cp's OEIS Frontend

Coordination sequence for infinite tree with valency 3.

Number of Hamiltonian cycles in K_3 X P_n.

Number of ternary words of length n avoiding aa, bb, cc.

For n > 0, row sums of A029635. - Paul Barry, Jan 30 2005

Binomial transform is {1, 4, 13, 40, 121, 364, ...}, see A003462. - Philippe Deléham, Jul 23 2005

Convolved with the Jacobsthal sequence A001045 = A001786: (1, 4, 12, 32, 80, ...). - Gary W. Adamson, May 23 2009

Equals (n+1)-th row sums of triangle A161175. - Gary W. Adamson, Jun 05 2009

a(n) written in base 2: a(0) = 1, a(n) for n >= 1: 11, 110, 11000, 110000, ..., i.e.: 2 times 1, (n-1) times 0 (see A003953(n)). - Jaroslav Krizek, Aug 17 2009

INVERTi transform of A003688. - Gary W. Adamson, Aug 05 2010

An elephant sequence, see A175655. For the central square four A[5] vectors, with decimal values 42, 138, 162 and 168, lead to this sequence. For the corner squares these vectors lead to the companion sequence A083329. - Johannes W. Meijer, Aug 15 2010

A216022(a(n)) != 2 and A216059(a(n)) != 3. - Reinhard Zumkeller, Sep 01 2012

Number of length-n strings of 3 letters with no two adjacent letters identical. The general case (strings of r letters) is the sequence with g.f. (1+x)/(1-(r-1)*x). - Joerg Arndt, Oct 11 2012

Sums of pairs of rows of Pascal's triangle A007318, T(2n,k)+T(2n+1,k); Sum_{n>=1} A000290(n)/a(n) = 4. - John Molokach, Sep 26 2013

From David W. Wilson: Numbers triangular, differences square.

To be precise, the differences are the squares of the powers of three with positive indices. Hence a(n+1) - a(n) = (A000244(n+1))^2 = A001019(n+1). [Added by Ant King, Jan 05 2011]

Partial sums of A001019. This is m-th triangular number, where m is partial sums of A000244. a(n) = A000217(A003462(n)). - Lekraj Beedassy, May 25 2004

With offset 0, binomial transform of A003951. - Philippe Deléham, Jul 22 2005

Numbers in base 9: 1, 11, 111, 1111, 11111, 111111, 1111111, etc. - Zerinvary Lajos, Apr 26 2009

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=9, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=10, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 2, a(n-1) = (-1)^n*charpoly(A,1). - Milan Janjic, Feb 21 2010

From Hieronymus Fischer, Jan 30 2013: (Start)

Least index k such that A052382(k) >= 10^(n-1), for n > 0.

Also index k such that A052382(k) = (10^n-1)/9, n > 0.

A052382(a(n)) is the least zerofree number with n digits, for n > 0.

For n > 1: A052382(a(n)-1) is the greatest zerofree number with n-1 digits. (End)

For n > 0, 4*a(n) is the total number of holes in a certain triangle fractal (start with 9 triangles, 4 holes) after n iterations. See illustration in links. - Kival Ngaokrajang, Feb 21 2015

For n > 0, a(n) is the sum of the numerators and denominators of the reduced fractions 0 < (b/3^(n-1)) < 1 plus 1. Example for n=3 gives fractions 1/9, 2/9, 1/3, 4/9, 5/9, 2/3, 7/9, and 8/9 plus 1 has sum of numerators and denominators +1 = a(3) = 91. - J. M. Bergot, Jul 11 2015

Except for 0 and 1, all terms are Brazilian repunits numbers in base 9, so belong to A125134. All these terms are composite because a(n) is the ((3^n - 1)/2)-th triangular number. - Bernard Schott, Apr 23 2017

These are also the second steps after the junctions of the Collatz trajectories of 2^(2k-1)-1 and 2^2k-1. - David Rabahy, Nov 01 2017

Coordination sequence for infinite tree with valency 8.

The n-th term of the coordination sequence of the infinite tree with valency 2m is the same as the number of reduced words of size n in the free group on m generators. In the five sequences A003946, A003948, A003950, A003952, A003954 m is 2, 3, 4, 5, 6 . - Avi Peretz (njk(AT)netvision.net.il), Feb 23 2001 and Ola Veshta (olaveshta(AT)my-deja.com), Mar 30 2001.

For n>=1, a(n) equals the number of words of length n on the alphabet {0,1,...,7} with no two adjacent letters identical. - Milan Janjic, Jan 31 2015 [Corrected by David Nacin, May 31 2017]

a(n) is the number of octonary sequences of length n such that no two consecutive terms have distance 4. - David Nacin, May 31 2017

Coordination sequence for infinite tree with valency 7.

For n >= 1, a(n+1) is equal to the number of functions f:{1,2,...,n+1}->{1,2,3,4,5,6,7} such that for fixed, different x_1, x_2,...,x_n in {1,2,...,n+1} and fixed y_1, y_2,...,y_n in {1,2,3,4,5,6,7} we have f(x_i)<>y_i, (i=1,2,...,n). - Milan Janjic, May 10 2007

For n >= 1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,2,3,5,6} with no two adjacent letters identical. - Milan Janjic, Jan 31 2015

Coordination sequence for infinite tree with valency 10.

The n-th term of the coordination sequence of the infinite tree with valency 2m is the same as the number of reduced words of size n in the free group on m generators. In the five sequences A003946, A003948, A003950, A003952, A003954 m is 2, 3, 4, 5, 6 . - Avi Peretz (njk(AT)netvision.net.il), Feb 23 2001 and Ola Veshta (olaveshta(AT)my-deja.com), Mar 30 2001

Except 1, all terms are in A033583. - Vincenzo Librandi, May 26 2014

For n>=1, a(n) equals the number of words of length n on alphabet {0,1,...,9} with no two adjacent letters identical. - Milan Janjic, Jan 31 2015 [Corrected by David Nacin, May 31 2017]

a(n) is the number of sequences over the alphabet {0,1,...,9} of length n such that no two consecutive terms have distance 5. - David Nacin, May 31 2017

The initial terms coincide with those of A003951, although the two sequences are eventually different.

Computed with MAGMA using commands similar to those used to compute A154638.

Consider the k-fold Cartesian products CP(n,k) of the vector A(n) = [1, 2, 3, ..., n].

An element of CP(n,k) is a n-tuple T_t of the form T_t = [i_1, i_2, i_3, ..., i_k] with t=1, .., n^k.

We count members T of CP(n,k) which satisfy some condition delta(T_t), so delta(.) is an indicator function which attains values of 1 or 0 depending on whether T_t is to be counted or not; the summation sum_{CP(n,k)} delta(T_t) over all elements T_t of CP produces the count.

For the triangle here we have delta(T_t) = 0 if for any two i_j, i_(j+1) in T_t one has i_j = i_(j+1): T(n,k) = Sum_{CP(n,k)} delta(T_t) = Sum_{CP(n,k)} delta(i_j = i_(j+1)).

The test on i_j > i_(j+1) generates A158498. One gets the Pascal triangle A007318 if the indicator function tests whether for any two i_j, i_(j+1) in T_t one has i_j >= i_(j+1).

Use of other indicator functions can also calculate the Bell numbers A000110, A000045 or A000108.

The n-th row is {T(n,0),T(n,1),...,T(n,n-1)}.

Let m denote the prime power p^e.

T(m,0) = A020478(m) = (p^(e+1) + p^e-1)*p^(2*e-1).

T(m,1) = A000056(m) = (p^2-1)*p^(3*e-2).

T(prime(n),1) = A127917(n).

Sum_{k=1..n-1} T(n,k) = A005353(n).

T(n,1) = n*A007434(n) for n>=1 because A000056(n) = n*Jordan_Function_J_2(n).

T(2^n,1) = A083233(n) = A164640(2n) for n>=1. Proof: a(n):=T(2^n,1); a(1)=6, a(n)=8*a(n-1); A083233(1)=6 and A083233(n) is a geometric series with ratio 8 (because of its g.f.), too; A164640 = {b(1)=1, b(2)=6, b(n)=8*b(n-2)}.

T(2^n,0) = A165148(n) for n>=0, because 2*T(2^n,0) = (3*2^n-1)*4^n.

T(2^e,2) = A003951(e) for 2 <= e. Proof: T(2^e,2) = 9*8^(e-1) is a series with ratio 8 and initial term 72, as A003951(2...inf) is.

Working with consecutive powers of a prime p, we need a definition (0 <= i < e):

N(p^e,i):=#{k: 0 < k < p^e, gcd(k,p^e) = p^i} = (p-1)*p^(e-1-i). We say that these k's belong to i (respect to p^e). Note that N(p^e,0) = EulerPhi(p^e), and if 0 < k < p^e then gcd(k,p^e) = gcd(k,p^(e+1)). Let T(p^e,[i]) denote the common value of T(p^e,k)'s, where k's belong to i (q.v.PROGRAM); for example, T(p^e,[0]) = T(p^e,1). The number of the 2 X 2 matrices over Z(p^e), T(p^e,0) + Sum_{i=0..e-1} T(p^e,[i])*N(p^e,i) = p^(4e) will be useful.

On the hexagon property: Let prime p be given and let T(p^e,[0]), T(p^e,[1]), T(p^e,[2]), ..., T(p^e,[e-2]), T(p^e,[e-1]) form the e-th row of a Pascal-like triangle, e>=1. Let denote X(r,s) an element of the triangle and its value T(p^r,[s]). Let positive integers a and b given, so that the entries A(m-a,n-b), B(m-a,n), C(m,n+a), D(m+b,n+a), E(m+b,n), F(m,n-b) of the triangle form a hexagon spaced around T(p^m,[n]); if a=b=1 then they surround it. If A*C*E = B*D*F, then we say that the triangle T(.,.) has the "hexagon property". (In the case of binomial coefficients X(r,s) = COMB(r,s), the "hexagon property" holds (see [Gupta]) and moreover gcd(A,C,E) = gcd(B,D,F) (see [Hitotumatu & Sato]).)

Corollary 2.2 in [Brent & McKay] says that, for the d X d matrices over Z(p^e), (mutatis mutandis) T_d(p^e,0) = K*(1-P(d+e-1)/P(e-1)) and T_d(p^e,[i]) = K*(q^e)*((1-q^d)/(1-q))*P(d+i-1)/P(i), where q=1/p, K=(p^e)^(d^2), P(t) = Product_{j=1..t} (1-q^j), P(0):=1. (For the case d=2, we have T(p^e,[i]) = (p+1)*(p^(i+1)-1)*p^(3*e-i-2).) Due to [Brent & McKay], it can be simply proved that for d X d matrices the "hexagon property" is true. The formulation implies an obvious generalization: For the entries A(r,u), B(r,v), C(s,w), D(t,w), E(t,v), F(s,u) of the T_d(.,.)-triangle, a hexagon-like property A*C*E = B*D*F holds. This is false in general for the COMB(.,.)-triangle.

Another (rotated-hexagon-like) property: for the entries A(m-b1,n), B(m-a1,n+c2), C(m+a2,n+c2), D(m+b2,n), E(m+a2,n-c1), F(m-a1,n-c1) of the T_d(.,.)-triangle, the property A*C*E = B*D*F holds, if and only if 2*(a1 + a2) = b1 + b2. This is also in general false for COMB(.,.)-triangle.

The initial terms coincide with those of A003951, although the two sequences are eventually different.

Computed with MAGMA using commands similar to those used to compute A154638.