A033114 -id:A033114 - 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

A043291 Every run length in base 2 is 2.

Original entry on oeis.org

3, 12, 51, 204, 819, 3276, 13107, 52428, 209715, 838860, 3355443, 13421772, 53687091, 214748364, 858993459, 3435973836, 13743895347, 54975581388, 219902325555, 879609302220, 3518437208883, 14073748835532, 56294995342131, 225179981368524, 900719925474099

Offset: 1

Clark Kimberling

a(n) is the number whose binary representation is A153435(n). - Omar E. Pol, Jan 18 2009

See A033001 and following for the analog in other bases and the variant with runs of length >= 2. - M. F. Hasler, Feb 01 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (4,1,-4).

Cf. A153435 (binary).
Bisections: A108020, A182512. Bisection of A077854.
Cf. A000975, A033001, A033114.

[Floor(4^(n+1)/5): n in [1..30]]; // Vincenzo Librandi, Jun 26 2011

seq(floor(4^(n+1)/5),n=1..25); # Mircea Merca, Dec 26 2010

f[n_] := Floor[4^(n + 1)/5]; Array[f, 23] (* or *)
a[1] = 3; a[2] = 12; a[3] = 51; a[n_] := a[n] = 4 a[n - 1] + a[n - 2] - 4 a[n - 3]; Array[a, 23] (* or *)
3 LinearRecurrence[{4, 1, -4}, {1, 4, 17}, 23] (* Robert G. Wilson v, Jul 01 2014 *)

A043291 = n->4^(n+1)\5 \\ M. F. Hasler, Feb 01 2014

def a(n): return int(''.join([['11', '00'][i%2] for i in range(n)]), 2)
print([a(n) for n in range(1, 26)]) # Michael S. Branicky, Mar 12 2021

a(n) = 4*a(n-1)+a(n-2)-4*a(n-3), n>3. - John W. Layman, Feb 01 2000
a(n) = floor(4^(n+1)/5). - Mircea Merca, Dec 26 2010
G.f.: 3*x / ( (x-1)*(4*x-1)*(1+x) ). - Joerg Arndt, Jan 08 2011
a(n) = 3*A033114(n). - R. J. Mathar, Jan 08 2011

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

A115255 "Correlation triangle" of central binomial coefficients A000984.

Original entry on oeis.org

1, 2, 2, 6, 5, 6, 20, 14, 14, 20, 70, 46, 41, 46, 70, 252, 160, 134, 134, 160, 252, 924, 574, 466, 441, 466, 574, 924, 3432, 2100, 1672, 1534, 1534, 1672, 2100, 3432, 12870, 7788, 6118, 5506, 5341, 5506, 6118, 7788, 12870, 48620, 29172, 22692, 20152, 19174

Offset: 0

Paul Barry, Jan 18 2006

Row sums are A033114. Diagonal sums are A115256. T(2n,n) is A115257. Corresponds to the triangle of antidiagonals of the correlation matrix of the sequence array for C(2n,n).

Let s=(1,2,6,20,...), (central binomial coefficients), and let T be the infinite square matrix whose n-th row is formed by putting n-1 zeros before the terms of s. Let T' be the transpose of T. Then A115255 represents the matrix product M=T'*T. M is the self-fusion matrix of s, as defined at A193722. See A203005 for characteristic polynomials of principal submatrices of M, with interlacing zeros. - Clark Kimberling, Dec 27 2011

			Triangle begins:
  1;
  2, 2;
  6, 5, 6;
  20, 14, 14, 20;
  70, 46, 41, 46, 70;
  252, 160, 134, 134, 160, 252;
Northwest corner (square format):
  1    2    6    20    70
  2    5    14   46    160
  6    14   41   134   466
  20   46   134  441   1534

Cf. A203004, A203001, A202453.

s[k_] := Binomial[2 k - 2, k - 1];
U = NestList[Most[Prepend[#, 0]] &, #, Length[#] - 1] &[Table[s[k], {k, 1, 15}]];
L = Transpose[U]; M = L.U; TableForm[M]
m[i_, j_] := M[[i]][[j]]; (* A115255 in square format *)
Flatten[Table[m[i, n + 1 - i], {n, 1, 12}, {i, 1, n}]]
f[n_] := Sum[m[i, n], {i, 1, n}] + Sum[m[n, j], {j, 1, n - 1}]; Table[f[n], {n, 1, 12}]
Table[Sqrt[f[n]], {n, 1, 12}]  (* A006134 *)
Table[m[1, j], {j, 1, 12}]     (* A000984 *)
Table[m[j, j], {j, 1, 12}]     (* A115257 *)
Table[m[j, j + 1], {j, 1, 12}] (* 2*A082578 *)
(* Clark Kimberling, Dec 27 2011 *)

G.f.: 1/(sqrt(1-4*x)*sqrt(1-4*x*y)*(1-x^2*y)) (format due to Christian G. Bower).

T(n, k) = Sum_{j=0..n} [j<=k]*C(2*k-2*j, k-j)*[j<=n-k]*C(2*n-2*k-2*j, n-k-j).

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

A097138 Convolution of 4^n and floor(n/2).

Original entry on oeis.org

0, 0, 1, 5, 22, 90, 363, 1455, 5824, 23300, 93205, 372825, 1491306, 5965230, 23860927, 95443715, 381774868, 1527099480, 6108397929, 24433591725, 97734366910, 390937467650, 1563749870611, 6254999482455, 25019997929832

Offset: 0

Paul Barry, Jul 29 2004

a(n+1) gives partial sums of A033114 and second partial sums of A015521.

Partial sums of 1/3*floor(4^n/5). - Mircea Merca, Dec 26 2010

			a(3) = 1/3*floor(4^0/5)+1/3*floor(4^1/5)+1/3*floor(4^2/5) +1/3*floor(4^3/5) = 0 + 0 + 1 + 4 = 5.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (5,-3,-5,4).

Column k=4 of A368296.

Cf. A015521, A033114.

[(4^(n+2)-30*n+9*(-1)^n-25)/180: n in [0..30]]; // Vincenzo Librandi, May 31 2011

A097138 := proc(n) (4^(n+2)-30*n+9*(-1)^n-25)/180 ; end proc: # R. J. Mathar, Jan 08 2011

LinearRecurrence[{5,-3,-5,4},{0,0,1,5},30] (* Harvey P. Dale, Sep 17 2017 *)

G.f.: x^2/((1-x)*(1-4*x)*(1-x^2)).
a(n) = Sum_{k=0..n} floor((n-k)/2)4^k = Sum_{k=0..n} floor(k/2)*4^(n-k).
a(n) = 5*a(n-1) - 3*a(n-2) - 5*a(n-3) + 4*a(n-4).
From Mircea Merca, Dec 26 2010: (Start)
3*a(n) = round((16*4^n-30*n-25)/60) = floor((8*4^n-15*n-8)/30) = ceiling((8*4^n-15*n-17)/30) = round((8*4^n-15*n-8)/30).
a(n) = a(n-2)+(4^(n-1)-1)/3, n>1. (End)
a(n) = (4^(n+2)-30*n+9*(-1)^n-25)/180. - Bruno Berselli, Dec 27 2010
a(n) = (floor(4^(n+1)/15) - floor((n+1)/2))/3. - Seiichi Manyama, Dec 22 2023

A117616 a(0)=0, a(n)=4a(n-1)+2 for n odd, a(n)=4a(n-1) for n even.

Original entry on oeis.org

0, 2, 8, 34, 136, 546, 2184, 8738, 34952, 139810, 559240, 2236962, 8947848, 35791394, 143165576, 572662306, 2290649224, 9162596898, 36650387592, 146601550370, 586406201480, 2345624805922, 9382499223688, 37529996894754

Offset: 0

Roger L. Bagula, Apr 07 2006

L. Rosenfeld, Nuclear Forces, section II, Interscience, New York, 1948, p 202

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,1,-4).

a:=proc(n) if n=0 then 0 elif n mod 2 = 1 then 4*a(n-1)+2 else 4*a(n-1) fi end: seq(a(n),n=0..23);

b[0] := 0 b[1] := 2 b[n_?EvenQ] := b[n] = 4*b[n - 1] b[n_?OddQ] := b[n] = 4*b[n - 1] + 2 a = Table[b[n], {n, 0, 25}]
nxt[{n_,a_}]:={n+1,If[EvenQ[n],4a+2,4a]}; NestList[nxt,{0,0},30][[;;,2]] (* or *) LinearRecurrence[{4,1,-4},{0,2,8},30] (* Harvey P. Dale, Mar 10 2023 *)

a(n) = (-5-3*(-1)^n+2^(3+2*n))/15. a(n) = 4*a(n-1)+a(n-2)-4*a(n-3). G.f.: 2*x / ((x-1)*(x+1)*(4*x-1)). [Colin Barker, Feb 17 2013]

a(n) = 2*A033114(n). - R. J. Mathar, Feb 27 2019

Edited by N. J. A. Sloane, Apr 16 2006

cp's OEIS Frontend

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

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A043291 Every run length in base 2 is 2.

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A056830 Alternate digits 1 and 0.

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A115255 "Correlation triangle" of central binomial coefficients A000984.

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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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A097138 Convolution of 4^n and floor(n/2).

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A117616 a(0)=0, a(n)=4a(n-1)+2 for n odd, a(n)=4a(n-1) for n even.

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