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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A214622 Triangle read by rows, matrix inverse of [x^(n-k)](skp(n,x)-skp(n,x-1)+x^n) where skp denotes the Swiss-Knife polynomials A153641.

Original entry on oeis.org

1, -1, 1, 3, -2, 1, -10, 9, -3, 1, 45, -40, 18, -4, 1, -256, 225, -100, 30, -5, 1, 1743, -1536, 675, -200, 45, -6, 1, -13840, 12201, -5376, 1575, -350, 63, -7, 1, 125625, -110720, 48804, -14336, 3150, -560, 84, -8, 1, -1282816, 1130625, -498240, 146412, -32256, 5670, -840, 108, -9, 1
Offset: 0

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Author

Peter Luschny, Jul 23 2012

Keywords

Examples

			Triangle begins:
     1;
    -1,     1;
     3,    -2,    1;
   -10,     9,   -3,    1;
    45,   -40,   18,   -4,  1;
  -256,   225, -100,   30, -5,  1;
  1743, -1536,  675, -200, 45, -6, 1;
  ...
		

Programs

  • Maple
    A214622_row := proc(n) local s,t,k;
      s := series(exp(z*x)/(sech(x)+tanh(x)),x,n+2);
      t := factorial(n)*coeff(s,x,n); seq(coeff(t,z,k), k=(0..n)) end:
    for n from 0 to 7 do A214622_row(n) od; # Peter Luschny, Aug 01 2012
  • Mathematica
    A214622row[n_] := Module[{s, t},
       s = Series[Exp[z*x]/(Sech[x] + Tanh[x]), {x, 0, n+2}];
       t = n!*Coefficient[s, x, n];
       Table[Coefficient[t, z, k], {k, 0, n}]];
    Table[A214622row[n], {n, 0, 9}] // Flatten (* Jean-François Alcover, May 25 2024, after Peter Luschny *)
  • Sage
    R = PolynomialRing(ZZ, 'x')
    @CachedFunction
    def skp(n, x) : # Swiss-Knife polynomials A153641.
        if n == 0 : return 1
        return add(skp(k, 0)*binomial(n, k)*(x^(n-k)-(n+1)%2) for k in range(n)[::2])
    def A109449_signed(n, k) : return 0 if k > n else R(skp(n, x)-skp(n, x-1)+x^n)[k]
    T = matrix(ZZ, 9, A109449_signed).inverse(); T

Formula

T(n,k) = matrix inverse of A109449(n,k)*(-1)^floor((k-n+5)/2).
T(n,0) = A003704(n+1).
E.g.f.: exp(x*z)/(sech(x)+tanh(x)). - Peter Luschny, Aug 01 2012

A000217 Triangular numbers: a(n) = binomial(n+1,2) = n*(n+1)/2 = 0 + 1 + 2 + ... + n.

Original entry on oeis.org

0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431
Offset: 0

Views

Author

Keywords

Comments

Also referred to as T(n) or C(n+1, 2) or binomial(n+1, 2) (preferred).
Also generalized hexagonal numbers: n*(2*n-1), n=0, +-1, +-2, +-3, ... Generalized k-gonal numbers are second k-gonal numbers and positive terms of k-gonal numbers interleaved, k >= 5. In this case k = 6. - Omar E. Pol, Sep 13 2011 and Aug 04 2012
Number of edges in complete graph of order n+1, K_{n+1}.
Number of legal ways to insert a pair of parentheses in a string of n letters. E.g., there are 6 ways for three letters: (a)bc, (ab)c, (abc), a(b)c, a(bc), ab(c). Proof: there are C(n+2,2) ways to choose where the parentheses might go, but n + 1 of them are illegal because the parentheses are adjacent. Cf. A002415.
For n >= 1, a(n) is also the genus of a nonsingular curve of degree n+2, such as the Fermat curve x^(n+2) + y^(n+2) = 1. - Ahmed Fares (ahmedfares(AT)my_deja.com), Feb 21 2001
From Harnack's theorem (1876), the number of branches of a nonsingular curve of order n is bounded by a(n-1)+1, and the bound can be achieved. See also A152947. - Benoit Cloitre, Aug 29 2002. Corrected by Robert McLachlan, Aug 19 2024
Number of tiles in the set of double-n dominoes. - Scott A. Brown, Sep 24 2002
Number of ways a chain of n non-identical links can be broken up. This is based on a similar problem in the field of proteomics: the number of ways a peptide of n amino acid residues can be broken up in a mass spectrometer. In general, each amino acid has a different mass, so AB and BC would have different masses. - James A. Raymond, Apr 08 2003
Triangular numbers - odd numbers = shifted triangular numbers; 1, 3, 6, 10, 15, 21, ... - 1, 3, 5, 7, 9, 11, ... = 0, 0, 1, 3, 6, 10, ... - Xavier Acloque, Oct 31 2003 [Corrected by Derek Orr, May 05 2015]
Centered polygonal numbers are the result of [number of sides * A000217 + 1]. E.g., centered pentagonal numbers (1,6,16,31,...) = 5 * (0,1,3,6,...) + 1. Centered heptagonal numbers (1,8,22,43,...) = 7 * (0,1,3,6,...) + 1. - Xavier Acloque, Oct 31 2003
Maximum number of lines formed by the intersection of n+1 planes. - Ron R. King, Mar 29 2004
Number of permutations of [n] which avoid the pattern 132 and have exactly 1 descent. - Mike Zabrocki, Aug 26 2004
Number of ternary words of length n-1 with subwords (0,1), (0,2) and (1,2) not allowed. - Olivier Gérard, Aug 28 2012
Number of ways two different numbers can be selected from the set {0,1,2,...,n} without repetition, or, number of ways two different numbers can be selected from the set {1,2,...,n} with repetition.
Conjecturally, 1, 6, 120 are the only numbers that are both triangular and factorial. - Christopher M. Tomaszewski (cmt1288(AT)comcast.net), Mar 30 2005
Binomial transform is {0, 1, 5, 18, 56, 160, 432, ...}, A001793 with one leading zero. - Philippe Deléham, Aug 02 2005
Each pair of neighboring terms adds to a perfect square. - Zak Seidov, Mar 21 2006
Number of transpositions in the symmetric group of n+1 letters, i.e., the number of permutations that leave all but two elements fixed. - Geoffrey Critzer, Jun 23 2006
With rho(n):=exp(i*2*Pi/n) (an n-th root of 1) one has, for n >= 1, rho(n)^a(n) = (-1)^(n+1). Just use the triviality a(2*k+1) == 0 (mod (2*k+1)) and a(2*k) == k (mod (2*k)).
a(n) is the number of terms in the expansion of (a_1 + a_2 + a_3)^(n-1). - Sergio Falcon, Feb 12 2007
a(n+1) is the number of terms in the complete homogeneous symmetric polynomial of degree n in 2 variables. - Richard Barnes, Sep 06 2017
The number of distinct handshakes in a room with n+1 people. - Mohammad K. Azarian, Apr 12 2007 [corrected, Joerg Arndt, Jan 18 2016]
Equal to the rank (minimal cardinality of a generating set) of the semigroup PT_n\S_n, where PT_n and S_n denote the partial transformation semigroup and symmetric group on [n]. - James East, May 03 2007
a(n) gives the total number of triangles found when cevians are drawn from a single vertex on a triangle to the side opposite that vertex, where n = the number of cevians drawn+1. For instance, with 1 cevian drawn, n = 1+1 = 2 and a(n)= 2*(2+1)/2 = 3 so there is a total of 3 triangles in the figure. If 2 cevians are drawn from one point to the opposite side, then n = 1+2 = 3 and a(n) = 3*(3+1)/2 = 6 so there is a total of 6 triangles in the figure. - Noah Priluck (npriluck(AT)gmail.com), Apr 30 2007
For n >= 1, a(n) is the number of ways in which n-1 can be written as a sum of three nonnegative integers if representations differing in the order of the terms are considered to be different. In other words, for n >= 1, a(n) is the number of nonnegative integral solutions of the equation x + y + z = n-1. - Amarnath Murthy, Apr 22 2001 (edited by Robert A. Beeler)
a(n) is the number of levels with energy n + 3/2 (in units of h*f0, with Planck's constant h and the oscillator frequency f0) of the three-dimensional isotropic harmonic quantum oscillator. See the comment by A. Murthy above: n = n1 + n2 + n3 with positive integers and ordered. Proof from the o.g.f. See the A. Messiah reference. - Wolfdieter Lang, Jun 29 2007
From Hieronymus Fischer, Aug 06 2007: (Start)
Numbers m >= 0 such that round(sqrt(2m+1)) - round(sqrt(2m)) = 1.
Numbers m >= 0 such that ceiling(2*sqrt(2m+1)) - 1 = 1 + floor(2*sqrt(2m)).
Numbers m >= 0 such that fract(sqrt(2m+1)) > 1/2 and fract(sqrt(2m)) < 1/2, where fract(x) is the fractional part of x (i.e., x - floor(x), x >= 0). (End)
If Y and Z are 3-blocks of an n-set X, then, for n >= 6, a(n-1) is the number of (n-2)-subsets of X intersecting both Y and Z. - Milan Janjic, Nov 09 2007
Equals row sums of triangle A143320, n > 0. - Gary W. Adamson, Aug 07 2008
a(n) is also an even perfect number in A000396 iff n is a Mersenne prime A000668. - Omar E. Pol, Sep 05 2008. Unnecessary assumption removed and clarified by Rick L. Shepherd, Apr 14 2025
Equals row sums of triangle A152204. - Gary W. Adamson, Nov 29 2008
The number of matches played in a round robin tournament: n*(n-1)/2 gives the number of matches needed for n players. Everyone plays against everyone else exactly once. - Georg Wrede (georg(AT)iki.fi), Dec 18 2008
-a(n+1) = E(2)*binomial(n+2,2) (n >= 0) where E(n) are the Euler numbers in the enumeration A122045. Viewed this way, a(n) is the special case k=2 in the sequence of diagonals in the triangle A153641. - Peter Luschny, Jan 06 2009
Equivalent to the first differences of successive tetrahedral numbers. See A000292. - Jeremy Cahill (jcahill(AT)inbox.com), Apr 15 2009
The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus a(k) = |2^(-3)(P(2,1)-(-1)^k P(2,2k+1))|. - Peter Luschny, Jul 12 2009
a(n) is the smallest number > a(n-1) such that gcd(n,a(n)) = gcd(n,a(n-1)). If n is odd this gcd is n; if n is even it is n/2. - Franklin T. Adams-Watters, Aug 06 2009
Partial sums of A001477. - Juri-Stepan Gerasimov, Jan 25 2010. [A-number corrected by Omar E. Pol, Jun 05 2012]
The numbers along the right edge of Floyd's triangle are 1, 3, 6, 10, 15, .... - Paul Muljadi, Jan 25 2010
From Charlie Marion, Dec 03 2010: (Start)
More generally, a(2k+1) == j*(2j-1) (mod 2k+2j+1) and
a(2k) == [-k + 2j*(j-1)] (mod 2k+2j).
Column sums of:
1 3 5 7 9 ...
1 3 5 ...
1 ...
...............
---------------
1 3 6 10 15 ...
Sum_{n>=1} 1/a(n)^2 = 4*Pi^2/3-12 = 12 less than the volume of a sphere with radius Pi^(1/3).
(End)
A004201(a(n)) = A000290(n); A004202(a(n)) = A002378(n). - Reinhard Zumkeller, Feb 12 2011
1/a(n+1), n >= 0, has e.g.f. -2*(1+x-exp(x))/x^2, and o.g.f. 2*(x+(1-x)*log(1-x))/x^2 (see the Stephen Crowley formula line). -1/(2*a(n+1)) is the z-sequence for the Sheffer triangle of the coefficients of the Bernoulli polynomials A196838/A196839. - Wolfdieter Lang, Oct 26 2011
From Charlie Marion, Feb 23 2012: (Start)
a(n) + a(A002315(k)*n + A001108(k+1)) = (A001653(k+1)*n + A001109(k+1))^2. For k=0 we obtain a(n) + a(n+1) = (n+1)^2 (identity added by N. J. A. Sloane on Feb 19 2004).
a(n) + a(A002315(k)*n - A055997(k+1)) = (A001653(k+1)*n - A001109(k))^2.
(End)
Plot the three points (0,0), (a(n), a(n+1)), (a(n+1), a(n+2)) to form a triangle. The area will be a(n+1)/2. - J. M. Bergot, May 04 2012
The sum of four consecutive triangular numbers, beginning with a(n)=n*(n+1)/2, minus 2 is 2*(n+2)^2. a(n)*a(n+2)/2 = a(a(n+1)-1). - J. M. Bergot, May 17 2012
(a(n)*a(n+3) - a(n+1)*a(n+2))*(a(n+1)*a(n+4) - a(n+2)*a(n+3))/8 = a((n^2+5*n+4)/2). - J. M. Bergot, May 18 2012
a(n)*a(n+1) + a(n+2)*a(n+3) + 3 = a(n^2 + 4*n + 6). - J. M. Bergot, May 22 2012
In general, a(n)*a(n+1) + a(n+k)*a(n+k+1) + a(k-1)*a(k) = a(n^2 + (k+2)*n + k*(k+1)). - Charlie Marion, Sep 11 2012
a(n)*a(n+3) + a(n+1)*a(n+2) = a(n^2 + 4*n + 2). - J. M. Bergot, May 22 2012
In general, a(n)*a(n+k) + a(n+1)*a(n+k-1) = a(n^2 + (k+1)*n + k-1). - Charlie Marion, Sep 11 2012
a(n)*a(n+2) + a(n+1)*a(n+3) = a(n^2 + 4*n + 3). - J. M. Bergot, May 22 2012
Three points (a(n),a(n+1)), (a(n+1),a(n)) and (a(n+2),a(n+3)) form a triangle with area 4*a(n+1). - J. M. Bergot, May 23 2012
a(n) + a(n+k) = (n+k)^2 - (k^2 + (2n-1)*k -2n)/2. For k=1 we obtain a(n) + a(n+1) = (n+1)^2 (see below). - Charlie Marion, Oct 02 2012
In n-space we can define a(n-1) nontrivial orthogonal projections. For example, in 3-space there are a(2)=3 (namely point onto line, point onto plane, line onto plane). - Douglas Latimer, Dec 17 2012
From James East, Jan 08 2013: (Start)
For n >= 1, a(n) is equal to the rank (minimal cardinality of a generating set) and idempotent rank (minimal cardinality of an idempotent generating set) of the semigroup P_n\S_n, where P_n and S_n denote the partition monoid and symmetric group on [n].
For n >= 3, a(n-1) is equal to the rank and idempotent rank of the semigroup T_n\S_n, where T_n and S_n denote the full transformation semigroup and symmetric group on [n].
(End)
For n >= 3, a(n) is equal to the rank and idempotent rank of the semigroup PT_n\S_n, where PT_n and S_n denote the partial transformation semigroup and symmetric group on [n]. - James East, Jan 15 2013
Conjecture: For n > 0, there is always a prime between A000217(n) and A000217(n+1). Sequence A065383 has the first 1000 of these primes. - Ivan N. Ianakiev, Mar 11 2013
The formula, a(n)*a(n+4k+2)/2 + a(k) = a(a(n+2k+1) - (k^2+(k+1)^2)), is a generalization of the formula a(n)*a(n+2)/2 = a(a(n+1)-1) in Bergot's comment dated May 17 2012. - Charlie Marion, Mar 28 2013
The series Sum_{k>=1} 1/a(k) = 2, given in a formula below by Jon Perry, Jul 13 2003, has partial sums 2*n/(n+1) (telescopic sum) = A022998(n)/A026741(n+1). - Wolfdieter Lang, Apr 09 2013
For odd m = 2k+1, we have the recurrence a(m*n + k) = m^2*a(n) + a(k). Corollary: If number T is in the sequence then so is 9*T+1. - Lekraj Beedassy, May 29 2013
Euler, in Section 87 of the Opera Postuma, shows that whenever T is a triangular number then 9*T + 1, 25*T + 3, 49*T + 6 and 81*T + 10 are also triangular numbers. In general, if T is a triangular number then (2*k + 1)^2*T + k*(k + 1)/2 is also a triangular number. - Peter Bala, Jan 05 2015
Using 1/b and 1/(b+2) will give a Pythagorean triangle with sides 2*b + 2, b^2 + 2*b, and b^2 + 2*b + 2. Set b=n-1 to give a triangle with sides of lengths 2*n,n^2-1, and n^2 + 1. One-fourth the perimeter = a(n) for n > 1. - J. M. Bergot, Jul 24 2013
a(n) = A028896(n)/6, where A028896(n) = s(n) - s(n-1) are the first differences of s(n) = n^3 + 3*n^2 + 2*n - 8. s(n) can be interpreted as the sum of the 12 edge lengths plus the sum of the 6 face areas plus the volume of an n X (n-1) X (n-2) rectangular prism. - J. M. Bergot, Aug 13 2013
Dimension of orthogonal group O(n+1). - Eric M. Schmidt, Sep 08 2013
Number of positive roots in the root system of type A_n (for n > 0). - Tom Edgar, Nov 05 2013
A formula for the r-th successive summation of k, for k = 1 to n, is binomial(n+r,r+1) [H. W. Gould]. - Gary Detlefs, Jan 02 2014
Also the alternating row sums of A095831. Also the alternating row sums of A055461, for n >= 1. - Omar E. Pol, Jan 26 2014
For n >= 3, a(n-2) is the number of permutations of 1,2,...,n with the distribution of up (1) - down (0) elements 0...011 (n-3 zeros), or, the same, a(n-2) is up-down coefficient {n,3} (see comment in A060351). - Vladimir Shevelev, Feb 14 2014
a(n) is the dimension of the vector space of symmetric n X n matrices. - Derek Orr, Mar 29 2014
Non-vanishing subdiagonal of A132440^2/2, aside from the initial zero. First subdiagonal of unsigned A238363. Cf. A130534 for relations to colored forests, disposition of flags on flagpoles, and colorings of the vertices of complete graphs. - Tom Copeland, Apr 05 2014
The number of Sidon subsets of {1,...,n+1} of size 2. - Carl Najafi, Apr 27 2014
Number of factors in the definition of the Vandermonde determinant V(x_1,x_2,...,x_n) = Product_{1 <= i < k <= n} x_i - x_k. - Tom Copeland, Apr 27 2014
Number of weak compositions of n into three parts. - Robert A. Beeler, May 20 2014
Suppose a bag contains a(n) red marbles and a(n+1) blue marbles, where a(n), a(n+1) are consecutive triangular numbers. Then, for n > 0, the probability of choosing two marbles at random and getting two red or two blue is 1/2. In general, for k > 2, let b(0) = 0, b(1) = 1 and, for n > 1, b(n) = (k-1)*b(n-1) - b(n-2) + 1. Suppose, for n > 0, a bag contains b(n) red marbles and b(n+1) blue marbles. Then the probability of choosing two marbles at random and getting two red or two blue is (k-1)/(k+1). See also A027941, A061278, A089817, A053142, A092521. - Charlie Marion, Nov 03 2014
Let O(n) be the oblong number n(n+1) = A002378 and S(n) the square number n^2 = A000290(n). Then a(4n) = O(3n) - O(n), a(4n+1) = S(3n+1) - S(n), a(4n+2) = S(3n+2) - S(n+1) and a(4n+3) = O(3n+2) - O(n). - Charlie Marion, Feb 21 2015
Consider the partition of the natural numbers into parts from the set S=(1,2,3,...,n). The length (order) of the signature of the resulting sequence is given by the triangular numbers. E.g., for n=10, the signature length is 55. - David Neil McGrath, May 05 2015
a(n) counts the partitions of (n-1) unlabeled objects into three (3) parts (labeled a,b,c), e.g., a(5)=15 for (n-1)=4. These are (aaaa),(bbbb),(cccc),(aaab),(aaac),(aabb),(aacc),(aabc),(abbc),(abcc),(abbb),(accc),(bbcc),(bccc),(bbbc). - David Neil McGrath, May 21 2015
Conjecture: the sequence is the genus/deficiency of the sinusoidal spirals of index n which are algebraic curves. The value 0 corresponds to the case of the Bernoulli Lemniscate n=2. So the formula conjectured is (n-1)(n-2)/2. - Wolfgang Tintemann, Aug 02 2015
Conjecture: Let m be any positive integer. Then, for each n = 1,2,3,... the set {Sum_{k=s..t} 1/k^m: 1 <= s <= t <= n} has cardinality a(n) = n*(n+1)/2; in other words, all the sums Sum_{k=s..t} 1/k^m with 1 <= s <= t are pairwise distinct. (I have checked this conjecture via a computer and found no counterexample.) - Zhi-Wei Sun, Sep 09 2015
The Pisano period lengths of reading the sequence modulo m seem to be A022998(m). - R. J. Mathar, Nov 29 2015
For n >= 1, a(n) is the number of compositions of n+4 into n parts avoiding the part 2. - Milan Janjic, Jan 07 2016
In this sequence only 3 is prime. - Fabian Kopp, Jan 09 2016
Suppose you are playing Bulgarian Solitaire (see A242424 and Chamberland's and Gardner's books) and, for n > 0, you are starting with a single pile of a(n) cards. Then the number of operations needed to reach the fixed state {n, n-1,...,1} is a(n-1). For example, {6}->{5,1}->{4,2}->{3,2,1}. - Charlie Marion, Jan 14 2016
Numbers k such that 8k + 1 is a square. - Juri-Stepan Gerasimov, Apr 09 2016
Every perfect cube is the difference of the squares of two consecutive triangular numbers. 1^2-0^2 = 1^3, 3^2-1^2 = 2^3, 6^2-3^2 = 3^3. - Miquel Cerda, Jun 26 2016
For n > 1, a(n) = tau_n(k*) where tau_n(k) is the number of ordered n-factorizations of k and k* is the square of a prime. For example, tau_3(4) = tau_3(9) = tau_3(25) = tau_3(49) = 6 (see A007425) since the number of divisors of 4, 9, 25, and 49's divisors is 6, and a(3) = 6. - Melvin Peralta, Aug 29 2016
In an (n+1)-dimensional hypercube, number of two-dimensional faces congruent with a vertex (see also A001788). - Stanislav Sykora, Oct 23 2016
Generalizations of the familiar formulas, a(n) + a(n+1) = (n+1)^2 (Feb 19 2004) and a(n)^2 + a(n+1)^2 = a((n+1)^2) (Nov 22 2006), follow: a(n) + a(n+2k-1) + 4a(k-1) = (n+k)^2 + 6a(k-1) and a(n)^2 + a(n+2k-1)^2 + (4a(k-1))^2 + 3a(k-1) = a((n+k)^2 + 6a(k-1)). - Charlie Marion, Nov 27 2016
a(n) is also the greatest possible number of diagonals in a polyhedron with n+4 vertices. - Vladimir Letsko, Dec 19 2016
For n > 0, 2^5 * (binomial(n+1,2))^2 represents the first integer in a sum of 2*(2*n + 1)^2 consecutive integers that equals (2*n + 1)^6. - Patrick J. McNab, Dec 25 2016
Does not satisfy Benford's law (cf. Ross, 2012). - N. J. A. Sloane, Feb 12 2017
Number of ordered triples (a,b,c) of positive integers not larger than n such that a+b+c = 2n+1. - Aviel Livay, Feb 13 2017
Number of inequivalent tetrahedral face colorings using at most n colors so that no color appears only once. - David Nacin, Feb 22 2017
Also the Wiener index of the complete graph K_{n+1}. - Eric W. Weisstein, Sep 07 2017
Number of intersections between the Bernstein polynomials of degree n. - Eric Desbiaux, Apr 01 2018
a(n) is the area of a triangle with vertices at (1,1), (n+1,n+2), and ((n+1)^2, (n+2)^2). - Art Baker, Dec 06 2018
For n > 0, a(n) is the smallest k > 0 such that n divides numerator of (1/a(1) + 1/a(2) + ... + 1/a(n-1) + 1/k). It should be noted that 1/1 + 1/3 + 1/6 + ... + 2/(n(n+1)) = 2n/(n+1). - Thomas Ordowski, Aug 04 2019
Upper bound of the number of lines in an n-homogeneous supersolvable line arrangement (see Theorem 1.1 in Dimca). - Stefano Spezia, Oct 04 2019
For n > 0, a(n+1) is the number of lattice points on a triangular grid with side length n. - Wesley Ivan Hurt, Aug 12 2020
From Michael Chu, May 04 2022: (Start)
Maximum number of distinct nonempty substrings of a string of length n.
Maximum cardinality of the sumset A+A, where A is a set of n numbers. (End)
a(n) is the number of parking functions of size n avoiding the patterns 123, 132, and 312. - Lara Pudwell, Apr 10 2023
Suppose two rows, each consisting of n evenly spaced dots, are drawn in parallel. Suppose we bijectively draw lines between the dots of the two rows. For n >= 1, a(n - 1) is the maximal possible number of intersections between the lines. Equivalently, the maximal number of inversions in a permutation of [n]. - Sela Fried, Apr 18 2023
The following equation complements the generalization in Bala's Comment (Jan 05 2015). (2k + 1)^2*a(n) + a(k) = a((2k + 1)*n + k). - Charlie Marion, Aug 28 2023
a(n) + a(n+k) + a(k-1) + (k-1)*n = (n+k)^2. For k = 1, we have a(n) + a(n+1) = (n+1)^2. - Charlie Marion, Nov 17 2023
a(n+1)/3 is the expected number of steps to escape from a linear row of n positions starting at a random location and randomly performing steps -1 or +1 with equal probability. - Hugo Pfoertner, Jul 22 2025
a(n+1) is the number of nonnegative integer solutions to p + q + r = n. By Sylvester's law of inertia, it is also the number of congruence classes of real symmetric n-by-n matrices or equivalently, the number of symmetric bilinear forms on a real n-dimensional vector space. - Paawan Jethva, Jul 24 2025

Examples

			G.f.: x + 3*x^2 + 6*x^3 + 10*x^4 + 15*x^5 + 21*x^6 + 28*x^7 + 36*x^8 + 45*x^9 + ...
When n=3, a(3) = 4*3/2 = 6.
Example(a(4)=10): ABCD where A, B, C and D are different links in a chain or different amino acids in a peptide possible fragments: A, B, C, D, AB, ABC, ABCD, BC, BCD, CD = 10.
a(2): hollyhock leaves on the Tokugawa Mon, a(4): points in Pythagorean tetractys, a(5): object balls in eight-ball billiards. - _Bradley Klee_, Aug 24 2015
From _Gus Wiseman_, Oct 28 2020: (Start)
The a(1) = 1 through a(5) = 15 ordered triples of positive integers summing to n + 2 [Beeler, McGrath above] are the following. These compositions are ranked by A014311.
  (111)  (112)  (113)  (114)  (115)
         (121)  (122)  (123)  (124)
         (211)  (131)  (132)  (133)
                (212)  (141)  (142)
                (221)  (213)  (151)
                (311)  (222)  (214)
                       (231)  (223)
                       (312)  (232)
                       (321)  (241)
                       (411)  (313)
                              (322)
                              (331)
                              (412)
                              (421)
                              (511)
The unordered version is A001399(n-3) = A069905(n), with Heinz numbers A014612.
The strict case is A001399(n-6)*6, ranked by A337453.
The unordered strict case is A001399(n-6), with Heinz numbers A007304.
(End)
		

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. 828.
  • C. Alsina and R. B. Nelson, Charming Proofs: A Journey into Elegant Mathematics, MAA, 2010. See Chapter 1.
  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 2.
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 109ff.
  • Marc Chamberland, Single Digits: In Praise of Small Numbers, Chapter 3, The Number Three, p. 72, Princeton University Press, 2015.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 33, 38, 40, 70.
  • J. M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 309 pp 46-196, Ellipses, Paris, 2004
  • 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. 1.
  • Martin Gardner, Colossal Book of Mathematics, Chapter 34, Bulgarian Solitaire and Other Seemingly Endless Tasks, pp. 455-467, W. W. Norton & Company, 2001.
  • James Gleick, The Information: A History, A Theory, A Flood, Pantheon, 2011. [On page 82 mentions a table of the first 19999 triangular numbers published by E. de Joncort in 1762.]
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §4.6 Mathematical Proof and §8.6 Figurate Numbers, pp. 158-159, 289-290.
  • Cay S. Horstmann, Scala for the Impatient. Upper Saddle River, New Jersey: Addison-Wesley (2012): 171.
  • Elemer Labos, On the number of RGB-colors we can distinguish. Partition Spectra. Lecture at 7th Hungarian Conference on Biometry and Biomathematics. Budapest. Jul 06 2005.
  • A. Messiah, Quantum Mechanics, Vol.1, North Holland, Amsterdam, 1965, p. 457.
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 52-53, 129-132, 274.
  • 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, pages 2-6, 13.
  • T. Trotter, Some Identities for the Triangular Numbers, Journal of Recreational Mathematics, Spring 1973, 6(2).
  • D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, pp. 91-93 Penguin Books 1987.

Crossrefs

The figurate numbers, with parameter k as in the second Python program: A001477 (k=0), this sequence (k=1), A000290 (k=2), A000326 (k=3), A000384 (k=4), A000566 (k=5), A000567 (k=6), A001106 (k=7), A001107 (k=8).
a(n) = A110449(n, 0).
a(n) = A110555(n+2, 2).
A diagonal of A008291.
Column 2 of A195152.
Numbers of the form n*t(n+k,h)-(n+k)*t(n,h), where t(i,h) = i*(i+2*h+1)/2 for any h (for A000217 is k=1): A005563, A067728, A140091, A140681, A212331.
Boustrophedon transforms: A000718, A000746.
Iterations: A007501 (start=2), A013589 (start=4), A050542 (start=5), A050548 (start=7), A050536 (start=8), A050909 (start=9).
Cf. A002817 (doubly triangular numbers), A075528 (solutions of a(n)=a(m)/2).
Cf. A104712 (first column, starting with a(1)).
Some generalized k-gonal numbers are A001318 (k=5), this sequence (k=6), A085787 (k=7), etc.
A001399(n-3) = A069905(n) = A211540(n+2) counts 3-part partitions.
A001399(n-6) = A069905(n-3) = A211540(n-1) counts 3-part strict partitions.
A011782 counts compositions of any length.
A337461 counts pairwise coprime triples, with unordered version A307719.

Programs

  • Haskell
    a000217 n = a000217_list !! n
    a000217_list = scanl1 (+) [0..] -- Reinhard Zumkeller, Sep 23 2011
    
  • J
    a000217=: *-:@>: NB. Stephen Makdisi, May 02 2018
    
  • Magma
    [n*(n+1)/2: n in [0..60]]; // Bruno Berselli, Jul 11 2014
    
  • Magma
    [n: n in [0..1500] | IsSquare(8*n+1)]; // Juri-Stepan Gerasimov, Apr 09 2016
    
  • Maple
    A000217 := proc(n) n*(n+1)/2; end;
    istriangular:=proc(n) local t1; t1:=floor(sqrt(2*n)); if n = t1*(t1+1)/2 then return true else return false; end if; end proc; # N. J. A. Sloane, May 25 2008
    ZL := [S, {S=Prod(B, B, B), B=Set(Z, 1 <= card)}, unlabeled]:
    seq(combstruct[count](ZL, size=n), n=2..55); # Zerinvary Lajos, Mar 24 2007
    isA000217 := proc(n)
        issqr(1+8*n) ;
    end proc: # R. J. Mathar, Nov 29 2015 [This is the recipe Leonhard Euler proposes in chapter VII of his "Vollständige Anleitung zur Algebra", 1765. Peter Luschny, Sep 02 2022]
  • Mathematica
    Array[ #*(# - 1)/2 &, 54] (* Zerinvary Lajos, Jul 10 2009 *)
    FoldList[#1 + #2 &, 0, Range@ 50] (* Robert G. Wilson v, Feb 02 2011 *)
    Accumulate[Range[0,70]] (* Harvey P. Dale, Sep 09 2012 *)
    CoefficientList[Series[x / (1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Jul 30 2014 *)
    (* For Mathematica 10.4+ *) Table[PolygonalNumber[n], {n, 0, 53}] (* Arkadiusz Wesolowski, Aug 27 2016 *)
    LinearRecurrence[{3, -3, 1}, {0, 1, 3}, 54] (* Robert G. Wilson v, Dec 04 2016 *)
    (* The following Mathematica program, courtesy of Steven J. Miller, is useful for testing if a sequence is Benford. To test a different sequence only one line needs to be changed. This strongly suggests that the triangular numbers are not Benford, since the second and third columns of the output disagree. - N. J. A. Sloane, Feb 12 2017 *)
    fd[x_] := Floor[10^Mod[Log[10, x], 1]]
    benfordtest[num_] := Module[{},
       For[d = 1, d <= 9, d++, digit[d] = 0];
       For[n = 1, n <= num, n++,
        {
         d = fd[n(n+1)/2];
         If[d != 0, digit[d] = digit[d] + 1];
         }];
       For[d = 1, d <= 9, d++, digit[d] = 1.0 digit[d]/num];
       For[d = 1, d <= 9, d++,
        Print[d, " ", 100.0 digit[d], " ", 100.0 Log[10, (d + 1)/d]]];
       ];
    benfordtest[20000]
    Table[Length[Join@@Permutations/@IntegerPartitions[n,{3}]],{n,0,15}] (* Gus Wiseman, Oct 28 2020 *)
  • PARI
    A000217(n) = n * (n + 1) / 2;
    
  • PARI
    is_A000217(n)=n*2==(1+n=sqrtint(2*n))*n \\ M. F. Hasler, May 24 2012
    
  • PARI
    is(n)=ispolygonal(n,3) \\ Charles R Greathouse IV, Feb 28 2014
    
  • PARI
    list(lim)=my(v=List(),n,t); while((t=n*n++/2)<=lim,listput(v,t)); Vec(v) \\ Charles R Greathouse IV, Jun 18 2021
    
  • Python
    for n in range(0,60): print(n*(n+1)//2, end=', ') # Stefano Spezia, Dec 06 2018
    
  • Python
    # Intended to compute the initial segment of the sequence, not
    # isolated terms. If in the iteration the line "x, y = x + y + 1, y + 1"
    # is replaced by "x, y = x + y + k, y + k" then the figurate numbers are obtained,
    # for k = 0 (natural A001477), k = 1 (triangular), k = 2 (squares), k = 3 (pentagonal), k = 4 (hexagonal), k = 5 (heptagonal), k = 6 (octagonal), etc.
    def aList():
        x, y = 1, 1
        yield 0
        while True:
            yield x
            x, y = x + y + 1, y + 1
    A000217 = aList()
    print([next(A000217) for i in range(54)]) # Peter Luschny, Aug 03 2019
  • SageMath
    [n*(n+1)/2 for n in (0..60)] # Bruno Berselli, Jul 11 2014
    
  • Scala
    (1 to 53).scanLeft(0)( + ) // Horstmann (2012), p. 171
    
  • Scheme
    (define (A000217 n) (/ (* n (+ n 1)) 2)) ;; Antti Karttunen, Jul 08 2017
    

Formula

G.f.: x/(1-x)^3. - Simon Plouffe in his 1992 dissertation
E.g.f.: exp(x)*(x+x^2/2).
a(n) = a(-1-n).
a(n) + a(n-1)*a(n+1) = a(n)^2. - Terrel Trotter, Jr., Apr 08 2002
a(n) = (-1)^n*Sum_{k=1..n} (-1)^k*k^2. - Benoit Cloitre, Aug 29 2002
a(n+1) = ((n+2)/n)*a(n), Sum_{n>=1} 1/a(n) = 2. - Jon Perry, Jul 13 2003
For n > 0, a(n) = A001109(n) - Sum_{k=0..n-1} (2*k+1)*A001652(n-1-k); e.g., 10 = 204 - (1*119 + 3*20 + 5*3 + 7*0). - Charlie Marion, Jul 18 2003
With interpolated zeros, this is n*(n+2)*(1+(-1)^n)/16. - Benoit Cloitre, Aug 19 2003
a(n+1) is the determinant of the n X n symmetric Pascal matrix M_(i, j) = binomial(i+j+1, i). - Benoit Cloitre, Aug 19 2003
a(n) = ((n+1)^3 - n^3 - 1)/6. - Xavier Acloque, Oct 24 2003
a(n) = a(n-1) + (1 + sqrt(1 + 8*a(n-1)))/2. This recursive relation is inverted when taking the negative branch of the square root, i.e., a(n) is transformed into a(n-1) rather than a(n+1). - Carl R. White, Nov 04 2003
a(n) = Sum_{k=1..n} phi(k)*floor(n/k) = Sum_{k=1..n} A000010(k)*A010766(n, k) (R. Dedekind). - Vladeta Jovovic, Feb 05 2004
a(n) + a(n+1) = (n+1)^2. - N. J. A. Sloane, Feb 19 2004
a(n) = a(n-2) + 2*n - 1. - Paul Barry, Jul 17 2004
a(n) = sqrt(Sum_{i=1..n} Sum_{j=1..n} (i*j)) = sqrt(A000537(n)). - Alexander Adamchuk, Oct 24 2004
a(n) = sqrt(sqrt(Sum_{i=1..n} Sum_{j=1..n} (i*j)^3)) = (Sum_{i=1..n} Sum_{j=1..n} Sum_{k=1..n} (i*j*k)^3)^(1/6). - Alexander Adamchuk, Oct 26 2004
a(n) == 1 (mod n+2) if n is odd and a(n) == n/2+2 (mod n+2) if n is even. - Jon Perry, Dec 16 2004
a(0) = 0, a(1) = 1, a(n) = 2*a(n-1) - a(n-2) + 1. - Miklos Kristof, Mar 09 2005
a(n) = a(n-1) + n. - Zak Seidov, Mar 06 2005
a(n) = A108299(n+3,4) = -A108299(n+4,5). - Reinhard Zumkeller, Jun 01 2005
a(n) = A111808(n,2) for n > 1. - Reinhard Zumkeller, Aug 17 2005
a(n)*a(n+1) = A006011(n+1) = (n+1)^2*(n^2+2)/4 = 3*A002415(n+1) = 1/2*a(n^2+2*n). a(n-1)*a(n) = (1/2)*a(n^2-1). - Alexander Adamchuk, Apr 13 2006 [Corrected and edited by Charlie Marion, Nov 26 2010]
a(n) = floor((2*n+1)^2/8). - Paul Barry, May 29 2006
For positive n, we have a(8*a(n))/a(n) = 4*(2*n+1)^2 = (4*n+2)^2, i.e., a(A033996(n))/a(n) = 4*A016754(n) = (A016825(n))^2 = A016826(n). - Lekraj Beedassy, Jul 29 2006
a(n)^2 + a(n+1)^2 = a((n+1)^2) [R B Nelsen, Math Mag 70 (2) (1997), p. 130]. - R. J. Mathar, Nov 22 2006
a(n) = A126890(n,0). - Reinhard Zumkeller, Dec 30 2006
a(n)*a(n+k)+a(n+1)*a(n+1+k) = a((n+1)*(n+1+k)). Generalizes previous formula dated Nov 22 2006 [and comments by J. M. Bergot dated May 22 2012]. - Charlie Marion, Feb 04 2011
(sqrt(8*a(n)+1)-1)/2 = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 26 2007
a(n) = A023896(n) + A067392(n). - Lekraj Beedassy, Mar 02 2007
Sum_{k=0..n} a(k)*A039599(n,k) = A002457(n-1), for n >= 1. - Philippe Deléham, Jun 10 2007
8*a(n)^3 + a(n)^2 = Y(n)^2, where Y(n) = n*(n+1)*(2*n+1)/2 = 3*A000330(n). - Mohamed Bouhamida, Nov 06 2007 [Edited by Derek Orr, May 05 2015]
A general formula for polygonal numbers is P(k,n) = (k-2)*(n-1)n/2 + n = n + (k-2)*A000217(n-1), for n >= 1, k >= 3. - Omar E. Pol, Apr 28 2008 and Mar 31 2013
a(3*n) = A081266(n), a(4*n) = A033585(n), a(5*n) = A144312(n), a(6*n) = A144314(n). - Reinhard Zumkeller, Sep 17 2008
a(n) = A022264(n) - A049450(n). - Reinhard Zumkeller, Oct 09 2008
If we define f(n,i,a) = Sum_{j=0..k-1} (binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j)), then a(n) = -f(n,n-1,1), for n >= 1. - Milan Janjic, Dec 20 2008
4*a(x) + 4*a(y) + 1 = (x+y+1)^2 + (x-y)^2. - Vladimir Shevelev, Jan 21 2009
a(n) = A000124(n-1) + n-1 for n >= 2. a(n) = A000124(n) - 1. - Jaroslav Krizek, Jun 16 2009
An exponential generating function for the inverse of this sequence is given by Sum_{m>=0} ((Pochhammer(1, m)*Pochhammer(1, m))*x^m/(Pochhammer(3, m)*factorial(m))) = ((2-2*x)*log(1-x)+2*x)/x^2, the n-th derivative of which has a closed form which must be evaluated by taking the limit as x->0. A000217(n+1) = (lim_{x->0} d^n/dx^n (((2-2*x)*log(1-x)+2*x)/x^2))^-1 = (lim_{x->0} (2*Gamma(n)*(-1/x)^n*(n*(x/(-1+x))^n*(-x+1+n)*LerchPhi(x/(-1+x), 1, n) + (-1+x)*(n+1)*(x/(-1+x))^n + n*(log(1-x)+log(-1/(-1+x)))*(-x+1+n))/x^2))^-1. - Stephen Crowley, Jun 28 2009
a(n) = A034856(n+1) - A005408(n) = A005843(n) + A000124(n) - A005408(n). - Jaroslav Krizek, Sep 05 2009
a(A006894(n)) = a(A072638(n-1)+1) = A072638(n) = A006894(n+1)-1 for n >= 1. For n=4, a(11) = 66. - Jaroslav Krizek, Sep 12 2009
With offset 1, a(n) = floor(n^3/(n+1))/2. - Gary Detlefs, Feb 14 2010
a(n) = 4*a(floor(n/2)) + (-1)^(n+1)*floor((n+1)/2). - Bruno Berselli, May 23 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=1. - Mark Dols, Aug 20 2010
From Charlie Marion, Oct 15 2010: (Start)
a(n) + 2*a(n-1) + a(n-2) = n^2 + (n-1)^2; and
a(n) + 3*a(n-1) + 3*a(n-2) + a(n-3) = n^2 + 2*(n-1)^2 + (n-2)^2.
In general, for n >= m > 2, Sum_{k=0..m} binomial(m,m-k)*a(n-k) = Sum_{k=0..m-1} binomial(m-1,m-1-k)*(n-k)^2.
a(n) - 2*a(n-1) + a(n-2) = 1, a(n) - 3*a(n-1) + 3*a(n-2) - a(n-3) = 0 and a(n) - 4*a(n-1) + 6*a(n-2) - 4*(a-3) + a(n-4) = 0.
In general, for n >= m > 2, Sum_{k=0..m} (-1)^k*binomial(m,m-k)*a(n-k) = 0.
(End)
a(n) = sqrt(A000537(n)). - Zak Seidov, Dec 07 2010
For n > 0, a(n) = 1/(Integral_{x=0..Pi/2} 4*(sin(x))^(2*n-1)*(cos(x))^3). - Francesco Daddi, Aug 02 2011
a(n) = A110654(n)*A008619(n). - Reinhard Zumkeller, Aug 24 2011
a(2*k-1) = A000384(k), a(2*k) = A014105(k), k > 0. - Omar E. Pol, Sep 13 2011
a(n) = A026741(n)*A026741(n+1). - Charles R Greathouse IV, Apr 01 2012
a(n) + a(a(n)) + 1 = a(a(n)+1). - J. M. Bergot, Apr 27 2012
a(n) = -s(n+1,n), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
a(n)*a(n+1) = a(Sum_{m=1..n} A005408(m))/2, for n >= 1. For example, if n=8, then a(8)*a(9) = a(80)/2 = 1620. - Ivan N. Ianakiev, May 27 2012
a(n) = A002378(n)/2 = (A001318(n) + A085787(n))/2. - Omar E. Pol, Jan 11 2013
G.f.: x * (1 + 3x + 6x^2 + ...) = x * Product_{j>=0} (1+x^(2^j))^3 = x * A(x) * A(x^2) * A(x^4) * ..., where A(x) = (1 + 3x + 3x^2 + x^3). - Gary W. Adamson, Jun 26 2012
G.f.: G(0) where G(k) = 1 + (2*k+3)*x/(2*k+1 - x*(k+2)*(2*k+1)/(x*(k+2) + (k+1)/G(k+1))); (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Nov 23 2012
a(n) = A002088(n) + A063985(n). - Reinhard Zumkeller, Jan 21 2013
G.f.: x + 3*x^2/(Q(0)-3*x) where Q(k) = 1 + k*(x+1) + 3*x - x*(k+1)*(k+4)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Mar 14 2013
a(n) + a(n+1) + a(n+2) + a(n+3) + n = a(2*n+4). - Ivan N. Ianakiev, Mar 16 2013
a(n) + a(n+1) + ... + a(n+8) + 6*n = a(3*n+15). - Charlie Marion, Mar 18 2013
a(n) + a(n+1) + ... + a(n+20) + 2*n^2 + 57*n = a(5*n+55). - Charlie Marion, Mar 18 2013
3*a(n) + a(n-1) = a(2*n), for n > 0. - Ivan N. Ianakiev, Apr 05 2013
In general, a(k*n) = (2*k-1)*a(n) + a((k-1)*n-1). - Charlie Marion, Apr 20 2015
Also, a(k*n) = a(k)*a(n) + a(k-1)*a(n-1). - Robert Israel, Apr 20 2015
a(n+1) = det(binomial(i+2,j+1), 1 <= i,j <= n). - Mircea Merca, Apr 06 2013
a(n) = floor(n/2) + ceiling(n^2/2) = n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
a(n) = floor((n+1)/(exp(2/(n+1))-1)). - Richard R. Forberg, Jun 22 2013
Sum_{n>=1} a(n)/n! = 3*exp(1)/2 by the e.g.f. Also see A067764 regarding ratios calculated this way for binomial coefficients in general. - Richard R. Forberg, Jul 15 2013
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2) - 2 = 0.7725887... . - Richard R. Forberg, Aug 11 2014
2/(Sum_{n>=m} 1/a(n)) = m, for m > 0. - Richard R. Forberg, Aug 12 2014
A228474(a(n))=n; A248952(a(n))=0; A248953(a(n))=a(n); A248961(a(n))=A000330(n). - Reinhard Zumkeller, Oct 20 2014
a(a(n)-1) + a(a(n+2)-1) + 1 = A000124(n+1)^2. - Charlie Marion, Nov 04 2014
a(n) = 2*A000292(n) - A000330(n). - Luciano Ancora, Mar 14 2015
a(n) = A007494(n-1) + A099392(n) for n > 0. - Bui Quang Tuan, Mar 27 2015
Sum_{k=0..n} k*a(k+1) = a(A000096(n+1)). - Charlie Marion, Jul 15 2015
Let O(n) be the oblong number n(n+1) = A002378(n) and S(n) the square number n^2 = A000290(n). Then a(n) + a(n+2k) = O(n+k) + S(k) and a(n) + a(n+2k+1) = S(n+k+1) + O(k). - Charlie Marion, Jul 16 2015
A generalization of the Nov 22 2006 formula, a(n)^2 + a(n+1)^2 = a((n+1)^2), follows. Let T(k,n) = a(n) + k. Then for all k, T(k,n)^2 + T(k,n+1)^2 = T(k,(n+1)^2 + 2*k) - 2*k. - Charlie Marion, Dec 10 2015
a(n)^2 + a(n+1)^2 = a(a(n) + a(n+1)). Deducible from N. J. A. Sloane's a(n) + a(n+1) = (n+1)^2 and R. B. Nelson's a(n)^2 + a(n+1)^2 = a((n+1)^2). - Ben Paul Thurston, Dec 28 2015
Dirichlet g.f.: (zeta(s-2) + zeta(s-1))/2. - Ilya Gutkovskiy, Jun 26 2016
a(n)^2 - a(n-1)^2 = n^3. - Miquel Cerda, Jun 29 2016
a(n) = A080851(0,n-1). - R. J. Mathar, Jul 28 2016
a(n) = A000290(n-1) - A034856(n-4). - Peter M. Chema, Sep 25 2016
a(n)^2 + a(n+3)^2 + 19 = a(n^2 + 4*n + 10). - Charlie Marion, Nov 23 2016
2*a(n)^2 + a(n) = a(n^2+n). - Charlie Marion, Nov 29 2016
G.f.: x/(1-x)^3 = (x * r(x) * r(x^3) * r(x^9) * r(x^27) * ...), where r(x) = (1 + x + x^2)^3 = (1 + 3*x + 6*x^2 + 7*x^3 + 6*x^4 + 3*x^5 + x^6). - Gary W. Adamson, Dec 03 2016
a(n) = sum of the elements of inverse of matrix Q(n), where Q(n) has elements q_i,j = 1/(1-4*(i-j)^2). So if e = appropriately sized vector consisting of 1's, then a(n) = e'.Q(n)^-1.e. - Michael Yukish, Mar 20 2017
a(n) = Sum_{k=1..n} ((2*k-1)!!*(2*n-2*k-1)!!)/((2*k-2)!!*(2*n-2*k)!!). - Michael Yukish, Mar 20 2017
Sum_{i=0..k-1} a(n+i) = (3*k*n^2 + 3*n*k^2 + k^3 - k)/6. - Christopher Hohl, Feb 23 2019
a(n) = A060544(n + 1) - A016754(n). - Ralf Steiner, Nov 09 2019
a(n) == 0 (mod n) iff n is odd (see De Koninck reference). - Bernard Schott, Jan 10 2020
8*a(k)*a(n) + ((a(k)-1)*n + a(k))^2 = ((a(k)+1)*n + a(k))^2. This formula reduces to the well-known formula, 8*a(n) + 1 = (2*n+1)^2, when k = 1. - Charlie Marion, Jul 23 2020
a(k)*a(n) = Sum_{i = 0..k-1} (-1)^i*a((k-i)*(n-i)). - Charlie Marion, Dec 04 2020
From Amiram Eldar, Jan 20 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(7)*Pi/2)/(2*Pi).
Product_{n>=2} (1 - 1/a(n)) = 1/3. (End)
a(n) = Sum_{k=1..2*n-1} (-1)^(k+1)*a(k)*a(2*n-k). For example, for n = 4, 1*28 - 3*21 + 6*15 - 10*10 + 15*6 - 21*3 + 28*1 = 10. - Charlie Marion, Mar 23 2022
2*a(n) = A000384(n) - n^2 + 2*n. In general, if P(k,n) = the n-th k-gonal number, then (j+1)*a(n) = P(5 + j, n) - n^2 + (j+1)*n. More generally, (j+1)*P(k,n) = P(2*k + (k-2)*(j-1),n) - n^2 + (j+1)*n. - Charlie Marion, Mar 14 2023
a(n) = A109613(n) * A004526(n+1). - Torlach Rush, Nov 10 2023
a(n) = (1/6)* Sum_{k = 0..3*n} (-1)^(n+k+1) * k*(k + 1) * binomial(3*n+k, 2*k). - Peter Bala, Nov 03 2024
From Peter Bala, Jul 05 2025: (Start)
The following series telescope: for k >= 0,
Sum_{n >= 1} a(n)*a(n+2)*...*a(n+2*k)/(a(n+1)*a(n+3)*...*a(n+2*k+3)) = 1/(2*k + 3);
Sum_{n >= 1} a(n+1)*a(n+3)*...*a(n+2*k+1)/(a(n)*a(n+2)*...*a(n+2*k+2)) = 2/(2*k + 3) * Sum_{i = 1..2*k+3} 1/i. (End)

Extensions

Edited by Derek Orr, May 05 2015

A046092 4 times triangular numbers: a(n) = 2*n*(n+1).

Original entry on oeis.org

0, 4, 12, 24, 40, 60, 84, 112, 144, 180, 220, 264, 312, 364, 420, 480, 544, 612, 684, 760, 840, 924, 1012, 1104, 1200, 1300, 1404, 1512, 1624, 1740, 1860, 1984, 2112, 2244, 2380, 2520, 2664, 2812, 2964, 3120, 3280, 3444, 3612, 3784, 3960, 4140, 4324
Offset: 0

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Comments

Consider all Pythagorean triples (X,Y,Z=Y+1) ordered by increasing Z; sequence gives Y values. X values are 1, 3, 5, 7, 9, ... (A005408), Z values are A001844.
In the triple (X, Y, Z) we have X^2=Y+Z. Actually, the triple is given by {x, (x^2 -+ 1)/2}, where x runs over the odd numbers (A005408) and x^2 over the odd squares (A016754). - Lekraj Beedassy, Jun 11 2004
a(n) is the number of edges in n X n square grid with all horizontal and vertical segments filled in. - Asher Auel, Jan 12 2000 [Corrected by Felix Huber, Apr 09 2024]
a(n) is the only number satisfying an inequality related to zeta(2) and zeta(3): Sum_{i>a(n)+1} 1/i^2 < Sum_{i>n} 1/i^3 < Sum_{i>a(n)} 1/i^2. - Benoit Cloitre, Nov 02 2001
Number of right triangles made from vertices of a regular n-gon when n is even. - Sen-Peng Eu, Apr 05 2001
Number of ways to change two non-identical letters in the word aabbccdd..., where there are n type of letters. - Zerinvary Lajos, Feb 15 2005
a(n) is the number of (n-1)-dimensional sides of an (n+1)-dimensional hypercube (e.g., squares have 4 corners, cubes have 12 edges, etc.). - Freek van Walderveen (freek_is(AT)vanwal.nl), Nov 11 2005
From Nikolaos Diamantis (nikos7am(AT)yahoo.com), May 23 2006: (Start)
Consider a triangle, a pentagon, a heptagon, ..., a k-gon where k is odd. We label a triangle with n=1, a pentagon with n=2, ..., a k-gon with n = floor(k/2). Imagine a player standing at each vertex of the k-gon.
Initially there are 2 frisbees, one held by each of two neighboring players. Every time they throw the frisbee to one of their two nearest neighbors with equal probability. Then a(n) gives the average number of steps needed so that the frisbees meet.
I verified this by simulating the processes with a computer program. For example, a(2) = 12 because in a pentagon that's the expected number of trials we need to perform. That is an exercise in Concrete Mathematics and it can be done using generating functions. (End)
A diagonal of A059056. - Zerinvary Lajos, Jun 18 2007
If X_1,...,X_n is a partition of a 2n-set X into 2-blocks then a(n-1) is equal to the number of 2-subsets of X containing none of X_i, (i=1,...,n). - Milan Janjic, Jul 16 2007
X values of solutions to the equation 2*X^3 + X^2 = Y^2. To find Y values: b(n) = 2n(n+1)(2n+1). - Mohamed Bouhamida, Nov 06 2007
Number of (n+1)-permutations of 3 objects u,v,w, with repetition allowed, containing n-1 u's. Example: a(1)=4 because we have vv, vw, wv and ww; a(2)=12 because we can place u in each of the previous four 2-permutations either in front, or in the middle, or at the end. - Zerinvary Lajos, Dec 27 2007
Sequence found by reading the line from 0, in the direction 0, 4, ... and the same line from 0, in the direction 0, 12, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, May 03 2008
a(n) is also the least weight of self-conjugate partitions having n different even parts. - Augustine O. Munagi, Dec 18 2008
From Peter Luschny, Jul 12 2009: (Start)
The general formula for alternating sums of powers of even integers is in terms of the Swiss-Knife polynomials P(n,x) A153641 (P(n,1)-(-1)^k P(n,2k+1))/2. Here n=2, thus
a(k) = |(P(2,1) - (-1)^k*P(2,2k+1))/2|. (End)
The sum of squares of n+1 consecutive numbers between a(n)-n and a(n) inclusive equals the sum of squares of n consecutive numbers following a(n). For example, for n = 2, a(2) = 12, and the corresponding equation is 10^2 + 11^2 + 12^2 = 13^2 + 14^2. - Tanya Khovanova, Jul 20 2009
Number of roots in the root system of type D_{n+1} (for n>2). - Tom Edgar, Nov 05 2013
Draw n ellipses in the plane (n>0), any 2 meeting in 4 points; sequence gives number of intersections of these ellipses (cf. A051890, A001844); a(n) = A051890(n+1) - 2 = A001844(n) - 1. - Jaroslav Krizek, Dec 27 2013
a(n) appears also as the second member of the quartet [p0(n), a(n), p2(n), p3(n)] of the square of [n, n+1, n+2, n+3] in the Clifford algebra Cl_2 for n >= 0. p0(n) = -A147973(n+3), p2(n) = A054000(n+1) and p3(n) = A139570(n). See a comment on A147973, also with a reference. - Wolfdieter Lang, Oct 15 2014
a(n) appears also as the third and fourth member of the quartet [p0(n), p0(n), a(n), a(n)] of the square of [n, n, n+1, n+1] in the Clifford algebra Cl_2 for n >= 0. p0(n) = A001105(n). - Wolfdieter Lang, Oct 16 2014
Consider two equal rectangles composed of unit squares. Then surround the 1st rectangle with 1-unit-wide layers to build larger rectangles, and surround the 2nd rectangle just to hide the previous layers. If r(n) and h(n) are the number of unit squares needed for n layers in the 1st case and the 2nd case, then for all rectangles, we have a(n) = r(n) - h(n) for n>=1. - Michel Marcus, Sep 28 2015
When greater than 4, a(n) is the perimeter of a Pythagorean triangle with an even short leg 2*n. - Agola Kisira Odero, Apr 26 2016
Also the number of minimum connected dominating sets in the (n+1)-cocktail party graph. - Eric W. Weisstein, Jun 29 2017
a(n+1) is the harmonic mean of A000384(n+2) and A014105(n+1). - Bob Andriesse, Apr 27 2019
Consider a circular cake from which wedges of equal center angle c are cut out in clockwise succession and turned around so that the bottom comes to the top. This goes on until the cake shows its initial surface again. An interesting case occurs if 360°/c is not an integer. Then, with n = floor(360°/c), the number of wedges which have to be cut out and turned equals a(n). (For the number of cutting line segments see A005408.) - According to Peter Winkler's book "Mathematical Mind-Benders", which presents the problem and its solution (see Winkler, pp. 111, 115) the problem seems to be of French origin but little is known about its history. - Manfred Boergens, Apr 05 2022
a(n-3) is the maximum irregularity over all maximal 2-degenerate graphs with n vertices. The extremal graphs are 2-stars (K_2 joined to n-2 independent vertices). (The irregularity of a graph is the sum of the differences between the degrees over all edges of the graph.) - Allan Bickle, May 29 2023
Number of ways of placing a domino on a (n+1)X(n+1) board of squares. - R. J. Mathar, Apr 24 2024
The sequence terms are the exponents in the expansion of (1/(1 + x)) * Sum_{n >= 0} x^n * Product_{k = 1..n} (1 - x^(2*k-1))/(1 + x^(2*k+1)) = 1 - x^4 + x^12 - x^24 + x^40 - x^60 + - ... (Andrews and Berndt, Entry 9.3.3, p. 229). Cf. A153140. - Peter Bala, Feb 15 2025
Number of edges in an (n+1)-dimensional orthoplex. 2D orthoplexes (diamonds) have 4 edges, 3D orthoplexes (octahedrons) have 12 edges, 4D orthoplexes (16-cell) have 24 edges, and so on. - Aaron Franke, Mar 23 2025

Examples

			a(7)=112 because 112 = 2*7*(7+1).
The first few triples are (1,0,1), (3,4,5), (5,12,13), (7,24,25), ...
The first such partitions, corresponding to a(n)=1,2,3,4, are 2+2, 4+4+2+2, 6+6+4+4+2+2, 8+8+6+6+4+4+2+2. - _Augustine O. Munagi_, Dec 18 2008
		

References

  • George E. Andrews and Bruce C. Berndt, Ramanujan's Lost Notebook, Part I, Springer, 2005.
  • Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.
  • Albert H. Beiler, Recreations in the Theory of Numbers. New York: Dover, p. 125, 1964.
  • Ronald L. Graham, D. E. Knuth and Oren Patashnik, Concrete Mathematics, Reading, Massachusetts: Addison-Wesley, 1994.
  • Peter Winkler, Mathematical Mind-Benders, Wellesley, Massachusetts: A K Peters, 2007.

Crossrefs

Main diagonal of array in A001477.
Equals A033996/2. Cf. A001844. - Augustine O. Munagi, Dec 18 2008
Cf. A078371, A141530 (see Librandi's comment in A078371).
Cf. similar sequences listed in A299645.
Cf. A005408.
Cf. A016754.
Cf. A002378, A046092, A028896 (irregularities of maximal k-degenerate graphs).

Programs

Formula

a(n) = A100345(n+1, n-1) for n>0.
a(n) = 2*A002378(n) = 4*A000217(n). - Lekraj Beedassy, May 25 2004
a(n) = C(2n, 2) - n = 4*C(n, 2). - Zerinvary Lajos, Feb 15 2005
From Lekraj Beedassy, Jun 04 2006: (Start)
a(n) - a(n-1)=4*n.
Let k=a(n). Then a(n+1) = k + 2*(1 + sqrt(2k + 1)). (End)
Array read by rows: row n gives A033586(n), A085250(n+1). - Omar E. Pol, May 03 2008
O.g.f.:4*x/(1-x)^3; e.g.f.: exp(x)*(2*x^2+4*x). - Geoffrey Critzer, May 17 2009
From Stephen Crowley, Jul 26 2009: (Start)
a(n) = 1/int(-(x*n+x-1)*(step((-1+x*n)/n)-1)*n*step((x*n+x-1)/(n+1)),x=0..1) where step(x)=piecewise(x<0,0,0<=x,1) is the Heaviside step function.
Sum_{n>=1} 1/a(n) = 1/2. (End)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=4, a(2)=12. - Harvey P. Dale, Jul 25 2011
For n > 0, a(n) = 1/(Integral_{x=0..Pi/2} (sin(x))^(2*n-1)*(cos(x))^3). - Francesco Daddi, Aug 02 2011
a(n) = A001844(n) - 1. - Omar E. Pol, Oct 03 2011
(a(n) - A000217(k))^2 = A000217(2n-k)*A000217(2n+1+k) - (A002378(n) - A000217(k)), for all k. See also A001105. - Charlie Marion, May 09 2013
From Ivan N. Ianakiev, Aug 30 2013: (Start)
a(n)*(2m+1)^2 + a(m) = a(n*(2m+1)+m), for any nonnegative integers n and m.
t(k)*a(n) + t(k-1)*a(n+1) = a((n+1)*(t(k)-t(k-1)-1)), where k>=2, n>=1, t(k)=A000217(k). (End)
a(n) = A245300(n,n). - Reinhard Zumkeller, Jul 17 2014
2*a(n)+1 = A016754(n) = A005408(n)^2, the odd squares. - M. F. Hasler, Oct 02 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2) - 1/2 = A187832. - Ilya Gutkovskiy, Mar 16 2017
a(n) = lcm(2*n,2*n+2). - Enrique Navarrete, Aug 30 2017
a(n)*a(n+k) + k^2 = m^2 (a perfect square), n >= 1, k >= 0. - Ezhilarasu Velayutham, May 13 2019
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = cosh(Pi/2)/(Pi/2).
Product_{n>=1} (1 - 1/a(n)) = -2*cos(sqrt(3)*Pi/2)/Pi. (End)
a(n) = A016754(n) - A001844(n). - Leo Tavares, Sep 20 2022

A001057 Canonical enumeration of integers: interleaved positive and negative integers with zero prepended.

Original entry on oeis.org

0, 1, -1, 2, -2, 3, -3, 4, -4, 5, -5, 6, -6, 7, -7, 8, -8, 9, -9, 10, -10, 11, -11, 12, -12, 13, -13, 14, -14, 15, -15, 16, -16, 17, -17, 18, -18, 19, -19, 20, -20, 21, -21, 22, -22, 23, -23, 24, -24, 25, -25, 26, -26, 27, -27, 28, -28, 29, -29, 30, -30, 31, -31
Offset: 0

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Comments

Go forwards and backwards with increasing step sizes. - Daniele Parisse and Franco Virga, Jun 06 2005
The partial sums of the divergent series 1 - 2 + 3 - 4 + ... give this sequence. Euler summed it to 1/4 which was one of the first examples of summing divergent series. - Michael Somos, May 22 2007
From Peter Luschny, Jul 12 2009: (Start)
The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus
a(k) = 2^(-2)(P(1,1)-(-1)^k P(1,2k+1)). (End)
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-1, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=4, a(n-3)=(-1)^(n-1)*coeff(charpoly(A,x),x). - Milan Janjic, Jan 26 2010
Cantor ordering of the integers producing a 1-1 and onto correspondence between the natural numbers and the integers showing that the set Z of integers has the same cardinality as the set N of natural numbers. The cardinal of N is the first transfinite cardinal aleph_null (or aleph_naught), which is the cardinality of a given infinite set if and only if it is countably infinite (denumerable), i.e., it can be put in 1-1 and onto correspondence (with a proper Cantor ordering) with the natural numbers. - Daniel Forgues, Jan 23 2010
a(n) is the determinant of the (n+2) X (n+2) (0,1)-Toeplitz matrix M satisfying: M(i,j)=0 iff i=j or i=j-1. The matrix M arises in the variation of ménage problem where not a round table, but one side of a rectangular table is considered (see comments of Vladimir Shevelev in A000271). Namely M(i,j) defines the class of permutations p of 1,2,...,n+2 such that p(i)<>i and p(i)<>i+1 for i=1,2,...,n+1, and p(n+2)<>n+2. And a(n) is also the difference between the number of even and odd such permutations. - Dmitry Efimov, Mar 02 2017

Examples

			G.f. = x - x^2 + 2*x^3 - 2*x^4 + 3*x^5 - 3*x^6 + 4*x^7 - 4*x^8 + 5*x^9 - 5*x^10 + ...
		

Crossrefs

Cf. A008619, A004526, A166711, A166871, A130472 (negation), A142150 (partial sums), A010551 (partial products for n > 0).
Alternating row sums of A104578 are a(n+1), for n >= 0.

Programs

  • Haskell
    a001057 n = (n' + m) * (-1) ^ (1 - m) where (n',m) = divMod n 2
    a001057_list = 0 : concatMap (\x -> [x,-x]) [1..]
    -- Reinhard Zumkeller, Apr 02 2012
    
  • Maple
    a := n -> (1-(-1)^n*(2*n+1))/4; # Peter Luschny, Jul 12 2009
  • Mathematica
    Join[{0},Riffle[Range[35],-Range[35]]] (* Harvey P. Dale, Sep 21 2011 *)
    a[ n_] := -(-1)^n Ceiling[n/2]; (* Michael Somos, Jun 05 2013 *)
    LinearRecurrence[{-1, 1, 1}, {0, 1, -1}, 63] (* Jean-François Alcover, Jan 07 2019 *)
  • PARI
    {a(n) = if( n%2, n\2 + 1, -n/2)}; /* Michael Somos, Jul 20 1999 */
    
  • Python
    def a(n): return n//2 + 1 if n%2 else -n//2
    print([a(n) for n in range(63)]) # Michael S. Branicky, Jul 14 2022

Formula

Euler transform of [-1, 2] is sequence a(n+1). - Michael Somos, Jun 11 2003
G.f.: x / ((1 + x) * (1 - x^2)). - Michael Somos, Jul 20 1999
E.g.f.: (exp(x) - (1 - 2*x) * exp(-x)) / 4. - Michael Somos, Jun 11 2003
a(n) = 1 - 2*a(n-1) -a(n-2); a(2*n) = -n, a(2*n+1) = n+1. - Michael Somos, Jul 20 1999
|a(n+1)| = A008619(n). |a(n-1)| = A004526(n). - Michael Somos, Jul 20 1999
a(n) = -a(n-1) + a(n-2) + a(n-3). a(n) = (-1)^(n+1) * floor((n+1) / 2). - Michael Somos, Jun 11 2003
a(1) = 1, a(n) = a(n-1)+n or a(n-1)-n whichever is closer to 0 on the number line. Or abs(a(n)) = min{abs(a(n-1)+n), abs(a(n-1)-n)}. - Amarnath Murthy, Jul 01 2003
a(n) = Sum_{k=0..n} k*(-1)^(k+1). - Paul Barry, Aug 20 2003
a(n) = (1-(2n+1)*(-1)^n)/4. - Paul Barry, Feb 02 2004
a(0) = 0; a(n) = (-1)^(n-1) * (n-|a(n-1)|) for n >= 1. - Rick L. Shepherd, Jul 14 2004
a(n) = a(n-1)-n*(-1)^n, a(0)=0; or a(n) = -a(n-1)+(1-(-1)^n)/2, a(0)=0. - Daniele Parisse and Franco Virga, Jun 06 2005
a(n) = ceiling(n/2) * (-1)^(n+1), n >= 0. - Franklin T. Adams-Watters, Nov 25 2011 (corrected by Daniel Forgues, Jul 21 2012)
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jun 05 2013
Sum_{n>=1} 1/a(n) = 0. - Jaume Oliver Lafont, Jul 14 2017

Extensions

Thanks to Michael Somos for helpful comments.
Name edited by Franklin T. Adams-Watters, Jan 30 2012

A001586 Generalized Euler numbers, or Springer numbers.

Original entry on oeis.org

1, 1, 3, 11, 57, 361, 2763, 24611, 250737, 2873041, 36581523, 512343611, 7828053417, 129570724921, 2309644635483, 44110959165011, 898621108880097, 19450718635716001, 445777636063460643, 10784052561125704811, 274613643571568682777, 7342627959965776406281
Offset: 0

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From Peter Bala, Feb 02 2011: (Start)
The Springer numbers were originally considered by Glaisher (see references). They are a type B analog of the zigzag numbers A000111 for the group of signed permutations.
COMBINATORIAL INTERPRETATIONS
Several combinatorial interpretations of the Springer numbers are known:
1) a(n) gives the number of Weyl chambers in the principal Springer cone of the Coxeter group B_n of symmetries of an n dimensional cube. An example can be found in [Arnold - The Calculus of snakes...].
2) Arnold found an alternative combinatorial interpretation of the Springer numbers in terms of snakes. Snakes are a generalization of alternating permutations to the group of signed permutations. A signed permutation is a sequence (x_1,x_2,...,x_n) of integers such that {|x_1|,|x_2|,...,|x_n|} = {1,2,...,n}. They form a group, the hyperoctahedral group of order 2^n*n! = A000165(n), isomorphic to the group of symmetries of the n dimensional cube. A snake of type B_n is a signed permutation (x_1,x_2,...,x_n) such that 0 < x_1 > x_2 < ... x_n. For example, (3,-4,-2,-5,1,-6) is a snake of type B_6. a(n) gives the number of snakes of type B_n [Arnold]. The cases n=2 and n=3 are given in the Example section below.
3) The Springer numbers also arise in the study of the critical points of functions; they count the topological types of odd functions with 2*n critical values [Arnold, Theorem 35].
4) Let F_n be the set of plane rooted forests satisfying the following conditions:
... each root has exactly one child, and each of the other internal nodes has exactly two (ordered) children,
... there are n nodes labeled by integers from 1 to n, but some leaves can be non-labeled (these are called empty leaves), and labels are increasing from each root down to the leaves. Then a(n) equals the cardinality of F_n. An example and proof are given in [Verges, Theorem 4.5].
OTHER APPEARANCES OF THE SPRINGER NUMBERS
1) Hoffman has given a connection between Springer numbers, snakes and the successive derivatives of the secant and tangent functions.
2) For integer N the quarter Gauss sums Q(N) are defined by ... Q(N) := Sum_{r = 0..floor(N/4)} exp(2*Pi*I*r^2/N). In the cases N = 1 (mod 4) and N = 3 (mod 4) an asymptotic series for Q(N) as N -> inf that involves the Springer numbers has been given by Evans et al., see 1.32 and 1.33.
For a sequence of polynomials related to the Springer numbers see A185417. For a table to recursively compute the Springer numbers see A185418.
(End)
Similar to the way in which the signed Euler numbers A122045 are 2^n times the value of the Euler polynomials at 1/2, the generalized signed Euler numbers A188458 can be seen as 2^n times the value of generalized Euler polynomials at 1/2. These are the Swiss-Knife polynomials A153641. A recursive definition of these polynomials is given in A081658. - Peter Luschny, Jul 19 2012
a(n) is the number of reverse-complementary updown permutations of [2n]. For example, the updown permutation 241635 is reverse-complementary because its complement is 536142, which is the same as its reverse, and a(2)=3 counts 1324, 2413, 3412. - David Callan, Nov 29 2012
a(n) = |2^n G(n,1/2;-1)|, a specialization of the Appell sequence of polynomials umbrally formed by G(n,x;t) = (G(.,0;t) + x)^n from the Grassmann polynomials G(n,0;t) of A046802 enumerating the cells of the positive Grassmannians. - Tom Copeland, Oct 14 2015
Named "Springer numbers" after the Dutch mathematician Tonny Albert Springer (1926-2011). - Amiram Eldar, Jun 13 2021

Examples

			a(2) = 3: The three snakes of type B_2 are
  (1,-2), (2,1), (2,-1).
a(3) = 11: The 11 snakes of type B_3 are
  (1,-2,3), (1,-3,2), (1,-3,-2),
  (2,1,3), (2,-1,3), (2,-3,1), (2,-3,-1),
  (3,1,2), (3,-1,2), (3,-2,1), (3,-2,-1).
		

References

  • V. I. Arnold, Springer numbers and Morsification spaces. J. Algebraic Geom., Vol. 1, No. 2 (1992), pp. 197-214.
  • J. W. L. Glaisher, "On the coefficients in the expansions of cos x/cos 2x and sin x/cos 2x", Quart. J. Pure and Applied Math., Vol. 45 (1914), pp. 187-222.
  • J. W. L. Glaisher, On the Bernoullian function, Q. J. Pure Appl. Math., Vol. 29 (1898), pp. 1-168.
  • Ulrike Sattler, Decidable classes of formal power series with nice closure properties, Diplomarbeit im Fach Informatik, Univ. Erlangen - Nürnberg, Jul 27 1994.
  • 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).
  • Tonny Albert Springer, Remarks on a combinatorial problem, Nieuw Arch. Wisk., Vol. 19, No. 3 (1971), pp. 30-36.

Crossrefs

Row 2 of A349271.
Bisections are A000281 and A000464. Overview in A349264.
Related polynomials are given in A098432, A081658 and A153641.
Cf. A046802.

Programs

  • Maple
    a := proc(n) local k; (-1)^iquo(n,2)*add(2^k*binomial(n,k)*euler(k),k=0..n) end; # Peter Luschny, Jul 08 2009
    a := n -> (-1)^(n+iquo(n,2))*2^(3*n+1)*(Zeta(0,-n,1/8) - Zeta(0,-n,5/8)):
    seq(a(n),n=0..21); # Peter Luschny, Mar 11 2015
  • Mathematica
    n=21; CoefficientList[Series[1/(Cos[x]-Sin[x]), {x, 0, n}], x] * Table[k!, {k, 0, n}] (* Jean-François Alcover, May 18 2011 *)
    Table[Abs[Numerator[EulerE[n,1/4]]],{n,0,35}] (* Harvey P. Dale, May 18 2011 *)
  • PARI
    {a(n) = if(n<0, 0, n! * polcoeff( 1 / (cos(x + x * O(x^n)) - sin(x + x * O(x^n))), n))}; /* Michael Somos, Feb 03 2004 */
    
  • PARI
    {a(n) = my(an); if(n<2, n>=0, an = vector(n+1, m, 1); for(m=2, n, an[m+1] = 2*an[m] + an[m-1] + sum(k=0, m-3, binomial(m-2, k) * (an[k+1] * an[m-1-k] + 2*an[k+2] * an[m-k] - an[k+3] * an[m-1-k]))); an[n+1])}; /* Michael Somos, Feb 03 2004 */
    
  • PARI
    /* Explicit formula by Peter Bala: */
    {a(n)=((1+I)/2)^n*sum(k=0,n,((1-I)/(1+I))^k*sum(j=0,k,(-1)^(k-j)*binomial(n+1,k-j)*(2*j+1)^n))}
    
  • Sage
    @CachedFunction
    def p(n,x) :
        if n == 0 : return 1
        w = -1 if n%2 == 0 else 0
        v =  1 if n%2 == 0 else -1
        return v*add(p(k,0)*binomial(n,k)*(x^(n-k)+w) for k in range(n)[::2])
    def A001586(n) : return abs(2^n*p(n, 1/2))
    [A001586(n) for n in (0..21)] # Peter Luschny, Jul 19 2012

Formula

E.g.f.: 1/(cos(x) - sin(x)).
Values at 1 of polynomials Q_n() defined in A104035. - N. J. A. Sloane, Nov 06 2009
a(n) = numerator of abs(Euler(n,1/4)). - N. J. A. Sloane, Nov 07 2009
Let B_n(x) = Sum_{k=0.. n*(n-1)/2} b(n,k)*x^k, where b(n,k) is number of n-node acyclic digraphs with k arcs, cf. A081064; then a(n) = |B_n(-2)|. - Vladeta Jovovic, Jan 25 2005
G.f. A(x) = y satisfies y'^2 = 2y^4 - y^2, y''y = y^2 + 2y'^2. - Michael Somos, Feb 03 2004
a(n) = (-1)^floor(n/2) Sum_{k=0..n} 2^k C(n,k) Euler(k). - Peter Luschny, Jul 08 2009
From Peter Bala, Feb 02 2011: (Start)
(1)... a(n) = ((1 + i)/2)^n*B(n,(1 - i)/(1 + i)), where i = sqrt(-1) and {B(n,x)}n>=0 = [1, 1 + x, 1 + 6*x + x^2, 1 + 23*x + 23*x^2 + x^3, ...] is the sequence of type B Eulerian polynomials - see A060187.
This yields the explicit formula
(2)... a(n) = ((1 + i)/2)^n*Sum_{k = 0..n} ((1 - i)/(1 + i))^k * Sum_{j = 0..k} (-1)^(k-j)*binomial(n+1,k-j)*(2*j + 1)^n.
The result (2) can be used to find congruences satisfied by the Springer numbers. For example, for odd prime p
(3)
... a(p) = 1 (mod p) when p = 4*n + 1
... a(p) = -1 (mod p) when p = 4*n + 3.
(End)
E.g.f.: 1/Q(0) where Q(k) = 1 - x/((2k+1)-x*(2k+1)/(x+(2k+2)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 19 2011
E.g.f.: 2/U(0) where U(k) = 1 + 1/(1 + x/(2*k + 1 -x - (2*k+1)/(2 - x/(x+ (2*k+2)/U(k+1))))); (continued fraction, 5-step). - Sergei N. Gladkovskii, Sep 24 2012
E.g.f.: 1/G(0) where G(k) = 1 - x/(4*k+1 - x*(4*k+1)/(4*k+2 + x + x*(4*k+2)/(4*k+3 - x - x*(4*k+3)/(x + (4*k+4)/G(k+1) )))); (continued fraction, 3rd kind, 5-step). - Sergei N. Gladkovskii, Oct 02 2012
G.f.: 1/G(0) where G(k) = 1 - x*(2*k+1) - 2*x^2*(k+1)*(k+1)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 11 2013
a(n) = | 2*4^n*lerchphi(-1, -n, 1/4) |. - Peter Luschny, Apr 27 2013
a(n) ~ 4 * n^(n+1/2) * (4/Pi)^n / (sqrt(Pi)*exp(n)). - Vaclav Kotesovec, Oct 07 2013
G.f.: T(0)/(1-x), where T(k) = 1 - 2*x^2*(k+1)^2/( 2*x^2*(k+1)^2 - (1-x-2*x*k)*(1-3*x-2*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 15 2013
a(n) = (-1)^C(n+1,2)*2^(3*n+1)*(Zeta(-n,1/8)-Zeta(-n,5/8)), where Zeta(a,z) is the generalized Riemann zeta function. - Peter Luschny, Mar 11 2015
E.g.f. A(x) satisfies: A(x) = exp( Integral A(x)/A(-x) dx ). - Paul D. Hanna, Feb 04 2017
E.g.f. A(x) satisfies: A'(x) = A(x)^2/A(-x). - Paul D. Hanna, Feb 04 2017

Extensions

More terms from Vladeta Jovovic, Jan 25 2005

A109449 Triangle read by rows, T(n,k) = binomial(n,k)*A000111(n-k), 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 3, 3, 1, 5, 8, 6, 4, 1, 16, 25, 20, 10, 5, 1, 61, 96, 75, 40, 15, 6, 1, 272, 427, 336, 175, 70, 21, 7, 1, 1385, 2176, 1708, 896, 350, 112, 28, 8, 1, 7936, 12465, 9792, 5124, 2016, 630, 168, 36, 9, 1, 50521, 79360, 62325, 32640, 12810, 4032, 1050, 240, 45, 10, 1
Offset: 0

Views

Author

Philippe Deléham, Aug 27 2005

Keywords

Comments

The boustrophedon transform {t} of a sequence {s} is given by t_n = Sum_{k=0..n} T(n,k)*s(k). Triangle may be called the boustrophedon triangle.
The 'signed version' of the triangle is the exponential Riordan array [sech(x) + tanh(x), x]. - Peter Luschny, Jan 24 2009
Up to signs, the matrix is self-inverse: T^(-1)(n,k) = (-1)^(n+k)*T(n,k). - R. J. Mathar, Mar 15 2013

Examples

			Triangle starts:
      1;
      1,     1;
      1,     2,     1;
      2,     3,     3,     1;
      5,     8,     6,     4,     1;
     16,    25,    20,    10,     5,    1;
     61,    96,    75,    40,    15,    6,    1;
    272,   427,   336,   175,    70,   21,    7,   1;
   1385,  2176,  1708,   896,   350,  112,   28,   8,  1;
   7936, 12465,  9792,  5124,  2016,  630,  168,  36,  9,  1;
  50521, 79360, 62325, 32640, 12810, 4032, 1050, 240, 45, 10, 1; ...
		

Crossrefs

Programs

  • Haskell
    a109449 n k = a109449_row n !! k
    a109449_row n = zipWith (*)
                    (a007318_row n) (reverse $ take (n + 1) a000111_list)
    a109449_tabl = map a109449_row [0..]
    -- Reinhard Zumkeller, Nov 02 2013
    
  • Magma
    f:= func< n,x | Evaluate(BernoulliPolynomial(n+1), x) >;
    A109449:= func< n,k | k eq n select 1 else 2^(2*n-2*k+1)*Binomial(n,k)*Abs(f(n-k,3/4) - f(n-k,1/4) + f(n-k,1) - f(n-k,1/2))/(n-k+1) >;
    [A109449(n,k): k in [0..n], n in [0..13]]; // G. C. Greubel, Jul 10 2025
  • Maple
    From Peter Luschny, Jul 10 2009, edited Jun 06 2022: (Start)
    A109449 := (n,k) -> binomial(n, k)*A000111(n-k):
    seq(print(seq(A109449(n, k), k=0..n)), n=0..9);
    B109449 := (n,k) -> 2^(n-k)*binomial(n, k)*abs(euler(n-k, 1/2)+euler(n-k, 1)) -`if`(n-k=0, 1, 0): seq(print(seq(B109449(n, k), k=0..n)), n=0..9);
    R109449 := proc(n, k) option remember; if k = 0 then A000111(n) else R109449(n-1, k-1)*n/k fi end: seq(print(seq(R109449(n, k), k=0..n)), n=0..9);
    E109449 := proc(n) add(binomial(n, k)*euler(k)*((x+1)^(n-k)+ x^(n-k)), k=0..n) -x^n end: seq(print(seq(abs(coeff(E109449(n), x, k)), k=0..n)), n=0..9);
    sigma := n -> ifelse(n=0, 1, [1,1,0,-1,-1,-1,0,1][n mod 8 + 1]/2^iquo(n-1,2)-1):
    L109449 := proc(n) add(add((-1)^v*binomial(k, v)*(x+v+1)^n*sigma(k), v=0..k), k=0..n) end: seq(print(seq(abs(coeff(L109449(n), x, k)), k=0..n)), n=0..9);
    X109449 := n -> n!*coeff(series(exp(x*t)*(sech(t)+tanh(t)), t, 24), t, n): seq(print(seq(abs(coeff(X109449(n), x, k)), k=0..n)), n=0..9);
    (End)
  • Mathematica
    lim = 10; s = CoefficientList[Series[(1 + Sin[x])/Cos[x], {x, 0, lim}], x] Table[k!, {k, 0, lim}]; Table[Binomial[n, k] s[[n - k + 1]], {n, 0, lim}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 24 2015, after Jean-François Alcover at A000111 *)
    T[n_, k_] := (n!/k!) SeriesCoefficient[(1 + Sin[x])/Cos[x], {x, 0, n - k}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 27 2019 *)
  • PARI
    A109449(n,k)=binomial(n,k)*if(n>k,2*abs(polylog(k-n,I)),1) \\ M. F. Hasler, Oct 05 2017
    
  • Sage
    R = PolynomialRing(ZZ, 'x')
    @CachedFunction
    def skp(n, x) :
        if n == 0 : return 1
        return add(skp(k, 0)*binomial(n, k)*(x^(n-k)-(n+1)%2) for k in range(n)[::2])
    def A109449_row(n):
        x = R.gen()
        return [abs(c) for c in list(skp(n,x)-skp(n,x-1)+x^n)]
    for n in (0..10) : print(A109449_row(n)) # Peter Luschny, Jul 22 2012
    

Formula

Sum_{k>=0} T(n, k) = A000667(n).
Sum_{k>=0} T(2n, 2k) = A000795(n).
Sum_{k>=0} T(2n, 2k+1) = A009747(n).
Sum_{k>=0} T(2n+1, 2k) = A003719(n).
Sum_{k>=0} T(2n+1, 2k+1) = A002084(n).
Sum_{k>=0} T(n, 2k) = A062272(n).
Sum_{k>=0} T(n, 2k+1) = A062161(n).
Sum_{k>=0} (-1)^(k)*T(n, k) = A062162(n). - Johannes W. Meijer, Apr 20 2011
E.g.f.: exp(x*y)*(sec(x)+tan(x)). - Vladeta Jovovic, May 20 2007
T(n,k) = 2^(n-k)*C(n,k)*|E(n-k,1/2) + E(n-k,1)| - [n=k] where C(n,k) is the binomial coefficient, E(m,x) are the Euler polynomials and [] the Iverson bracket. - Peter Luschny, Jan 24 2009
From Reikku Kulon, Feb 26 2009: (Start)
A109449(n, 0) = A000111(n), approx. round(2^(n + 2) * n! / Pi^(n + 1)).
A109449(n, n - 1) = n.
A109449(n, n) = 1.
For n > 0, k > 0: A109449(n, k) = A109449(n - 1, k - 1) * n / k. (End)
From Peter Luschny, Jul 10 2009: (Start)
Let p_n(x) = Sum_{k=0..n} Sum_{v=0..k} (-1)^v C(k,v)*F(k)*(x+v+1)^n, where F(0)=1 and for k>0 F(k)=-1 + s_k 2^floor((k-1)/2), s_k is 0 if k mod 8 in {2,6}, 1 if k mod 8 in {0,1,7} and otherwise -1. T(n,k) are the absolute values of the coefficients of these polynomials.
Another way to express the polynomials p_n(x) is
p_n(x) = -x^n + Sum_{k=0..n} binomial(n,k)*Euler(k)((x+1)^(n-k) + x^(n-k)). (End)
From Peter Bala, Jan 26 2011: (Start)
An explicit formula for the n-th row polynomial is
x^n + i*Sum_{k=1..n}((1+i)/2)^(k-1)*Sum_{j=0..k} (-1)^j*binomial(k,j)*(x+i*j)^n, where i = sqrt(-1). This is the triangle of connection constants between the polynomial sequences {Z(n,x+1)} and {Z(n,x)}, where Z(n,x) denotes the zigzag polynomials described in A147309.
Denote the present array by M. The first column of the array (I-x*M)^-1 is a sequence of rational functions in x whose numerator polynomials are the row polynomials of A145876 - the generalized Eulerian numbers associated with the zigzag numbers. (End)
Let skp{n}(x) denote the Swiss-Knife polynomials A153641. Then
T(n,k) = [x^(n-k)] |skp{n}(x) - skp{n}(x-1) + x^n|. - Peter Luschny, Jul 22 2012
T(n,k) = A007318(n,k) * A000111(n - k), k = 0..n. - Reinhard Zumkeller, Nov 02 2013
T(n,k) = abs(A247453(n,k)). - Reinhard Zumkeller, Sep 17 2014

Extensions

Edited, formula corrected, typo T(9,4)=2016 (before 2816) fixed by Peter Luschny, Jul 10 2009

A077221 a(0) = 0 and then alternately even and odd numbers in increasing order such that the sum of any two successive terms is a square.

Original entry on oeis.org

0, 1, 8, 17, 32, 49, 72, 97, 128, 161, 200, 241, 288, 337, 392, 449, 512, 577, 648, 721, 800, 881, 968, 1057, 1152, 1249, 1352, 1457, 1568, 1681, 1800, 1921, 2048, 2177, 2312, 2449, 2592, 2737, 2888, 3041, 3200, 3361, 3528, 3697, 3872, 4049, 4232
Offset: 0

Views

Author

Amarnath Murthy, Nov 03 2002

Keywords

Comments

This sequence arises from reading the line from 0, in the direction 0, 1, ... and the same line from 0, in the direction 0, 8, ..., in the square spiral whose vertices are the triangular numbers A000217. Cf. A139591, etc. - Omar E. Pol, May 03 2008
The general formula for alternating sums of powers of odd integers is in terms of the Swiss-Knife polynomials P(n,x) A153641 (P(n,0)-(-1)^k*P(n,2*k))/2. Here n=2, thus a(k) = |(P(2,0)-(-1)^k*P(2,2*k))/2|. - Peter Luschny, Jul 12 2009
Axis perpendicular to A046092 in the square spiral whose vertices are the triangular numbers A000217. See the comment above. - Omar E. Pol, Sep 14 2011
Column 8 of A195040. - Omar E. Pol, Sep 28 2011
Concentric octagonal numbers. A139098 and A069129 interleaved. - Omar E. Pol, Sep 17 2011
Subsequence of A194274. - Bruno Berselli, Sep 22 2011
Partial sums of A047522. - Reinhard Zumkeller, Jan 07 2012
Alternating sum of the first n odd squares in decreasing order, n >= 1. Also number of "ON" cells at n-th stage in simple 2-dimensional cellular automaton. The rules are: on the infinite square grid, start with all cells OFF, so a(0) = 0. Turn a single cell to the ON state, so a(1) = 1. At each subsequent step, the neighbor cells of each cell of the old generation are turned ON, and the cells of the old generation are turned OFF. Here "neighbor" refers to the eight adjacent cells of each ON cell. See example. - Omar E. Pol, Feb 16 2014

Examples

			From _Omar E. Pol_, Feb 16 2014: (Start)
Illustration of initial terms as a cellular automaton:
.
.                                   O O O O O O O
.                     O O O O O     O           O
.           O O O     O       O     O   O O O   O
.     O     O   O     O   O   O     O   O   O   O
.           O O O     O       O     O   O O O   O
.                     O O O O O     O           O
.                                   O O O O O O O
.
.     1       8           17              32
.
(End)
		

Crossrefs

Programs

Formula

a(2n) = 8*n^2, a(2n+1) = 8*n(n+1) + 1.
From Ralf Stephan, Mar 31 2003: (Start)
a(n) = 2*n^2 + 4*n + 1 [+1 if n is odd] with a(0)=1.
G.f.: x*(x^2+6*x+1)/(1-x)^3/(1+x). (End)
Row sums of triangle A131925; binomial transform of (1, 7, 2, 4, -8, 16, -32, ...). - Gary W. Adamson, Jul 29 2007
a(n) = a(-n); a(n+1) = A195605(n) - (-1)^n. - Bruno Berselli, Sep 22 2011
a(n) = 2*n^2 + ((-1)^n-1)/2. - Omar E. Pol, Sep 28 2011
Sum_{n>=1} 1/a(n) = Pi^2/48 + tan(Pi/(2*sqrt(2)))*Pi /(4*sqrt(2)). - Amiram Eldar, Jan 16 2023

Extensions

Extended by Ralf Stephan, Mar 31 2003

A062392 a(n) = n^4 - (n-1)^4 + (n-2)^4 - ... 0^4.

Original entry on oeis.org

0, 1, 15, 66, 190, 435, 861, 1540, 2556, 4005, 5995, 8646, 12090, 16471, 21945, 28680, 36856, 46665, 58311, 72010, 87990, 106491, 127765, 152076, 179700, 210925, 246051, 285390, 329266, 378015, 431985, 491536, 557040, 628881, 707455, 793170, 886446, 987715
Offset: 0

Views

Author

Henry Bottomley, Jun 21 2001

Keywords

Comments

Number of edges in the join of two complete graphs of order n^2 and n, K_n^2 * K_n. - Roberto E. Martinez II, Jan 07 2002
The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus a(k) = |2^(-5)(P(4,1)-(-1)^k P(4,2k+1))|. - Peter Luschny, Jul 12 2009
Define an infinite symmetric array by T(n,m) = n*(n-1) + m for 0 <= m <= n and T(n,m) = T(m,n), n >= 0. Then a(n) is the sum of terms in the top left (n+1) X (n+1) subarray: a(n) = Sum_{r=0..n} Sum_{c=0..n} T(r,c). - J. M. Bergot, Jul 05 2013
a(n) is the sum of all positive numbers less than A002378(n). - J. M. Bergot, Aug 30 2013
Except the first term, these are triangular numbers that remain triangular when divided by their index, e.g., 66 divided by 11 gives 6. - Waldemar Puszkarz, Sep 14 2017
a(n) is the semiperimeter of the unique primitive Pythagorean triple such that (a-b+c)/2 = T(n) = A000217(n). Its long leg and hypotenuse are consecutive natural numbers and the triple is (2*T(n) - 1, 2*T(n)*(T(n) - 1), 2*T(n)*(T(n) - 1) + 1). - Miguel-Ángel Pérez García-Ortega, May 27 2025

Examples

			From _Bruno Berselli_, Oct 30 2017: (Start)
After 0:
1   =                 -(1) + (2);
15  =             -(1 + 2) + (3 + 4 + 5 + 2*3);
66  =         -(1 + 2 + 3) + (4 + 5 + 6 + 7 + ... + 11 + 3*4);
190 =     -(1 + 2 + 3 + 4) + (5 + 6 + 7 + 8 + ... + 19 + 4*5);
435 = -(1 + 2 + 3 + 4 + 5) + (6 + 7 + 8 + 9 + ... + 29 + 5*6), etc. (End)
		

References

  • T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.

Crossrefs

Cf. A000538, A000583. A062393 provides the result for 5th powers, A011934 for cubes, A000217 for squares, A001057 (unsigned) for nonnegative integers, A000035 (offset) for 0th powers.
Cf. A236770 (see crossrefs).

Programs

  • Maple
    a := n -> (2*n^2+n^3-1)*n/2; # Peter Luschny, Jul 12 2009
  • Mathematica
    Table[n (n + 1) (n^2 + n - 1)/2, {n, 0, 40}] (* Harvey P. Dale, Oct 19 2011 *)
  • PARI
    { a=0; for (n=0, 1000, write("b062392.txt", n, " ", a=n^4 - a) ) } \\ Harry J. Smith, Aug 07 2009

Formula

a(n) = n*(n+1)*(n^2 + n - 1)/2 = n^4 - a(n-1) = A000583(n) - a(n-1) = A000217(A028387(n-1)) = A000217(n)*A028387(n-1).
a(n) = Sum_{i=0..n} A007588(i) for n > 0. - Jonathan Vos Post, Mar 15 2006
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 4. - Harvey P. Dale, Oct 19 2011
G.f.: x*(x*(x + 10) + 1)/(1 - x)^5. - Harvey P. Dale, Oct 19 2011
a(n) = A000384(A000217(n)). - Bruno Berselli, Jan 31 2014
a(n) = A110450(n) - A002378(n). - Gionata Neri, May 13 2015
Sum_{n>=1} 1/a(n) = tan(sqrt(5)*Pi/2)*2*Pi/sqrt(5). - Amiram Eldar, Jan 22 2024
a(n) = sqrt(144*A288876(n-2) + 72*A006542(n+2) + A000537(n)). - Yasser Arath Chavez Reyes, Jul 22 2024
E.g.f.: exp(x)*x*(2 + 13*x + 8*x^2 + x^3)/2. - Stefano Spezia, Apr 27 2025
a(n) = A000217(n)*(2*A000217(n)-1). - Miguel-Ángel Pérez García-Ortega, May 27 2025

A119879 Exponential Riordan array (sech(x),x).

Original entry on oeis.org

1, 0, 1, -1, 0, 1, 0, -3, 0, 1, 5, 0, -6, 0, 1, 0, 25, 0, -10, 0, 1, -61, 0, 75, 0, -15, 0, 1, 0, -427, 0, 175, 0, -21, 0, 1, 1385, 0, -1708, 0, 350, 0, -28, 0, 1, 0, 12465, 0, -5124, 0, 630, 0, -36, 0, 1, -50521, 0, 62325, 0, -12810, 0, 1050, 0, -45, 0, 1
Offset: 0

Views

Author

Paul Barry, May 26 2006

Keywords

Comments

Row sums have e.g.f. exp(x)*sech(x) (signed version of A009006). Inverse of masked Pascal triangle A119467. Transforms the sequence with e.g.f. g(x) to the sequence with e.g.f. g(x)*sech(x).
Coefficients of the Swiss-Knife polynomials for the computation of Euler, tangent and Bernoulli number (triangle read by rows). Another version in A153641. - Philippe Deléham, Oct 26 2013
Relations to Green functions and raising/creation and lowering/annihilation/destruction operators are presented in Hodges and Sukumar and in Copeland's discussion of this sequence and 2020 pdf. - Tom Copeland, Jul 24 2020

Examples

			Triangle begins:
     1;
     0,    1;
    -1,    0,     1;
     0,   -3,     0,   1;
     5,    0,    -6,   0,   1;
     0,   25,     0, -10,   0,   1;
   -61,    0,    75,   0, -15,   0,   1;
     0, -427,     0, 175,   0, -21,   0,  1;
  1385,    0, -1708,   0, 350,   0, -28,  0,  1;
		

Crossrefs

Row sums are A155585. - Johannes W. Meijer, Apr 20 2011
Rows reversed: A081658.

Programs

  • Maple
    T := (n,k) -> binomial(n,k)*2^(n-k)*euler(n-k,1/2): # Peter Luschny, Jan 25 2009
  • Mathematica
    T[n_, k_] := Binomial[n, k] 2^(n-k) EulerE[n-k, 1/2];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 20 2018, after Peter Luschny *)
  • PARI
    {T(n,k) = binomial(n,k)*2^(n-k)*(2/(n-k+1))*(subst(bernpol(n-k+1, x), x, 1/2) - 2^(n-k+1)*subst(bernpol(n-k+1, x), x, 1/4))};
    for(n=0,5, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Feb 25 2019
  • Sage
    @CachedFunction
    def A119879_poly(n, x) :
        return 1 if n == 0  else add(A119879_poly(k, 0)*binomial(n, k)*(x^(n-k)-1+n%2) for k in range(n)[::2])
    def A119879_row(n) :
        R = PolynomialRing(ZZ, 'x')
        return R(A119879_poly(n,x)).coeffs()  # Peter Luschny, Jul 16 2012
    # Alternatively:
    
  • Sage
    # uses[riordan_array from A256893]
    riordan_array(sech(x), x, 9, exp=true) # Peter Luschny, Apr 19 2015
    

Formula

Number triangle whose k-th column has e.g.f. sech(x)*x^k/k!.
T(n,k) = C(n,k)*2^(n-k)*E_{n-k}(1/2) where C(n,k) is the binomial coefficient and E_{m}(x) are the Euler polynomials. - Peter Luschny, Jan 25 2009
The coefficients in ascending order of x^i of the polynomials p{0}(x) = 1 and p{n}(x) = Sum_{k=0..n-1; k even} binomial(n,k)*p{k}(0)*((n mod 2) - 1 + x^(n-k)). - Peter Luschny, Jul 16 2012
E.g.f.: exp(x*z)/cosh(x). - Peter Luschny, Aug 01 2012
Sum_{k=0..n} T(n,k)*x^k = A122045(n), A155585(n), A119880(n), A119881(n) for x = 0, 1, 2, 3 respectively. - Philippe Deléham, Oct 27 2013
With all offsets 0, let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of A046802 (the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of A248727 (the face polynomials of the stellahedra), by f_n(x) = A_n(1 + x;1); the Swiss-knife polynomials of this entry, A119879, by Sw_n(x) = A_n(-1;1 + x); and the row polynomials of the Worpitsky triangle (A130850), by w_n(x) = A(1 + x;0). Other specializations of A_n(x;y) give A090582 (the f-polynomials of the permutohedra, cf. also A019538) and A028246 (another version of the Worpitsky triangle). - Tom Copeland, Jan 24 2020
Triangle equals P*((I + P^2)/2)^(-1), where P denotes Pascal's triangle A007318. - Peter Bala, Mar 07 2024

A000281 Expansion of cos(x)/cos(2x).

Original entry on oeis.org

1, 3, 57, 2763, 250737, 36581523, 7828053417, 2309644635483, 898621108880097, 445777636063460643, 274613643571568682777, 205676334188681975553003, 184053312545818735778213457, 193944394596325636374396208563
Offset: 0

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Author

Keywords

Comments

a(n) is (2n)! times the coefficient of x^(2n) in the Taylor series for cos(x)/cos(2x).

Examples

			cos x / cos 2*x = 1 + 3*x^2/2 + 19*x^4/8 + 307*x^6/80 + ...
		

References

  • J. W. L. Glaisher, "On the coefficients in the expansions of cos x / cos 2x and sin x / cos 2x", Quart. J. Pure and Applied Math., 45 (1914), 187-222.
  • I. J. Schwatt, Intro. to Operations with Series, Chelsea, p. 278.
  • 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

Programs

  • Maple
    a := n -> (-1)^n*2^(6*n+1)*(Zeta(0,-2*n,1/8)-Zeta(0,-2*n,5/8)):
    seq(a(n), n=0..13); # Peter Luschny, Mar 11 2015
  • Mathematica
    With[{nn=30},Take[CoefficientList[Series[Cos[x]/Cos[2x],{x,0,nn}],x] Range[0,nn]!,{1,-1,2}]] (* Harvey P. Dale, Oct 06 2011 *)
  • PARI
    {a(n) = if( n<0, 0, n*=2; n! * polcoeff( cos(x + x * O(x^n)) / cos(2*x + x * O(x^n)), n))}; /* Michael Somos, Feb 09 2006 */

Formula

a(n) = Sum_{k=0..n} (-1)^k*binomial(2n, 2k)*A000364(n-k)*4^(n-k). - Philippe Deléham, Jan 26 2004
E.g.f.: Sum_{k>=0} a(k)x^(2k)/(2k)! = cos(x)/cos(2x).
a(n-1) is approximately 2^(4*n-3)*(2*n-1)!*sqrt(2)/((Pi^(2*n-1))*(2*n-1)). The approximation is quite good a(250) is of the order of 10^1181 and this formula is accurate to 238 digits. - Simon Plouffe, Jan 31 2007
G.f.: 1 / (1 - 1*3*x / (1 - 4*4*x / (1 - 5*7*x / (1 - 8*8*x / (1 - 9*11*x / ... ))))). - Michael Somos, May 12 2012
G.f.: 1/E(0) where E(k) = 1 - 3*x - 16*x*k*(2*k+1) - 16*x^2*(k+1)^2*(4*k+1)*(4*k+3)/E(k+1) (continued fraction, 1-step). - Sergei N. Gladkovskii, Sep 17 2012
G.f.: T(0)/(1-3*x), where T(k) = 1 - 16*x^2*(4*k+1)*(4*k+3)*(k+1)^2/( 16*x^2*(4*k+1)*(4*k+3)*(k+1)^2 - (32*x*k^2+16*x*k+3*x-1 )*(32*x*k^2+80*x*k+51*x -1)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 11 2013
From Peter Bala, Mar 09 2015: (Start)
a(n) = (-1)^n*4^(2*n)*E(2*n,1/4), where E(n,x) denotes the n-th Euler polynomial.
O.g.f.: Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 + x*(4*k + 1)^2) = 1 + 3*x + 57*x^2 + 2763*x^3 + ....
We appear to have the asymptotic expansion Pi/(2*sqrt(2)) - Sum {k = 0..n - 1} (-1)^floor(k/2)/(2*k + 1) ~ 1/(2*n) - 3/(2*n)^3 + 57/(2*n)^5 - 2763/(2*n)^7 + .... See A093954.
Bisection of A001586. See also A188458 and A212435. Second row of A235605 (read as a square array).
The expansion of exp( Sum_{n >= 1} a(n)*x^n/n ) appears to have integer coefficients. See A255883. (End)
From Peter Luschny, Mar 11 2015: (Start)
a(n) = ((-64)^n/((n+1/2)))*(B(2*n+1,7/8)-B(2*n+1,3/8)), B(n,x) Bernoulli polynomials.
a(n) = 2*(-16)^n*LerchPhi(-1, -2*n, 1/4).
a(n) = (-1)^n*Sum_{0..2*n} 2^k*C(2*n,k)*E(k), E(n) the Euler secant numbers A122045.
a(n) = (-4)^n*SKP(2*n,1/2) where SKP are the Swiss-Knife polynomials A153641.
a(n) = (-1)^n*2^(6*n+1)*(Zeta(-2*n,1/8) - Zeta(-2*n,5/8)), where Zeta(a,z) is the generalized Riemann zeta function. (End)
From Peter Bala, May 13 2017: (Start)
G.f.: 1/(1 + x - 4*x/(1 - 12*x/(1 + x - 40*x/(1 - 56*x/(1 + x - ... - 4*n(4*n - 3)*x/(1 - 4*n(4*n - 1)*x/(1 + x - ...
G.f.: 1/(1 + 9*x - 12*x/(1 - 4*x/(1 + 9*x - 56*x/(1 - 40*x/(1 + 9*x - ... - 4*n(4*n - 1)*x/(1 - 4*n(4*n - 3)*x/(1 + 9*x - .... (End)
From Peter Bala, Nov 08 2019: (Start)
a(n) = sqrt(2)*4^n*Integral_{x = 0..inf} x^(2*n)*cosh(Pi*x/2)/cosh(Pi*x) dx. Cf. A002437.
The L-series 1 + 1/3^(2*n+1) - 1/5^(2*n+1) - 1/7^(2*n+1) + + - - ... = sqrt(2)*(Pi/4)^(2*n+1)*a(n)/(2*n)! (see Shanks), which gives a(n) ~ (1/sqrt(2))*(2*n)!*(4/Pi)^(2*n+1). (End)
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