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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A244420 Numerators of coefficient triangle for expansion of x^n in terms of polynomials Todd(k, x) = T(2*k+1, sqrt(x))/sqrt(x) (A084930), with the Chebyshev polynomials of the first kind (type T).

Original entry on oeis.org

1, 3, 1, 5, 5, 1, 35, 21, 7, 1, 63, 21, 9, 9, 1, 231, 165, 165, 55, 11, 1, 429, 1287, 715, 143, 39, 13, 1, 6435, 5005, 3003, 1365, 455, 105, 15, 1, 12155, 2431, 1547, 1547, 595, 85, 17, 17, 1, 46189, 37791, 12597, 6783, 2907, 969, 969, 171, 19, 1, 88179, 146965, 101745, 14535, 6783, 20349, 5985, 665, 105, 21, 1
Offset: 0

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Author

Wolfdieter Lang, Aug 04 2014

Keywords

Comments

This expansion is due to the Riordan property of the triangle A084930. The inverse of the lower triangular matrix built by A084930 is therefore also a (rational) Riordan triangle, namely ((2 - c(z/4))/(1-z), -1 + c(z/4)) in the standard notation, where c is the o.g.f. of A000108 (Catalan).
For the denominators of this triangle see A244421.
The expansion is x^n = sum(R(n,m)*Todd(m, x), m=0..n), n >= 0, with the rational triangle with entries R(n,m) = a(n, m)/b(n, m) with b(n, m) = A244421(n, m).
If one uses instead the expansion of (4*x)^n one finds the integer triangle A111418: (4*x)^n = sum(A111418(n,k) * Todd(k, x), k=0..n).
The row sums of the rational triangle R(n,m) are identically 1. The alternating row sums have o.g.f. 1/sqrt(1-x) which generates A001790(n)/A046161(n) (see a Michael Somos comment on A046161), namely 1, 1/2, 3/8, 5/16, 35/128, 63/256, 231/1024, 429/2048, ...
From Wolfdieter Lang, Jun 13 2016: (Start)
R(n,m) = a(n, m)/ A244421(n, m) is also the rational triangle for the expansion cos^{2*n+1}(x) = Sum_{m=0..n} R(n, m)*cos((2*m+1)*x), n >= 0, m = 0..n. Compare with the odd numbered rows of A273496. In terms of Chebyshev T-polynomials (A053120) this is the identity x^(2*n+1) = Sum_{m=0..n} R(n, m)*T(2*m+1, x).
S(n,m) = (-1)^m*a(n, m)/ A244421(n, m) is the rational triangle for the expansion sin^{2*n+1}(x) = Sum_{m=0..n} S(n, m)*sin((2*m+1)*x), n >= 0, m = 0..n. In terms of Chebyshev S-polynomials (A049310) this is equivalent to the identity (4 - x^2)*n = Sum_{m=0..n} (-1)^m * binomial(n, n-m)*S(2*m,x), n >= 0.
(End)

Examples

			The numerator triangle a(n,m) begins:
  n\m      0      1      2     3    4     5    6   7   8   9
  0:       1
  1:       3      1
  2:       5      5      1
  3:      35     21      7     1
  4:      63     21      9     9    1
  5:     231    165    165    55   11     1
  6:     429   1287    715   143   39    13    1
  7:    6435   5005   3003  1365  455   105   15   1
  8:   12155   2431   1547  1547  595    85   17  17   1
  9:   46189  37791  12597  6783 2907   969  969 171  19   1
  ...
The rational triangle R(n,m) begins:
  n\m       0        1         2       3        4        5
  0:        1
  1:      3/4      1/4
  2:      5/8     5/16      1/16
  3:    35/64    21/64      7/64    1/64
  4:   63/128    21/64      9/64   9/256    1/256
  5:  231/512  165/512  165/1024 55/1024  11/1024   1/1024
  ...
The next rows are:
  n=6: 429/1024, 1287/4096, 715/4096, 143/2048, 39/2048, 13/4096, 1/4096,
  n=7: 6435/16384, 5005/16384, 3003/16384, 1365/16384, 455/16384, 105/16384, 15/16384, 1/16384,
  n=8: 12155/32768, 2431/8192, 1547/8192, 1547/16384, 595/16384, 85/8192, 17/8192, 17/65536, 1/65536,
  n=9: 46189/131072, 37791/131072, 12597/65536, 6783/65536, 2907/65536, 969/65536, 969/262144, 171/262144, 19/262144, 1/262144,
  ...
Expansions:
x^2 = 5/8 * Todd(0,x) + 5/16 * Todd(1,x) + 1/16 * Todd(2,x) = 5/8 + (5/16)*(-3 + 4*x) +(1/16)*(5 -20*x + 16*x^2).
x^3 = (35*Todd(0, x) + 21*Todd(1, x) + 7*Todd(2, x) + 1*Todd(3, x))/64 = (35 + 21*(-3+4*x) + 7*( 5-20*x+16*x^2) + (-7+56*x-112*x^2+64*x^3))/64.
For the Todd polynomials see the coefficient table A084930.
		

Crossrefs

Formula

a(n, m) = numerator(R(n, m)) with the rationals Riordan matrix elements R(n, m)= [x^m]R(n, x), with the row polynomials R(n, x) generated by ((2 - c(z/4))/(1-z))/(1 - x*(-1 + c(z/4))) = 2*((1+x)*(z-1) + (1-x)*sqrt(1-z))/((1-z)*((1+x)^2*z - 4*x)), where c(x) is the o.g.f. of the Catalan numbers A000108.
The rationals R(n, m) = binomial(2*n+1, m)/2^(2*n). - Wolfdieter Lang, Jun 12 2016

A244421 Denominators of coefficient triangle for expansion of x^n in terms of polynomials Todd(k,x) = T(2*k+1, sqrt(x))/sqrt(x) (A084930), with the Chebyshev T-polynomials.

Original entry on oeis.org

1, 4, 4, 8, 16, 16, 64, 64, 64, 64, 128, 64, 64, 256, 256, 512, 512, 1024, 1024, 1024, 1024, 1024, 4096, 4096, 2048, 2048, 4096, 4096, 16384, 16384, 16384, 16384, 16384, 16384, 16384, 16384, 32768, 8192, 8192, 16384, 16384, 8192, 8192, 65536, 65536, 131072, 131072, 65536, 65536, 65536, 65536, 262144, 262144, 262144, 262144, 262144, 524288, 524288, 131072, 131072, 1048576, 1048576, 524288, 524288, 1048576, 1048576
Offset: 0

Views

Author

Wolfdieter Lang, Aug 04 2014

Keywords

Comments

For the numerator triangle see A244420, also for comments, and the rational entries R(n,m) of the lower triangular Riordan matrix denoted in standard fashion by ((2 - c(z/4))/(1-z), -1 + c(z/4)) with c the o.g.f. of the Catalan numbers A000108.

Examples

			The triangle a(n,m) begins:
  n\m   0     1     2     3     4     5     6 ...
  0:    1
  1:    4     4
  2:    8    16    16
  3:   64    64    64    64
  4:  128    64    64   256   256
  5:  512   512  1024  1024  1024  1024
  6: 1024  4096  4096  2048  2048  4096  4096
  ...
For more rows see the link.
For the rational triangle R(n,m) see the example section of A244420.
Expansion: x^3 = (35*Todd(0, x) + 21*Todd(1, x) + 7*Todd(2, x) + 1*Todd(3, x))/64 = (35 + 21*(-3+4*x) + 7*( 5-20*x+16*x^2) + (-7+56*x-112*x^2+64*x^3))/64. For the Todd polynomials see A084930.
		

Crossrefs

Formula

a(n,m) = denominator(R(n,m)) with the rationals Riordan matrix elements R(n,m)= [x^m]R(n,x), with the row polynomials R(n,x) generated by ((2 - c(z/4))/(1-z))/(1 - x*(-1 + c(z/4))) = 2*((1+x)*(z-1) + (1-x)*sqrt(1-z))/((1-z)*((1+x)^2*z - 4*x)), where c(x) is the o.g.f. of the Catalan numbers A000108.

A000330 Square pyramidal numbers: a(n) = 0^2 + 1^2 + 2^2 + ... + n^2 = n*(n+1)*(2*n+1)/6.

Original entry on oeis.org

0, 1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819, 1015, 1240, 1496, 1785, 2109, 2470, 2870, 3311, 3795, 4324, 4900, 5525, 6201, 6930, 7714, 8555, 9455, 10416, 11440, 12529, 13685, 14910, 16206, 17575, 19019, 20540, 22140, 23821, 25585, 27434, 29370
Offset: 0

Views

Author

Keywords

Comments

The sequence contains exactly one square greater than 1, namely 4900 (according to Gardner). - Jud McCranie, Mar 19 2001, Mar 22 2007 [This is a result from Watson. - Charles R Greathouse IV, Jun 21 2013] [See A351830 for further related comments and references.]
Number of rhombi in an n X n rhombus. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000
Number of acute triangles made from the vertices of a regular n-polygon when n is odd (cf. A007290). - Sen-Peng Eu, Apr 05 2001
Gives number of squares with sides parallel to the axes formed from an n X n square. In a 1 X 1 square, one is formed. In a 2 X 2 square, five squares are formed. In a 3 X 3 square, 14 squares are formed and so on. - Kristie Smith (kristie10spud(AT)hotmail.com), Apr 16 2002; edited by Eric W. Weisstein, Mar 05 2025
a(n-1) = B_3(n)/3, where B_3(x) = x(x-1)(x-1/2) is the third Bernoulli polynomial. - Michael Somos, Mar 13 2004
Number of permutations avoiding 13-2 that contain the pattern 32-1 exactly once.
Since 3*r = (r+1) + r + (r-1) = T(r+1) - T(r-2), where T(r) = r-th triangular number r*(r+1)/2, we have 3*r^2 = r*(T(r+1) - T(r-2)) = f(r+1) - f(r-1) ... (i), where f(r) = (r-1)*T(r) = (r+1)*T(r-1). Summing over n, the right hand side of relation (i) telescopes to f(n+1) + f(n) = T(n)*((n+2) + (n-1)), whence the result Sum_{r=1..n} r^2 = n*(n+1)*(2*n+1)/6 immediately follows. - Lekraj Beedassy, Aug 06 2004
Also as a(n) = (1/6)*(2*n^3 + 3*n^2 + n), n > 0: structured trigonal diamond numbers (vertex structure 5) (cf. A006003 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Number of triples of integers from {1, 2, ..., n} whose last component is greater than or equal to the others.
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Jun 12 2005
Sum of the first n positive squares. - Cino Hilliard, Jun 18 2007
Maximal number of cubes of side 1 in a right pyramid with a square base of side n and height n. - Pasquale CUTOLO (p.cutolo(AT)inwind.it), Jul 09 2007
If a 2-set Y and an (n-2)-set Z are disjoint subsets of an n-set X then a(n-3) is the number of 4-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 19 2007
We also have the identity 1 + (1+4) + (1+4+9) + ... + (1+4+9+16+ ... + n^2) = n(n+1)(n+2)(n+(n+1)+(n+2))/36; ... and in general the k-fold nested sum of squares can be expressed as n(n+1)...(n+k)(n+(n+1)+...+(n+k))/((k+2)!(k+1)/2). - Alexander R. Povolotsky, Nov 21 2007
The terms of this sequence are coefficients of the Engel expansion of the following converging sum: 1/(1^2) + (1/1^2)*(1/(1^2+2^2)) + (1/1^2)*(1/(1^2+2^2))*(1/(1^2+2^2+3^2)) + ... - Alexander R. Povolotsky, Dec 10 2007
Convolution of A000290 with A000012. - Sergio Falcon, Feb 05 2008
Hankel transform of binomial(2*n-3, n-1) is -a(n). - Paul Barry, Feb 12 2008
Starting (1, 5, 14, 30, ...) = binomial transform of [1, 4, 5, 2, 0, 0, 0, ...]. - Gary W. Adamson, Jun 13 2008
Starting (1,5,14,30,...) = second partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = Sum_{i=0..n} binomial(n+2,i+2)*b(i), where b(i)=1,2,0,0,0,... - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
Convolution of A001477 with A005408: a(n) = Sum_{k=0..n} (2*k+1)*(n-k). - Reinhard Zumkeller, Mar 07 2009
Sequence of the absolute values of the z^1 coefficients of the polynomials in the GF1 denominators of A156921. See A157702 for background information. - Johannes W. Meijer, Mar 07 2009
The sequence is related to A000217 by a(n) = n*A000217(n) - Sum_{i=0..n-1} A000217(i) and this is the case d = 1 in the identity n^2*(d*n-d+2)/2 - Sum_{i=0..n-1} i*(d*i-d+2)/2 = n*(n+1)(2*d*n-2*d+3)/6, or also the case d = 0 in n^2*(n+2*d+1)/2 - Sum_{i=0..n-1} i*(i+2*d+1)/2 = n*(n+1)*(2*n+3*d+1)/6. - Bruno Berselli, Apr 21 2010, Apr 03 2012
a(n)/n = k^2 (k = integer) for n = 337; a(337) = 12814425, a(n)/n = 38025, k = 195, i.e., the number k = 195 is the quadratic mean (root mean square) of the first 337 positive integers. There are other such numbers -- see A084231 and A084232. - Jaroslav Krizek, May 23 2010
Also the number of moves to solve the "alternate coins game": given 2n+1 coins (n+1 Black, n White) set alternately in a row (BWBW...BWB) translate (not rotate) a pair of adjacent coins at a time (1 B and 1 W) so that at the end the arrangement shall be BBBBB..BW...WWWWW (Blacks separated by Whites). Isolated coins cannot be moved. - Carmine Suriano, Sep 10 2010
From J. M. Bergot, Aug 23 2011: (Start)
Using four consecutive numbers n, n+1, n+2, and n+3 take all possible pairs (n, n+1), (n, n+2), (n, n+3), (n+1, n+2), (n+1, n+3), (n+2, n+3) to create unreduced Pythagorean triangles. The sum of all six areas is 60*a(n+1).
Using three consecutive odd numbers j, k, m, (j+k+m)^3 - (j^3 + k^3 + m^3) equals 576*a(n) = 24^2*a(n) where n = (j+1)/2. (End)
From Ant King, Oct 17 2012: (Start)
For n > 0, the digital roots of this sequence A010888(a(n)) form the purely periodic 27-cycle {1, 5, 5, 3, 1, 1, 5, 6, 6, 7, 2, 2, 9, 7, 7, 2, 3, 3, 4, 8, 8, 6, 4, 4, 8, 9, 9}.
For n > 0, the units' digits of this sequence A010879(a(n)) form the purely periodic 20-cycle {1, 5, 4, 0, 5, 1, 0, 4, 5, 5, 6, 0, 9, 5, 0, 6, 5, 9, 0, 0}. (End)
Length of the Pisano period of this sequence mod n, n>=1: 1, 4, 9, 8, 5, 36, 7, 16, 27, 20, 11, 72, 13, 28, 45, 32, 17, 108, 19, 40, ... . - R. J. Mathar, Oct 17 2012
Sum of entries of n X n square matrix with elements min(i,j). - Enrique Pérez Herrero, Jan 16 2013
The number of intersections of diagonals in the interior of regular n-gon for odd n > 1 divided by n is a square pyramidal number; that is, A006561(2*n+1)/(2*n+1) = A000330(n-1) = (1/6)*n*(n-1)*(2*n-1). - Martin Renner, Mar 06 2013
For n > 1, a(n)/(2n+1) = A024702(m), for n such that 2n+1 = prime, which results in 2n+1 = A000040(m). For example, for n = 8, 2n+1 = 17 = A000040(7), a(8) = 204, 204/17 = 12 = A024702(7). - Richard R. Forberg, Aug 20 2013
A formula for the r-th successive summation of k^2, for k = 1 to n, is (2*n+r)*(n+r)!/((r+2)!*(n-1)!) (H. W. Gould). - Gary Detlefs, Jan 02 2014
The n-th square pyramidal number = the n-th triangular dipyramidal number (Johnson 12), which is the sum of the n-th + (n-1)-st tetrahedral numbers. E.g., the 3rd tetrahedral number is 10 = 1+3+6, the 2nd is 4 = 1+3. In triangular "dipyramidal form" these numbers can be written as 1+3+6+3+1 = 14. For "square pyramidal form", rebracket as 1+(1+3)+(3+6) = 14. - John F. Richardson, Mar 27 2014
Beukers and Top prove that no square pyramidal number > 1 equals a tetrahedral number A000292. - Jonathan Sondow, Jun 21 2014
Odd numbered entries are related to dissections of polygons through A100157. - Tom Copeland, Oct 05 2014
From Bui Quang Tuan, Apr 03 2015: (Start)
We construct a number triangle from the integers 1, 2, 3, ..., n as follows. The first column contains 2*n-1 integers 1. The second column contains 2*n-3 integers 2, ... The last column contains only one integer n. The sum of all the numbers in the triangle is a(n).
Here is an example with n = 5:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1
(End)
The Catalan number series A000108(n+3), offset 0, gives Hankel transform revealing the square pyramidal numbers starting at 5, A000330(n+2), offset 0 (empirical observation). - Tony Foster III, Sep 05 2016; see Dougherty et al. link p. 2. - Andrey Zabolotskiy, Oct 13 2016
Number of floating point additions in the factorization of an (n+1) X (n+1) real matrix by Gaussian elimination as e.g. implemented in LINPACK subroutines sgefa.f or dgefa.f. The number of multiplications is given by A007290. - Hugo Pfoertner, Mar 28 2018
The Jacobi polynomial P(n-1,-n+2,2,3) or equivalently the sum of dot products of vectors from the first n rows of Pascal's triangle (A007318) with the up-diagonal Chebyshev T coefficient vector (1,3,2,0,...) (A053120) or down-diagonal vector (1,-7,32,-120,400,...) (A001794). a(5) = 1 + (1,1).(1,3) + (1,2,1).(1,3,2) + (1,3,3,1).(1,3,2,0) + (1,4,6,4,1).(1,3,2,0,0) = (1 + (1,1).(1,-7) + (1,2,1).(1,-7,32) + (1,3,3,1).(1,-7,32,-120) + (1,4,6,4,1).(1,-7,32,-120,400))*(-1)^(n-1) = 55. - Richard Turk, Jul 03 2018
Coefficients in the terminating series identity 1 - 5*n/(n + 4) + 14*n*(n - 1)/((n + 4)*(n + 5)) - 30*n*(n - 1)*(n - 2)/((n + 4)*(n + 5)*(n + 6)) + ... = 0 for n = 1,2,3,.... Cf. A002415 and A108674. - Peter Bala, Feb 12 2019
n divides a(n) iff n == +- 1 (mod 6) (see A007310). (See De Koninck reference.) Examples: a(11) = 506 = 11 * 46, and a(13) = 819 = 13 * 63. - Bernard Schott, Jan 10 2020
For n > 0, a(n) is the number of ternary words of length n+2 having 3 letters equal to 2 and 0 only occurring as the last letter. For example, for n=2, the length 4 words are 2221,2212,2122,1222,2220. - Milan Janjic, Jan 28 2020
Conjecture: Every integer can be represented as a sum of three generalized square pyramidal numbers. A related conjecture is given in A336205 corresponding to pentagonal case. A stronger version of these conjectures is that every integer can be expressed as a sum of three generalized r-gonal pyramidal numbers for all r >= 3. In here "generalized" means negative indices are included. - Altug Alkan, Jul 30 2020
The natural number y is a term if and only if y = a(floor((3 * y)^(1/3))). - Robert Israel, Dec 04 2024
Also the number of directed bishop moves on an n X n chessboard, where two moves are considered the same if one can be obtained from the other by a rotation of the board. Reflections are ignored. Equivalently, number of directed bishop moves on an n X n chessboard, where two moves are considered the same if one can be obtained from the other by an axial reflection of the board (horizontal or vertical). Rotations and diagonal reflections are ignored. - Hilko Koning, Aug 22 2025

Examples

			G.f. = x + 5*x^2 + 14*x^3 + 30*x^4 + 55*x^5 + 91*x^6 + 140*x^7 + 204*x^8 + ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 813.
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover Publications, NY, 1964, p. 194.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 215,223.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 122, see #19 (3(1)), I(n); p. 155.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 47-49.
  • H. S. M. Coxeter, Polyhedral numbers, pp. 25-35 of R. S. Cohen, J. J. Stachel and M. W. Wartofsky, eds., For Dirk Struik: Scientific, historical and political essays in honor of Dirk J. Struik, Reidel, Dordrecht, 1974.
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p.165).
  • J. M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 310, pp. 46-196, Ellipses, Paris, 2004.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 2.
  • M. Gardner, Fractal Music, Hypercards and More, Freeman, NY, 1991, p. 293.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, p. 293.
  • M. Holt, Math puzzles and games, Walker Publishing Company, 1977, p. 2 and p. 89.
  • Simon Singh, The Simpsons and Their Mathematical Secrets. London: Bloomsbury Publishing PLC (2013): 188.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 126.

Crossrefs

Sums of 2 consecutive terms give A005900.
Column 0 of triangle A094414.
Column 1 of triangle A008955.
Right side of triangle A082652.
Row 2 of array A103438.
Partial sums of A000290.
Cf. similar sequences listed in A237616 and A254142.
Cf. |A084930(n, 1)|.
Cf. A253903 (characteristic function).
Cf. A034705 (differences of any two terms).

Programs

  • GAP
    List([0..30], n-> n*(n+1)*(2*n+1)/6); # G. C. Greubel, Dec 31 2019
  • Haskell
    a000330 n = n * (n + 1) * (2 * n + 1) `div` 6
    a000330_list = scanl1 (+) a000290_list
    -- Reinhard Zumkeller, Nov 11 2012, Feb 03 2012
    
  • Magma
    [n*(n+1)*(2*n+1)/6: n in [0..50]]; // Wesley Ivan Hurt, Jun 28 2014
    
  • Magma
    [0] cat [((2*n+3)*Binomial(n+2,2))/3: n in [0..40]]; // Vincenzo Librandi, Jul 30 2014
    
  • Maple
    A000330 := n -> n*(n+1)*(2*n+1)/6;
    a := n->(1/6)*n*(n+1)*(2*n+1): seq(a(n),n=0..53); # Emeric Deutsch
    with(combstruct): ZL:=[st, {st=Prod(left, right), left=Set(U, card=r), right=Set(U, card=r), U=Sequence(Z, card>=1)}, unlabeled]: subs(r=1, stack): seq(count(subs(r=2, ZL), size=m*2), m=1..45) ; # Zerinvary Lajos, Jan 02 2008
    nmax := 44; for n from 0 to nmax do fz(n) := product( (1-(2*m-1)*z)^(n+1-m) , m=1..n); c(n) := abs(coeff(fz(n),z,1)); end do: a := n-> c(n): seq(a(n), n=0..nmax); # Johannes W. Meijer, Mar 07 2009
  • Mathematica
    Table[Binomial[w+2, 3] + Binomial[w+1, 3], {w, 0, 30}]
    CoefficientList[Series[x(1+x)/(1-x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Jul 30 2014 *)
    Accumulate[Range[0,50]^2] (* Harvey P. Dale, Sep 25 2014 *)
  • Maxima
    A000330(n):=binomial(n+2,3)+binomial(n+1,3)$
    makelist(A000330(n),n,0,20); /* Martin Ettl, Nov 12 2012 */
    
  • PARI
    {a(n) = n * (n+1) * (2*n+1) / 6};
    
  • PARI
    upto(n) = [x*(x+1)*(2*x+1)/6 | x<-[0..n]] \\ Cino Hilliard, Jun 18 2007, edited by M. F. Hasler, Jan 02 2024
    
  • Python
    a=lambda n: (n*(n+1)*(2*n+1))//6 # Indranil Ghosh, Jan 04 2017
    
  • Sage
    [n*(n+1)*(2*n+1)/6 for n in (0..30)] # G. C. Greubel, Dec 31 2019
    

Formula

G.f.: x*(1+x)/(1-x)^4. - Simon Plouffe (in his 1992 dissertation: generating function for sequence starting at a(1))
E.g.f.: (x + 3*x^2/2 + x^3/3)*exp(x).
a(n) = n*(n+1)*(2*n+1)/6 = binomial(n+2, 3) + binomial(n+1, 3).
2*a(n) = A006331(n). - N. J. A. Sloane, Dec 11 1999
Can be extended to Z with a(n) = -a(-1-n) for all n in Z.
a(n) = A002492(n)/4. - Paul Barry, Jul 19 2003
a(n) = (((n+1)^4 - n^4) - ((n+1)^2 - n^2))/12. - Xavier Acloque, Oct 16 2003
From Alexander Adamchuk, Oct 26 2004: (Start)
a(n) = sqrt(A271535(n)).
a(n) = (Sum_{k=1..n} Sum_{j=1..n} Sum_{i=1..n} (i*j*k)^2)^(1/3). (End)
a(n) = Sum_{i=1..n} i*(2*n-2*i+1); sum of squares gives 1 + (1+3) + (1+3+5) + ... - Jon Perry, Dec 08 2004
a(n+1) = A000217(n+1) + 2*A000292(n). - Creighton Dement, Mar 10 2005
Sum_{n>=1} 1/a(n) = 6*(3-4*log(2)); Sum_{n>=1} (-1)^(n+1)*1/a(n) = 6*(Pi-3). - Philippe Deléham, May 31 2005
Sum of two consecutive tetrahedral (or pyramidal) numbers a(n) = A000292(n-1) + A000292(n). - Alexander Adamchuk, May 17 2006
Euler transform of length-2 sequence [ 5, -1 ]. - Michael Somos, Sep 04 2006
a(n) = a(n-1) + n^2. - Rolf Pleisch, Jul 22 2007
a(n) = A132121(n,0). - Reinhard Zumkeller, Aug 12 2007
a(n) = binomial(n, 2) + 2*binomial(n, 3). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009, corrected by M. F. Hasler, Jan 02 2024
a(n) = A168559(n) + 1 for n > 0. - Reinhard Zumkeller, Feb 03 2012
a(n) = Sum_{i=1..n} J_2(i)*floor(n/i), where J_2 is A007434. - Enrique Pérez Herrero, Feb 26 2012
a(n) = s(n+1, n)^2 - 2*s(n+1, n-1), where s(n, k) are Stirling numbers of the first kind, A048994. - Mircea Merca, Apr 03 2012
a(n) = A001477(n) + A000217(n) + A007290(n+2) + 1. - J. M. Bergot, May 31 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 2. - Ant King, Oct 17 2012
a(n) = Sum_{i = 1..n} Sum_{j = 1..n} min(i,j). - Enrique Pérez Herrero, Jan 15 2013
a(n) = A000217(n) + A007290(n+1). - Ivan N. Ianakiev, May 10 2013
a(n) = (A047486(n+2)^3 - A047486(n+2))/24. - Richard R. Forberg, Dec 25 2013
a(n) = Sum_{i=0..n-1} (n-i)*(2*i+1), with a(0) = 0. After 0, row sums of the triangle in A101447. - Bruno Berselli, Feb 10 2014
a(n) = n + 1 + Sum_{i=1..n+1} (i^2 - 2i). - Wesley Ivan Hurt, Feb 25 2014
a(n) = A000578(n+1) - A002412(n+1). - Wesley Ivan Hurt, Jun 28 2014
a(n) = Sum_{i = 1..n} Sum_{j = i..n} max(i,j). - Enrique Pérez Herrero, Dec 03 2014
a(n) = A055112(n)/6, see Singh (2013). - Alonso del Arte, Feb 20 2015
For n >= 2, a(n) = A028347(n+1) + A101986(n-2). - Bui Quang Tuan, Apr 03 2015
For n > 0: a(n) = A258708(n+3,n-1). - Reinhard Zumkeller, Jun 23 2015
a(n) = A175254(n) + A072481(n), n >= 1. - Omar E. Pol, Aug 12 2015
a(n) = A000332(n+3) - A000332(n+1). - Antal Pinter, Dec 27 2015
Dirichlet g.f.: zeta(s-3)/3 + zeta(s-2)/2 + zeta(s-1)/6. - Ilya Gutkovskiy, Jun 26 2016
a(n) = A080851(2,n-1). - R. J. Mathar, Jul 28 2016
a(n) = (A005408(n) * A046092(n))/12 = (2*n+1)*(2*n*(n+1))/12. - Bruce J. Nicholson, May 18 2017
12*a(n) = (n+1)*A001105(n) + n*A001105(n+1). - Bruno Berselli, Jul 03 2017
a(n) = binomial(n-1, 1) + binomial(n-1, 2) + binomial(n, 3) + binomial(n+1, 2) + binomial(n+1, 3). - Tony Foster III, Aug 24 2018
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Nathan Fox, Dec 04 2019
Let T(n) = A000217(n), the n-th triangular number. Then a(n) = (T(n)+1)^2 + (T(n)+2)^2 + ... + (T(n)+n)^2 - (n+2)*T(n)^2. - Charlie Marion, Dec 31 2019
a(n) = 2*n - 1 - a(n-2) + 2*a(n-1). - Boštjan Gec, Nov 09 2023
a(n) = 2/(2*n)! * Sum_{j = 1..n} (-1)^(n+j) * j^(2*n+2) * binomial(2*n, n-j). Cf. A060493. - Peter Bala, Mar 31 2025

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A111418 Right-hand side of odd-numbered rows of Pascal's triangle.

Original entry on oeis.org

1, 3, 1, 10, 5, 1, 35, 21, 7, 1, 126, 84, 36, 9, 1, 462, 330, 165, 55, 11, 1, 1716, 1287, 715, 286, 78, 13, 1, 6435, 5005, 3003, 1365, 455, 105, 15, 1, 24310, 19448, 12376, 6188, 2380, 680, 136, 17, 1, 92378, 75582, 50388
Offset: 0

Views

Author

Philippe Deléham, Nov 13 2005

Keywords

Comments

Riordan array (c(x)/sqrt(1-4*x),x*c(x)^2) where c(x) is g.f. of A000108. Unsigned version of A113187. Diagonal sums are A014301(n+1).
Triangle T(n,k),0<=k<=n, read by rows defined by :T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=3*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+2*T(n-1,k)+T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 22 2007
Reversal of A122366. - Philippe Deléham, Mar 22 2007
Column k has e.g.f. exp(2x)(Bessel_I(k,2x)+Bessel_I(k+1,2x)). - Paul Barry, Jun 06 2007
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1 . Other triangles arise by choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0)-> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007
Diagonal sums are A014301(n+1). - Paul Barry, Mar 08 2011
This triangle T(n,k) appears in the expansion of odd powers of Fibonacci numbers F=A000045 in terms of F-numbers with multiples of odd numbers as indices. See the Ozeki reference, p. 108, Lemma 2. The formula is: F_l^(2*n+1) = sum(T(n,k)*(-1)^((n-k)*(l+1))* F_{(2*k+1)*l}, k=0..n)/5^n, n >= 0, l >= 0. - Wolfdieter Lang, Aug 24 2012
Central terms give A052203. - Reinhard Zumkeller, Mar 14 2014
This triangle appears in the expansion of (4*x)^n in terms of the polynomials Todd(n, x):= T(2*n+1, sqrt(x))/sqrt(x) = sum(A084930(n,m)*x^m), n >= 0. This follows from the inversion of the lower triangular Riordan matrix built from A084930 and comparing the g.f. of the row polynomials. - Wolfdieter Lang, Aug 05 2014
From Wolfdieter Lang, Aug 15 2014: (Start)
This triangle is the inverse of the signed Riordan triangle (-1)^(n-m)*A111125(n,m).
This triangle T(n,k) appears in the expansion of x^n in terms of the polynomials todd(k, x):= T(2*k+1, sqrt(x)/2)/(sqrt(x)/2) = S(k, x-2) - S(k-1, x-2) with the row polynomials T and S for the triangles A053120 and A049310, respectively: x^n = sum(T(n,k)*todd(k, x), k=0..n). Compare this with the preceding comment.
The A- and Z-sequences for this Riordan triangle are [1, 2, 1, repeated 0] and [3, 1, repeated 0]. For A- and Z-sequences for Riordan triangles see the W. Lang link under A006232. This corresponds to the recurrences given in the Philippe Deléham, Mar 22 2007 comment above. (End)

Examples

			From _Wolfdieter Lang_, Aug 05 2014: (Start)
The triangle T(n,k) begins:
n\k      0      1      2      3     4     5    6    7   8  9  10 ...
0:       1
1:       3      1
2:      10      5      1
3:      35     21      7      1
4:     126     84     36      9     1
5:     462    330    165     55    11     1
6:    1716   1287    715    286    78    13    1
7:    6435   5005   3003   1365   455   105   15    1
8:   24310  19448  12376   6188  2380   680  136   17   1
9:   92378  75582  50388  27132 11628  3876  969  171  19  1
10: 352716 293930 203490 116280 54264 20349 5985 1330 210 21   1
...
Expansion examples (for the Todd polynomials see A084930 and a comment above):
(4*x)^2 = 10*Todd(n,  0) + 5*Todd(n, 1) + 1*Todd(n, 2) = 10*1 + 5*(-3 + 4*x) + 1*(5 - 20*x + 16*x^2).
(4*x)^3 =  35*1 + 21*(-3 + 4*x) + 7*(5 - 20*x + 16*x^2) + (-7 + 56*x - 112*x^2 +64*x^3)*1. (End)
---------------------------------------------------------------------
Production matrix is
3, 1,
1, 2, 1,
0, 1, 2, 1,
0, 0, 1, 2, 1,
0, 0, 0, 1, 2, 1,
0, 0, 0, 0, 1, 2, 1,
0, 0, 0, 0, 0, 1, 2, 1,
0, 0, 0, 0, 0, 0, 1, 2, 1,
0, 0, 0, 0, 0, 0, 0, 1, 2, 1
- _Paul Barry_, Mar 08 2011
Application to odd powers of Fibonacci numbers F, row n=2:
F_l^5 = (10*(-1)^(2*(l+1))*F_l + 5*(-1)^(1*(l+1))*F_{3*l} + 1*F_{5*l})/5^2, l >= 0. - _Wolfdieter Lang_, Aug 24 2012
		

Crossrefs

Programs

  • Haskell
    a111418 n k = a111418_tabl !! n !! k
    a111418_row n = a111418_tabl !! n
    a111418_tabl = map reverse a122366_tabl
    -- Reinhard Zumkeller, Mar 14 2014
  • Mathematica
    Table[Binomial[2*n+1, n-k], {n,0,10}, {k,0,n}] (* G. C. Greubel, May 22 2017 *)
    T[0, 0, x_, y_] := 1; T[n_, 0, x_, y_] := x*T[n - 1, 0, x, y] + T[n - 1, 1, x, y]; T[n_, k_, x_, y_] := T[n, k, x, y] = If[k < 0 || k > n, 0,
    T[n - 1, k - 1, x, y] + y*T[n - 1, k, x, y] + T[n - 1, k + 1, x, y]];
    Table[T[n, k, 3, 2], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, May 22 2017 *)

Formula

T(n, k) = C(2*n+1, n-k).
Sum_{k=0..n} T(n, k) = 4^n.
Sum_{k, 0<=k<=n}(-1)^k *T(n,k) = binomial(2*n,n) = A000984(n). - Philippe Deléham, Mar 22 2007
T(n,k) = sum{j=k..n, C(n,j)*2^(n-j)*C(j,floor((j-k)/2))}. - Paul Barry, Jun 06 2007
Sum_{k, k>=0} T(m,k)*T(n,k) = T(m+n,0)= A001700(m+n). - Philippe Deléham, Nov 22 2009
G.f. row polynomials: ((1+x) - (1-x)/sqrt(1-4*z))/(2*(x - (1+x)^2*z))
(see the Riordan property mentioned in a comment above). - Wolfdieter Lang, Aug 05 2014

A050486 a(n) = binomial(n+6,6)*(2n+7)/7.

Original entry on oeis.org

1, 9, 44, 156, 450, 1122, 2508, 5148, 9867, 17875, 30888, 51272, 82212, 127908, 193800, 286824, 415701, 591261, 826804, 1138500, 1545830, 2072070, 2744820, 3596580, 4665375, 5995431, 7637904, 9651664, 12104136, 15072200, 18643152, 22915728, 28001193
Offset: 0

Views

Author

Barry E. Williams, Dec 26 1999

Keywords

Comments

If a 2-set Y and an (n-3)-set Z are disjoint subsets of an n-set X then a(n-8) is the number of 8-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 08 2007
7-dimensional square numbers, sixth partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = Sum_{i=0..n} C(n+6,i+6)*b(i), where b(i) = [1,2,0,0,0,...]. - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
2*a(n) is number of ways to place 6 queens on an (n+6) X (n+6) chessboard so that they diagonally attack each other exactly 15 times. The maximal possible attack number, p=binomial(k,2)=15 for k=6 queens, is achievable only when all queens are on the same diagonal. In graph-theory representation they thus form a corresponding complete graph. - Antal Pinter, Dec 27 2015
Coefficients in the terminating series identity 1 - 9*n/(n + 8) + 44*n*(n - 1)/((n + 8)*(n + 9)) - 156*n*(n - 1)*(n - 2)/((n + 8)*(n + 9)*(n + 10)) + ... = 0 for n = 1,2,3,.... Cf. A005585 and A053347. - Peter Bala, Feb 18 2019

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Crossrefs

Partial sums of A040977, A005585.
Fourth column (s=3, without leading zeros) of A111125. - Wolfdieter Lang, Oct 18 2012
Cf. A084960 (unsigned fourth column divided by 64). - Wolfdieter Lang, Aug 04 2014

Programs

  • Magma
    [Binomial(n+6, 6) + 2*Binomial(n+6, 7): n in [0..35]]; // Vincenzo Librandi, Jun 09 2013
    
  • Maple
    A050486:=n->binomial(n+6,6)*(2*n+7)/7: seq(A050486(n), n=0..50); # Wesley Ivan Hurt, Jan 01 2016
  • Mathematica
    CoefficientList[Series[(1 + x) / (1 - x)^8, {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)
    Table[SeriesCoefficient[(1 + x)/(1 - x)^8, {x, 0, n}], {n, 0, 28}] (* or *)
    Table[Binomial[n + 6, 6] (2 n + 7)/7, {n, 0, 30}] (* Michael De Vlieger, Dec 31 2015 *)
  • PARI
    a(n)=binomial(n+6,6)*(2*n+7)/7 \\ Charles R Greathouse IV, Sep 24 2015
    
  • Python
    A050486_list, m = [], [2]+[1]*7
    for _ in range(10**2):
        A050486_list.append(m[-1])
        for i in range(7):
            m[i+1] += m[i] # Chai Wah Wu, Jan 24 2016

Formula

a(n) = (-1)^n*A053120(2*n+7, 7)/64 (1/64 of eighth unsigned column of Chebyshev T-triangle, zeros omitted).
G.f.: (1+x)/(1-x)^8.
a(n) = 2*C(n+7, 7)-C(n+6, 6). - Paul Barry, Mar 04 2003
a(n) = C(n+6,6)+2*C(n+6,7). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
a(n) = (-1)^n*A084930(n+3, 3)/64. Compare with the first line above. - Wolfdieter Lang, Aug 04 2014
a(n) = 8*a(n-1)-28*a(n-2)+56*a(n-3)-70*a(n-4)+56*a(n-5)-28*a(n-6)+8*a(n-7)-a(n-8) for n>7. - Wesley Ivan Hurt, Jan 01 2016
From Amiram Eldar, Jan 25 2022: (Start)
Sum_{n>=0} 1/a(n) = 24871/25 - 7168*log(2)/5.
Sum_{n>=0} (-1)^n/a(n) = 1792*Pi/5 - 28126/25. (End)

A098597 Numerator of Catalan(n)/2^(2n+1). Also, numerators of (2n-1)!!/(n+1)!. Odd part of the n-th Catalan number.

Original entry on oeis.org

1, 1, 1, 5, 7, 21, 33, 429, 715, 2431, 4199, 29393, 52003, 185725, 334305, 9694845, 17678835, 64822395, 119409675, 883631595, 1641030105, 6116566755, 11435320455, 171529806825, 322476036831, 1215486600363, 2295919134019, 17383387729001, 32968493968795, 125280277081421
Offset: 0

Views

Author

Michael Somos, Sep 15 2004

Keywords

Comments

Also numerators of g.f. c(x/2) = (1-sqrt(1-2x))/x where c(x) = g.f. of A000108. - Paul Barry, Sep 04 2007
Also numerator of x(n)=Sum(x(k)*x(n-k-1):0<=kA086117(n). - Reinhard Zumkeller, Feb 06 2008
Also numerator of (1/Pi)*int(x^n*sqrt((1-x)/x), x=0..1). - Groux Roland, Mar 17 2011
The negative of this sequence appears in the A-sequence of the Riordan triangle A084930 as numerators 4, -2, -seq(a(n-1), n >= 2). The denominators look like 1, seq(A120777(n-1), n >= 1). - Wolfdieter Lang, Aug 04 2014
The series of a(n)/A046161(n+1) is absolutely convergent to 1. - Ralf Steiner, Feb 09 2017

Examples

			1/(1 + sqrt(1-x)) = 1/2 + 1/8*x + 1/16*x^2 + 5/128*x^3 + 7/256*x^4 + ...
		

Crossrefs

Cf. Equals A000265(A000108(n)).
Essentially the absolute values of A002596. Cf. A000108, A001795.

Programs

  • Magma
    [Numerator(Catalan(n)/2^(2*n+1)):n in [0..30]]; // Vincenzo Librandi, Jan 14 2016
  • Maple
    a:= n-> abs(numer(binomial(1/2, n+1))): seq(a(n), n=0..50); # Alois P. Heinz, Apr 10 2009
  • Mathematica
    Table[Numerator[CatalanNumber[n]/2^(2n+1)],{n,0,30}] (* Harvey P. Dale, Jul 27 2011 *)
    A098597[n_] := With[{c = CatalanNumber[n]}, c / 2^IntegerExponent[c, 2]];
    Table[A098597[n], {n, 0, 29}]  (* Peter Luschny, Apr 16 2024 *)
  • PARI
    {a(n) = if( n < 0, 0, numerator(polcoeff(1 / (1 + sqrt(1 - x + x * O(x^n))), n)))};
    

Formula

Numerators of g.f.: 1/(1 + sqrt(1-x)).
a(n) = A000108(n) / 2^A048881(n).

Extensions

Edited by Ralf Stephan, Dec 28 2004

A082985 Coefficient table for Chebyshev's U(2*n,x) polynomial expanded in decreasing powers of (-4*(1-x^2)).

Original entry on oeis.org

1, 1, 3, 1, 5, 5, 1, 7, 14, 7, 1, 9, 27, 30, 9, 1, 11, 44, 77, 55, 11, 1, 13, 65, 156, 182, 91, 13, 1, 15, 90, 275, 450, 378, 140, 15, 1, 17, 119, 442, 935, 1122, 714, 204, 17, 1, 19, 152, 665, 1729, 2717, 2508, 1254, 285, 19
Offset: 0

Views

Author

Gary W. Adamson, May 29 2003

Keywords

Comments

Sum of row #n = A000204(2n+1), i.e., A002878(n).
Row #n has the unsigned coefficients of a polynomial whose roots are 2 sin(2*Pi*k/(2n+1)) [for k=1 to 2n].
The positive roots are the diagonal lengths of a regular (2n+1)-gon, inscribed in the unit circle.
Polynomial of row #n = Sum_{m=0..n} (-1)^m T(n,m) x^(2n-2m).
This is also the unsigned coefficient table of Chebyshev's 2*T(2*n+1,x) polynomials expanded in decreasing odd powers of 2*x. - Wolfdieter Lang, Mar 07 2007
The n-th row are the coefficients of the polynomial S(n) where S(0)=1, S(1)=x+3, and S(n) = (x+2)*S(n-1) - S(n-2) (see Sun link). - Michel Marcus, Mar 07 2016

Examples

			Expansion of polynomials:
  x^0;
  x^2  -  3*x^0;
  x^4  -  5*x^2 +  5*x^0;
  x^6  -  7*x^4 + 14*x^2 -  7*x^0;
  x^8  -  9*x^6 + 27*x^4 - 30*x^2 +  9*x^0;
  x^10 - 11*x^8 + 44*x^6 - 77*x^4 + 55*x^2 - 11*x^0; ...
Polynomial #4 has 8 roots: 2*sin(2*Pi*k/9) for k=1 to 8.
Coefficients (with signs removed) are
  1;
  1,  3;
  1,  5,  5;
  1,  7, 14,  7;
  1,  9, 27, 30,  9;
  1, 11, 44, 77, 55, 11;
  ...
		

References

  • J. D'Angelo, Several Complex Variables and the Geometry of Real Hypersurfaces, CRC Press, 1992; see pp. 151,175.
  • Stephen Eberhart, "Mathematical-Physical Correspondence," Number 37-38, Jan 08, 1982.
  • Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990.

Crossrefs

Cf. companion triangle A084534.

Programs

  • GAP
    Flat(List([0..10], n-> List([0..n], k-> Binomial(2*n-k,k)*(2*n+1)/(2*n-2*k+1) ))); # G. C. Greubel, Dec 30 2019
  • Magma
    [Binomial(2*n-k,k)*(2*n+1)/(2*n-2*k+1): k in [0..n], n in [0..10]]; // G. C. Greubel, Dec 30 2019
    
  • Maple
    A082985 := proc(n,m)
        binomial(2*n-m,m)*(2*n+1)/(2*n-2*m+1) ;
    end proc: # R. J. Mathar, Sep 08 2013
  • Mathematica
    T[n_, m_]:= Binomial[2*n-m, m]*(2*n+1)/(2*n-2*m+1); Table[T[n, m], {n, 0, 9}, {m, 0, n}]//Flatten (* Jean-François Alcover, Oct 08 2013, after R. J. Mathar *)
  • PARI
    T(n,k)=binomial(2*n-k,k)*(2*n+1)/(2*n-2*k+1); \\ G. C. Greubel, Dec 30 2019
    
  • Sage
    [[binomial(2*n-k,k)*(2*n+1)/(2*n-2*k+1) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Dec 30 2019
    

Formula

Triangle read by rows: row #n has n+1 terms. T(n,0)=1, T(n,n)=2n+1, T(n,m) = T(n-1,m-1) + Sum_{k=0..m} T(n-1-k, m-k).
T(k, s) = ((2k+1)/(2s+1))*binomial(k+s, 2s), 0 <= s <= k; then transpose the triangle. - Gary W. Adamson, May 29 2003
From Wolfdieter Lang, Mar 07 2007: (Start)
Signed version: a(n,m)=0 if n < m, otherwise a(n,m) = ((-1)^m)*binomial(2*n+1-m,m)*(2*n+1)/(2*n+1-m). From the Rivlin reference, p. 37, eq.(1.92), using the differential eq. for T(2*n+1,x). Also from Waring's formula.
Signed version: a(n,m)=0 if n < m, otherwise a(n,m) = ((-1)^m)*(Sum_{k=0..n-m} binomial(m+k,k)*binomial(2*n+1,2*(k+m)))/2^(2*(n-m)). Proof: De Moivre's formula for cos((2*n+1)*phi) rewritten in terms of odd powers of cos(phi). Cf. Rivlin reference p. 4, eq.(1.10).
Signed version: a(n,m) = A084930(n,n-m)/2^(2*(n-m)) (scaled coefficients of Chebyshev's T(2*n+1,x), decreasing odd powers).
Unsigned version: a(n,m)=0 if n < m, otherwise a(n,m) = binomial(2*n-m,m)*(2*n+1)/(2*(n-m)+1). From the differential eq. for U(2*n,x). (End)
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-2). - Philippe Deléham, Feb 24 2012
Sum_{i>=0} T(n-i,n-2*i) = A003945(n). - Philippe Deléham, Feb 24 2012
Sum_{i>=0} T(n-i, n-2*i)*4^i = 3^n = A000244(n). - Philippe Deléham, Feb 24 2012
From Paul Weisenhorn Nov 25 2019: (Start)
T(r,k) = binomial(2*r-k,k-1) + binomial(2*r-1-k,k-2) with 1 <= r and 1 <= k <= r.
For a given n, one gets r = floor((1+sqrt(8*n))/2), k = n-(r^2-r)/2, a(n) = binomial(2*r-k,k-1) + binomial(2*r-1-k,k-2). (End)

Extensions

Edited by Anne Donovan (anned3005(AT)aol.com), Jun 11 2003
Re-edited by Don Reble, Nov 12 2005

A127674 Coefficient table for Chebyshev polynomials T(2*n,x) (increasing even powers x, without zeros).

Original entry on oeis.org

1, -1, 2, 1, -8, 8, -1, 18, -48, 32, 1, -32, 160, -256, 128, -1, 50, -400, 1120, -1280, 512, 1, -72, 840, -3584, 6912, -6144, 2048, -1, 98, -1568, 9408, -26880, 39424, -28672, 8192, 1, -128, 2688, -21504, 84480, -180224, 212992, -131072, 32768, -1, 162, -4320, 44352, -228096, 658944, -1118208
Offset: 0

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Author

Wolfdieter Lang Mar 07 2007

Keywords

Comments

Let C_n be the root lattice generated as a monoid by {+-2*e_i: 1 <= i <= n; +-e_i +- e_j: 1 <= i not equal to j <= n}. Let P(C_n) be the polytope formed by the convex hull of this generating set. Then the rows of (the signless version of) this array are the f-vectors of a unimodular triangulation of P(C_n) [Ardila et al.]. See A086645 for the corresponding array of h-vectors for these type C_n polytopes. See A063007 for the array of f-vectors for type A_n polytopes and A108556 for the array of f-vectors associated with type D_n polytopes. - Peter Bala, Oct 23 2008

Examples

			[1];
[-1,2];
[1,-8,8];
[-1,18,-48,32];
[1,-32,160,-256,128];
...
See a link for the row polynomials.
The T-polynomial for row n=3, [-1,18,-48,32], is T(2*3,x) =  -1*x^0 + 18*x^2 - 48*x^4 + 32*x^6.
		

References

  • Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990. p. 37, eq.(1.96) and p. 4. eq.(1.10).

Crossrefs

Cf. A075733 (different signs and offset). A084930 (coefficients of odd-indexed T-polynomials).
Cf. A053120 (coefficients of T-polynomials, with interspersed zeros).

Formula

a(n,m) = 0 if n < m, a(0,0) = 1; otherwise a(n,m) = ((-1)^(n-m))*(2^(2*m-1))*binomial(n+m,2*m)*(2*n)/(n+m).
O.g.f.: (1 + z*(1 - 2*x))/((1 + z)^2 - 4*x*z) = 1 + (-1 + 2*x)*z + (1 - 8*x + 8*x^2)*z^2 + ... . [Peter Bala, Oct 23 2008] For the t-polynomials actually with x -> x^2. - Wolfdieter Lang, Aug 02 2014
Denoting the row polynomials by R(n,x) we have exp( Sum_{n >= 1} R(n,x)*z^n/n ) = 1/sqrt( (1 + z)^2 - 4*x*z ) = 1 + (-1 + 2*x)*z + (1 - 6*x + 6*x^2)*z^2 + ..., the o.g.f. for a signed version of A063007. - Peter Bala, Sep 02 2015
The n-th row polynomial equals T(n, 2*x - 1). - Peter Bala, Jul 09 2023

A127675 Coefficient table for Chebyshev's U(2*n,x) polynomials in decreasing powers of (1-x^2).

Original entry on oeis.org

1, -4, 3, 16, -20, 5, -64, 112, -56, 7, 256, -576, 432, -120, 9, -1024, 2816, -2816, 1232, -220, 11, 4096, -13312, 16640, -9984, 2912, -364, 13, -16384, 61440, -92160, 70400, -28800, 6048, -560, 15, 65536, -278528, 487424, -452608, 239360, -71808, 11424, -816, 17, -262144, 1245184
Offset: 0

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Author

Wolfdieter Lang, Mar 07 2007

Keywords

Comments

This table gives therefore sin((2*n+1)*phi) in terms of falling odd powers of sin(phi).
The unsigned triangle with reversed rows is A084930 (the signs differ).

Examples

			[1];[ -4,3];[16,-20,5];[ -64,112,-56,7];[256,-576,432,-120,9]; ...
Row n=3: -64*(1-x^2)^3+ 112*(1-x^2)^2 -56*(1-x^2)^1 + 7 = 64*x^6 - 80*x^4 + 24* x^2 -1 =U(6,x).
Row n=3: sin(7*phi)=-64*sin(phi)^7 + 112*sin(phi)^5 - 56*sin(phi)^3 + 7*sin(phi).
		

Crossrefs

Row sums (signed) A033999(n)=(-1)^n. Row sums (unsigned) A002315(n).
Cf. A082985 (scaled coefficient table).

Formula

a(n,m)=0 if n < m else a(n,m) = ((-4)^(n-m))*binomial(2n-m,m)*(2*n+1)/(2*(n-m)+1), n >= m >= 0. (Proof from the differential eq. for U(2*n,x): (1-x^2)*(d^2/dx^2)U(2*n,x) - 3*x*(d/dx)U(2*n,x) + 4*n*(n+1)*U(2*n,x) = 0.)
a(n,m)=0 if n < m else a(n,m) = Sum_{k=0..n-m} (binomial(m+k,k)*binomial(2*n+1,2*(m+k))*(-1)^(n-m)) (from de Moivre's formula for sin((2*n+1)*phi) after replacing cos(phi)^2 with 1-sin(phi)^2).
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