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Fractal - deleting the first occurrence of each integer leaves the original sequence. Also, deleting all the 1's leaves the original sequence plus 1. New values occur at square indices. 1's occur at indices m^2+1 and m^2+m+1. Ordinal transform of A122196.

Except for its initial 1, A122197 is the natural fractal sequence of A002620; that is, A122197(n+1) is the number of the row of A194061 that contains n. See A194029 for definition of natural fractal sequence. - Clark Kimberling, Aug 12 2011

From Johannes W. Meijer, Sep 09 2013: (Start)

Triangle read by rows formed from antidiagonals of triangle A002260.

The row sums equal A008805(n-1) and the antidiagonal sums equal A211534(n+5). (End)

From Vladimir Shevelev, Apr 23 2011: (Start)

Also number of non-equivalent necklaces of 5 beads each of them painted by one of n colors.

The sequence solves the so-called Reis problem about convex k-gons in case k=5. The full solution was given by H. Gupta (1979); I gave a short proof of Gupta's result and showed an equivalence of this problem and every one of the following problems: enumerating the bracelets of n beads of 2 colors, k of them black, and enumerating the necklaces of k beads each of them painted by one of n colors.

a(n) is an essentially unimprovable upper estimate for the number of distinct values of the permanent in (0,1)-circulants of order n with five 1's in every row. (End)

a(n+5) is the number of symmetry-allowed, linearly-independent terms at n-th order in the series expansion of the T_1 X h vibronic perturbation matrix, H(Q) (cf. Dunn & Bates). - Bradley Klee, Jul 20 2015

From Petros Hadjicostas, Jul 17 2018: (Start)

Let (c(n): n >= 1) be a sequence of nonnegative integers and let C(x) = Sum_{n>=1} c(n)*x^n be its g.f. Let k be a positive integer. Let a_k = (a_k(n): n >= 1) be the output sequence of the DIK[k] transform of sequence (c(n): n >= 1), and let A_k(x) = Sum_{n>=1} a_k(n)*x^n be its g.f. See Christian G. Bower's web link below. It can be proved that, when k is odd, A_k(x) = ((1/k)*Sum_{d|k} phi(d)*C(x^d)^(k/d) + C(x^2)^((k-1)/2)*C(x))/2.

For this sequence, k = 5, c(n) = 1 for all n >= 1, and C(x) = x/(1-x). Thus, a(n) = a_5(n) for all n >= 1. Since a_k(n) = 0 for 1 <= n <= k-1, the offset of this sequence is n = k = 5. Applying the formula for the g.f. of DIK[5] of (c(n): n >= 1) with C(x) = x/(1-x) and k = 5, we get A(x) = A_5(x) = x^5*((1/5)*Sum_{d|5} phi(d)*(1-x^d)^(-5/d) + (1+x)/(1-x^2)^3)/2, which obviously equals the g.f. in the formula section below.

The g.f. is also a special case of Herbert Kociemba's formula that is valid for both even and odd k: A_k(x) = x^k*((1/k)*Sum_{d|k} phi(d)*(1-x^d)^(-k/d) + (1+x)/(1-x^2)^Floor[(k+2)/2])/2.

Here, a(n) is defined to be the number of n-bead bracelets of two colors with 5 black beads and n-5 white beads. But it is also the number of dihedral compositions of n with 5 positive parts. (This statement is equivalent to Vladimir Shevelev's statement above that a(n) is the "number of non-equivalent necklaces of 5 beads each of them painted by one of n colors." By "necklaces" he means "turnover necklaces". See paragraph (2) of Section 2 in his 2004 paper in the Indian Journal of Pure and Applied Mathematics.)

Two cyclic compositions of n (with k = 5 parts) belong to the same equivalence class corresponding to a dihedral composition of n if and only if one can be obtained from the other by a rotation or reversal of order. (End)

Same as A097364 without the 0's.

Also number of partitions of n such that the difference between the largest and smallest parts is k (see A097364). Example: T(6,2)=3 because we have [4,2],[3,2,1] and [3,1,1,1].

Row 1 has one term; row n (n>=2) has n-1 terms.

Row sums yield the partition numbers (A000041).

T(n,0)=A000005(n) (number of divisors of n).

T(n,1)=A049820(n) (n minus number of divisors of n).

T(n,2)=A008805(n-4) for n>=4.

Sum(k*T(n,k),k=0..n-2)=A116686

This is the row reversed version of A180870. See also A157751 and A228565.

The Dirichlet kernel is D(n,x) = Sum_{k=-n..n} exp(i*k*x) = 1 + 2*Sum_{k=1..n} T(n,x) = S(n, 2*y) + S(n-1, 2*y) = S(2*n, sqrt(2*(1+y))) with y = cos(x), n >= 0, with the Chebyshev polynomials T (A053120) and S (A049310). This triangle T(n, k) gives in row n the coefficients of the polynomial Dir(n,y) = D(n,x=arccos(y)) = Sum_{m=0..n} T(n,m)*y^m. See A180870, especially the Peter Bala comments and formulas.

This is the Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)) due to the o.g.f. for Dir(n,y) given by (1+z)/(1 - 2*y*z + z^2) = G(z)/(1 - y*F(z)) with G(z) = (1+z)/(1+z^2) and F(z) = 2*z/(1+z^2) (see the Peter Bala formula under A180870). For Riordan triangles and references see the W. Lang link 'Sheffer a- and z- sequences' under A006232.

The A- and Z- sequences of this Riordan triangle are (see the mentioned W. Lang link in the preceding comment also for the references): The A-sequence has o.g.f. 1+sqrt(1-x^2) and is given by A(2*k+1) = 0 and A(2*k) [2, -1/2, -1/8, -1/16, -5/128, -7/256, -21/1024, -33/2048, -429/32768, -715/65536, ...], k >= 0. (See A098597 and A046161.)

The Z-sequence has o.g.f. sqrt((1-x)/(1+x)) and is given by

[1, -1, 1/2, -1/2, 3/8, -3/8, 5/16, -5/16, 35/128, -35/128, ...]. (See A001790 and A046161.)

The column sequences are A057077, 2*(A004526 with even numbers signed), 4*A008805 (signed), 8*A058187 (signed), 16*A189976 (signed), 32*A189980 (signed) for m = 0, 1, ..., 5.

The row sums give A005408 (from the o.g.f. due to the Riordan property), and the alternating row sums give A033999.

The row polynomials Dir(n, x), n >= 0, give solutions to the diophantine equation (a + 1)*X^2 - (a - 1)*Y^2 = 2 by virtue of the identity (a + 1)*Dir(n, -a)^2 - (a - 1)*Dir(n, a)^2 = 2, which is easily proved inductively using the recurrence Dir(n, a) = (1 + a)*(-1)^(n-1)*Dir(n-1, -a) + a*Dir(n-1, a) given below by Wolfdieter Lang. - Peter Bala, May 08 2025

Conjecture 1. If n>1 is even then F(n,x) has no real roots.

Conjecture 2. If n>0 is odd then F(n,x) has exactly one real root, r,

and if n>4 then 0 < -r < n.

Conjectures 1 and 2 are true. [From Alain Thiery (Alain.Thiery(AT)math.u-bordeaux1.fr), May 14 2010]

Cayley (1876) states "We, in fact, find 1 + sin u = 1 + x, 1 - sin 3u = (1 + x)(1 - 2x)^2, 1 + sin 5u = (1 + x)(1 + 2x - 4x^2)^2, 1 - sin 7u = (1 + x)(1 - 4x - 4x^2 + 8x^3)^2, &c.". - Michael Somos, Jun 19 2012

Appears to be the unsigned row reverse of A180870 and A228565. - Peter Bala, Feb 17 2014

From Wolfdieter Lang, Jul 29 2014: (Start)

This triangle is the Riordan triangle ((1+z)/(1-z^2), 2*z/(1-z^2)). For Riordan triangles see the W. Lang link 'Sheffer a-and z-sequences' under A006232, also for references. The o.g.f. given by Peter Bala in the formula section refers to the row reversed triangle. The usual information on this triangle, like o.g.f. for the columns, the row sums, the alternating row sums, the recurrences using A- and Z-sequences, etc. follows from this Riordan property. The Riordan proof follows from the given o.g.f. by Peter Bala, call it Grev(x,z), by row reversion: G(x,z) = Grev(1/x,x*z) = (1+z)/(1- 2*x*z - z^2) = G(z)*(1/(1 - x*F(z))) with G(z) = (1+z)/(1-z^2) and F(z) = z*2/(1-z^2). See A244419 for the discussion for a signed version of this triangle.

(End)

Equals row sums of triangle A177994. - Gary W. Adamson, May 16 2010

From Ralf Stephan, Mar 09 2014: (Start)

Write the positive integers in a skewed triangle:

1, 2;

0, 3, 4, 5;

0, 0, 6, 7, 8, 9;

0, 0, 0, 10, 11, 12, 13, 14;

...

Sequence consists of the first number in each column. (End)

In a regular k-polygon draw lines connecting all the vertices. Select a triangle that tiles the polygon into k pieces. This triangle contains two adjacent polygon vertices. The third vertex is for even k the center of the polygon and for odd k one of the vertices of the central k-polygon (which is not included in the tiling). Count all lines connecting vertices in the original k-polygon that passes through the interior of the tiling triangle. That count is a(k-5). (See illustrations below.) - Lars Blomberg, Feb 20 2020

a(n) is the smallest number which has n+1 as a part in any of its maximally refined strict partitions. The first such are:(1),(2),(1,3),(1,4),(1,2,5),(1,2,6),(1,2,3,7),(1,2,3,8),(1,2,3,4,9) etc. - Sigurd Kittilsen, Oct 18 2024

Length of n-th row is A272651(n) + 1, where A272651(n) is the maximal number of points that can be placed under the condition mentioned.

Rotations and reflections of placements are not counted. If they are to be counted, see A279445.

For condition "no more than 2 points on a straight line at any angle", see A235453.

For n >= 9, a(n-1) is the number of incongruent two-color bracelets of n beads, 9 from them are black (A032281), having a diameter of symmetry.

Refer to arrangement same as A005169: "A fountain is formed by starting with a row of coins, then stacking additional coins on top so that each new coin touches two in the previous row". The 3 curves coins patterns consist of a part of each coin circumference and forms a continuous area. There are total 4 distinct patterns. For selected pattern, I would like to call "3c3s" type as it cover 3 coins and symmetry. When packing 3c3s into fountain of coins base n, the total number of 3c3s is A008805, the coins left is A008795 and voids left is a(n). See illustration in links.

The triangle T(n, k) is irregularly shaped: 1 <= k <= 2n. First row corresponds to n = 1.

Without the restriction "non-equivalent (mod D_4)" the numbers are given by triangle A194193. (But this one is read by antidiagonals!)

T(n, 2n) = A000769(n).

2n is an upper bound on the number of points that can be placed on the grid. For large n, it is conjectured that this bound is not reached (see MathWorld link).