cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A193091 Augmentation of the triangular array A158405. See Comments.

Original entry on oeis.org

1, 1, 3, 1, 6, 14, 1, 9, 37, 79, 1, 12, 69, 242, 494, 1, 15, 110, 516, 1658, 3294, 1, 18, 160, 928, 3870, 11764, 22952, 1, 21, 219, 1505, 7589, 29307, 85741, 165127, 1, 24, 287, 2274, 13355, 61332, 224357, 638250, 1217270, 1, 27, 364, 3262, 21789, 115003
Offset: 0

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Author

Clark Kimberling, Jul 30 2011

Keywords

Comments

Suppose that P is an infinite triangular array of numbers:
p(0,0)
p(1,0)...p(1,1)
p(2,0)...p(2,1)...p(2,2)
p(3,0)...p(3,1)...p(3,2)...p(3,3)...
...
Let w(0,0)=1, w(1,0)=p(1,0), w(1,1)=p(1,1), and define
W(n)=(w(n,0), w(n,1), w(n,2),...w(n,n-1), w(n,n)) recursively by W(n)=W(n-1)*PP(n), where PP(n) is the n X (n+1) matrix given by
...
row 0 ... p(n,0) ... p(n,1) ...... p(n,n-1) ... p(n,n)
row 1 ... 0 ..... p(n-1,0) ..... p(n-1,n-2) .. p(n-1,n-1)
row 2 ... 0 ..... 0 ............ p(n-2,n-3) .. p(n-2,n-2)
...
row n-1 . 0 ..... 0 ............. p(2,1) ..... p(2,2)
row n ... 0 ..... 0 ............. p(1,0) ..... p(1,1)
...
The augmentation of P is here introduced as the triangular array whose n-th row is W(n), for n>=0. The array P may be represented as a sequence of polynomials; viz., row n is then the vector of coefficients: p(n,0), p(n,1),...,p(n,n), from p(n,0)*x^n+p(n,1)*x^(n-1)+...+p(n,n). For example, (C(n,k)) is represented by ((x+1)^n); using this choice of P (that is, Pascal's triangle), the augmentation of P is calculated one row at a time, either by the above matrix products or by polynomial substitutions in the following manner:
...
row 0 of W: 1, by decree
row 1 of W: 1 augments to 1,1
...polynomial version: 1 -> x+1
row 2 of W: 1,1 augments to 1,3,2
...polynomial version: x+1 -> (x^2+2x+1)+(x+1)=x^2+3x+2
row 3 to W: 1,3,2 augments to 1,6,11,6
...polynomial version:
x^2+3x+2 -> (x+1)^3+3(x+1)^2+2(x+1)=(x+1)(x+2)(x+3)
...
Examples of augmented triangular arrays:
(p(n,k)=1) augments to A009766, Catalan triangle.
Catalan triangle augments to A193560.
Pascal triangle augments to A094638, Stirling triangle.
A002260=((k+1)) augments to A023531.
A154325 augments to A033878.
A158405 augments to A193091.
((k!)) augments to A193092.
A094727 augments to A193093.
A130296 augments to A193094.
A004736 augments to A193561.
...
Regarding the specific augmentation W=A193091: w(n,n)=A003169.
From Peter Bala, Aug 02 2012: (Start)
This is the table of g(n,k) in the notation of Carlitz (p. 124). The triangle enumerates two-line arrays of positive integers
............a_1 a_2 ... a_n..........
............b_1 b_2 ... b_n..........
such that
1) max(a_i, b_i) <= min(a_(i+1), b_(i+1)) for 1 <= i <= n-1
2) max(a_i, b_i) <= i for 1 <= i <= n
3) max(a_n, b_n) = k.
See A071948 and A211788 for other two-line array enumerations.
(End)

Examples

			The triangle P, at A158405, is given by rows
  1
  1...3
  1...3...5
  1...3...5...7
  1...3...5...7...9...
The augmentation of P is the array W starts with w(0,0)=1, by definition of W.
Successive polynomials (rows of W) arise from P as shown here:
  ...
  1->x+3, so that W has (row 1)=(1,3);
  ...
  x+3->(x^2+3x+5)+3*(x+3), so that W has (row 2)=(1,6,14);
  ...
  x^2+6x+14->(x^3+3x^2+5x+7)+6(x^2+3x+5)+14(x+3), so that (row 3)=(1,9,37,79).
  ...
First 7 rows of W:
  1
  1    3
  1    6    14
  1    9    37    79
  1   12    69   242    494
  1   15   110   516   1658    3294
  1   18   160   928   3870   11764   22952
		

Crossrefs

Programs

  • Mathematica
    p[n_, k_] := 2 k + 1
    Table[p[n, k], {n, 0, 5}, {k, 0, n}] (* A158405 *)
    m[n_] := Table[If[i <= j, p[n + 1 - i, j - i], 0], {i, n}, {j, n + 1}]
    TableForm[m[4]]
    w[0, 0] = 1; w[1, 0] = p[1, 0]; w[1, 1] = p[1, 1];
    v[0] = w[0, 0]; v[1] = {w[1, 0], w[1, 1]};
    v[n_] := v[n - 1].m[n]
    TableForm[Table[v[n], {n, 0, 6}]] (* A193091 *)
    Flatten[Table[v[n], {n, 0, 9}]]

Formula

From Peter Bala, Aug 02 2012: (Start)
T(n,k) = (n-k+1)/n*Sum_{i=0..k} C(n+1,n-k+i+1)*C(2*n+i+1,i) for 0 <= k <= n.
Recurrence equation: T(n,k) = Sum_{i=0..k} (2*k-2*i+1)*T(n-1,i).
(End)

A166866 Interleave the fractals A158405, A002260.

Original entry on oeis.org

1, 1, 1, 3, 1, 2, 1, 3, 5, 1, 2, 3, 1, 3, 5, 7, 1, 2, 3, 4, 1, 3, 5, 7, 9, 1, 2, 3, 4, 5, 1, 3, 5, 7, 9, 11, 1, 2, 3, 4, 5, 6, 1, 3, 5, 7, 9, 11, 13, 1, 2, 3, 4, 5, 6, 7, 1, 3, 5, 7, 9, 11, 13, 15, 1, 2, 3, 4, 5, 6, 7, 8, 1, 3, 5, 7, 9, 11, 13, 15, 17, 1, 2, 3, 4, 5, 6, 7, 8, 9
Offset: 0

Views

Author

Paul Curtz, Oct 22 2009

Keywords

Comments

See A166494. Consider a(n+1) sum by 1,1,2,2,3,3=A008619 terms: 1; 1; 3+1=4; 2+1=3; 3+5+1=9; 2+3+1=6; 3+5+7+1=16; 2+3+4+1=10; this is A123596(n+2).Via A001105(n+1),A123596 is linked to (extended) Mendeleyev-Moseley-Seaborg table of the elements.

Crossrefs

A003215 Hex (or centered hexagonal) numbers: 3*n*(n+1)+1 (crystal ball sequence for hexagonal lattice).

Original entry on oeis.org

1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, 469, 547, 631, 721, 817, 919, 1027, 1141, 1261, 1387, 1519, 1657, 1801, 1951, 2107, 2269, 2437, 2611, 2791, 2977, 3169, 3367, 3571, 3781, 3997, 4219, 4447, 4681, 4921, 5167, 5419, 5677, 5941, 6211, 6487, 6769
Offset: 0

Views

Author

Keywords

Comments

The hexagonal lattice is the familiar 2-dimensional lattice in which each point has 6 neighbors. This is sometimes called the triangular lattice.
Crystal ball sequence for A_2 lattice. - Michael Somos, Jun 03 2012
Sixth spoke of hexagonal spiral (cf. A056105-A056109).
Number of ordered integer triples (a,b,c), -n <= a,b,c <= n, such that a+b+c=0. - Benoit Cloitre, Jun 14 2003
Also the number of partitions of 6n into at most 3 parts, A001399(6n). - R. K. Guy, Oct 20 2003
Also, a(n) is the number of partitions of 6(n+1) into exactly 3 distinct parts. - William J. Keith, Jul 01 2004
Number of dots in a centered hexagonal figure with n+1 dots on each side.
Values of second Bessel polynomial y_2(n) (see A001498).
First differences of cubes (A000578). - Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004
Final digits of Hex numbers (hex(n) mod 10) are periodic with palindromic period of length 5 {1, 7, 9, 7, 1}. Last two digits of Hex numbers (hex(n) mod 100) are periodic with palindromic period of length 100. - Alexander Adamchuk, Aug 11 2006
All divisors of a(n) are congruent to 1, modulo 6. Proof: If p is an odd prime different from 3 then 3n^2 + 3n + 1 = 0 (mod p) implies 9(2n + 1)^2 = -3 (mod p), whence p = 1 (mod 6). - Nick Hobson, Nov 13 2006
For n>=1, a(n) is the side of Outer Napoleon Triangle whose reference triangle is a right triangle with legs (3a(n))^(1/2) and 3n(a(n))^(1/2). - Tom Schicker (tschicke(AT)email.smith.edu), Apr 25 2007
Number of triples (a,b,c) where 0<=(a,b)<=n and c=n (at least once the term n). E.g., for n = 1: (0,0,1), (0,1,0), (1,0,0), (0,1,1), (1,0,1), (1,1,0), (1,1,1), so a(1)=7. - Philippe Lallouet (philip.lallouet(AT)wanadoo.fr), Aug 20 2007
Equals the triangular numbers convolved with [1, 4, 1, 0, 0, 0, ...]. - Gary W. Adamson and Alexander R. Povolotsky, May 29 2009
From Terry Stickels, Dec 07 2009: (Start)
Also the maximum number of viewable cubes from any one static point while viewing a cube stack of identical cubes of varying magnitude.
For example, viewing a 2 X 2 X 2 stack will yield 7 maximum viewable cubes.
If the stack is 3 X 3 X 3, the maximum number of viewable cubes from any one static position is 19, and so on.
The number of cubes in the stack must always be the same number for width, length, height (at true regular cubic stack) and the maximum number of visible cubes can always be found by taking any cubic number and subtracting the number of the cube that is one less.
Examples: 125 - 64 = 61, 64 - 27 = 37, 27 - 8 = 19. (End)
The sequence of digital roots of the a(n) is period 3: repeat [1,7,1]. - Ant King, Jun 17 2012
The average of the first n (n>0) centered hexagonal numbers is the n-th square. - Philippe Deléham, Feb 04 2013
A002024 is the following array A read along antidiagonals:
1, 2, 3, 4, 5, 6, ...
2, 3, 4, 5, 6, 7, ...
3, 4, 5, 6, 7, 8, ...
4, 5, 6, 7, 8, 9, ...
5, 6, 7, 8, 9, 10, ...
6, 7, 8, 9, 10, 11, ...
and a(n) is the hook sum Sum_{k=0..n} A(n,k) + Sum_{r=0..n-1} A(r,n). - R. J. Mathar, Jun 30 2013
a(n) is the sum of the terms in the n+1 X n+1 matrices minus those in n X n matrices in an array formed by considering A158405 an array (the beginning terms in each row are 1,3,5,7,9,11,...). - J. M. Bergot, Jul 05 2013
The formula also equals the product of the three distinct combinations of two consecutive numbers: n^2, (n+1)^2, and n*(n+1). - J. M. Bergot, Mar 28 2014
The sides of any triangle ABC are divided into 2n + 1 equal segments by 2n points: A_1, A_2, ..., A_2n in side a, and also on the sides b and c cyclically. If A'B'C' is the triangle delimited by AA_n, BB_n and CC_n cevians, we have (ABC)/(A'B'C') = a(n) (see Java applet link). - Ignacio Larrosa Cañestro, Jan 02 2015
a(n) is the maximal number of parts into which (n+1) triangles can intersect one another. - Ivan N. Ianakiev, Feb 18 2015
((2^m-1)n)^t mod a(n) = ((2^m-1)(n+1))^t mod a(n) = ((2^m-1)(2n+1))^t mod a(n), where m any positive integer, and t = 0(mod 6). - Alzhekeyev Ascar M, Oct 07 2016
((2^m-1)n)^t mod a(n) = ((2^m-1)(n+1))^t mod a(n) = a(n) - (((2^m-1)(2n+1))^t mod a(n)), where m any positive integer, and t = 3(mod 6). - Alzhekeyev Ascar M, Oct 07 2016
(3n+1)^(a(n)-1) mod a(n) = (3n+2)^(a(n)-1) mod a(n) = 1. If a(n) not prime, then always strong pseudoprime. - Alzhekeyev Ascar M, Oct 07 2016
Every positive integer is the sum of 8 hex numbers (zero included), at most 3 of which are greater than 1. - Mauro Fiorentini, Jan 01 2018
Area enclosed by the segment of Archimedean spiral between n*Pi/2 and (n+1)*Pi/2 in Pi^3/48 units. - Carmine Suriano, Apr 10 2018
This sequence contains all numbers k such that 12*k - 3 is a square. - Klaus Purath, Oct 19 2021
The continued fraction expansion of sqrt(3*a(n)) is [3n+1; {1, 1, 2n, 1, 1, 6n+2}]. For n = 0, this collapses to [1; {1, 2}]. - Magus K. Chu, Sep 12 2022

Examples

			G.f. = 1 + 7*x + 19*x^2 + 37*x^3 + 61*x^4 + 91*x^5 + 127*x^6 + 169*x^7 + 217*x^8 + ...
From _Omar E. Pol_, Aug 21 2011: (Start)
Illustration of initial terms:
.
.                                 o o o o
.                   o o o        o o o o o
.         o o      o o o o      o o o o o o
.   o    o o o    o o o o o    o o o o o o o
.         o o      o o o o      o o o o o o
.                   o o o        o o o o o
.                                 o o o o
.
.   1      7          19             37
.
(End)
From _Klaus Purath_, Dec 03 2021: (Start)
(1) a(19) is not a prime number, because besides a(19) = a(9) + P(29), a(19) = a(15) + P(20) = a(2) + P(33) is also true.
(2) a(25) is prime, because except for a(25) = a(12) + P(38) there is no other equation of this pattern. (End)
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 81.
  • M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 18.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Column k=3 of A080853, and column k=2 of A047969.
See also A220083 for a list of numbers of the form n*P(s,n)-(n-1)*P(s,n-1), where P(s,n) is the n-th polygonal number with s sides.
Cf. A287326(A000124(n), 1).
Cf. A008292.
Cf. A154105.

Programs

Formula

a(n) = 3*n*(n+1) + 1, n >= 0 (see the name).
a(n) = (n+1)^3 - n^3 = a(-1-n).
G.f.: (1 + 4*x + x^2) / (1 - x)^3. - Simon Plouffe in his 1992 dissertation
a(n) = 6*A000217(n) + 1.
a(n) = a(n-1) + 6*n = 2a(n-1) - a(n-2) + 6 = 3*a(n-1) - 3*a(n-2) + a(n-3) = A056105(n) + 5n = A056106(n) + 4*n = A056107(n) + 3*n = A056108(n) + 2*n = A056108(n) + n.
n-th partial arithmetic mean is n^2. - Amarnath Murthy, May 27 2003
a(n) = 1 + Sum_{j=0..n} (6*j). E.g., a(2)=19 because 1+ 6*0 + 6*1 + 6*2 = 19. - Xavier Acloque, Oct 06 2003
The sum of the first n hexagonal numbers is n^3. That is, Sum_{n>=1} (3*n*(n-1) + 1) = n^3. - Edward Weed (eweed(AT)gdrs.com), Oct 23 2003
a(n) = right term in M^n * [1 1 1], where M = the 3 X 3 matrix [1 0 0 / 2 1 0 / 3 3 1]. M^n * [1 1 1] = [1 2n+1 a(n)]. E.g., a(4) = 61, right term in M^4 * [1 1 1], since M^4 * [1 1 1] = [1 9 61] = [1 2n+1 a(4)]. - Gary W. Adamson, Dec 22 2004
Row sums of triangle A130298. - Gary W. Adamson, Jun 07 2007
a(n) = 3*n^2 + 3*n + 1. Proof: 1) If n occurs once, it may be in 3 positions; for the two other ones, n terms are independently possible, then we have 3*n^2 different triples. 2) If the term n occurs twice, the third one may be placed in 3 positions and have n possible values, then we have 3*n more different triples. 3) The term n may occurs 3 times in one way only that gives the formula. - Philippe Lallouet (philip.lallouet(AT)wanadoo.fr), Aug 20 2007
Binomial transform of [1, 6, 6, 0, 0, 0, ...]; Narayana transform (A001263) of [1, 6, 0, 0, 0, ...]. - Gary W. Adamson, Dec 29 2007
a(n) = (n-1)*A000166(n) + (n-2)*A000166(n-1) = (n-1)floor(n!*e^(-1)+1) + (n-2)*floor((n-1)!*e^(-1)+1) (with offset 0). - Gary Detlefs, Dec 06 2009
a(n) = A028896(n) + 1. - Omar E. Pol, Oct 03 2011
a(n) = integral( (sin((n+1/2)x)/sin(x/2))^3, x=0..Pi)/Pi. - Yalcin Aktar, Dec 03 2011
Sum_{n>=0} 1/a(n) = Pi/sqrt(3)*tanh(Pi/(2*sqrt(3))) = 1.305284153013581... - Ant King, Jun 17 2012
a(n) = A000290(n) + A000217(2n+1). - Ivan N. Ianakiev, Sep 24 2013
a(n) = A002378(n+1) + A056220(n) = A005408(n) + 2*A005449(n) = 6*A000217(n) + 1. - Ivan N. Ianakiev, Sep 26 2013
a(n) = 6*A000124(n) - 5. - Ivan N. Ianakiev, Oct 13 2013
a(n) = A239426(n+1) / A239449(n+1) = A215630(2*n+1,n+1). - Reinhard Zumkeller, Mar 19 2014
a(n) = A243201(n) / A002061(n + 1). - Mathew Englander, Jun 03 2014
a(n) = A101321(6,n). - R. J. Mathar, Jul 28 2016
E.g.f.: (1 + 6*x + 3*x^2)*exp(x). - Ilya Gutkovskiy, Jul 28 2016
a(n) = (A001844(n) + A016754(n))/2. - Bruce J. Nicholson, Aug 06 2017
a(n) = A045943(2n+1). - Miquel Cerda, Jan 22 2018
a(n) = 3*Integral_{x=n..n+1} x^2 dx. - Carmine Suriano, Apr 10 2018
a(n) = A287326(A000124(n), 1). - Kolosov Petro, Oct 22 2018
From Amiram Eldar, Jun 20 2020: (Start)
Sum_{n>=0} a(n)/n! = 10*e.
Sum_{n>=0} (-1)^(n+1)*a(n)/n! = 2/e. (End)
G.f.: polylog(-3, x)*(1-x)/x. See the Simon Plouffe formula above, and the g.f. of the rows of A008292 by Vladeta Jovovic, Sep 02 2002. - Wolfdieter Lang, May 08 2021
a(n) = T(n-1)^2 - 2*T(n)^2 + T(n+1)^2, n >= 1, T = triangular number A000217. - Klaus Purath, Oct 11 2021
a(n) = 1 + 2*Sum_{j=n..2n} j. - Klaus Purath, Oct 19 2021
a(n) = A069099(n+1) - A000217(n). - Klaus Purath, Nov 03 2021
From Leo Tavares, Dec 03 2021: (Start)
a(n) = A005448(n) + A140091(n);
a(n) = A001844(n) + A002378(n);
a(n) = A005891(n) + A000217(n);
a(n) = A000290(n) + A000384(n+1);
a(n) = A060544(n-1) + 3*A000217(n);
a(n) = A060544(n-1) + A045943(n).
a(2*n+1) = A154105(n).
(End)

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A180662 The Golden Triangle: T(n,k) = A001654(k) for n>=0 and 0<=k<=n.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 2, 6, 0, 1, 2, 6, 15, 0, 1, 2, 6, 15, 40, 0, 1, 2, 6, 15, 40, 104, 0, 1, 2, 6, 15, 40, 104, 273, 0, 1, 2, 6, 15, 40, 104, 273, 714, 0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 4895
Offset: 0

Views

Author

Johannes W. Meijer, Sep 21 2010

Keywords

Comments

The terms in the n-th row of the Golden Triangle are the first (n+1) golden rectangle numbers. The golden rectangle numbers are A001654(n)=F(n)*F(n+1), with F(n) the Fibonacci numbers. The mirror image of the Golden Triangle is A180663.
We define below 24 mostly new triangle sums. The Row1 and Row2 sums are the ordinary and alternating row sums respectively and the Kn11 and Kn12 sums are commonly known as antidiagonal sums. Each of the names of these sums, except for the row sums, comes from a (fairy) chess piece that moves in its own peculiar way over a chessboard, see Hooper and Whyld. All pieces are leapers: knight (sqrt(5) or 1,2), fil (sqrt(8) or 2,2), camel (sqrt(10) or 3,1), giraffe (sqrt(17) or 4,1) and zebra (sqrt(13) or 3,2). Information about the origin of these chess sums can be found in "Famous numbers on a chessboard", see Meijer.
Each triangle or chess sum formula adds up numbers on a chessboard using the moves of its namesake. Converting a number triangle to a square array of numbers shows this most clearly (use the table button!). The formulas given below are for number triangles.
The chess sums of the Golden Triangle lead to six different sequences, see the crossrefs. As could be expected all these sums are related to the golden rectangle numbers.
Some triangles with complete sets of triangle sums are: A002260 (Natural Numbers), A007318 (Pascal), A008288 (Delannoy) A013609 (Pell-Jacobsthal), A036561 (Nicomachus), A104763 (Fibonacci(n)), A158405 (Odd Numbers) and of course A180662 (Golden Triangle).
#..Name....Type..Code....Definition of triangle sums.
1. Row......1....Row1.. a(n) = Sum_{k=0..n} T(n, k).
2. Row Alt..2....Row2.. a(n) = Sum_{k=0..n} (-1)^(n+k)*T(n, k).
3. Knight...1....Kn11.. a(n) = Sum_{k=0..floor(n/2)} T(n-k, k).
4. Knight...1....Kn12.. a(n) = Sum_{k=0..floor(n/2)} T(n-k+1, k+1).
5. Knight...1....Kn13.. a(n) = Sum_{k=0..floor(n/2)} T(n-k+2, k+2).
6. Knight...2....Kn21.. a(n) = Sum_{k=0..floor(n/2)} T(n-k, n-2*k).
7. Knight...2....Kn22.. a(n) = Sum_{k=0..floor(n/2)} T(n-k+1, n-2*k).
8. Knight...2....Kn23.. a(n) = Sum_{k=0..floor(n/2)} T(n-k+2, n-2*k).
9. Knight...3....Kn3... a(n) = Sum_{k=0..n} T(n+k, 2*k).
10. Knight...4....Kn4... a(n) = Sum_{k=0..n} T(n+k, n-k).
11. Fil......1....Fi1... a(n) = Sum_{k=0..floor(n/2)} T(n, 2*k).
12. Fil......2....Fi2... a(n) = Sum_{k=0..floor(n/2)} T(n, n-2*k).
13. Camel....1....Ca1... a(n) = Sum_{k=0..floor(n/3)} T(n-2*k, k).
14. Camel....2....Ca2... a(n) = Sum_{k=0..floor(n/3)} T(n-2*k, n-3*k).
15. Camel....3....Ca3... a(n) = Sum_{k=0..n} T(n+2*k, 3*k).
16. Camel....4....Ca4... a(n) = Sum_{k=0..n} T(n+2*k, n-k).
17. Giraffe..1....Gi1... a(n) = Sum_{k=0..floor(n/4)} T(n-3*k, k).
18. Giraffe..2....Gi2... a(n) = Sum_{k=0..floor(n/4)} T(n-3*k, n-4*k).
19. Giraffe..3....Gi3... a(n) = Sum_{k=0..n} T(n+3*k, 4*k).
20. Giraffe..4....Gi4... a(n) = Sum_{k=0..n} T(n+3*k, n-k).
21. Zebra....1....Ze1... a(n) = Sum_{k=0..floor(n/2)} T(n+k, 3*k).
22. Zebra....2....Ze2... a(n) = Sum_{k=0..floor(n/2)} T(n+k, n-2*k).
23. Zebra....3....Ze3... a(n) = Sum_{k=0..floor(n/3)} T(n-k, 2*k).
24. Zebra....4....Ze4... a(n) = Sum_{k=0..floor(n/3)} T(n-k, n-3*k).

Examples

			The first few rows of the Golden Triangle are:
  0;
  0, 1;
  0, 1, 2;
  0, 1, 2, 6;
  0, 1, 2, 6, 15;
  0, 1, 2, 6, 15, 40;
		

References

  • David Hooper and Kenneth Whyld, The Oxford Companion to Chess, p. 221, 1992.

Crossrefs

Cf. A180663 (Mirror), A001654 (Golden Rectangle), A000045 (F(n)).
Triangle sums: A064831 (Row1, Kn11, Kn12, Kn4, Ca1, Ca4, Gi1, Gi4), A077916 (Row2), A180664 (Kn13), A180665 (Kn21, Kn22, Kn23, Fi2, Ze2), A180665(2*n) (Kn3, Fi1, Ze3), A115730(n+1) (Ca2, Ze4), A115730(3*n+1) (Ca3, Ze1), A180666 (Gi2), A180666(4*n) (Gi3).

Programs

  • Haskell
    import Data.List (inits)
    a180662 n k = a180662_tabl !! n !! k
    a180662_row n = a180662_tabl !! n
    a180662_tabl = tail $ inits a001654_list
    -- Reinhard Zumkeller, Jun 08 2013
    
  • Magma
    [Fibonacci(k)*Fibonacci(k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, May 25 2021
    
  • Maple
    F:= combinat[fibonacci]:
    T:= (n, k)-> F(k)*F(k+1):
    seq(seq(T(n, k), k=0..n), n=0..10); # revised Johannes W. Meijer, Sep 13 2012
  • Mathematica
    Table[Times @@ Fibonacci@ {k, k + 1}, {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Aug 18 2016 *)
    Module[{nn=20,f},f=Times@@@Partition[Fibonacci[Range[0,nn]],2,1];Table[Take[f,n],{n,nn}]]//Flatten (* Harvey P. Dale, Nov 26 2022 *)
  • PARI
    T(n,k)=fibonacci(k)*fibonacci(k+1) \\ Charles R Greathouse IV, Nov 07 2016
    
  • Sage
    flatten([[fibonacci(k)*fibonacci(k+1) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 25 2021

Formula

T(n, k) = F(k)*F(k+1) with F(n) = A000045(n), for n>=0 and 0<=k<=n.
From Johannes W. Meijer, Jun 22 2015: (Start)
Kn1p(n) = Sum_{k=0..floor(n/2)} T(n-k+p-1, k+p-1), p >= 1.
Kn1p(n) = Kn11(n+2*p-2) - Sum_{k=0..p-2} T(n-k+2*p-2, k), p >= 2.
Kn2p(n) = Sum_{k=0..floor(n/2)} T(n-k+p-1, n-2*k), p >= 1.
Kn2p(n) = Kn21(n+2*p-2) - Sum_{k=0..p-2} T(n+k+p, n+2*k+2), p >= 2. (End)
G.f. as triangle: xy/((1-x)(1+xy)(1-3xy+x^2 y^2)). - Robert Israel, Sep 06 2015

A258708 Triangle read by rows: T(i,j) = integer part of binomial(i+j, i-j)/(2*j+1) for i >= 1 and j = 0..i-1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 5, 7, 4, 1, 1, 7, 14, 12, 5, 1, 1, 9, 25, 30, 18, 6, 1, 1, 12, 42, 66, 55, 26, 7, 1, 1, 15, 66, 132, 143, 91, 35, 8, 1, 1, 18, 99, 245, 334, 273, 140, 45, 9, 1, 1, 22, 143, 429, 715, 728, 476, 204, 57, 10, 1
Offset: 1

Views

Author

N. J. A. Sloane, Jun 12 2015

Keywords

Comments

In the Loh-Shannon-Horadam paper, Table 3 contains a typo (see Extensions lines).
T(n,k) = round(A258993(n,k)/(2*k+1)). - Reinhard Zumkeller, Jun 22 2015
From Reinhard Zumkeller, Jun 23 2015: (Start)
(using tables 4 and 5 of the Loh-Shannon-Horadam paper, p. 8f).
T(n, n-1) = 1;
T(n, n-2) = n for n > 1;
T(n, n-3) = A000969(n-3) for n > 2;
T(n, n-4) = A000330(n-3) for n > 3;
T(n, n-5) = T(2*n-7, 2) = A000970(n) for n > 4;
T(n, n-6) = A000971(n) for n > 5;
T(n, n-7) = A000972(n) for n > 6;
T(n, n-8) = A000973(n) for n > 7;
T(n, 1) = A001840(n-1) for n > 1;
T(2*n, n) = A001764(n);
T(3*n-1, 1) = A000326(n);
T(3*n, 2*n) = A002294(n);
T(4*n, 3*n) = A002296(n). (End)

Examples

			Triangle T(i, j) (with rows i >= 1 and columns j >= 0) begins as follows:
  1;
  1,  1;
  1,  2,  1;
  1,  3,  3,   1;
  1,  5,  7,   4,   1;
  1,  7, 14,  12,   5,   1;
  1,  9, 25,  30,  18,   6,   1;
  1, 12, 42,  66,  55,  26,   7,  1;
  1, 15, 66, 132, 143,  91,  35,  8, 1;
  1, 18, 99, 245, 334, 273, 140, 45, 9, 1;
  ...
		

Crossrefs

Programs

  • Haskell
    a258708 n k = a258708_tabl !! (n-1) !! k
    a258708_row n = a258708_tabl !! (n-1)
    a258708_tabl = zipWith (zipWith ((round .) . ((/) `on` fromIntegral)))
                           a258993_tabl a158405_tabl
    -- Reinhard Zumkeller, Jun 22 2015, Jun 16 2015

Extensions

Corrected T(8,5) = 26 from Reinhard Zumkeller, Jun 13 2015

A099375 Sequence matrix for odd numbers.

Original entry on oeis.org

1, 3, 1, 5, 3, 1, 7, 5, 3, 1, 9, 7, 5, 3, 1, 11, 9, 7, 5, 3, 1, 13, 11, 9, 7, 5, 3, 1, 15, 13, 11, 9, 7, 5, 3, 1, 17, 15, 13, 11, 9, 7, 5, 3, 1, 19, 17, 15, 13, 11, 9, 7, 5, 3, 1, 21, 19, 17, 15, 13, 11, 9, 7, 5, 3, 1, 23, 21, 19, 17, 15, 13, 11, 9, 7, 5, 3, 1, 25, 23, 21, 19, 17, 15, 13, 11, 9
Offset: 0

Views

Author

Paul Barry, Jan 22 2005

Keywords

Comments

Riordan array ((1+x)/(1-x)^2, x).
Inverse matrix is A101038.
Row sums yield (n+1)^2.
Diagonal sums yield sum{k=0..floor(n/2),2(n-2k)+1}=C(n+2,2)=A000217(n+1). Note that sum{k=0..n,2(n-2k)+1}=n+1.
From Paul Curtz, Sep 25 2011. (Start)
Consider from A187870(n-2) and A171080(n)
1 + 1/3 - 4/45 + 44/945 - 428/14175 =1/(1 -1/3 +1/5 -1/7 ..= Pi/4)=4/Pi.
For c(0)=-1, c(1)=1/3, c(2)=4/45, c(3)=44/945, c(4)=428/14175,
c(0)/3 + c(1)=0,
c(0)/5 + c(1)/3 + c(2)=0,
c(0)/7 + c(1)/5 + c(2)/3 + c(3)=0.
Hence a(n+1). Numbers are
-1/3 + 1/3, 1=1,
-1/5 + 1/9 + 4/45, 4=9-5,
-1/7 + 1/15 + 4/135 + 44/945 44=135-63-28. (End)
T(n,k) = A158405(n+1,n+1-k), 1<=k<=n. [Reinhard Zumkeller, Mar 31 2012]
From Peter Bala, Jul 22 2014: (Start)
Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/
having the k x k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite matrix product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to A208904. (End)

Examples

			Rows start
1;
3,1;
5,3,1;
7,5,3,1;
9,7,5,3,1;
11,9,7,5,3,1;
13,11,9,7,5,3,1;
		

Crossrefs

Programs

  • Haskell
    a099375 n k = a099375_row n !! k
    a099375_row n = a099375_tabl !! n
    a099375_tabl = iterate (\xs -> (head xs + 2) : xs) [1]
    -- Reinhard Zumkeller, Mar 31 2012

Formula

Number triangle T(n, k)=if(k<=n, 2(n-k)+1, 0)=binomial(2(n-k)+1, 2(n-k))
a(n)=2*A004736(n)-1; a(n)=2*((t*t+3*t+4)/2-n)-1, where t=floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Feb 08 2013

A255437 In positive integers: replace k^2 with the first k odd numbers.

Original entry on oeis.org

1, 2, 3, 1, 3, 5, 6, 7, 8, 1, 3, 5, 10, 11, 12, 13, 14, 15, 1, 3, 5, 7, 17, 18, 19, 20, 21, 22, 23, 24, 1, 3, 5, 7, 9, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 1, 3, 5, 7, 9, 11, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 1, 3, 5, 7, 9, 11, 13, 50, 51
Offset: 1

Views

Author

Reinhard Zumkeller, Mar 23 2015

Keywords

Comments

a(A005448(n)) = 1;
conjecture: a(A068722(n)) = (2*n+1)^2, i.e. A068722(n) = gives the position of the first occurrence of n-th odd square;
A164514(n) = a(A255527(n)) and a(m) < A164514(n) for m < A255527(n).

Examples

			.  A000290 | 1,    4,          9,                      16,         . . .
.  A000027 | _,2,3,___,5,6,7,8,_____,10,11,12,13,14,15,_______,17,18,...
.  A158405 | 1,    1,3,        1,3,5,                  1,3,5,7,
.  --------+-------------------------------------------------------------
.     a(n) | 1,2,3,1,3,5,6,7,8,1,3,5,10,11,12,13,14,15,1,3,5,7,17,18,19 .
		

Crossrefs

Cf. A256188, A000290, A000037, A158405, A016742, A164514, A255527, A005448, A255507 (first differences), A255508 (partial sums).

Programs

  • Haskell
    a255437 n = a255437_list !! (n-1)
    a255437_list = f 0 [1..] a158405_tabl where
       f k xs (zs:zss) = us ++ zs ++ f (k + 2) vs zss
                         where (us, v:vs) = splitAt k xs

A152204 Triangle read by rows: T(n,k) = 2*n-4*k+5 (n >= 0, 1 <= k <= 1+floor(n/2)).

Original entry on oeis.org

1, 3, 5, 1, 7, 3, 9, 5, 1, 11, 7, 3, 13, 9, 5, 1, 15, 11, 7, 3, 17, 13, 9, 5, 1, 19, 15, 11, 7, 3, 21, 17, 13, 9, 5, 1, 23, 19, 15, 11, 7, 3, 25, 21, 17, 13, 9, 5, 1, 27, 23, 19, 15, 11, 7, 3, 29, 25, 21, 17, 13, 9, 5, 1, 31, 27, 23, 19, 15, 11, 7, 3, 33
Offset: 0

Views

Author

Gary W. Adamson, Nov 29 2008

Keywords

Comments

All terms are odd, decreasing across rows. Row sums = A000217, the triangular numbers.
From Johannes W. Meijer, Sep 08 2013: (Start)
Triangle read by rows formed from the antidiagonals of triangle A099375.
The alternating row sums equal A098181(n). (End)

Examples

			First few rows of the triangle:
  1
  3
  5  1
  7  3
  9  5  1
  11 7  3
  13 9  5  1
  15 11 7  3
  17 13 9  5 1
  19 15 11 7 3
  21 17 13 9 5 1
  ...
		

Crossrefs

Cf. A000217.

Programs

  • Maple
    T := proc(n,k) return 2*n-4*k+5: end: seq(seq(T(n,k), k=1..1+floor(n/2)), n=0..20); # Nathaniel Johnston, May 01 2011

Formula

By columns, odd terms in every column, n-th column starts at row (2*n).
From Johannes W. Meijer, Sep 08 2013: (Start)
T(n, k) = A099375(n-k+1, k-1), n >= 0 and 1 <= k <= 1+floor(n/2).
T(n, k) = A158405(n+1, n-2*k+2). (End)

Extensions

Edited by N. J. A. Sloane, Sep 25 2010, following a suggestion from Emeric Deutsch
Offset corrected by Johannes W. Meijer, Sep 07 2013

A208057 Triangle by rows, generated from the odd integers and related to A000165.

Original entry on oeis.org

1, 1, 1, 4, 3, 1, 24, 18, 5, 1, 192, 144, 40, 7, 1, 1920, 1440, 400, 70, 9, 1, 23040, 17280, 4800, 840, 108, 11, 1, 322560, 241920, 67200, 11760, 1512, 154, 13, 1, 5160960, 3870720, 1075200, 188160, 24192, 2464, 208, 15, 1
Offset: 0

Views

Author

Gary W. Adamson, Feb 22 2012

Keywords

Comments

Row sums = A000165, the double factorial numbers: (1, 2, 8, 48, 384,...).
Left border = A002866 and the eigensequence of the odd integers prefaced with a 1.

Examples

			First few rows of the triangle:
  1;
  1, 1;
  4, 3, 1;
  24, 18, 5, 1;
  192, 144, 40, 7, 1;
  1920, 1440, 400, 70, 9, 1;
  23040, 17280, 4800, 840, 108, 11, 1;
  322560, 241920, 67200, 11760, 1512, 154, 13, 1;
  ...
		

Crossrefs

Programs

  • Maple
    T:= proc(n) option remember; local M;
          M:= (Matrix(n+1, (i, j)-> `if`(i=j, 1, `if`(i>j, -2*j+1, 0)))^(-1));
          seq(M[n+1, k], k=1..n+1)
        end:
    seq(T(n), n=0..10);  # Alois P. Heinz, Feb 27 2012
  • Mathematica
    T[n_] := T[n] = Module[{M}, M = Table[If[i == j, 1, If[i>j, -2*j+1, 0]], {i, 1, n+1 }, {j, 1, n+1}] // Inverse; M[[n+1]]]; Table[T[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Mar 09 2015, after Alois P. Heinz *)

Formula

Eigentriangle of triangle A158405 (odd integers in every row: (1, 3, 5,...)); the inverse of:
1;
-1, 1;
-1, -3, 1;
-1, -3, -5, 1;
-1, -3, -5, -7, 1;
...

Extensions

Typo in term 17 corrected by Alois P. Heinz, Dec 06 2012

A104716 Triangle T(n,k) = (2k-3+4n)*(k-1-n)*(k-2-n)/6, 1<=k<=n.

Original entry on oeis.org

1, 7, 3, 22, 13, 5, 50, 34, 19, 7, 95, 70, 46, 25, 9, 161, 125, 90, 58, 31, 11, 252, 203, 155, 110, 70, 37, 13, 372, 308, 245, 185, 130, 82, 43, 15, 525, 444, 364, 287, 215, 150, 94, 49, 17, 715, 615, 516, 420, 329, 245, 170, 106, 55, 19, 946, 825, 705, 588, 476, 371, 275, 190, 118, 61, 21
Offset: 1

Views

Author

Gary W. Adamson, Mar 20 2005

Keywords

Comments

The triangle is created by the matrix product A158405 * A004736, regarding both as infinite lower triangular matrices, rest of the terms filled in with zeros.
Apparently, row n contains the initial terms of row 2n-2 of A177877. - R. J. Mathar, Aug 31 2011

Examples

			First few rows are:
1;
7, 3;
22, 13, 5;
50, 34, 19, 7;
95, 70, 46, 25, 9;
...
		

Crossrefs

Cf. A104715, A002419 (row sums).

Programs

  • Maple
    A104716 := proc(n,k) (2*k-3+4*n)*(k-1-n)*(k-2-n)/6 ; end proc:
    seq(seq(A104716(n,k),k=1..n),n=1..15) ; # R. J. Mathar, Aug 31 2011

Extensions

Closed-form definition by R. J. Mathar, Aug 31 2011
Showing 1-10 of 18 results. Next