cp's OEIS Frontend

Essentially the same as A078039, A141015, and A141683.

Named after a 14th-century Indian mathematician. [The sequence first appeared in the book "Ganita Kaumudi" (1356) by the Indian mathematician Narayana Pandita (c. 1340 - c. 1400). - Amiram Eldar, Apr 15 2021]

Number of compositions of n into parts 1 and 3. - Joerg Arndt, Jun 25 2011

A Lamé sequence of higher order.

Could have begun 1,0,0,1,1,1,2,3,4,6,9,... (A078012) but that would spoil many nice properties.

Number of tilings of a 3 X n rectangle with straight trominoes.

Number of ways to arrange n-1 tatami mats in a 2 X (n-1) room such that no 4 meet at a point. For example, there are 6 ways to cover a 2 X 5 room, described by 11111, 2111, 1211, 1121, 1112, 212.

Equivalently, number of compositions (ordered partitions) of n-1 into parts 1 and 2 with no two 2's adjacent. E.g., there are 6 such ways to partition 5, namely 11111, 2111, 1211, 1121, 1112, 212, so a(6) = 6. [Minor edit by Keyang Li, Oct 10 2020]

This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n-1) + a(n-m), with a(n) = 1 for n = 0...m-1. The generating function is 1/(1-x-x^m). Also a(n) = Sum_{i=0..floor(n/m)} binomial(n-(m-1)*i, i). This family of binomial summations or recurrences gives the number of ways to cover (without overlapping) a linear lattice of n sites with molecules that are m sites wide. Special case: m=1: A000079; m=4: A003269; m=5: A003520; m=6: A005708; m=7: A005709; m=8: A005710.

a(n+2) is the number of n-bit 0-1 sequences that avoid both 00 and 010. - David Callan, Mar 25 2004 [This can easily be proved by the Cluster Method - see for example the Noonan-Zeilberger article. - N. J. A. Sloane, Aug 29 2013]

a(n-4) is the number of n-bit sequences that start and end with 0 but avoid both 00 and 010. For n >= 6, such a sequence necessarily starts 011 and ends 110; deleting these 6 bits is a bijection to the preceding item. - David Callan, Mar 25 2004

Also number of compositions of n+1 into parts congruent to 1 mod m. Here m=3, A003269 for m=4, etc. - Vladeta Jovovic, Feb 09 2005

Row sums of Riordan array (1/(1-x^3), x/(1-x^3)). - Paul Barry, Feb 25 2005

Row sums of Riordan array (1,x(1+x^2)). - Paul Barry, Jan 12 2006

Starting with offset 1 = row sums of triangle A145580. - Gary W. Adamson, Oct 13 2008

Number of digits in A061582. - Dmitry Kamenetsky, Jan 17 2009

From Jon Perry, Nov 15 2010: (Start)

The family a(n) = a(n-1) + a(n-m) with a(n)=1 for n=0..m-1 can be generated by considering the sums (A102547):

1 1 1 1 1 1 1 1 1 1 1 1 1

1 2 3 4 5 6 7 8 9 10

1 3 6 10 15 21 28

1 4 10 20

1

------------------------------

1 1 1 2 3 4 6 9 13 19 28 41 60

with (in this case 3) leading zeros added to each row.

(End)

Number of pairs of rabbits existing at period n generated by 1 pair. All pairs become fertile after 3 periods and generate thereafter a new pair at all following periods. - Carmine Suriano, Mar 20 2011

The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=3, 2*a(n-3) equals the number of 2-colored compositions of n with all parts >= 3, such that no adjacent parts have the same color. - Milan Janjic, Nov 27 2011

For n>=2, row sums of Pascal's triangle (A007318) with triplicated diagonals. - Vladimir Shevelev, Apr 12 2012

Pisano period lengths of the sequence read mod m, m >= 1: 1, 7, 8, 14, 31, 56, 57, 28, 24, 217, 60, 56, 168, ... (A271953) If m=3, for example, the remainder sequence becomes 1, 1, 1, 2, 0, 1, 0, 0, 1, 1, 1, 2, 0, 1, 0, 0, 1, 1, 1, 2, 0, 1, 0, 0, 1, 1, 1, 2, 0, 1, 0, 0, 1, 1, ... with a period of length 8. - R. J. Mathar, Oct 18 2012

Diagonal sums of triangle A011973. - John Molokach, Jul 06 2013

"In how many ways can a kangaroo jump through all points of the integer interval [1,n+1] starting at 1 and ending at n+1, while making hops that are restricted to {-1,1,2}? (The OGF is the rational function 1/(1 - z - z^3) corresponding to A000930.)" [Flajolet and Sedgewick, p. 373] - N. J. A. Sloane, Aug 29 2013

a(n) is the number of length n binary words in which the length of every maximal run of consecutive 0's is a multiple of 3. a(5) = 4 because we have: 00011, 10001, 11000, 11111. - Geoffrey Critzer, Jan 07 2014

a(n) is the top left entry of the n-th power of the 3X3 matrix [1, 0, 1; 1, 0, 0; 0, 1, 0] or of the 3 X 3 matrix [1, 1, 0; 0, 0, 1; 1, 0, 0]. - R. J. Mathar, Feb 03 2014

a(n-3) is the top left entry of the n-th power of any of the 3 X 3 matrices [0, 1, 0; 0, 1, 1; 1, 0, 0], [0, 0, 1; 1, 1, 0; 0, 1, 0], [0, 1, 0; 0, 0, 1; 1, 0, 1] or [0, 0, 1; 1, 0, 0; 0, 1, 1]. - R. J. Mathar, Feb 03 2014

Counts closed walks of length (n+3) on a unidirectional triangle, containing a loop at one of remaining vertices. - David Neil McGrath, Sep 15 2014

a(n+2) equals the number of binary words of length n, having at least two zeros between every two successive ones. - Milan Janjic, Feb 07 2015

a(n+1)/a(n) tends to x = 1.465571... (decimal expansion given in A092526) in the limit n -> infinity. This is the real solution of x^3 - x^2 -1 = 0. See also the formula by Benoit Cloitre, Nov 30 2002. - Wolfdieter Lang, Apr 24 2015

a(n+2) equals the number of subsets of {1,2,..,n} in which any two elements differ by at least 3. - Robert FERREOL, Feb 17 2016

Let T* be the infinite tree with root 0 generated by these rules: if p is in T*, then p+1 is in T* and x*p is in T*. Let g(n) be the set of nodes in the n-th generation, so that g(0) = {0}, g(1) = {1}, g(2) = {2,x}, g(3) = {3,2x,x+1,x^2}, etc. Let T(r) be the tree obtained by substituting r for x. If a positive integer N such that r = N^(1/3) is not an integer, then the number of (not necessarily distinct) integers in g(n) is A000930(n), for n >= 1. (See A274142.) - Clark Kimberling, Jun 13 2016

a(n-3) is the number of compositions of n excluding 1 and 2, n >= 3. - Gregory L. Simay, Jul 12 2016

Antidiagonal sums of array A277627. - Paul Curtz, May 16 2019

a(n+1) is the number of multus bitstrings of length n with no runs of 3 ones. - Steven Finch, Mar 25 2020

Suppose we have a(n) samples, exactly one of which is positive. Assume the cost for testing a mix of k samples is 3 if one of the samples is positive (but you will not know which sample was positive if you test more than 1) and 1 if none of the samples is positive. Then the cheapest strategy for finding the positive sample is to have a(n-3) undergo the first test and then continue with testing either a(n-4) if none were positive or with a(n-6) otherwise. The total cost of the tests will be n. - Ruediger Jehn, Dec 24 2020

From Petros Hadjicostas, Jun 12 2019: (Start)

This is a mirror image of the triangular array A140997. The current array has index of asymmetry s = 2 and index of obliqueness (obliquity) e = 1. Array A140997 has the same index of asymmetry, but has index of obliqueness e = 0. (In other related sequences, the author uses the letter y for the index of asymmetry and the letter z for the index of obliqueness, but on the stone slab that appears over a tomb in a picture that he posted in those sequences, the letters s and e are used instead. See, for example, the documentation for sequences A140998, A141065, A141066, and A141067.)

In general, if the index of asymmetry (from the Pascal triangle A007318) is s, then the order of the recurrence is s + 2 (because the recurrence of the Pascal triangle has order 2). There are also s + 2 infinite sets of initial conditions (as opposed to the Pascal triangle, which has only 2 infinite sets of initial conditions, namely, G(n, 0) = G(n+1, n+1) = 1 for n >= 0).

Pascal's triangle A007318 has s = 0 and is symmetric, arrays A140998 and A140993 have s = 1 (with e = 0 and e = 1, respectively), and arrays A140996 and A140995 have s = 3 (with e = 0 and e = 1, respectively).

If A(x,y) = Sum_{n,k >= 0} G(n, k)*x^n*y^k is the bivariate g.f. for this array (with G(n, k) = 0 for 0 <= n < k) and B(x, y) = Sum_{n, k} A140997(n, k)*x^n*y^k, then A(x, y) = B(x*y, y^(-1)). This can be proved using formal manipulation of double series expansions and the fact G(n, k) = A140997(n, n-k) for 0 <= k <= n.

If we let b(k) = lim_{n -> infinity} G(n, k) for k >= 0, then b(0) = 1, b(1) = 2, b(2) = 4, and b(k) = b(k-1) + 2*b(k-2) + b(k-3) for k >= 3. (The existence of the limit can be proved by induction on k.) It follows that b(k) = A141015(k) for k >= 0.

(End)

From Petros Hadjicostas, Jun 12 2019: (Start)

This is a mirror image of the triangular array A140994. The current array has index of asymmetry s = 2 and index of obliqueness (obliquity) e = 0. Array A140994 has the same index of asymmetry, but has index of obliqueness e = 1. (In other related sequences, the author uses the letter y for the index of asymmetry and the letter z for the index of obliqueness, but on the stone slab that appears over a tomb in a picture that he posted in those sequences, the letters s and e are used instead. See, for example, the documentation for sequences A140998, A141065, A141066, and A141067.)

In general, if the index of asymmetry (from the Pascal triangle A007318) is s, then the order of the recurrence is s + 2 (because the recurrence of the Pascal triangle has order 2). There are also s + 2 infinite sets of initial conditions (as opposed to the Pascal triangle, which has only 2 infinite sets of initial conditions, namely, G(n, 0) = G(n+1, n+1) = 1 for n >= 0).

Pascal's triangle A007318 has s = 0 and is symmetric, arrays A140998 and A140993 have s = 1 (with e = 0 and e = 1, respectively), and arrays A140996 and A140995 have s = 3 (with e = 0 and e = 1, respectively).

(End)

Row sums are A028859. Diagonal sums are A141015(n+1). Inverse is A154930. Product of A030528 and A007318.

Transforms sequence m^n with g.f. 1/(1-m*x) to the sequence with g.f. (1+x)/(1-(m+1)x-(m+1)x^2).

Subtriangle of triangle T(n,k), given by (0, 2, -1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. This triangle is the Riordan array (1, x(1+x)/(1-x-x^2)). - Philippe Deléham, Jan 25 2012

Essentially the same as A141015. - R. J. Mathar, Sep 14 2008

In the attached photograph, we see that the index of asymmetry is denoted by s and the index of obliqueness by e. The general recurrence is G(n+s+2, k) = G(n+1, k-e*s+e-1) + Sum_{1 <= m <= s+1} G(n+m, k-e*s+m*e-2*e) for n >= 0 with k = 1..(n+1) when e = 0 and k = (s+1)..(n+s+1) when e = 1. The initial conditions are G(n+x+1, n-e*n+e*x-e+1) = 2^x for x=0..s and n >= 0. There is one more initial condition, namely, G(n, e*n) = 1 for n >= 0.

For s = 0, we get Pascal's triangle A007318. For s = 1, we get A140998 (e = 0) and A140993 (e = 1). For s = 2, we get A140997 (e = 0) and A140994 (e = 1). For s = 3, we get A140996 (e = 0) and A140995 (e = 1). For s = 4, we have arrays A141020 (with e = 0) and A141021 (with e = 1). In some of these arrays, the indices n and k are shifted.

For the triangular array G(n, k) = A141021(n, k), we have G(n+6, k) = G(n+1, k-4) + G(n+1, k-5) + G(n+2, k-4) + G(n+3, k-3) + G(n+4, k-2) + G(n+5, k-1) for n >= 0 and k = 5..(n+5) with G(n+x+1, x) = 2^x for x = 0..4 and n >= 0.

With G(n, k) = A141021(n, k), the current sequence (a(k): k >= 0) is defined by a(k) = lim_{n -> infinity} G(n, k) for k >= 0. Then a(k) = a(k-5) + 2*a(k-4) + a(k-3) + a(k-2) + a(k-1) for k >= 5 with a(x) = 2^x for x = 0..4.