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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A000384 Hexagonal numbers: a(n) = n*(2*n-1).

Original entry on oeis.org

0, 1, 6, 15, 28, 45, 66, 91, 120, 153, 190, 231, 276, 325, 378, 435, 496, 561, 630, 703, 780, 861, 946, 1035, 1128, 1225, 1326, 1431, 1540, 1653, 1770, 1891, 2016, 2145, 2278, 2415, 2556, 2701, 2850, 3003, 3160, 3321, 3486, 3655, 3828, 4005, 4186, 4371, 4560
Offset: 0

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Author

Keywords

Comments

Number of edges in the join of two complete graphs, each of order n, K_n * K_n. - Roberto E. Martinez II, Jan 07 2002
The power series expansion of the entropy function H(x) = (1+x)log(1+x) + (1-x)log(1-x) has 1/a_i as the coefficient of x^(2i) (the odd terms being zero). - Tommaso Toffoli (tt(AT)bu.edu), May 06 2002
Partial sums of A016813 (4n+1). Also with offset = 0, a(n) = (2n+1)(n+1) = A005408 * A000027 = 2n^2 + 3n + 1, i.e., a(0) = 1. - Jeremy Gardiner, Sep 29 2002
Sequence also gives the greatest semiperimeter of primitive Pythagorean triangles having inradius n-1. Such a triangle has consecutive longer sides, with short leg 2n-1, hypotenuse a(n) - (n-1) = A001844(n), and area (n-1)*a(n) = 6*A000330(n-1). - Lekraj Beedassy, Apr 23 2003
Number of divisors of 12^(n-1), i.e., A000005(A001021(n-1)). - Henry Bottomley, Oct 22 2001
More generally, if p1 and p2 are two arbitrarily chosen distinct primes then a(n) is the number of divisors of (p1^2*p2)^(n-1) or equivalently of any member of A054753^(n-1). - Ant King, Aug 29 2011
Number of standard tableaux of shape (2n-1,1,1) (n>=1). - Emeric Deutsch, May 30 2004
It is well known that for n>0, A014105(n) [0,3,10,21,...] is the first of 2n+1 consecutive integers such that the sum of the squares of the first n+1 such integers is equal to the sum of the squares of the last n; e.g., 10^2 + 11^2 + 12^2 = 13^2 + 14^2.
Less well known is that for n>1, a(n) [0,1,6,15,28,...] is the first of 2n consecutive integers such that sum of the squares of the first n such integers is equal to the sum of the squares of the last n-1 plus n^2; e.g., 15^2 + 16^2 + 17^2 = 19^2 + 20^2 + 3^2. - Charlie Marion, Dec 16 2006
a(n) is also a perfect number A000396 when n is an even superperfect number A061652. - Omar E. Pol, Sep 05 2008
Sequence found by reading the line from 0, in the direction 0, 6, ... and the line from 1, in the direction 1, 15, ..., in the square spiral whose vertices are the generalized hexagonal numbers A000217. - Omar E. Pol, Jan 09 2009
For n>=1, 1/a(n) = Sum_{k=0..2*n-1} ((-1)^(k+1)*binomial(2*n-1,k)*binomial(2*n-1+k,k)*H(k)/(k+1)) with H(k) harmonic number of order k.
The number of possible distinct colorings of any 2 colors chosen from n colors of a square divided into quadrants. - Paul Cleary, Dec 21 2010
Central terms of the triangle in A051173. - Reinhard Zumkeller, Apr 23 2011
For n>0, a(n-1) is the number of triples (w,x,y) with all terms in {0,...,n} and max(|w-x|,|x-y|) = |w-y|. - Clark Kimberling, Jun 12 2012
a(n) is the number of positions of one domino in an even pyramidal board with base 2n. - César Eliud Lozada, Sep 26 2012
Partial sums give A002412. - Omar E. Pol, Jan 12 2013
Let a triangle have T(0,0) = 0 and T(r,c) = |r^2 - c^2|. The sum of the differences of the terms in row(n) and row(n-1) is a(n). - J. M. Bergot, Jun 17 2013
With T_(i+1,i)=a(i+1) and all other elements of the lower triangular matrix T zero, T is the infinitesimal generator for A176230, analogous to A132440 for the Pascal matrix. - Tom Copeland, Dec 11 2013
a(n) is the number of length 2n binary sequences that have exactly two 1's. a(2) = 6 because we have: {0,0,1,1}, {0,1,0,1}, {0,1,1,0}, {1,0,0,1}, {1,0,1,0}, {1,1,0,0}. The ordinary generating function with interpolated zeros is: (x^2 + 3*x^4)/(1-x^2)^3. - Geoffrey Critzer, Jan 02 2014
For n > 0, a(n) is the largest integer k such that k^2 + n^2 is a multiple of k + n. More generally, for m > 0 and n > 0, the largest integer k such that k^(2*m) + n^(2*m) is a multiple of k + n is given by k = 2*n^(2*m) - n. - Derek Orr, Sep 04 2014
Binomial transform of (0, 1, 4, 0, 0, 0, ...) and second partial sum of (0, 1, 4, 4, 4, ...). - Gary W. Adamson, Oct 05 2015
a(n) also gives the dimension of the simple Lie algebras D_n, for n >= 4. - Wolfdieter Lang, Oct 21 2015
For n > 0, a(n) equals the number of compositions of n+11 into n parts avoiding parts 2, 3, 4. - Milan Janjic, Jan 07 2016
Also the number of minimum dominating sets and maximal irredundant sets in the n-cocktail party graph. - Eric W. Weisstein, Jun 29 and Aug 17 2017
As Beedassy's formula shows, this Hexagonal number sequence is the odd bisection of the Triangle number sequence. Both of these sequences are figurative number sequences. For A000384, a(n) can be found by multiplying its triangle number by its hexagonal number. For example let's use the number 153. 153 is said to be the 17th triangle number but is also said to be the 9th hexagonal number. Triangle(17) Hexagonal(9). 17*9=153. Because the Hexagonal number sequence is a subset of the Triangle number sequence, the Hexagonal number sequence will always have both a triangle number and a hexagonal number. n* (2*n-1) because (2*n-1) renders the triangle number. - Bruce J. Nicholson, Nov 05 2017
Also numbers k with the property that in the symmetric representation of sigma(k) the smallest Dyck path has a central valley and the largest Dyck path has a central peak, n >= 1. Thus all hexagonal numbers > 0 have middle divisors. (Cf. A237593.) - Omar E. Pol, Aug 28 2018
k^a(n-1) mod n = 1 for prime n and k=2..n-1. - Joseph M. Shunia, Feb 10 2019
Consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z: a(n+1) gives the semiperimeter of related triangles; A005408, A046092 and A001844 give the X, Y and Z values. - Ralf Steiner, Feb 25 2020
See A002939(n) = 2*a(n) for the corresponding perimeters. - M. F. Hasler, Mar 09 2020
It appears that these are the numbers k with the property that the smallest subpart in the symmetric representation of sigma(k) is 1. - Omar E. Pol, Aug 28 2021
The above conjecture is true. See A280851 for a proof. - Hartmut F. W. Hoft, Feb 02 2022
The n-th hexagonal number equals the sum of the n consecutive integers with the same parity starting at n; for example, 1, 2+4, 3+5+7, 4+6+8+10, etc. In general, the n-th 2k-gonal number is the sum of the n consecutive integers with the same parity starting at (k-2)*n - (k-3). When k = 1 and 2, this result generates the positive integers, A000027, and the squares, A000290, respectively. - Charlie Marion, Mar 02 2022
Conjecture: For n>0, min{k such that there exist subsets A,B of {0,1,2,...,a(n)} such that |A|=|B|=k and A+B={0,1,2,...,2*a(n)}} = 2*n. - Michael Chu, Mar 09 2022

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, pp. 77-78. (In the integral formula on p. 77 a left bracket is missing for the cosine argument.)
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 38.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
  • 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.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 53-54, 129-130, 132.
  • 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).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 21.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See pp. 122-123.

Crossrefs

a(n)= A093561(n+1, 2), (4, 1)-Pascal column.
a(n) = A100345(n, n-1) for n>0.
Cf. A002939 (twice a(n): sums of Pythagorean triples (X, Y, Z=Y+1)).
Cf. A280851.

Programs

  • Haskell
    a000384 n = n * (2 * n - 1)
    a000384_list = scanl (+) 0 a016813_list
    -- Reinhard Zumkeller, Dec 16 2012
    
  • Maple
    A000384:=n->n*(2*n-1); seq(A000384(k), k=0..100); # Wesley Ivan Hurt, Sep 27 2013
  • Mathematica
    Table[n*(2 n - 1), {n, 0, 100}] (* Wesley Ivan Hurt, Sep 27 2013 *)
    LinearRecurrence[{3, -3, 1}, {0, 1, 6}, 50] (* Harvey P. Dale, Sep 10 2015 *)
    Join[{0}, Accumulate[Range[1, 312, 4]]] (* Harvey P. Dale, Mar 26 2016 *)
    (* For Mathematica 10.4+ *) Table[PolygonalNumber[RegularPolygon[6], n], {n, 0, 48}] (* Arkadiusz Wesolowski, Aug 27 2016 *)
    PolygonalNumber[6, Range[0, 20]] (* Eric W. Weisstein, Aug 17 2017 *)
    CoefficientList[Series[x*(1 + 3*x)/(1 - x)^3 , {x, 0, 100}], x] (* Stefano Spezia, Sep 02 2018 *)
  • PARI
    a(n)=n*(2*n-1)
    
  • PARI
    a(n) = binomial(2*n,2) \\ Altug Alkan, Oct 06 2015
    
  • Python
    # Intended to compute the initial segment of the sequence, not isolated terms.
    def aList():
         x, y = 1, 1
         yield 0
         while True:
             yield x
             x, y = x + y + 4, y + 4
    A000384 = aList()
    print([next(A000384) for i in range(49)]) # Peter Luschny, Aug 04 2019

Formula

a(n) = Sum_{k=1..n} tan^2((k - 1/2)*Pi/(2n)). - Ignacio Larrosa Cañestro, Apr 17 2001
E.g.f.: exp(x)*(x+2x^2). - Paul Barry, Jun 09 2003
G.f.: x*(1+3*x)/(1-x)^3. - Simon Plouffe in his 1992 dissertation, dropping the initial zero
a(n) = A000217(2*n-1) = A014105(-n).
a(n) = 4*A000217(n-1) + n. - Lekraj Beedassy, Jun 03 2004
a(n) = right term of M^n * [1,0,0], where M = the 3 X 3 matrix [1,0,0; 1,1,0; 1,4,1]. Example: a(5) = 45 since M^5 *[1,0,0] = [1,5,45]. - Gary W. Adamson, Dec 24 2006
Row sums of triangle A131914. - Gary W. Adamson, Jul 27 2007
Row sums of n-th row, triangle A134234 starting (1, 6, 15, 28, ...). - Gary W. Adamson, Oct 14 2007
Starting with offset 1, = binomial transform of [1, 5, 4, 0, 0, 0, ...]. Also, A004736 * [1, 4, 4, 4, ...]. - Gary W. Adamson, Oct 25 2007
a(n)^2 + (a(n)+1)^2 + ... + (a(n)+n-1)^2 = (a(n)+n+1)^2 + ... + (a(n)+2n-1)^2 + n^2; e.g., 6^2 + 7^2 = 9^2 + 2^2; 28^2 + 29^2 + 30^2 + 31^2 = 33^2 + 34^2 + 35^2 + 4^2. - Charlie Marion, Nov 10 2007
a(n) = binomial(n+1,2) + 3*binomial(n,2).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0)=0, a(1)=1, a(2)=6. - Jaume Oliver Lafont, Dec 02 2008
a(n) = T(n) + 3*T(n-1), where T(n) is the n-th triangular number. - Vincenzo Librandi, Nov 10 2010
a(n) = a(n-1) + 4*n - 3 (with a(0)=0). - Vincenzo Librandi, Nov 20 2010
a(n) = A007606(A000290(n)). - Reinhard Zumkeller, Feb 12 2011
a(n) = 2*a(n-1) - a(n-2) + 4. - Ant King, Aug 26 2011
a(n+1) = A045896(2*n). - Reinhard Zumkeller, Dec 12 2011
a(2^n) = 2^(2n+1) - 2^n. - Ivan N. Ianakiev, Apr 13 2013
a(n) = binomial(2*n,2). - Gary Detlefs, Jul 28 2013
a(n+1) = A128918(2*n+1). - Reinhard Zumkeller, Oct 13 2013
a(4*a(n)+7*n+1) = a(4*a(n)+7*n) + a(4*n+1). - Vladimir Shevelev, Jan 24 2014
Sum_{n>=1} 1/a(n) = 2*log(2) = 1.38629436111989...= A016627. - Vaclav Kotesovec, Apr 27 2016
Sum_{n>=1} (-1)^n/a(n) = log(2) - Pi/2. - Vaclav Kotesovec, Apr 20 2018
a(n+1) = trinomial(2*n+1, 2) = trinomial(2*n+1, 4*n), for n >= 0, with the trinomial irregular triangle A027907. a(n+1) = (n+1)*(2*n+1) = (1/Pi)*Integral_{x=0..2} (1/sqrt(4 - x^2))*(x^2 - 1)^(2*n+1)*R(4*n-2, x) with the R polynomial coefficients given in A127672. [Comtet, p. 77, the integral formula for q=3, n -> 2*n+1, k = 2, rewritten with x = 2*cos(phi)]. - Wolfdieter Lang, Apr 19 2018
Sum_{n>=1} 1/(a(n))^2 = 2*Pi^2/3-8*log(2) = 1.0345588... = 10*A182448 - A257872. - R. J. Mathar, Sep 12 2019
a(n) = (A005408(n-1) + A046092(n-1) + A001844(n-1))/2. - Ralf Steiner, Feb 27 2020
Product_{n>=2} (1 - 1/a(n)) = 2/3. - Amiram Eldar, Jan 21 2021
a(n) = floor(Sum_{k=(n-1)^2..n^2} sqrt(k)), for n >= 1. - Amrit Awasthi, Jun 13 2021
a(n+1) = A084265(2*n), n>=0. - Hartmut F. W. Hoft, Feb 02 2022
a(n) = A000290(n) + A002378(n-1). - Charles Kusniec, Sep 11 2022

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A016813 a(n) = 4*n + 1.

Original entry on oeis.org

1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 57, 61, 65, 69, 73, 77, 81, 85, 89, 93, 97, 101, 105, 109, 113, 117, 121, 125, 129, 133, 137, 141, 145, 149, 153, 157, 161, 165, 169, 173, 177, 181, 185, 189, 193, 197, 201, 205, 209, 213, 217, 221, 225, 229, 233, 237
Offset: 0

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Keywords

Comments

Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0( 23 ).
Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 64 ).
Numbers k such that k and (k+1) have the same binary digital sum. - Benoit Cloitre, Jun 05 2002
Numbers k such that (1 + sqrt(k))/2 is an algebraic integer. - Alonso del Arte, Jun 04 2012
Numbers k such that 2 is the only prime p that satisfies the relationship p XOR k = p + k. - Brad Clardy, Jul 22 2012
This may also be interpreted as the array T(n,k) = A001844(n+k) + A008586(k) read by antidiagonals:
1, 9, 21, 37, 57, 81, ...
5, 17, 33, 53, 77, 105, ...
13, 29, 49, 73, 101, 133, ...
25, 45, 69, 97, 129, 165, ...
41, 65, 93, 125, 161, 201, ...
61, 89, 121, 157, 197, 241, ...
...
- R. J. Mathar, Jul 10 2013
With leading term 2 instead of 1, 1/a(n) is the largest tolerance of form 1/k, where k is a positive integer, so that the nearest integer to (n - 1/k)^2 and to (n + 1/k)^2 is n^2. In other words, if interval arithmetic is used to square [n - 1/k, n + 1/k], every value in the resulting interval of length 4n/k rounds to n^2 if and only if k >= a(n). - Rick L. Shepherd, Jan 20 2014
Odd numbers for which the number of prime factors congruent to 3 (mod 4) is even. - Daniel Forgues, Sep 20 2014
For the Collatz conjecture, we identify two types of odd numbers. This sequence contains all the descenders: where (3*a(n) + 1) / 2 is even and requires additional divisions by 2. See A004767 for the ascenders. - Fred Daniel Kline, Nov 29 2014 [corrected by Jaroslav Krizek, Jul 29 2016]
a(n-1), n >= 1, is also the complex dimension of the manifold M(S), the set of all conjugacy classes of irreducible representations of the fundamental group pi_1(X,x_0) of rank 2, where S = {a_1, ..., a_{n}, a_{n+1} = oo}, a subset of P^1 = C U {oo}, X = X(S) = P^1 \ S, and x_0 a base point in X. See the Iwasaki et al. reference, Proposition 2.1.4. p. 150. - Wolfdieter Lang, Apr 22 2016
For n > 3, also the number of (not necessarily maximal) cliques in the n-sunlet graph. - Eric W. Weisstein, Nov 29 2017
For integers k with absolute value in A047202, also exponents of the powers of k having the same unit digit of k in base 10. - Stefano Spezia, Feb 23 2021
Starting with a(1) = 5, numbers ending with 01 in base 2. - John Keith, May 09 2022

Examples

			From _Leo Tavares_, Jul 02 2021: (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
(End)
		

References

  • K. Iwasaki, H. Kimura, S. Shimomura and M. Yoshida, From Gauss to Painlevé, Vieweg, 1991. p. 150.

Crossrefs

Subsequence of A042963 and of A079523.
a(n) = A093561(n+1, 1), (4, 1)-Pascal column.
Cf. A004772 (complement).
Cf. A017557.

Programs

Formula

a(n) = A005408(2*n).
Sum_{n>=0} (-1)^n/a(n) = (1/(4*sqrt(2)))*(Pi+2*log(sqrt(2)+1)) = A181048 [Jolley]. - Benoit Cloitre, Apr 05 2002 [corrected by Amiram Eldar, Jul 30 2023]
G.f.: (1+3*x)/(1-x)^2. - Paul Barry, Feb 27 2003 [corrected for offset 0 by Wolfdieter Lang, Oct 03 2014]
(1 + 5*x + 9*x^2 + 13*x^3 + ...) = (1 + 2*x + 3*x^2 + ...) / (1 - 3*x + 9*x^2 - 27*x^3 + ...). - Gary W. Adamson, Jul 03 2003
a(n) = A001969(n) + A000069(n). - Philippe Deléham, Feb 04 2004
a(n) = A004766(n-1). - R. J. Mathar, Oct 26 2008
a(n) = 2*a(n-1) - a(n-2); a(0)=1, a(1)=5. a(n) = 4 + a(n-1). - Philippe Deléham, Nov 03 2008
A056753(a(n)) = 3. - Reinhard Zumkeller, Aug 23 2009
A179821(a(n)) = a(A179821(n)). - Reinhard Zumkeller, Jul 31 2010
a(n) = 8*n - 2 - a(n-1) for n > 0, a(0) = 1. - Vincenzo Librandi, Nov 20 2010
The identity (4*n+1)^2 - (4*n^2+2*n)*(2)^2 = 1 can be written as a(n)^2 - A002943(n)*2^2 = 1. - Vincenzo Librandi, Mar 11 2009 - Nov 25 2012
A089911(6*a(n)) = 8. - Reinhard Zumkeller, Jul 05 2013
a(n) = A004767(n) - 2. - Jean-Bernard François, Sep 27 2013
a(n) = A058281(3n+1). - Eli Jaffe, Jun 07 2016
From Ilya Gutkovskiy, Jul 29 2016: (Start)
E.g.f.: (1 + 4*x)*exp(x).
a(n) = Sum_{k = 0..n} A123932(k).
a(A005098(k)) = x^2 + y^2.
Inverse binomial transform of A014480. (End)
Dirichlet g.f.: 4*Zeta(-1 + s) + Zeta(s). - Stefano Spezia, Nov 02 2018

A002417 4-dimensional figurate numbers: a(n) = n*binomial(n+2, 3).

Original entry on oeis.org

1, 8, 30, 80, 175, 336, 588, 960, 1485, 2200, 3146, 4368, 5915, 7840, 10200, 13056, 16473, 20520, 25270, 30800, 37191, 44528, 52900, 62400, 73125, 85176, 98658, 113680, 130355, 148800, 169136, 191488, 215985, 242760, 271950, 303696, 338143, 375440, 415740
Offset: 1

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Author

Keywords

Comments

a(n) is 1/6 the number of colorings of a 2 X 2 hexagonal array with n+2 colors. - R. H. Hardin, Feb 23 2002
a(n) is the sum of all numbers that cannot be written as t*(n+1) + u*(n+2) for nonnegative integers t,u (see Schuh). - Floor van Lamoen, Oct 09 2002
a(n) is the total number of rectangles (including squares) contained in a stepped pyramid of n rows (or of base 2n-1) of squares. A stepped pyramid of squares of base 2*6 - 1 = 11, for instance, has the following vertices:
..........X.X
........X.X.X.X
......X.X.X.X.X.X
....X.X.X.X.X.X.X.X
..X.X.X.X.X.X.X.X.X.X
X.X.X.X.X.X.X.X.X.X.X.X
X.X.X.X.X.X.X.X.X.X.X.X - Lekraj Beedassy, Sep 02 2003
Partial sums of A002412. - Jonathan Vos Post, Mar 16 2006
a(n) equals -1 times the coefficient of x^3 of the characteristic polynomial of the (n + 2) X (n + 2) matrix with 2's along the main diagonal and 1's everywhere else (see Mathematica code below). - John M. Campbell, May 28 2011
a(n) is the n-th antidiagonal sum of the convolution array A213750. - Clark Kimberling, Jun 20 2012
Convolution of A000027 with A000384 (excluding 0). - Bruno Berselli, Dec 06 2012
The sequence is the binomial transform of (1, 7, 15, 13, 4, 0, 0, 0, ...). - Gary W. Adamson, Jul 31 2015
Also the number of 3-cycles in the (n+2)-triangular graph. - Eric W. Weisstein, Aug 14 2017

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 195.
  • K. -W. Lau, Solution to Problem 2495, Journal of Recreational Mathematics 2002-3 31(1) 79-80.
  • Fred. Schuh, Vragen betreffende een onbepaalde vergelijking, Nieuw Tijdschrift voor Wiskunde, 52 (1964-1965) 193-198.
  • 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).

Crossrefs

Bisection of A002624.
a(n) = A093561(n+3, 4).
Cf. A220212 for a list of sequences produced by the convolution of the natural numbers with the k-gonal numbers.
Cf. A151974 (number of 4-cycles in the triangular graph), A290939 (5-cycles), A290940 (6-cycles).

Programs

  • GAP
    List([1..40], n-> n^2*(n+1)*(n+2)/6 ); # G. C. Greubel, Jul 03 2019
  • Magma
    /* A000027 convolved with A000384 (excluding 0): */ A000384:=func; [&+[(n-i+1)*A000384(i): i in [1..n]]: n in [1..40]]; // Bruno Berselli, Dec 06 2012
    
  • Magma
    [n*Binomial(n+2,3):n in [1..40]]; // Vincenzo Librandi, Aug 02 2015
    
  • Maple
    seq(n^2*(n+1)*(n+2)/6, n=1..50);
  • Mathematica
    Table[n Binomial[n+2, 3], {n, 40}]
    Table[-Coefficient[CharacteristicPolynomial[Array[KroneckerDelta[#1, #2] + 1 &, {n+2, n+2}], x], x^3], {n, 40}] (* John M. Campbell, May 28 2011 *)
    Nest[Accumulate, Range[1, 170, 4], 3] (* Vladimir Joseph Stephan Orlovsky, Jan 21 2012 *)
    LinearRecurrence[{5, -10, 10, -5, 1}, {1, 8, 30, 80, 175}, 40] (* Harvey P. Dale, Jan 11 2014 *)
    Table[n Pochhammer[n, 3]/6, {n, 40}] (* or *) CoefficientList[Series[ (1+3x)/(1-x)^5, {x,0,40}], x] (* Eric W. Weisstein, Aug 14 2017 *)
  • PARI
    a(n)=n^2*(n+1)*(n+2)/6 \\ Charles R Greathouse IV, Jun 10 2011
    
  • Sage
    [n*binomial(n+2,3) for n in (1..40)] # G. C. Greubel, Jul 03 2019
    

Formula

a(n) = n^2*(n+1)*(n+2)/6.
G.f.: x*(1+3*x)/(1-x)^5. - Simon Plouffe in his 1992 dissertation
a(n) = C(n+2, 2)*n^2/3. - Paul Barry, Jun 26 2003
a(n) = C(n+3, n)*C(n+1, 1). - Zerinvary Lajos, Apr 27 2005
a(n) = (binomial(n+3,n-1) - binomial(n+2,n-2))*(binomial(n+1,n-1) - binomial(n,n-2)). - Zerinvary Lajos, May 12 2006
a(n) = 5*a(n-1) -10*a(n-2) +10*a(n-3) -5*a(n-4) +a(n-5), n>5. - Wesley Ivan Hurt, Aug 01 2015
G.f.: x*hypergeometric2F1(2,4;1;x). - R. J. Mathar, Aug 09 2015
a(n) = A080852(4,n-1). - R. J. Mathar, Jul 28 2016
Sum_{n>=1} 1/a(n) = Pi^2/2 - 15/4. - Jaume Oliver Lafont, Jul 13 2017
E.g.f.: x*(6 + 18*x + 9*x^2 + x^3)*exp(x)/3!. - G. C. Greubel, Jul 03 2019
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi^2 + 27 - 48*log(2))/4. - Amiram Eldar, Jun 28 2020
a(n) = A000332(n+3) + 3*A000332(n+2). - Mircea Dan Rus, Jul 29 2020

Extensions

Edited and extended by Floor van Lamoen, Oct 09 2002

A002412 Hexagonal pyramidal numbers, or greengrocer's numbers.

Original entry on oeis.org

0, 1, 7, 22, 50, 95, 161, 252, 372, 525, 715, 946, 1222, 1547, 1925, 2360, 2856, 3417, 4047, 4750, 5530, 6391, 7337, 8372, 9500, 10725, 12051, 13482, 15022, 16675, 18445, 20336, 22352, 24497, 26775, 29190, 31746, 34447, 37297, 40300
Offset: 0

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Comments

Binomial transform of (1, 6, 9, 4, 0, 0, 0, ...). - Gary W. Adamson, Oct 16 2007
a(n) is the sum of the maximum(m,n) over {(m,n): m,n in positive integers, m<=n}. - Geoffrey Critzer, Oct 11 2009
We obtain these numbers for d=2 in the identity n*(n*(d*n-d+2)/2)-sum(k*(d*k-d+2)/2, k=0..n-1) = n*(n+1)*(2*d*n-2*d+3)/6 (see Klaus Strassburger in Formula lines). - Bruno Berselli, Apr 21 2010, Nov 16 2010
q^a(n) is the Hankel transform of the q-Catalan numbers. - Paul Barry, Dec 15 2010
Row 1 of the convolution array A213835. - Clark Kimberling, Jul 04 2012
From Ant King, Oct 24 2012: (Start)
For n>0, the digital roots of this sequence A010888(A002412(n)) form the purely periodic 27-cycle {1,7,4,5,5,8,9,3,3,4,1,7,8,8,2,3,6,6,7,4,1,2,2,5,6,9,9}.
For n>0, the units' digits of this sequence A010879(A002412(n)) form the purely periodic 20-cycle {1,7,2,0,5,1,2,2,5,5,6,2,7,5,0,6,7,7,0,0}.
(End)
Partial sums of A000384. - Omar E. Pol, Jan 12 2013
Row sums of A094728. - J. M. Bergot, Jun 14 2013
Number of orbits of Aut(Z^7) as function of the infinity norm (n+1) of the representative integer lattice point of the orbit, when the cardinality of the orbit is equal to 40320. - Philippe A.J.G. Chevalier, Dec 28 2015
Coefficients in the hypergeometric series identity 1 - 7*(x - 1)/(3*x + 1) + 22*(x - 1)*(x - 2)/((3*x + 1)*(3*x + 2)) - 50*(x - 1)*(x - 2)*(x - 3)/((3*x + 1)*(3*x + 2)*(3*x + 3)) + ... = 0, valid for Re(x) > 1. Cf. A000326 and A002418. Column 3 of A103450. - Peter Bala, Mar 14 2019

Examples

			Let n=5, 2*n=10. Since 10 = 1 + 9 = 2 + 8 = 3 + 7 = 4 + 6 = 5 + 5, a(5) = 1*9 + 2*8 + 3*7 + 4*6 + 5*5 = 95. - _Vladimir Shevelev_, May 11 2012
		

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 194.
  • 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.
  • T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.
  • I. Siap, Linear codes over F_2 + u*F_2 and their complete weight enumerators, in Codes and Designs (Ohio State, May 18, 2000), pp. 259-271. De Gruyter, 2002.
  • 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).

Crossrefs

Bisection of A002623. Equals A000578(n) - A000330(n-1).
a(n) = A093561(n+2, 3), (4, 1)-Pascal column.
Cf. A220084 for a list of numbers of the form n*P(k,n)-(n-1)*P(k,n-1), where P(k,n) is the n-th k-gonal pyramidal number (see Adamson's formula).
Cf. similar sequences listed in A237616.
Orbits of Aut(Z^7) as function of the infinity norm A000579, A154286, A102860, A002412, A045943, A115067, A008585, A005843, A001477, A000217.

Programs

  • GAP
    List([0..40],n->n*(n+1)*(4*n-1)/6); # Muniru A Asiru, Mar 18 2019
    
  • Magma
    [n*(n+1)*(4*n-1)/6: n in [0..40]]; // Vincenzo Librandi, Nov 28 2015
    
  • Maple
    seq(sum(i*(2*k-i), i=1..k), k=0..100); # Wesley Ivan Hurt, Sep 25 2013
  • Mathematica
    Figurate[ ngon_, rank_, dim_] := Binomial[rank + dim - 2, dim - 1] ((rank - 1)*(ngon - 2) + dim)/dim; Table[ Figurate[6, r, 3], {r, 0, 40}] (* Robert G. Wilson v, Aug 22 2010 *)
    Table[n(n+1)(4n-1)/6, {n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1}, {0,1,7,22}, 40] (* Harvey P. Dale, Jul 16 2011 *)
  • Maxima
    A002412(n):=n*(n+1)*(4*n-1)/6$ makelist(A002412(n),n,0,20); /* Martin Ettl, Dec 12 2012 */
    
  • PARI
    v=vector(40,i,(i*(i+1))\2); s=0; print1(s","); forstep(i=1,40,2,s+=v[i]; print1(s","))
    
  • Python
    print([n*(n+1)*(4*n-1)//6 for n in range(40)]) # Michael S. Branicky, Mar 28 2022

Formula

a(n) = n(n + 1)(4n - 1)/6.
G.f.: x*(1+3*x)/(1-x)^4. - Simon Plouffe in his 1992 dissertation.
a(n) = n^3 - Sum_{i=1..n-1} i^2. - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de)
Partial sums of n odd-indexed triangular numbers, e.g., a(3) = t(1)+t(3)+t(5) = 1+6+15 = 22. - Jon Perry, Jul 23 2003
a(n) = Sum_{i=0..n-1} (n - i)*(n + i). - Jon Perry, Sep 26 2004
a(n) = n*A000292(n) - (n-1)*A000292(n-1) = n*binomial((n+2),3) - (n-1)*binomial((n+1),3); e.g., a(5) = 95 = 5*35 - 4*20. - Gary W. Adamson, Dec 28 2007
a(n) = Sum_{i=0..n} (2i^2 + 3i + 1), for n >= 0 (Omits the leading 0). - William A. Tedeschi, Aug 25 2010
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4), with a(0)=0, a(1)=1, a(2)=7, a(3)=22. - Harvey P. Dale, Jul 16 2011
a(n) = sum a*b, where the summing is over all unordered partitions 2*n = a+b. - Vladimir Shevelev, May 11 2012
From Ant King, Oct 24 2012: (Start)
a(n) = a(n-1) + n*(2*n-1).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 4.
a(n) = (n+1)*(2*A000384(n) + n)/6 = (4*n-1)*A000217(n)/3.
a(n) = A000292(n) + 3*A000292(n-1) = A002411(n) + A000292(n-1).
a(n) = binomial(n+2,3) + 3*binomial(n+1,3) = (4*n-1)*binomial(n+1,2)/3.
Sum_{n>=1} 1/a(n) = 6*(12*log(2)-2*Pi-1)/5 = 1.2414...
(End)
a(n) = Sum_{i=1..n} Sum_{j=1..n} max(i,j) = Sum_{i=1..n} i*(2*n-i). - Enrique Pérez Herrero, Jan 15 2013
a(n) = A005900(n+1) - A000326(n+1) = Octahedral - Pentagonal Numbers. - Richard R. Forberg, Aug 07 2013
a(n) = n*A000217(n) + Sum_{i=0..n-1} A000217(i). - Bruno Berselli, Dec 18 2013
a(n) = 2n * A000217(n) - A000330(n). - J. M. Bergot, Apr 05 2014
a(n) = A080851(4,n-1). - R. J. Mathar, Jul 28 2016
E.g.f.: x*(6 + 15*x + 4*x^2)*exp(x)/6. - Ilya Gutkovskiy, May 12 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = 6*(1 + 2*sqrt(2)*Pi - 2*(3+sqrt(2))*log(2) + 4*sqrt(2)*log(2-sqrt(2)))/5. - Amiram Eldar, Jan 04 2022

A020714 a(n) = 5 * 2^n.

Original entry on oeis.org

5, 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120, 10240, 20480, 40960, 81920, 163840, 327680, 655360, 1310720, 2621440, 5242880, 10485760, 20971520, 41943040, 83886080, 167772160, 335544320, 671088640, 1342177280, 2684354560, 5368709120, 10737418240
Offset: 0

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Comments

Same as Pisot sequences E(5,10), L(5,10), P(5,10), T(5,10). See A008776 for definitions of Pisot sequences.
The first differences are the sequence itself. - Alexandre Wajnberg & Eric Angelini, Sep 07 2005
5 times powers of 2. - Omar E. Pol, Dec 16 2008
Subsequence of A051916. - Reinhard Zumkeller, Mar 20 2010
With the addition of "2, 3," at the beginning, this sequence gives terms (n + 3) through the first term greater than 2^n, for n odd, of the negabinary Keith sequence for 2^n, thus proving that with the exception of 2 itself, no odd-indexed power of 2 is a negabinary Keith number (see A188381). - Alonso del Arte, Feb 02 2012
Let b(0) = 5 and b(n+1) = the smallest number not in the sequence such that b(n+1) - Product_{i=0..n} b(i) divides b(n+1) - Sum_{i=0..n} b(i). Then b(n+2) = a(n) for n > 0. - Derek Orr, Jan 15 2015

Crossrefs

Row sums of (4, 1)-Pascal triangle A093561.
Row sums of (9, 1)-Pascal triangle A093644.
Row sums of (1, 4)-Pascal triangle A095666 (with leading 4).

Programs

Formula

a(n) = 5*2^n. a(n) = 2*a(n-1).
G.f.: 5/(1-2*x).
If m is a term greater than 5 of this sequence then m = 5*phi(phi(m)). - Farideh Firoozbakht, Aug 16 2005
a(n) = A118416(n+1,3) for n>2. - Reinhard Zumkeller, Apr 27 2006
a(n) = A000079(n)*5. - Omar E. Pol, Dec 16 2008
a(n) = A173786(n+2,n) for n > 1. - Reinhard Zumkeller, Feb 28 2010
a(n) = A001045(n+4) - A001045(n). - Paul Curtz, Nov 08 2012
Sum_{n>=1} 1/a(n) = 2/5. - Amiram Eldar, Oct 28 2020
E.g.f.: 5*exp(2*x). - Stefano Spezia, May 15 2021

A029653 Numbers in (2,1)-Pascal triangle (by row).

Original entry on oeis.org

1, 2, 1, 2, 3, 1, 2, 5, 4, 1, 2, 7, 9, 5, 1, 2, 9, 16, 14, 6, 1, 2, 11, 25, 30, 20, 7, 1, 2, 13, 36, 55, 50, 27, 8, 1, 2, 15, 49, 91, 105, 77, 35, 9, 1, 2, 17, 64, 140, 196, 182, 112, 44, 10, 1, 2, 19, 81, 204, 336, 378, 294, 156, 54, 11, 1, 2, 21, 100, 285
Offset: 0

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Reverse of A029635. Row sums are A003945. Diagonal sums are Fibonacci(n+2) = Sum_{k=0..floor(n/2)} (2n-3k)*C(n-k,n-2k)/(n-k). - Paul Barry, Jan 30 2005
Riordan array ((1+x)/(1-x), x/(1-x)). The signed triangle (-1)^(n-k)T(n,k) or ((1-x)/(1+x), x/(1+x)) is the inverse of A055248. Row sums are A003945. Diagonal sums are F(n+2). - Paul Barry, Feb 03 2005
Row sums = A003945: (1, 3, 6, 12, 24, 48, 96, ...) = (1, 3, 7, 15, 31, 63, 127, ...) - (0, 0, 1, 3, 7, 15, 31, ...); where (1, 3, 7, 15, ...) = A000225. - Gary W. Adamson, Apr 22 2007
Triangle T(n,k), read by rows, given by (2,-1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 17 2011
A029653 is jointly generated with A208510 as an array of coefficients of polynomials v(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=u(n-1,x)+x*v(n-1)x and v(n,x)=u(n-1,x)+x*v(n-1,x)+1. See the Mathematica section. - Clark Kimberling, Feb 28 2012
For a closed-form formula for arbitrary left and right borders of Pascal like triangle, see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle, see A228576. - Boris Putievskiy, Sep 04 2013
The n-th row polynomial is (2 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Feb 25 2018

Examples

			The triangle T(n,k) begins:
n\k 0  1  2   3   4   5   6   7  8  9 10 ...
0:  1
1:  2  1
2:  2  3  1
3:  2  5  4   1
4:  2  7  9   5   1
5:  2  9 16  14   6   1
6:  2 11 25  30  20   7   1
7:  2 13 36  55  50  27   8   1
8:  2 15 49  91 105  77  35   9  1
9:  2 17 64 140 196 182 112  44 10  1
10: 2 19 81 204 336 378 294 156 54 11  1
... Reformatted. - _Wolfdieter Lang_, Jan 09 2015
With the array M(k) as defined in the Formula section, the infinite product M(0)*M(1)*M(2)*... begins
/1        \/1         \/1        \      /1        \
|2 1      ||0 1       ||0 1      |      |2 1      |
|2 1 1    ||0 2 1     ||0 0 1    |... = |2 3 1    |
|2 1 1 1  ||0 2 1 1   ||0 0 2 1  |      |2 5 4 1  |
|2 1 1 1 1||0 2 1 1 1 ||0 0 2 1 1|      |2 7 9 5 1|
|...      ||...       ||...      |      |...      |
- _Peter Bala_, Dec 27 2014
		

References

  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.

Crossrefs

(d, 1) Pascal triangles: A007318(d=1), A093560(3), A093561(4), A093562(5), A093563(6), A093564(7), A093565(8), A093644(9), A093645(10).

Programs

  • Haskell
    a029653 n k = a029653_tabl !! n !! k
    a029653_row n = a029653_tabl !! n
    a029653_tabl = [1] : iterate
                   (\xs -> zipWith (+) ([0] ++ xs) (xs ++ [0])) [2, 1]
    -- Reinhard Zumkeller, Dec 16 2013
    
  • Maple
    A029653 :=  proc(n,k)
    if n = 0 then
      1;
    else
      binomial(n-1, k)+binomial(n, k)
    fi
    end proc: # R. J. Mathar, Jun 30 2013
  • Mathematica
    u[1, x_] := 1; v[1, x_] := 1; z = 16;
    u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
    v[n_, x_] := u[n - 1, x] + x*v[n - 1, x] + 1;
    Table[Expand[u[n, x]], {n, 1, z/2}]
    Table[Expand[v[n, x]], {n, 1, z/2}]
    cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
    TableForm[cu]
    Flatten[%]  (* A208510 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]  (* A029653 *)
    (* Clark Kimberling, Feb 28 2012 *)
  • Python
    from sympy import Poly
    from sympy.abc import x
    def u(n, x): return 1 if n==1 else u(n - 1, x) + x*v(n - 1, x)
    def v(n, x): return 1 if n==1 else u(n - 1, x) + x*v(n - 1, x) + 1
    def a(n): return Poly(v(n, x), x).all_coeffs()[::-1]
    for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 27 2017
    
  • Python
    from math import comb, isqrt
    def A029653(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*((r<<1)-a)//r if n else 1 # Chai Wah Wu, Nov 12 2024

Formula

T(n, k) = C(n-2, k-1) + C(n-2, k) + C(n-1, k-1) + C(n-1, k) except for n=0.
G.f.: (1 + x + y + xy)/(1 - y - xy). - Ralf Stephan, May 17 2004
T(n, k) = (2n-k)*binomial(n, n-k)/n, n, k > 0. - Paul Barry, Jan 30 2005
Sum_{k=0..n} T(n, k)*x^k gives A003945-A003954 for x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. - Philippe Deléham, Jul 10 2005
T(n, k) = C(n-1, k) + C(n, k). - Philippe Deléham, Jul 10 2005
Equals A097806 * A007318, i.e., the pairwise operator * Pascal's Triangle as infinite lower triangular matrices. - Gary W. Adamson, Apr 22 2007
From Peter Bala, Dec 27 2014: (Start)
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(2 + 5*x + 4*x^2/2! + x^3/3!) = 2 + 7*x + 16*x^2/2! + 30*x^3/3! + 50*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ).
Let M denote the lower unit triangular array with 1's on the main diagonal and 1's everywhere else below the main diagonal except for the first column which consists of the sequence [1,2,2,2,...]. 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. Then the present triangle equals the infinite product M(0)*M(1)*M(2)*... (which is clearly well-defined). See the Example section. (End)

Extensions

More terms from James Sellers

A000285 a(0) = 1, a(1) = 4, and a(n) = a(n-1) + a(n-2) for n >= 2.

Original entry on oeis.org

1, 4, 5, 9, 14, 23, 37, 60, 97, 157, 254, 411, 665, 1076, 1741, 2817, 4558, 7375, 11933, 19308, 31241, 50549, 81790, 132339, 214129, 346468, 560597, 907065, 1467662, 2374727, 3842389, 6217116, 10059505, 16276621, 26336126, 42612747, 68948873, 111561620, 180510493, 292072113, 472582606
Offset: 0

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Comments

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(4;n-1-k,k), n >= 1, with a(-1)=3. These are the sums over the SW-NE diagonals in P(4;n,k), the (4,1) Pascal triangle A093561. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also SW-NE diagonal sums in the Pascal (1,3) triangle A095660.
In general, for a Fibonacci sequence beginning with 1,b we have a(n) = (2^(-1-n)*((1-sqrt(5))^n*(1+sqrt(5)-2b) + (1+sqrt(5))^n*(-1+sqrt(5)+2b)))/sqrt(5). In this case we have b=4. - Herbert Kociemba, Dec 18 2011
Pisano period lengths: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 5, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... - R. J. Mathar, Aug 10 2012
a(n) = number of independent vertex subsets (i.e., the Merrifield-Simmons index) of the tree obtained from the path tree P_{n-1} by attaching two pendant edges to one of its endpoints (n >= 2). Example: if n=3, then we have the star tree with edges ab, ac, ad; it has 9 independent vertex subsets: empty, a, b, c, d, bc, cd, bd, bcd.
For n >= 2, the number a(n-1) is the dimension of a commutative Hecke algebra of type D_n with independent parameters. See Theorem 1.4 and Corollary 1.5 in the link "Hecke algebras with independent parameters". - Jia Huang, Jan 20 2019
For n >= 1, a(n) is the number of edge covers of the tadpole graph T_{3,n-1} with T_{3,0} interpreted as just the cycle C_3. Example: If n=2, we have C_3 and P_1 joined by a bridge, which is just the triangle with a pendant, and this graph has 5 edge covers. In general, because of the path portion of the graph, the number of edge covers of T{3,n-1} satisfies the same recurrence as Fibonacci sequence and it starts with 4,5. - Feryal Alayont, Aug 27 2023
Eswarathasan (1978) called these numbers "pseudo-Fibonacci numbers", and proved that 1, 4, and 9 are the only squares in this sequence. If the recurrence is extended to negative indices, then there is only one more square, a(-9) = 81. Eswarathasan (1979) proved that none of the terms (even with negative indices) are twice a square. - Amiram Eldar, Mar 09 2024
For n>2, a(n) + (-1)^ceiling(n/2) is the number of ways to tile this strip of length n-1, with a central staircase, using unit squares and dominoes:
_
||
|||_|||_|||_|||. - Greg Dresden and Runhe Zhang, Sep 07 2024

Examples

			G.f. = 1 + 4*x + 5*x^2 + 9*x^3 + 14*x^4 + 23*x^5 + 37*x^6 + 60*x^7 + ...
		

References

  • Richard E. Merrifield and Howard E. Simmons, Topological Methods in Chemistry, Wiley, New York, 1989. pp. 131.
  • Joe Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 224.
  • 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).

Crossrefs

Essentially the same as A104449, which only has A104449(0)=3 prefixed.
Cf. A090888, A101220, A109754, A091157 (subsequence of primes).

Programs

  • GAP
    F:=Fibonacci;; List([0..40], n-> F(n+2) +2*F(n) ); # G. C. Greubel, Nov 08 2019
  • Haskell
    a000285 n = a000285_list !! n
    a000285_list = 1 : 4 : zipWith (+) a000285_list (tail a000285_list)
    -- Reinhard Zumkeller, Apr 28 2011
    
  • Magma
    a0:=1; a1:=4; [GeneralizedFibonacciNumber(a0,a1,n): n in [0..30]]; // Bruno Berselli, Feb 12 2013
    
  • Maple
    with(combinat):a:=n->2*fibonacci(n)+fibonacci(n+2): seq(a(n), n=0..34);
  • Mathematica
    LinearRecurrence[{1,1},{1,4},40] (* or *) Table[(3*LucasL[n]- Fibonacci[n])/2,{n,40}] (* Harvey P. Dale, Jul 18 2011 *)
    a[ n_]:= Fibonacci[n] + LucasL[n+1]; (* Michael Somos, May 28 2014 *)
  • Maxima
    a[0]:1$ a[1]:4$ a[n]:=a[n-1]+a[n-2]$ makelist(a[n],n,0,30); /* Martin Ettl, Oct 25 2012 */
    
  • PARI
    Vec((1+3*x)/(1-x-x^2)+O(x^40)) \\ Charles R Greathouse IV, Nov 20 2012
    
  • Sage
    f=fibonacci; [f(n+2) +2*f(n) for n in (0..40)] # G. C. Greubel, Nov 08 2019
    

Formula

G.f.: (1+3*x)/(1-x-x^2). - Simon Plouffe in his 1992 dissertation
Row sums of A131775 starting (1, 4, 5, 9, 14, 23, ...). - Gary W. Adamson, Jul 14 2007
a(n) = 2*Fibonacci(n) + Fibonacci(n+2). - Zerinvary Lajos, Oct 05 2007
a(n) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)) + (3/2)* ((1+sqrt(5))^(n-1) - (1-sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). Offset 1. a(3)=5. - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
a(n) = 3*Fibonacci(n+2) - 2*Fibonacci(n+1). - Gary Detlefs, Dec 21 2010
a(n) = A104449(n+1). - Michael Somos, Apr 07 2012
From Michael Somos, May 28 2014: (Start)
a(n) = A101220(3, 0, n+1).
a(n) = A109754(3, n+1).
a(k) = A090888(2, k-1), for k > 0.
a(-1 - n) = (-1)^n * A013655(n).
a(n) = Fibonacci(n) + Lucas(n+1), see Mathematica field. (End)
11*Fibonacci(n+1) = a(n+3) - a(n-2) = 3*a(n-1) + 2*a(n). - Manfred Arens and Michel Marcus, Jul 14 2014
a(n) = (9*F(n) + F(n-3))/2. - J. M. Bergot, Jul 15 2017
a(n-1) = 3 * A000045(n) + A000045(n+1). - R. J. Mathar, Feb 14 2024
E.g.f.: (cosh(x/2) + sinh(x/2))*(5*cosh(sqrt(5)*x/2) + 7*sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Dec 31 2024

A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

Original entry on oeis.org

0, 1, 2, 4, 3, 4, 9, 7, 7, 8, 16, 16, 14, 15, 16, 25, 32, 30, 29, 31, 32, 36, 57, 62, 59, 60, 63, 64, 49, 93, 119, 121, 119, 123, 127, 128, 64, 142, 212, 240, 240, 242, 250, 255, 256, 81, 206, 354, 452, 480, 482, 492, 505, 511, 512, 100, 287, 560, 806, 932, 962, 974, 997, 1016, 1023, 1024
Offset: 1

Views

Author

Boris Putievskiy, Aug 15 2013

Keywords

Comments

The third row is (n^4 - n^2 + 24*n + 24)/12.
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013

Examples

			The start of the sequence as a triangular array read by rows:
   0;
   1,  2;
   4,  3,  4;
   9,  7,  7,  8;
  16, 16, 14, 15, 16;
  25, 32, 30, 29, 31, 32;
  36, 57, 62, 59, 60, 63, 64;
		

Crossrefs

Cf. We denote Pascal-like triangle with L(n) on the left border and R(n) on the right border by (L(n),R(n)). A007318 (1,1), A008949 (1,2^n), A029600 (2,3), A029618 (3,2), A029635 (1,2), A029653 (2,1), A037027 (Fibonacci(n),1), A051601 (n,n) n>=0, A051597 (n,n) n>0, A051666 (n^2,n^2), A071919 (1,0), A074829 (Fibonacci(n), Fibonacci(n)), A074909 (1,n), A093560 (3,1), A093561 (4,1), A093562 (5,1), A093563 (6,1), A093564 (7,1), A093565 (8,1), A093644 (9,1), A093645 (10,1), A095660 (1,3), A095666 (1,4), A096940 (1,5), A096956 (1,6), A106516 (3^n,1), A108561(1,(-1)^n), A132200 (4,4), A134636 (2n+1,2n+1), A137688 (2^n,2^n), A160760 (3^(n-1),1), A164844(1,10^n), A164847 (100^n,1), A164855 (101*100^n,1), A164866 (101^n,1), A172171 (1,9), A172185 (9,11), A172283 (-9,11), A177954 (int(n/2),1), A193820 (1,2^n), A214292 (n,-n), A227074 (4^n,4^n), A227075 (3^n,3^n), A227076 (5^n,5^n), A227550 (n!,n!), A228053 ((-1)^n,(-1)^n), A228074 (Fibonacci(n), n).
Cf. A000290 (row 1), A153056 (row 2), A000079 (column 1), A000225 (column 2), A132753 (column 3), A118885 (row sums of triangle array + 1), A228576 (generalized Pascal's triangle).

Programs

  • GAP
    T:= function(n,k)
        if k=0 then return n^2;
        elif k=n then return 2^n;
        else return T(n-1,k-1) + T(n-1,k);
        fi;
      end;
    Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 12 2019
  • Maple
    T:= proc(n, k) option remember;
          if k=0 then n^2
        elif k=n then 2^k
        else T(n-1, k-1) + T(n-1, k)
          fi
        end:
    seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Nov 12 2019
  • Mathematica
    T[n_, k_]:= T[n, k] = If[k==0, n^2, If[k==n, 2^k, T[n-1, k-1] + T[n-1, k]]]; Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 12 2019 *)
    Flatten[Table[Sum[i^2 Binomial[n-1-i, n-k-i], {i,1,n-k}] + Sum[2^i Binomial[n-1-i, k-i], {i,1,k}], {n,0,10}, {k,0,n}]] (* Greg Dresden, Aug 06 2022 *)
  • PARI
    T(n,k) = if(k==0, n^2, if(k==n, 2^k, T(n-1, k-1) + T(n-1, k) )); \\ G. C. Greubel, Nov 12 2019
    
  • Python
    def funcL(n):
       q = n**2
       return q
    def funcR(n):
       q = 2**n
       return q
    for n in range (1,9871):
       t=int((math.sqrt(8*n-7) - 1)/ 2)
       i=n-t*(t+1)/2-1
       j=(t*t+3*t+4)/2-n-1
       sum1=0
       sum2=0
       for m1 in range (1,i+1):
          sum1=sum1+funcR(m1)*binomial(i+j-m1-1,i-m1)
       for m2 in range (1,j+1):
          sum2=sum2+funcL(m2)*binomial(i+j-m2-1,j-m2)
       sum=sum1+sum2
    
  • Sage
    @CachedFunction
    def T(n, k):
        if (k==0): return n^2
        elif (k==n): return 2^n
        else: return T(n-1, k-1) + T(n-1, k)
    [[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 12 2019
    

Formula

T(n,0) = n^2, n>0; T(0,k) = 2^k; T(n, k) = T(n-1, k-1) + T(n-1, k) for n,k > 0. [corrected by G. C. Greubel, Nov 12 2019]
Closed-form formula for general case. Let L(m) and R(m) be the left border and the right border of Pascal like triangle, respectively. We denote binomial(n,k) by C(n,k).
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} R(m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} L(m2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} R(m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} L(m2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2); n>0.
Some special cases. If L(m)={b,b,b...} b*A000012, then the second sum takes form b*C(n+k-1,j). If L(m) is {0,b,2b,...} b*A001477, then the second sum takes form b*C(n+k,n-1). Similarly for R(m) and the first sum.
For this sequence L(m)=m^2 and R(m)=2^m.
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} (2^m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} (m2^2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} (2^m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} (m2^2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2).
As a triangular array read by rows, T(n,k) = Sum_{i=1..n-k} i^2*C(n-1-i, n-k-i) + Sum_{i=1..k} 2^i*C(n-1-i, k-i); n,k >=0. - Greg Dresden, Aug 06 2022

Extensions

Cross-references corrected and extended by Philippe Deléham, Dec 27 2013

A093560 (3,1) Pascal triangle.

Original entry on oeis.org

1, 3, 1, 3, 4, 1, 3, 7, 5, 1, 3, 10, 12, 6, 1, 3, 13, 22, 18, 7, 1, 3, 16, 35, 40, 25, 8, 1, 3, 19, 51, 75, 65, 33, 9, 1, 3, 22, 70, 126, 140, 98, 42, 10, 1, 3, 25, 92, 196, 266, 238, 140, 52, 11, 1, 3, 28, 117, 288, 462, 504, 378, 192, 63, 12, 1, 3, 31, 145, 405, 750, 966, 882, 570, 255, 75, 13, 1
Offset: 0

Views

Author

Wolfdieter Lang, Apr 22 2004

Keywords

Comments

The array F(3;n,m) gives in the columns m >= 1 the figurate numbers based on A016777, including the pentagonal numbers A000326 (see the W. Lang link).
This is the third member, d=3, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal (d=1), A029653 (d=2).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=g(z)/(1-x*z*f(z)). Here: g(x)=(1+2*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(1+2*z)/(1-(1+x)*z).
The SW-NE diagonals give the Lucas numbers A000032: L(n) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with L(0)=2. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
Triangle T(n,k), read by rows, given by [3,-2,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 17 2009
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
From Wolfdieter Lang, Jan 09 2015: (Start)
The signed lower triangular matrix (-1)^(n-1)*a(n,m) is the inverse of the Riordan matrix A106516; that is Riordan ((1-2*x)/(1+x),x/(1+x)).
See the Peter Bala comment from Dec 23 2014 in A106516 for general Riordan triangles of the type (g(x), x/(1-x)): exp(x)*r(n,x) = d(n,x) with the e.g.f. r(n,x) of row n and the e.g.f. of diagonal n.
Similarly, for general Riordan triangles of the type (g(x), x/(1+x)): exp(x)*r(n,-x) = d(n,x). (End)
The n-th row polynomial is (3 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Mar 02 2018
Binomial(n-2,k)+2*Binomial(n-3,k) is also the number of permutations avoiding both 123 and 132 with k double descents, i.e., positions with w[i]>w[i+1]>w[i+2]. - Lara Pudwell, Dec 19 2018

Examples

			Triangle begins
  1,
  3,  1,
  3,  4,  1,
  3,  7,  5,   1,
  3, 10, 12,   6,   1,
  3, 13, 22,  18,   7,   1,
  3, 16, 35,  40,  25,   8,   1,
  3, 19, 51,  75,  65,  33,   9,  1,
  3, 22, 70, 126, 140,  98,  42, 10,  1,
  3, 25, 92, 196, 266, 238, 140, 52, 11, 1,
		

References

  • Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
  • Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Crossrefs

Cf. Column sequences for m=1..9: A016777, A000326 (pentagonal), A002411, A001296, A051836, A051923, A050494, A053367, A053310;
A007318 (Pascal's triangle), A029653 ((2,1) Pascal triangle), A093561 ((4,1) Pascal triangle), A228196, A228576.

Programs

  • GAP
    Concatenation([1],Flat(List([1..11],n->List([0..n],k->Binomial(n,k)+2*Binomial(n-1,k))))); # Muniru A Asiru, Dec 20 2018
    
  • Haskell
    a093560 n k = a093560_tabl !! n !! k
    a093560_row n = a093560_tabl !! n
    a093560_tabl = [1] : iterate
                   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [3, 1]
    -- Reinhard Zumkeller, Aug 31 2014
    
  • Python
    from math import comb, isqrt
    def A093560(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+(r-a<<1))//r if n else 1 # Chai Wah Wu, Nov 12 2024

Formula

a(n, m)=F(3;n-m, m) for 0<= m <= n, otherwise 0, with F(3;0, 0)=1, F(3;n, 0)=3 if n>=1 and F(3;n, m):=(3*n+m)*binomial(n+m-1, m-1)/m if m>=1.
G.f. column m (without leading zeros): (1+2*x)/(1-x)^(m+1), m>=0.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=3 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
T(n, k) = C(n, k) + 2*C(n-1, k). - Philippe Deléham, Aug 28 2005
Equals M * A007318, where M = an infinite triangular matrix with all 1's in the main diagonal and all 2's in the subdiagonal. - Gary W. Adamson, Dec 01 2007
Sum_{k=0..n} T(n,k) = A151821(n+1). - Philippe Deléham, Sep 17 2009
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(3 + 7*x + 5*x^2/2! + x^3/3!) = 3 + 10*x + 22*x^2/2! + 40*x^3/3! + 65*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014
G.f.: (-1-2*x)/(-1+x+x*y). - R. J. Mathar, Aug 11 2015

Extensions

Incorrect connection with A046055 deleted by N. J. A. Sloane, Jul 08 2009

A034263 a(n) = binomial(n+4,4)*(4*n+5)/5.

Original entry on oeis.org

1, 9, 39, 119, 294, 630, 1218, 2178, 3663, 5863, 9009, 13377, 19292, 27132, 37332, 50388, 66861, 87381, 112651, 143451, 180642, 225170, 278070, 340470, 413595, 498771, 597429, 711109, 841464, 990264, 1159400, 1350888, 1566873, 1809633, 2081583, 2385279
Offset: 0

Views

Author

Keywords

Comments

Kekulé numbers for certain benzenoids. - Emeric Deutsch, Nov 18 2005
5-dimensional form of hexagonal-based pyramid numbers. - Ben Creech (mathroxmysox(AT)yahoo.com), Nov 17 2005
Convolution of triangular numbers (A000217) and hexagonal numbers (A000384). - Bruno Berselli, Jun 27 2013

Examples

			By the third comment: A000217(1..6) and A000384(1..6) give the term a(5) = 1*21+5*15+12*10+22*6+35*3+51*1 = 630. - _Bruno Berselli_, Jun 27 2013
		

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
  • Herbert John Ryser, Combinatorial Mathematics, "The Carus Mathematical Monographs", No. 14, John Wiley and Sons, 1963, pp. 1-8.
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (pp. 167-169, Table 10.5/II/4).

Crossrefs

Partial sums of A002417.
Cf. similar sequences listed in A254142.

Programs

  • GAP
    List([0..35], n-> (4*n+5)*Binomial(n+4,4)/5); # G. C. Greubel, Aug 28 2019
  • Magma
    [(4*n+5)*Binomial(n+4,4)/5: n in [0..35]]; // G. C. Greubel, Aug 28 2019
    
  • Maple
    a:=n->(n+1)*(n+2)*(n+3)*(n+4)*(4*n+5)/120: seq(a(n),n=0..35); # Emeric Deutsch, Nov 18 2005
  • Mathematica
    Table[Binomial[n+4, 4]*(4*n+5)/5, {n,0,35}] (* Vladimir Joseph Stephan Orlovsky, Jan 26 2012 *)
    a[n_] := (1+n)(2+n)(3+n)(4+n)(4n+5)/120; Array[a, 36, 0] (* or *)
    LinearRecurrence[{6, -15, 20, -15, 6, -1}, {1, 9, 39, 119, 294, 630}, 36] (* or *)
    CoefficientList[ Series[(1+3*x)/(1-x)^6, {x, 0, 35}], x] (* Robert G. Wilson v, Feb 26 2015 *)
    Table[Sum[-x^2 + y^2 + z^2, {x, 0, g}, {y, x, g}, {z, y, g}], {g, 1, 30}]/4 (* Horst H. Manninger, Jun 19 2025 *)
  • PARI
    a(n)=(n+1)*(n+2)*(n+3)*(n+4)*(4*n+5)/120 \\ Charles R Greathouse IV, Sep 24 2015, corrected by Altug Alkan, Aug 15 2017
    
  • Sage
    [(4*n+5)*binomial(n+4,4)/5 for n in (0..35)] # G. C. Greubel, Aug 28 2019
    

Formula

a(n) = A093561(n+5, 5).
a(n) = A034261(n+1, 3).
G.f.: (1+3*x)/(1-x)^6.
a(n) = (n+1)*(n+2)*(n+3)*(n+4)*(4*n+5)/120. - Emeric Deutsch and Ben Creech (mathroxmysox(AT)yahoo.com), Nov 17 2005, corrected by Eric Rowland, Aug 15 2017
a(-n-4) = -A059599(n). - Bruno Berselli, Aug 23 2011
a(n) = Sum_{i=1..n+1} i*A000292(i). - Bruno Berselli, Jan 23 2015
Sum_{n>=0} 1/a(n) = 28300/231 - 1280*Pi/77 - 7680*log(2)/77. - Amiram Eldar, Feb 15 2022

Extensions

Corrected and extended by N. J. A. Sloane, Apr 21 2000
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