A131708 -id:A131708 - cp's OEIS frontend

A111927 Expansion of x^3 / ((x-1)(2x-1)*(x^2-x+1)).

0, 0, 0, 1, 4, 10, 21, 42, 84, 169, 340, 682, 1365, 2730, 5460, 10921, 21844, 43690, 87381, 174762, 349524, 699049, 1398100, 2796202, 5592405, 11184810, 22369620, 44739241, 89478484, 178956970, 357913941, 715827882, 1431655764, 2863311529, 5726623060

Offset: 0

Creighton Dement, Aug 21 2005

Binomial transform of sequence (0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0). Note: the binomial transform of the sequence (0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0) is A111926; the binomial transform of the sequence (0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0) is A024495 (disregarding first two terms, which are both zero).

The sequence relates the calculation of the logarithm of the Twin Prime Constants of order 3 to the sequence of prime zeta functions, see definition 7 in arXiv:0903.2514. - R. J. Mathar, Mar 28 2009

Colin Barker, Table of n, a(n) for n = 0..1000
Antoine-Augustin Cournot, Solution d'un problème d'analyse combinatoire, Bulletin des Sciences Mathématiques, Physiques et Chimiques, item 34, volume 11, 1829, pages 93-97. Also at Google Books. Page 97 case p=3 formula y^(0) = a(n). (But misprint "- (2/3)*cos" should be "+ (2/3)*cos".)
R. J. Mathar, Hardy-Littlewood constants embedded into infinite products over all positive integers, arXiv:0903.2514 [math.NT], 2009-2011.
Christian Ramus, Solution générale d'un problème d'analyse combinatoire, Journal für die Reine und Angewandte Mathematik (Crelle's journal), volume 11, 1834, pages 353-355. Page 353 case p=3 formula y^(0) = a(n). (But misprint "+ (1/3)*cos" should be "+ (2/3)*cos", per the general case equation A page 354.)
Kevin Ryde, Iterations of the Terdragon Curve, section Lines, quantity Lines_k(2) = a(k+1).
Index entries for linear recurrences with constant coefficients, signature (4,-6,5,-2).

Cf. A000295, A111926, A024495, A131708.

seq(sum(binomial(n, k*3), k=1..n), n=0..33); # Zerinvary Lajos, Oct 23 2007

LinearRecurrence[{4,-6,5,-2},{0,0,0,1},40] (* Harvey P. Dale, Jul 04 2017 *)

concat(vector(3), Vec(x^3/((x-1)*(2*x-1)*(x^2-x+1)) + O(x^40))) \\ Colin Barker, Feb 10 2017

a(n+2) - a(n+1) + a(n) = A000225(n).
a(n) - a(n-1) = A024495(n-1).
From Colin Barker, Feb 10 2017: (Start)
a(n) = 2^n/3 + 2*cos((Pi*n)/3)/3 - 1. [Cournot]
a(n) = 4*a(n-1) - 6*a(n-2) + 5*a(n-3) - 2*a(n-4) for n > 3. (End)
a(n) = (2^n+A087204(n))/3 - 1. - R. J. Mathar, Aug 07 2017
a(n) = (1/3)*Sum_{k=0..n-1} binomial(n, 3*floor(k/3)+3). - Taras Goy, Jan 26 2025
E.g.f.: (exp(x)*(exp(x) - 3) + 2*exp(x/2)*cos(sqrt(3)*x/2))/3. - Stefano Spezia, Feb 06 2025

A081374 Size of "uniform" Hamming covers of distance 1, that is, Hamming covers in which all vectors of equal weight are treated the same, included or excluded from the cover together.

Original entry on oeis.org

1, 2, 2, 5, 10, 22, 43, 86, 170, 341, 682, 1366, 2731, 5462, 10922, 21845, 43690, 87382, 174763, 349526, 699050, 1398101, 2796202, 5592406, 11184811, 22369622, 44739242, 89478485, 178956970, 357913942, 715827883, 1431655766, 2863311530, 5726623061

Offset: 1

David Applegate, Aug 22 2003

nonn

Motivation: consideration of the "hats" problem (which boils down to normal hamming covering codes) in the case when the people are indistinguishable or unlabeled.
From Paul Curtz, May 26 2011: (Start)
If we add a(0)=1 in front and build the table of a(n) and iterated differences in further rows we get:
1,    1,  2,  2,  5, 10,
0,    1,  0,  3,  5, 12,
1,   -1,  3,  2,  7,  9,
-2,   4, -1,  5,  2, 13,
6,   -5,  6, -3, 11,  6
-11, 11, -9, 14, -5, 21.
The first column is the inverse binomial transform, which is 1,0 followed by (-1)^n*A083322(n-1), n>=2.
The main diagonal in the table above is A001045, the adjacent upper diagonals are A078008, A048573 and A062092. (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,-1,2).

Cf. A083322.

I:=[1,2,2,5]; [n le 4 select I[n] else 2*Self(n-1)-Self(n-3)+2*Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jul 08 2016

hatwork := proc(n,i,covered) local val, val2; options remember;
# computes the minimum cover of the i-bit through n-bit words.
# if covered is true the i-bit words are already covered (by the (i-1)-bit words)
if (i>n or (i = n and covered)) then 0; elif (i = n and not covered) then 1; else
# one choice is to include the i-bit words in the cover
val := hatwork(n, i+1, true) + binomial(n,i);
# the other choice is not to include the i-bit words in the cover
if (covered) then val2 := hatwork (n, i+1, false); if (val2 < val) then val := val2; fi; else
# if the i-bit words were not covered by (i-1), they must be covered by the (i+1)-bit words
if (i <= n) then val2 := hatwork (n, i+2, true) + binomial(n,i+1); if (val2 < val) then val := val2; fi; fi; fi; val; fi; end proc;
A081374 := proc (n) hatwork(n, 0, false); end proc;

LinearRecurrence[{2,0,-1,2},{1,2,2,5},40] (* Harvey P. Dale, Feb 11 2015 *)

If (n mod 6 = 5) then sum(binomial(n, 3*i+1), i=0..n/3); elif (n mod 6 = 2) then sum(binomial(n, 3*i), i=0..n/3)+1; else sum(binomial(n, 3*i), i=0..n/3); fi;
G.f.: x*(2*x^3-2*x^2+1)/( (1-2*x)*(1+x)*(1-x+x^2) ).
a(n)=2*a(n-1)-a(n-3)+2*a(n-4).
From Paul Curtz, May 26 2011: (Start)
a(n+1) - 2*a(n) has period length 6: repeat 0, -2, 1, 0, 2, -1 (see A080425).
a(n) - A083322(n-1) = A010892(n-1) has period length 6.
a(n) + a(n+3) = 3*2^n = A007283(n).
a(n+6)-a(n) = 21*2^n = A175805(n).
a(n) - A131708(n) = -A131531(n).  (End)

A131090 First differences of A131666.

Original entry on oeis.org

0, 1, 0, 1, 1, 4, 7, 15, 28, 57, 113, 228, 455, 911, 1820, 3641, 7281, 14564, 29127, 58255, 116508, 233017, 466033, 932068, 1864135, 3728271, 7456540, 14913081, 29826161, 59652324, 119304647, 238609295, 477218588, 954437177, 1908874353

Offset: 0

Paul Curtz, Sep 24 2007

The first differences b(n)=a(n+1)-a(n) obey the recurrence b(n+1)-2b(n) = (-3,3,-2,3,-3,2), continued with period 6.
The 2nd differences c(n)=b(n+1)-b(n) obey the recurrence c(n+1)-2c(n) = (6,-5,5,-6,5,-5), periodically continued with period 6.
The hexaperiodic coefficients in these recurrences for A113405, A131666 and their higher order differences define a table,
0, 0, 1, 0, 0, -1 <- A113405
0, 1, -1, 0, -1, 1 <- A131666
1, -2, 1, -1, 2, -1 <- a(n)
-3, 3, -2, 3, -3, 2 <- b(n)
6, -5, 5, -6, 5, -5 <- c(n)
-11,10,-11, 11,-10, 11
21,-21,22,-21, 21,-22
...
in which the first three columns are A024495, A131708 and A024493, multiplied by a checkerboard pattern of signs.

Index entries for linear recurrences with constant coefficients, signature (2,0,-1,2).

LinearRecurrence[{2,0,-1,2},{0,1,0,1},40] (* Harvey P. Dale, Jan 15 2016 *)

a(n) = A131666(n+1)-A131666(n).
a(n+1)-2a(n) = A131556(n), a sequence with period length 6.
G.f.: -(x-1)^2*x / ((x+1)*(2*x-1)*(x^2-x+1)). - Colin Barker, Mar 04 2013

Edited by R. J. Mathar, Jun 28 2008

A167613 Array T(n,k) read by antidiagonals: the k-th term of the n-th difference of A131531.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 0, -1, -2, -3, 0, 0, 1, 3, 6, -1, -1, -1, -2, -5, -11, 0, 1, 2, 3, 5, 10, 21, 0, 0, -1, -3, -6, -11, -21, -42, 1, 1, 1, 2, 5, 11, 22, 43, 85, 0, -1, -2, -3, -5, -10, -21, -43, -86, -171, 0, 0, 1, 3, 6, 11, 21, 42, 85, 171, 342, -1, -1, -1, -2, -5, -11, -22, -43, -85, -170, -341, -683, 0, 1, 2, 3, 5, 10, 21, 43, 86, 171, 341, 682, 1365

Offset: 0

Paul Curtz, Nov 07 2009

The array contains A131708(0) in diagonal 0, then -A024495(0..1) in diagonal 1, then A024493(0..2) in diagonal 2, then -A131708(0..3), then A024495(0..4), then -A024493(0..5).

			The table starts in row n=0 with columns k >= 0 as:
0, 0, 1, 0, 0, -1, 0, 0, 1, 0, 0, -1, 0, 0, 1, 0, 0, -1, 0, 0, 1, 0, 0 A131531
0, 1, -1, 0, -1, 1, 0, 1, -1, 0, -1, 1, 0, 1, -1, 0, -1, 1, 0, 1, -1 A092220
1, -2, 1, -1, 2, -1, 1, -2, 1, -1, 2, -1, 1, -2, 1, -1, 2, -1, 1, -2 A131556
-3, 3, -2, 3, -3, 2, -3, 3, -2, 3, -3, 2, -3, 3, -2, 3, -3, 2, -3 A164359
6, -5, 5, -6, 5, -5, 6, -5, 5, -6, 5, -5, 6, -5, 5, -6, 5, -5, 6, -5
-11, 10, -11, 11, -10, 11, -11, 10, -11, 11, -10, 11, -11, 10, -11
21, -21, 22, -21, 21, -22, 21, -21, 22, -21, 21, -22, 21, -21, 22

Cf. A167617 (antidiagonal sums).

A131531 := proc(n) op((n mod 6)+1,[0,0,1,0,0,-1]) ; end proc:
A167613 := proc(n,k) option remember; if n= 0 then A131531(k); else procname(n-1,k+1)-procname(n-1,k) ; end if;end proc: # R. J. Mathar, Dec 17 2010

nmax = 13;
A131531 = Table[{0, 0, 1, 0, 0, -1}, {nmax}] // Flatten;
T[n_] := T[n] = Differences[A131531, n];
T[n_, k_] := T[n][[k]];
Table[T[n-k, k], {n, 1, nmax}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Oct 20 2023 *)

T(0,k) = A131531(k). T(n,k) = T(n-1,k+1) - T(n-1,k), n > 0.
T(n,n) = A001045(n). T(n,n+1) = -A001045(n). T(n,n+2) = A078008(n).
T(n,0) = -T(n,3) = (-1)^(n+1)*A024495(n).
T(n,1) = (-1)^(n+1)*A131708(n).
T(n,2) = (-1)^n*A024493(n).
T(n,k+6) = T(n,k).
a(n) = A131708(0), -A024495(0,1), A024493(0,1,2), -A131708(0,1,2,3), A024495(0,1,2,3,4), -A024493(0,1,2,3,4,5).

A138635 a(n) =3a(n-3)-3a(n-6)+2*a(n-9).

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 1, 2, 1, 3, 3, 2, 6, 5, 5, 11, 10, 11, 21, 21, 22, 42, 43, 43, 85, 86, 85, 171, 171, 170, 342, 341, 341, 683, 682, 683, 1365, 1365, 1366, 2730, 2731, 2731, 5461, 5462, 5461, 10923, 10923, 10922, 21846, 21845, 21845, 43691, 43690, 43691, 87381

Offset: 0

Paul Curtz, May 14 2008

As the recurrence shows, these are three interleaved sequences which obey recurrences b(n)=3*b(n-1)-3*b(n-2)+2*b(n-3), indicating that the b(n) equal their third differences.
These three sequences are A024495, A024494 (or A131708) and A024493 (or A130781).
Their starting "vectors" b(0,1,2) are 0,0,1 and 0,1,2 and 1,1,1, respectively, therefore linearly independent, such that other sequences with the same recursion as b(n) can be written as linear combinations of these.

Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-3,0,0,2).

Cf. A024493, A024494, A024495, A131708, A130781.

a(18*n) = 21*A133853(n).

G.f.: -x^2*(1+x^2-2*x^3+x^4-x^5+x^6)/((2*x^3-1)*(x^6-x^3+1)). - R. J. Mathar, May 17 2009

Edited by R. J. Mathar, May 17 2009

A290285 Determinant of circulant matrix of order 3 with entries in the first row (-1)^j * Sum_{k>=0} binomial(n,3*k+j), j=0,1,2.

Original entry on oeis.org

1, 0, 0, 62, 666, 5292, 39754, 307062, 2456244, 19825910, 159305994, 1274445900, 10184391946, 81430393590, 651443132340, 5212260963062, 41700950994186, 333607607822412, 2668815050206474, 21350337149539062, 170802697195263924, 1366424509598012150

Offset: 0

Vladimir Shevelev, Jul 26 2017

nonn

In the Shevelev link the author proved that, for even N>=2 and every n>=1, the determinant of circulant matrix of order N with entries in the first row being (-1)^j*Sum_{k>=0} binomial(n,N*k+j), j=0..N-1, is 0. This sequence shows what happens for the first odd N>2.

Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
Wikipedia, Circulant matrix

Cf. A024493, A024495, A131708, A290286.

a:= n-> LinearAlgebra[Determinant](Matrix(3, shape=Circulant[seq(
        (-1)^j*add(binomial(n, 3*k+j), k=0..(n-j)/3), j=0..2)])):
seq(a(n), n=0..25);  # Alois P. Heinz, Jul 27 2017

ro[n_] := Table[(-1)^j Sum[Binomial[n, 3k+j], {k, 0, n/3}], {j, 0, 2}];
M[n_] := Table[RotateRight[ro[n], m], {m, 0, 2}];
a[n_] := Det[M[n]];
Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Aug 09 2018 *)

mj(j,n) = (-1)^j*sum(k=0, n\3, binomial(n, 3*k+j));
a(n) = {m = matrix(3, 3); for (j=1, 3, m[1, j] = mj(j-1,n)); for (j=2, 3, m[2, j] = m[1, j-1]); m[2, 1] = m[1, 3]; for (j=2, 3, m[3, j] = m[2, j-1]); m[3, 1] = m[2, 3]; matdet(m);} \\ Michel Marcus, Jul 26 2017

from sympy.matrices import Matrix
from sympy import binomial
def mj(j, n):
    return (-1)**j*sum(binomial(n, 3*k + j) for k in range(n//3 + 1))
def a(n):
    m=Matrix(3, 3, [0]*9)
    for j in range(3):m[0, j]=mj(j, n)
    for j in range(1, 3):m[1, j]=m[0, j - 1]
    m[1, 0]=m[0, 2]
    for j in range(1, 3):m[2, j] = m[1, j - 1]
    m[2, 0]=m[1, 2]
    return m.det()
print([a(n) for n in range(22)]) # Indranil Ghosh, Jul 31 2017

G.f.: (1-12*x+48*x^2-73*x^3+6*x^4-60*x^5+736*x^6-576*x^7)/((1+x)*(-1+2*x)*(-1+8*x)* (1-x+x^2)*(1+2*x+4*x^2)*(1-4*x+16*x^2)). - Peter J. C. Moses, Jul 26 2017

More terms from Peter J. C. Moses, Jul 26 2017

A086953 Binomial transform of (-1)^mod(n,3) (A257075).

Original entry on oeis.org

1, 0, 0, 2, 6, 12, 22, 42, 84, 170, 342, 684, 1366, 2730, 5460, 10922, 21846, 43692, 87382, 174762, 349524, 699050, 1398102, 2796204, 5592406, 11184810, 22369620, 44739242, 89478486, 178956972, 357913942, 715827882, 1431655764, 2863311530, 5726623062

Offset: 0

Paul Barry, Jul 25 2003

Index entries for linear recurrences with constant coefficients, signature (3,-3,2).

Cf. A024493, A024495, A131370, A131708, A257075.

Join[{1, a = 0, b = 0}, Table[c = 2^n - a + b; a = b; b = c, {n, 1, 100}]] (* Vladimir Joseph Stephan Orlovsky, Jun 28 2011 *)
LinearRecurrence[{3,-3,2},{1,0,0},40] (* Harvey P. Dale, Aug 02 2017 *)

a(n+3)/2 = A024495(n+2). - corrected by Vladimir Shevelev, Aug 08 2017
a(n) = 0^n + Sum{k=0..floor((n-1)/3)} C(n-1, 3*k+2).
a(n) = Sum{k=0..n} C(n, k)(-1)^mod(k, 3).
G.f.: (1 - 3*x + 3*x^2)/((1 - 2*x)*(1 - x + x^2)). - Paul Barry, Dec 14 2004
From Vladimir Shevelev, Aug 02 2017: (Start)
a(n) = A024493(n) - A131708(n) + A024495(n);
a(n) = A024495(n) if and only if n == 1 (mod 3);
a(n) = A024495(n) - 1 if and only if n == 2 or 3 (mod 6);
a(n) = A024495(n) + 1 if and only if n == 0 or 5 (mod 6);
a(3*k+1) = 2*A024495(3*k). (End)
a(n) = A131370(n+1)/2. - Rick L. Shepherd, Aug 02 2017
3*a(n) = 2^n + 2*A057079(n+2). - R. J. Mathar, Aug 04 2017

A227430 Expansion of x^2(1-x)^3/((1-2x)(1-x+x^2)(1-3*x+3x^2)).

Original entry on oeis.org

0, 0, 1, 3, 6, 10, 15, 21, 29, 45, 90, 220, 561, 1365, 3095, 6555, 13110, 25126, 46971, 87381, 164921, 320001, 640002, 1309528, 2707629, 5592405, 11450531, 23166783, 46333566, 91869970, 181348455, 357913941, 708653429, 1410132405, 2820264810, 5662052980

Offset: 0

Paul Curtz, Jul 11 2013

Consider the binomial transform of 0, 0, 0, 0, 0, 1 (period 6) with its differences:
0, 0, 0, 0, 0,  1,  6, 21, 56, 126,...    d(n): after 0, it is A192080.
0, 0, 0, 0, 1,  5, 15, 35, 70, 126,...    e(n)
0, 0, 0, 1, 4, 10, 20, 35, 56,  85,...    f(n)
0, 0, 1, 3, 6, 10, 15, 21, 29,  45,...    a(n)
0, 1, 2, 3, 4,  5,  6,  8, 16,  45,...    b(n)
1, 1, 1, 1, 1,  1,  2,  8, 29,  85,...    c(n)
0, 0, 0, 0, 0,  1,  6, 21, 56, 126,...    d(n).
a(n) + d(n) = A024495(n),
b(n) + e(n) = A131708(n),
c(n) + f(n) = A024493(n).
a(n) - d(n) =  0,  0,  1,  3,  6,  9,   9,   0,...    A057083(n-2)
b(n) - e(n) =  0,  1,  2,  3,  3,  0,  -9, -27,...    A057682(n)
c(n) - f(n) =  1,  1,  1,  0, -3, -9, -18, -27,...    A057681(n)
d(n) - a(n) =  0,  0, -1, -3, -6, -9,  -9,   0,...   -A057083(n-2)
e(n) - b(n) =  0, -1, -2, -3, -3,  0,   9,  27,...   -A057682(n)
f(n) - c(n) = -1, -1, -1,  0,  3,  9,  18,  27,...   -A057681(n).
The first column is A131531(n).
The first two trisections are multiples of 3. Is the third (1, 10, 29,...) mod 9  A029898(n)?

			a(6)=6*10-15*6+20*3-15*1+6*0=15, a(7)=90-150+120-45+6=21.

Seiichi Manyama, Table of n, a(n) for n = 0..3000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6).

Join[{0},LinearRecurrence[{6,-15,20,-15,6},{0,1,3,6,10},40]] (* Harvey P. Dale, Dec 17 2014 *)

{a(n) = sum(k=0, n\6, binomial(n, 6*k+2))} \\ Seiichi Manyama, Mar 23 2019

a(n) = 6*a(n-1) -15*a(n-2) +20*a(n-3) -15*a(n-4) +6*a(n-5) for n>5, a(0)=a(1)=0, a(2)=1, a(3)=3, a(4)=6, a(5)=10.
a(n) = A024495(n) - A192080(n-5) for n>4.
G.f.: -(x^5 - 3*x^4 + 3*x^3 - x^2)/((1-2*x)*(1-x+x^2)*(1-3*x+3*x^2)). - Ralf Stephan, Jul 13 2013
a(n) = Sum_{k=0..floor(n/6)} binomial(n,6*k+2). - Seiichi Manyama, Mar 23 2019

Definition uses the g.f. of Ralf Stephan.

More terms from Harvey P. Dale, Dec 17 2014

cp's OEIS Frontend

A111927 Expansion of x^3 / ((x-1)*(2*x-1)*(x^2-x+1)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A081374 Size of "uniform" Hamming covers of distance 1, that is, Hamming covers in which all vectors of equal weight are treated the same, included or excluded from the cover together.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

A131090 First differences of A131666.

Views

Author

Keywords

Comments

Links

Programs

Mathematica

Formula

Extensions

A167613 Array T(n,k) read by antidiagonals: the k-th term of the n-th difference of A131531.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

Formula

A138635 a(n) =3*a(n-3)-3*a(n-6)+2*a(n-9).

Views

Author

Keywords

Comments

Links

Crossrefs

Formula

Extensions

A290285 Determinant of circulant matrix of order 3 with entries in the first row (-1)^j * Sum_{k>=0} binomial(n,3*k+j), j=0,1,2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Formula

Extensions

A086953 Binomial transform of (-1)^mod(n,3) (A257075).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A227430 Expansion of x^2*(1-x)^3/((1-2*x)*(1-x+x^2)*(1-3*x+3x^2)).

Views

A111927 Expansion of x^3 / ((x-1)(2x-1)*(x^2-x+1)).

A138635 a(n) =3a(n-3)-3a(n-6)+2*a(n-9).

A227430 Expansion of x^2(1-x)^3/((1-2x)(1-x+x^2)(1-3*x+3x^2)).