cp's OEIS Frontend

Previous name: Triangular array made of three copies of Pascal's triangle.

Computing each term modulo 2 also gives A047999, i.e., a(n) mod 2 = A007318(n) mod 2 for all n. (The triangle is paritywise isomorphic to Pascal's Triangle.) - Antti Karttunen

5th row/column gives entries of A000217 (triangular numbers C(n+1,2)) repeated twice and every other entry in 6th row/column form A000217. 7th row/column gives entries of A000292 (Tetrahedral (or pyramidal) nos: C(n+3,3)) repeated twice and every other entry in 8th row/column form A000292. 9th row/column gives entries of A000332 (binomial coefficients binomial(n,4)) repeated twice and every other entry in 10th row/column form A000332. 11th row/column gives entries of A000389 (binomial coefficients C(n,5)) repeated twice and every other entry in 12th row/column form A000389. - Gerald McGarvey, Aug 21 2004

If Sum_{k=0..n} A(k)*T(n,k) = B(n), the sequence B is the S-D transform of the sequence A. - Philippe Deléham, Aug 02 2006

Number of n-bead black-white reversible strings with k black beads; also binary grids; string is palindromic. - Yosu Yurramendi, Aug 07 2008

Row sums give A016116(n+1). - Yosu Yurramendi, Aug 07 2008 [corrected by Petros Hadjicostas, Nov 04 2017]

Coefficients in expansion of (x + y)^n where x and y anticommute (y x = -x y), that is, q-binomial coefficients when q = -1. - Michael Somos, Feb 16 2009

The sequence of coefficients of a general polynomial recursion that links at w=2 to the Pascal triangle is here w=0. Row sums are {1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 32, 64, ...}. - Roger L. Bagula and Gary W. Adamson, Dec 04 2009

T(n,k) is the number of palindromic compositions of n+1 with exactly k+1 parts. T(6,4) = 3 because we have the following compositions of n+1=7 with length k+1=5: 1+1+3+1+1, 2+1+1+1+2, 1+2+1+2+1. - Geoffrey Critzer, Mar 15 2014 [corrected by Petros Hadjicostas, Nov 03 2017]

Let P(n,k) be the number of palindromic compositions of n with exactly k parts. MacMahon (1893) was the first to prove that P(n,k) = T(n-1,k-1), where T(n,k) are the numbers in this sequence (see the comment above by G. Critzer). He actually proved that, for 1 <= s <= m, we have P(2*m,2*s) = P(2*m,2*s-1) = P(2*m-1, 2*s-1) = bin(m-1, s-1), but P(2*m-1, 2*s) = 0. For the current sequence, this can be translated into T(2*m-1, 2*s-1) = T(2*m-1,2*s-2) = T(2*m-2, 2*s-2) = bin(m-1,s-1), but T(2m-2, 2*s-1) = 0 (valid again for 1 <= s <= m). - Petros Hadjicostas, Nov 03 2017

T is the infinite lower triangular matrix for this sequence; define two others, U and V; let U(n,k)=e_k(-1,2,-3,...,(-1)^n n), where e_k is the k-th elementary symmetric polynomial, and let V be the diagonal matrix A057077 (periodic sequence 1,1,-1,-1). Clearly V^-1 = V. Conjecture: U = U^-1, T = U . V, T^-1 = V . U, and |T| = |U|. - George Beck, Dec 16 2017

Let T*(n,k)=T(n,k) except when n is odd and k=(n+1)/2, where T*(n,k) = T(n,k)+2^((n-1)/2). Thus, T*(n,k) is the number of non-isomorphic symmetric stairs with n cells and k steps, i.e., k-1 changes of direction. See A016116. - Christian Barrientos and Sarah Minion, Jul 29 2018

Number of periodic palindromes of length n using a maximum of k different symbols.

From Petros Hadjicostas, Sep 02 2018: (Start)

According to Christian Bower's theory of transforms, we have boxes of different sizes and different colors. The size of a box is determined by the number of balls it can hold. In this case, we assume all balls are the same and are unlabeled. Assume also that the number of possible colors a box with m balls can have is given by c(m). We place the boxes on a circle at equal distances from each other. Two configurations of boxes on the circle are considered equivalent if one can be obtained from the other by rotation. We are interested about circular configurations of boxes that are circular palindromes (i.e., necklaces with boxes that are the same when turned over). Let b(n) be the number of circularly palindromic configurations of boxes on a circle when the total number of balls in the boxes is n (and each box contains at least one ball).

Bower calls the sequence (b(n): n >= 1), the CPAL ("circular palindrome") transform of the input sequence (c(m): m >= 1). If the g.f. of the input sequence (c(m): m >= 1) is C(x) = Sum_{m>=1} c(m)*x^m, then the g.f. of the output sequence (b(n): n >= 1) is B(x) = Sum_{n >= 1} b(n)*x^n = (1 + C(x))^2/(2*(1 - C(x^2))) - 1/2.

In the current sequence, each box holds only one ball but can have one of k colors. Hence, c(1) = k but c(m) = 0 for m >= 2. Thus, C(x) = k*x. Then, for fixed k, the output sequence is (b(n): n >= 1) = (T(n, k): n >= 1), where T(n, k) = number of necklaces with n beads and k colors that are the same when turned over. If we let B_k(x) = Sum_{n>=1} T(n, k)*x^n, then B_k(x) = (1 + k*x)^2/(2*(1 - k*x^2)) - 1/2. From this, we can easily prove the formulae below.

Note that T(n, k=2) - 1 is the total number of Sommerville symmetric cyclic compositions of n. See pp. 301-304 in his paper in the links below. To see why this is the case, we use MacMahon's method of representing a cyclic composition of n with a necklace of 2 colors (see p. 273 in Sommerville's paper where the two "colors" are an x and a dot . rather than B and W). Given a Sommerville symmetrical composition b_1 + ... + b_r of n (with b_i >= 1 for all i and 1 <= r <= n), create the following circularly palindromic necklace with n beads of 2 colors: Start with a B bead somewhere on the circle and place b_1 - 1 W beads to the right of it; place a B bead to the right of the W beads (if any) followed by b_2 - 1 W beads; and so on. At the end, place a B bead followed with b_r - 1 W beads. (If b_i = 1 for some i, then a B bead follows a B bead since there are 0 W beads between them.) We thus get a circularly palindromic necklace with n beads of two colors. (The only necklace we cannot get with this method is the one than has all n beads colored W.)

It is interesting that the representation of a necklace of length n, say s_1, s_2, ..., s_n, as a periodic sequence (..., s_{-2}, s_{-1}, s_0, s_1, s_2, ...) with the property s_i = s_{i+n} for all i, as was done by Marks R. Nester in Chapter 2 of his 1999 PhD thesis, was considered by Sommerville in his 1909 paper (in the very first paragraph of his paper). (End)

Binomial transform of A056487. Unsigned version of A152174.

Number of words of length n over the alphabet {1,2,3,4} such that no odd letter is followed by an odd letter. - Armend Shabani, Feb 18 2017

From Sean A. Irvine, Jun 06 2025: (Start)

Also, the number of walks of length n starting at 0 in the following graph:

1---2

|\ /|

| 0 |

|/ \|

4---3. (End)

From Petros Hadjicostas, Sep 01 2018: (Start)

The DIK transform of the sequence (c(n): n >= 1), with g.f. C(x) = Sum_{n >= 1} c(n)*x^n, has g.f. -(1/2)*Sum_{m >= 1} (phi(m)/m))*log(1-C(x^m)) + (1 + C(x))^2/(4*(1-C(x^2))) - 1/4.

Here, c(1) = 5 and c(n) = 0 for n >= 2, and thus, C(x) = 5*x. Substituting this to the above g.f., we get that the g.f. of the current sequence is A(x) = Sum_{n >= 1} a(n)*x^n = -(1/2)*Sum_{m >= 1} (phi(m)/m))*log(1-5*x^m) + (1 + 5*x)^2/(4*(1-5*x^2)) - 1/4. This agrees with Herbert Kociemba's g.f. below except for an extra 1 because (1 + (1+5*x+10*x^2)/(1-5*x^2))/2 = 1 + (1 + 5*x)^2/(4*(1-5*x^2)) - 1/4.

(End)

Binomial transform is A163062, second binomial transform is A163063, third binomial transform is A098648 without initial 1, fourth binomial transform is A163064, fifth binomial transform is A163065.

Apparently identical to A056487! Is this a theorem? - Klaus Brockhaus, Jul 21 2009

Number of chiral bracelets of n beads using up to five different colors.

a(n) ~ c*(sqrt(5)+1)^n, where c = (sqrt(5)+3)/10.

The inverse binomial transform is 1,1,3,5,... (1 followed by A056487). Partial sum of 1,1,4,12,..., i.e., 1 plus n-th partial sum of A087206. [R. J. Mathar, Oct 04 2010]

From R. J. Mathar, Oct 12 2010: (Start)

Apparently the row n=4 of an array which counts walks with k steps on an n X n board, starting at a corner, each step to one of the <= 4 adjacent squares:

1,2,4,8,16,32,64,128,256,512,1024,2048,4096,

1,2,6,16,48,128,384,1024,3072,8192,24576,65536,196608,

1,2,6,18,58,186,602,1946,6298,20378,65946,213402,690586,

1,2,6,18,60,198,684,2322,8100,27702,96876,331938,1161540,

1,2,6,18,60,200,698,2432,8658,30762,110374,395428,1422916,

1,2,6,18,60,200,700,2448,8800,31552,115104,418176,1537536,

1,2,6,18,60,200,700,2450,8818,31730,116182,425172,1573416,

1,2,6,18,60,200,700,2450,8820,31750,116400,426600,1583400,

1,2,6,18,60,200,700,2450,8820,31752,116422,426862,1585246,

1,2,6,18,60,200,700,2450,8820,31752,116424,426886,1585556,

1,2,6,18,60,200,700,2450,8820,31752,116424,426888,1585582,

(End)

Decomposing rook walks of length=n on a 4 X 4 board into combinations of independent vertical and horizontal walks in 4-wide corridors leads to an exponential convolution of the Fibonacci numbers, cf. A052899. [David Scambler, Oct 17 2010]

Number of aperiodic necklaces with five colors that are the same when turned over and hence have reflectional symmetry but no rotational symmetry. - Herbert Kociemba, Nov 29 2016