A120987 -id:A120987

Search: a120987 -id:a120987

Displaying 1-3 of 3 results found.

page 1

A099464

Trisection of tribonacci numbers.

+10
4

0, 1, 7, 44, 274, 1705, 10609, 66012, 410744, 2555757, 15902591, 98950096, 615693474, 3831006429, 23837527729, 148323355432, 922906855808, 5742568741225, 35731770264967, 222332455004452, 1383410902447554, 8607945812375585, 53560898629395777, 333269972246340068

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,3

COMMENTS

Let A = [1,1,1; 2,4,3; 1,2,2]. a(n) is given by the (1,2) term in A^n.

LINKS

Table of n, a(n) for n=0..23.

Index entries for linear recurrences with constant coefficients, signature (7,-5,1).

FORMULA

G.f.: x/(1-7*x+5*x^2-x^3).

a(n) = 7a(n-1) -5a(n-2) +a(n-3).

a(n) = A000073(3n).

a(n) = Sum_{i>=n-1} A120987(i,n-1) for n>0. - Alois P. Heinz, Dec 11 2015

MAPLE

a:= n-> (<<0|1|0>, <0|0|1>, <1|-5|7>>^n)[3, 1]:

seq(a(n), n=0..30); # Alois P. Heinz, Dec 11 2015

MATHEMATICA

LinearRecurrence[{7, -5, 1}, {0, 1, 7}, 30] (* Harvey P. Dale, Jan 14 2016 *)

CROSSREFS

Cf. A009943, A120987.

KEYWORD

easy,nonn

AUTHOR

Paul Barry, Oct 16 2004

STATUS

approved

A157241

Expansion of x / ((1-x)*(4*x^2-2*x+1)).

+10
4

0, 1, 3, 3, -5, -21, -21, 43, 171, 171, -341, -1365, -1365, 2731, 10923, 10923, -21845, -87381, -87381, 174763, 699051, 699051, -1398101, -5592405, -5592405, 11184811, 44739243, 44739243, -89478485, -357913941

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,3

COMMENTS

Generating floretion is Y = .5('i + 'j + 'k + i' + j' + k') + ee. ("ibasek"). This is the same floretion which generates A157240.

LINKS

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (3,-6,4).

FORMULA

a(n+1) - a(n) = A088138(n+1).

a(n+1) = Sum_{k=0..n} A120987(n,k)*(-1)^(n-k). - Philippe Deléham, Oct 25 2011

G.f.: 2*x-2*x/(G(0) + 1) where G(k)= 1 + 2*(2*k+3)*x/(2*k+1 - 2*x*(k+2)*(2*k+1)/(2*x*(k+2) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2012

a(n) = 1/9*(3 + (-1)^floor((n-2)/3)*2^(4+3*floor((n-2)/3)) + (-1)^floor((n-1)/3)*2^(3+3*floor((n-1)/3))). - John M. Campbell, Dec 23 2016

From Colin Barker, May 22 2019: (Start)

a(n) = (2 - (1+i*sqrt(3))^(1+n) + i*(1-i*sqrt(3))^n*(i+sqrt(3))) / 6 where i=sqrt(-1).

a(n) = 3*a(n-1) - 6*a(n-2) + 4*a(n-3) for n>2.

(End)

MATHEMATICA

CoefficientList[Series[x/((1-x)(4x^2-2x+1)), {x, 0, 40}], x] (* or *) LinearRecurrence[{3, -6, 4}, {0, 1, 3}, 40] (* Harvey P. Dale, Oct 27 2013 *)

Table[1/9 (3 + (-1)^Floor[1/3 (-2 + n)] 2^(4 + 3 Floor[1/3 (-2 + n)]) + (-1)^Floor[1/3 (-1 + n)] 2^(3 + 3 Floor[1/3 (-1 + n)])), {n, 0, 500}] (* John M. Campbell, Dec 23 2016 *)

PROG

(PARI) concat(0, Vec(x / ((1 - x)*(1 - 2*x + 4*x^2)) + O(x^40))) \\ Colin Barker, May 22 2019

CROSSREFS

Cf. A128018, A157240.

KEYWORD

easy,sign

AUTHOR

Creighton Dement, Feb 25 2009

STATUS

approved

A265644

Triangle read by rows: T(n,m) is the number of quaternary words of length n with m strictly increasing runs (0 <= m <= n).

+10
1

1, 0, 4, 0, 6, 10, 0, 4, 40, 20, 0, 1, 65, 155, 35, 0, 0, 56, 456, 456, 56, 0, 0, 28, 728, 2128, 1128, 84, 0, 0, 8, 728, 5328, 7728, 2472, 120, 0, 0, 1, 486, 8451, 27876, 23607, 4950, 165

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

0,3

COMMENTS

In the following description the alphabet {0..r} is taken as a basis, with r = 3 in this case.

For example, the quaternary word 2|03|123|3 of length n=7, has m=4 strictly increasing runs.

The empty word has n = 0 and m = 0, and T(0, 0) = 1.

T(n, 0) = 0 for n >= 1.

T(n, m) <> 0 for m <= n <= m*(r+1). T(m*(r+1), m) = 1.

T(n,m) is a partition, based on m, of all the words of length n, so Sum_{k=0..n} T(n,k) = (r+1)^n.

REFERENCES

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, p. 24, p. 154.

LINKS

Table of n, a(n) for n=0..44.

MathPages, Balls In Bins With Limited Capacity

FORMULA

Refer to comment to A120987 concerning formulas for general values of r and considerations.

Therefrom we get

T(n, m) = Qsc(3, n, m) =

Nb(4*m-n, 3, n+1) = Nb(4*(n-m)+3, 3, n+1) =

Sum_{j=0..n+1} (-1)^j*Cb(n+1, j)*Cb(4*(m-j), 4*(m-j)-n) =

Sum_{j=0..m} (-1)^(m-j)*Cb(n+1, m-j)*Cb(4*j, n) =

(in this last version Cb(n,m) can be replaced by binomial(n,m))

Sum_{j=0..m} (-1)^(m-j)*binomial(n+1, m-j)*binomial(4*j, n) = [z^n, t^m](1-t)/(1-t(1+(1-t)z)^4) where [x^n]F(x) denotes the coefficient of x^n in the formal power series expansion of F(x),

Nb(s,r,n) denotes the (r+1)-nomial coefficient [x^s](1+x+..+x^r)^n,(Nb(s,3,n) = A008287(n,s)).

Cb(x,m) denotes the binomial coefficient in its extended falling factorial notation (Cb(x,m)= x^_m/m! iff m is a nonnegative integer, 0 otherwise), as defined in the Graham et al. reference.

The diagonal T(n, n) = Nb(3, 3, n+1) = Sum_{j=0..n} (-1)^(n-j)*Cb(n+1, n-j)*Cb(4*j, n) = Cb(n+3, 3) = binomial(n+3, 3) = A000292(n+1).

EXAMPLE

Triangle starts:

0, 4;

0, 6, 10;

0, 4, 40, 20;

0, 1, 65, 155, 35;

0, 0, 56, 456, 456, 56;

T(3,2) = 40, which accounts for the following words:

[0 <= a <= 0, 1 | 0 <= b <= 1] = 2

[0 <= a <= 1, 2 | 0 <= b <= 2] = 6

[0 <= a <= 2, 3 | 0 <= b <= 3] = 12

[0 <= a <= 3 | 0, 1 <= b <= 3] = 12

[1 <= a <= 3 | 1, 2 <= b <= 3] = 6

[2 <= a <= 3 | 2, 3 <= b <= 3] = 2

PROG

(MuPAD) T:=(n, m)->_plus((-1)^(m-j)*binomial(n+1, m-j)*binomial(4*j, n)$j=0..m):

(PARI) T(n, k) = sum(j=0, k, (-1)^(k-j)*binomial(n+1, k-j)*binomial(4*j, n));

tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print()); \\ Michel Marcus, Feb 09 2016

CROSSREFS

Cf. A119900 (r=1, binary words), A120987 (r=2, ternary words), A008287 (quadrinomial coefficients).

Row sums give A000302.

Cf. A000292.

KEYWORD

tabl,nonn

AUTHOR

Giuliano Cabrele, Dec 13 2015

STATUS

approved

page 1

Search completed in 0.005 seconds