jul 14 2019

a(n) = +10*a(n-1) + a(n-2) - 10*a(n-3).
a(n) = floor(10^(n+1)/(10^2-1)) = a(n-2)+10^(n-1) = 10*a(n-1) + (1 - (-1)^n)/2.
From Paul Barry, Nov 12 2003: (Start)
a(n+1) = Sum_{k=0..floor(n/2)} 10^(n-2*k).
a(n+1) = Sum_{k=0..n} Sum_{j=0..k} (-1)^(j+k)*10^j.
G.f.: x/((1-x)*(1+x)*(1-10*x)).
a(n) = 9*a(n-1) + 10*a(n-2) + 1.
a(n) = 10^(n+1)/99 - (-1)^n/22 - 1/18. (End)
a(n) = A007088(A107909(A104161(n))) = A007088(A000975(n)). - Reinhard Zumkeller, May 28 2005
a(n) = round((20*10^n-11)/198) = floor((10*10^n-1)/99) = ceiling((10*10^n-10)/99) = round((10*10^n-10)/99). - Mircea Merca, Dec 27 2010
From Daniel Forgues, Sep 20 2018: (Start)
If a(n) is interpreted in base 2:
a(2n) = Sum_{k=1..n} 2^(2n-1), n >= 0; a(2n-1) = a(2n)/2, n >= 1.
a(2n) = A020988(n), n >= 0.
a(0) = 0; a(2n) = 4*a(2n-2) + 2, n >= 1. (End)

More terms from Casey Mongoven, Mar 19 2006

A059848 As a square table by antidiagonals, the n-digit number which in base k starts 1010101...

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 2, 2, 0, 0, 1, 3, 5, 2, 1, 0, 1, 4, 10, 10, 3, 0, 0, 1, 5, 17, 30, 21, 3, 1, 0, 1, 6, 26, 68, 91, 42, 4, 0, 0, 1, 7, 37, 130, 273, 273, 85, 4, 1, 0, 1, 8, 50, 222, 651, 1092, 820, 170, 5, 0, 0, 1, 9, 65, 350, 1333, 3255, 4369, 2460, 341, 5, 1, 0, 1, 10

Offset: 0

Henry Bottomley, Feb 26 2001

			T(5,3)=10101 base 3=81+9+1=91; T(4,6)=1010 base 6=216+6=222. Table starts {0,0,0,0,...}, {1,1,1,1,...}, {0,1,2,3,...}, {1,2,5,10,...}, ...

Rows include A000004, A000012, A001477, A002522, A034262, A059826, A059830, A059839. Columns include A000035, A008619, A000975, A033113, A033114, A033115, A033116, A033117, A033118, A033119, A056830.

T(n, k)=floor[k^(n+1)/(k^2-1)] =T(n-2, k)+k^(n-1) =k*T(n-1, k)-((-1)^n-1)/2

A097139 Convolution of 5^n and floor(n/2).

Original entry on oeis.org

0, 0, 1, 6, 32, 162, 813, 4068, 20344, 101724, 508625, 2543130, 12715656, 63578286, 317891437, 1589457192, 7947285968, 39736429848, 198682149249, 993410746254, 4967053731280, 24835268656410, 124176343282061, 620881716410316

Offset: 0

Paul Barry, Jul 29 2004

a(n+1) gives partial sums of A033115 and second partial sums of A015531.

Partial sums of (1/4)*floor(5^n/6) = (1/3)*floor(5^n/8). - Mircea Merca, Dec 27 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-4,-6,5).

Column k=5 of A368296.

[5^(n+1)/96 -n/8 -3/32 +(-1)^n/24: n in [0..30]]; // Vincenzo Librandi, Jun 25 2011

A097139 := proc(n) 5^(n+1)/96 -n/8 -3/32 +(-1)^n/24 ; end proc: # R. J. Mathar, Jan 08 2011

f[n_] := Floor[5^n/6]/4; Accumulate@ Array[f, 24, 0]
a[n_] := a[n] = 6 a[n - 1] - 4 a[n - 2] - 6 a[n - 3] + 5 a[n - 4]; a[0] = a[1] = 0; a[2] = 1; a[3] = 6; Array[a, 24, 0]
CoefficientList[ Series[x^2/((1 - x) (1 - 5 x) (1 - x^2)), {x, 0, 23}], x] (* Robert G. Wilson v, Jan 02 2011 *)
LinearRecurrence[{6,-4,-6,5},{0,0,1,6},30] (* Harvey P. Dale, Mar 16 2019 *)

a(n) = 5^(n+1)/96 -n/8 -3/32 +(-1)^n/24. - R. J. Mathar, Jan 08 2011
G.f.: x^2/((1-x)*(1-5*x)*(1-x^2)).
a(n) = 6*a(n-1) - 4*a(n-2) - 6*a(n-3) + 5*a(n-4).
a(n) = Sum_{k=0..n} floor((n-k)/2)*4^k = Sum_{k=0..n} floor(k/2)*4^(n-k).
From Mircea Merca, Dec 27 2010: (Start)
4*a(n) = round((5*5^n-12*n-9)/24) = floor((5*5^n-12*n-5)/24) = ceiling((5*5^n-12*n-13)/24) = round((5*5^n-12*n-5)/24).
a(n) = a(n-2) + (5^(n-1)-1)/4, n>1. (End)
a(n) = (floor(5^(n+1)/24) - floor((n+1)/2))/4. - Seiichi Manyama, Dec 22 2023

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A128174 Transform, (1,0,1,...) in every column.

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A015531 Linear 2nd order recurrence: a(n) = 4*a(n-1) + 5*a(n-2).

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A056830 Alternate digits 1 and 0.

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A059848 As a square table by antidiagonals, the n-digit number which in base k starts 1010101...

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A097139 Convolution of 5^n and floor(n/2).

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A015531 Linear 2nd order recurrence: a(n) = 4a(n-1) + 5a(n-2).