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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A171218 a(n) = Sum_{k=0..n} A109613(k)*A005843(n-k).

Original entry on oeis.org

0, 2, 6, 16, 32, 58, 94, 144, 208, 290, 390, 512, 656, 826, 1022, 1248, 1504, 1794, 2118, 2480, 2880, 3322, 3806, 4336, 4912, 5538, 6214, 6944, 7728, 8570, 9470, 10432, 11456, 12546, 13702, 14928, 16224, 17594, 19038, 20560, 22160, 23842, 25606
Offset: 0

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Author

Reinhard Zumkeller, Dec 05 2009

Keywords

Comments

a(n) is the number of triples (w,x,y) with all terms in {0,...,n} and 2|w-x|Clark Kimberling, Jun 11 2012]

Programs

  • Magma
    [&+[(2*k+(-1)^k+1)*(n-k): k in [0..n]]: n in [0..42]]; // Bruno Berselli, Nov 16 2011
  • Mathematica
    CoefficientList[Series[2x (1+x^2)/((1+x)(1-x)^4),{x,0,50}],x] (* or *) LinearRecurrence[ {3,-2,-2,3,-1},{0,2,6,16,32},50] (* Harvey P. Dale, Jan 22 2023 *)

Formula

a(n+1) - a(n) = A137928(n+1).
From Bruno Berselli, Nov 16 2011: (Start)
G.f.: 2*x*(1+x^2)/((1+x)*(1-x)^4).
a(n) = 2*A131941(n) = (2*n*(2*n^2+3*n+4)-3*(-1)^n+3)/12.
a(n) = -a(-n-1) = 3*a(n-1)-2*a(n-2)-2*a(n-3)+3*a(n-4)-a(n-5). (End)

A305198 Number of set partitions of [2n+1] with symmetric block size list of length A109613(n).

Original entry on oeis.org

1, 1, 7, 56, 470, 10299, 91925, 3939653, 36298007, 2571177913, 24158837489, 2557117944391, 24350208829581, 3601150175699409, 34626777577615921, 6820331445080882282, 66066554102006208712, 16719951521837764142510, 162903256982698962545956
Offset: 0

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Author

Alois P. Heinz, May 27 2018

Keywords

Crossrefs

Bisection (odd part) of A305197.

Programs

  • Maple
    b:= proc(n, s) option remember; expand(`if`(n>s,
          binomial(n-1, n-s-1)*x, 1)+add(binomial(n-1, j-1)*
          b(n-j, s+j)*binomial(s+j-1, j-1), j=1..(n-s)/2)*x^2)
        end:
    a:= n-> coeff(b(2*n+1, 0), x, n+irem(n+1, 2)):
    seq(a(n), n=0..20);
  • Mathematica
    b[n_, s_] := b[n, s] = Expand[If[n > s, Binomial[n - 1, n - s - 1] x, 1] + Sum[Binomial[n - 1, j - 1] b[n - j, s + j] Binomial[s + j - 1, j - 1], {j, 1, (n - s)/2}] x^2];
    a[n_] := Coefficient[b[2n + 1, 0], x, n + Mod[n + 1, 2]];
    a /@ Range[0, 20] (* Jean-François Alcover, Dec 08 2020, after Alois P. Heinz *)

Formula

a(n) = A275281(2n+1,A109613(n)).

A180116 A008619(n-1)-fold concatenation of A109613(n).

Original entry on oeis.org

1, 3, 33, 55, 555, 777, 7777, 9999, 99999, 1111111111, 111111111111, 131313131313, 13131313131313, 15151515151515, 1515151515151515, 1717171717171717, 171717171717171717, 191919191919191919, 19191919191919191919, 21212121212121212121, 2121212121212121212121
Offset: 1

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Author

Mark Dols, Aug 10 2010

Keywords

Comments

Written underneath, the first terms have increasing lengths filling a triangular shape:
1,
3,
33,
55,
555,
777,
...

Crossrefs

Programs

  • Maple
    cat2 := proc(a,b) a*10^(max(1,1+ilog10(b)))+b ; end proc:
    A008619 := proc(n) 1+floor(n/2) ; end proc:
    A109613 := proc(n) 2*floor(n/2)+1 ; end proc:
    A180116 := proc(n) a := A109613(n) ; for t from 2 to A008619(n-1) do a := cat2(a,A109613(n)) ; end do: a ; end proc:
    seq(A180116(n),n=1..24) ; # R. J. Mathar, Sep 19 2010

Extensions

Edited by R. J. Mathar, Sep 19 2010

A008866 Prime(A052928(n+1)) + (-1)^n* prime(A109613(n)).

Original entry on oeis.org

1, 8, 2, 18, 2, 30, 2, 42, 6, 60, 6, 78, 2, 90, 6, 112, 2, 128, 4, 144, 6, 162, 6, 186, 4, 204, 4, 216, 4, 240, 4, 268, 2, 288, 2, 308, 6, 330, 6, 352, 2, 372, 2, 390, 2, 410, 12, 450, 2, 462, 6, 480, 10, 508, 6, 532
Offset: 1

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Author

Keywords

References

  • Problem #1210, Series Formation, by Charles W. Trigg, J. Rec. Math., 15 (1982), 221-222.

Programs

  • Maple
    ithprime(2*ceil(n/2))+(-1)^n*ithprime(2*floor(n/2)+1);

A180115 A109613(n)-fold concatenation of A008619(n).

Original entry on oeis.org

1, 111, 222, 22222, 33333, 3333333, 4444444, 444444444, 555555555, 55555555555, 66666666666, 6666666666666, 7777777777777, 777777777777777, 888888888888888
Offset: 1

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Author

Mark Dols, Aug 10 2010

Keywords

Crossrefs

Extensions

Definition rephrased, kewyord:base,less added - R. J. Mathar, Aug 19 2010

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

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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

A005408 The odd numbers: a(n) = 2*n + 1.

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131
Offset: 0

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Leibniz's series: Pi/4 = Sum_{n>=0} (-1)^n/(2n+1) (cf. A072172).
Beginning of the ordering of the natural numbers used in Sharkovski's theorem - see the Cielsielski-Pogoda paper.
The Sharkovski ordering begins with the odd numbers >= 3, then twice these numbers, then 4 times them, then 8 times them, etc., ending with the powers of 2 in decreasing order, ending with 2^0 = 1.
Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0(6).
Also continued fraction for coth(1) (A073747 is decimal expansion). - Rick L. Shepherd, Aug 07 2002
a(1) = 1; a(n) is the smallest number such that a(n) + a(i) is composite for all i = 1 to n-1. - Amarnath Murthy, Jul 14 2003
Smallest number greater than n, not a multiple of n, but containing it in binary representation. - Reinhard Zumkeller, Oct 06 2003
Numbers n such that phi(2n) = phi(n), where phi is Euler's totient (A000010). - Lekraj Beedassy, Aug 27 2004
Pi*sqrt(2)/4 = Sum_{n>=0} (-1)^floor(n/2)/(2n+1) = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 ... [since periodic f(x)=x over -Pi < x < Pi = 2(sin(x)/1 - sin(2x)/2 + sin(3x)/3 - ...) using x = Pi/4 (Maor)]. - Gerald McGarvey, Feb 04 2005
For n > 1, numbers having 2 as an anti-divisor. - Alexandre Wajnberg, Oct 02 2005
a(n) = shortest side a of all integer-sided triangles with sides a <= b <= c and inradius n >= 1.
First differences of squares (A000290). - Lekraj Beedassy, Jul 15 2006
The odd numbers are the solution to the simplest recursion arising when assuming that the algorithm "merge sort" could merge in constant unit time, i.e., T(1):= 1, T(n):= T(floor(n/2)) + T(ceiling(n/2)) + 1. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 14 2006
2n-5 counts the permutations in S_n which have zero occurrences of the pattern 312 and one occurrence of the pattern 123. - David Hoek (david.hok(AT)telia.com), Feb 28 2007
For n > 0: number of divisors of (n-1)th power of any squarefree semiprime: a(n) = A000005(A001248(k)^(n-1)); a(n) = A000005(A000302(n-1)) = A000005(A001019(n-1)) = A000005(A009969(n-1)) = A000005(A087752(n-1)). - Reinhard Zumkeller, Mar 04 2007
For n > 2, a(n-1) is the least integer not the sum of < n n-gonal numbers (0 allowed). - Jonathan Sondow, Jul 01 2007
A134451(a(n)) = abs(A134452(a(n))) = 1; union of A134453 and A134454. - Reinhard Zumkeller, Oct 27 2007
Numbers n such that sigma(2n) = 3*sigma(n). - Farideh Firoozbakht, Feb 26 2008
a(n) = A139391(A016825(n)) = A006370(A016825(n)). - Reinhard Zumkeller, Apr 17 2008
Number of divisors of 4^(n-1) for n > 0. - J. Lowell, Aug 30 2008
Equals INVERT transform of A078050 (signed - cf. comments); and row sums of triangle A144106. - Gary W. Adamson, Sep 11 2008
Odd numbers(n) = 2*n+1 = square pyramidal number(3*n+1) / triangular number(3*n+1). - Pierre CAMI, Sep 27 2008
A000035(a(n))=1, A059841(a(n))=0. - Reinhard Zumkeller, Sep 29 2008
Multiplicative closure of A065091. - Reinhard Zumkeller, Oct 14 2008
a(n) is also the maximum number of triangles that n+2 points in the same plane can determine. 3 points determine max 1 triangle; 4 points can give 3 triangles; 5 points can give 5; 6 points can give 7 etc. - Carmine Suriano, Jun 08 2009
Binomial transform of A130706, inverse binomial transform of A001787(without the initial 0). - Philippe Deléham, Sep 17 2009
Also the 3-rough numbers: positive integers that have no prime factors less than 3. - Michael B. Porter, Oct 08 2009
Or n without 2 as prime factor. - Juri-Stepan Gerasimov, Nov 19 2009
Given an L(2,1) labeling l of a graph G, let k be the maximum label assigned by l. The minimum k possible over all L(2,1) labelings of G is denoted by lambda(G). For n > 0, this sequence gives lambda(K_{n+1}) where K_{n+1} is the complete graph on n+1 vertices. - K.V.Iyer, Dec 19 2009
A176271 = odd numbers seen as a triangle read by rows: a(n) = A176271(A002024(n+1), A002260(n+1)). - Reinhard Zumkeller, Apr 13 2010
For n >= 1, a(n-1) = numbers k such that arithmetic mean of the first k positive integers is an integer. A040001(a(n-1)) = 1. See A145051 and A040001. - Jaroslav Krizek, May 28 2010
Union of A179084 and A179085. - Reinhard Zumkeller, Jun 28 2010
For n>0, continued fraction [1,1,n] = (n+1)/a(n); e.g., [1,1,7] = 8/15. - Gary W. Adamson, Jul 15 2010
Numbers that are the sum of two sequential integers. - Dominick Cancilla, Aug 09 2010
Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n + (h-4)*(-1)^n - h)/4 (h and n in A000027), therefore ((2*h*n + (h-4)*(-1)^n - h)/4)^2 - 1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 4). Also a(n)^2 - 1 == 0 (mod 8). - Bruno Berselli, Nov 17 2010
A004767 = a(a(n)). - Reinhard Zumkeller, Jun 27 2011
A001227(a(n)) = A000005(a(n)); A048272(a(n)) < 0. - Reinhard Zumkeller, Jan 21 2012
a(n) is the minimum number of tosses of a fair coin needed so that the probability of more than n heads is at least 1/2. In fact, Sum_{k=n+1..2n+1} Pr(k heads|2n+1 tosses) = 1/2. - Dennis P. Walsh, Apr 04 2012
A007814(a(n)) = 0; A037227(a(n)) = 1. - Reinhard Zumkeller, Jun 30 2012
1/N (i.e., 1/1, 1/2, 1/3, ...) = Sum_{j=1,3,5,...,infinity} k^j, where k is the infinite set of constants 1/exp.ArcSinh(N/2) = convergents to barover(N). The convergent to barover(1) or [1,1,1,...] = 1/phi = 0.6180339..., whereas c.f. barover(2) converges to 0.414213..., and so on. Thus, with k = 1/phi we obtain 1 = k^1 + k^3 + k^5 + ..., and with k = 0.414213... = (sqrt(2) - 1) we get 1/2 = k^1 + k^3 + k^5 + .... Likewise, with the convergent to barover(3) = 0.302775... = k, we get 1/3 = k^1 + k^3 + k^5 + ..., etc. - Gary W. Adamson, Jul 01 2012
Conjecture on primes with one coach (A216371) relating to the odd integers: iff an integer is in A216371 (primes with one coach either of the form 4q-1 or 4q+1, (q > 0)); the top row of its coach is composed of a permutation of the first q odd integers. Example: prime 19 (q = 5), has 5 terms in each row of its coach: 19: [1, 9, 5, 7, 3] ... [1, 1, 1, 2, 4]. This is interpreted: (19 - 1) = (2^1 * 9), (19 - 9) = (2^1 * 5), (19 - 5) = (2^1 - 7), (19 - 7) = (2^2 * 3), (19 - 3) = (2^4 * 1). - Gary W. Adamson, Sep 09 2012
A005408 is the numerator 2n-1 of the term (1/m^2 - 1/n^2) = (2n-1)/(mn)^2, n = m+1, m > 0 in the Rydberg formula, while A035287 is the denominator (mn)^2. So the quotient a(A005408)/a(A035287) simulates the Hydrogen spectral series of all hydrogen-like elements. - Freimut Marschner, Aug 10 2013
This sequence has unique factorization. The primitive elements are the odd primes (A065091). (Each term of the sequence can be expressed as a product of terms of the sequence. Primitive elements have only the trivial factorization. If the products of terms of the sequence are always in the sequence, and there is a unique factorization of each element into primitive elements, we say that the sequence has unique factorization. So, e.g., the composite numbers do not have unique factorization, because for example 36 = 4*9 = 6*6 has two distinct factorizations.) - Franklin T. Adams-Watters, Sep 28 2013
These are also numbers k such that (k^k+1)/(k+1) is an integer. - Derek Orr, May 22 2014
a(n-1) gives the number of distinct sums in the direct sum {1,2,3,..,n} + {1,2,3,..,n}. For example, {1} + {1} has only one possible sum so a(0) = 1. {1,2} + {1,2} has three distinct possible sums {2,3,4} so a(1) = 3. {1,2,3} + {1,2,3} has 5 distinct possible sums {2,3,4,5,6} so a(2) = 5. - Derek Orr, Nov 22 2014
The number of partitions of 4*n into at most 2 parts. - Colin Barker, Mar 31 2015
a(n) is representable as a sum of two but no fewer consecutive nonnegative integers, e.g., 1 = 0 + 1, 3 = 1 + 2, 5 = 2 + 3, etc. (see A138591). - Martin Renner, Mar 14 2016
Unique solution a( ) of the complementary equation a(n) = a(n-1)^2 - a(n-2)*b(n-1), where a(0) = 1, a(1) = 3, and a( ) and b( ) are increasing complementary sequences. - Clark Kimberling, Nov 21 2017
Also the number of maximal and maximum cliques in the n-centipede graph. - Eric W. Weisstein, Dec 01 2017
Lexicographically earliest sequence of distinct positive integers such that the average of any number of consecutive terms is always an integer. (For opposite property see A042963.) - Ivan Neretin, Dec 21 2017
Maximum number of non-intersecting line segments between vertices of a convex (n+2)-gon. - Christoph B. Kassir, Oct 21 2022
a(n) is the number of parking functions of size n+1 avoiding the patterns 123, 132, and 231. - Lara Pudwell, Apr 10 2023

Examples

			G.f. = q + 3*q^3 + 5*q^5 + 7*q^7 + 9*q^9 + 11*q^11 + 13*q^13 + 15*q^15 + ...
		

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 2.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 28.
  • T. Dantzig, The Language of Science, 4th Edition (1954) page 276.
  • H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 73.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.1 Terminology, p. 264.
  • D. Hök, Parvisa mönster i permutationer [Swedish], (2007).
  • E. Maor, Trigonometric Delights, Princeton University Press, NJ, 1998, pp. 203-205.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

See A120062 for sequences related to integer-sided triangles with integer inradius n.
Cf. A001651 (n=1 or 2 mod 3), A047209 (n=1 or 4 mod 5).
Cf. A003558, A216371, A179480 (relating to the Coach theorem).
Cf. A000754 (boustrophedon transform).

Programs

Formula

a(n) = 2*n + 1. a(-1 - n) = -a(n). a(n+1) = a(n) + 2.
G.f.: (1 + x) / (1 - x)^2.
E.g.f.: (1 + 2*x) * exp(x).
G.f. with interpolated zeros: (x^3+x)/((1-x)^2 * (1+x)^2); e.g.f. with interpolated zeros: x*(exp(x)+exp(-x))/2. - Geoffrey Critzer, Aug 25 2012
a(n) = L(n,-2)*(-1)^n, where L is defined as in A108299. - Reinhard Zumkeller, Jun 01 2005
Euler transform of length 2 sequence [3, -1]. - Michael Somos, Mar 30 2007
G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = v * (1 + 2*u) * (1 - 2*u + 16*v) - (u - 4*v)^2 * (1 + 2*u + 2*u^2). - Michael Somos, Mar 30 2007
a(n) = b(2*n + 1) where b(n) = n if n is odd is multiplicative. [This seems to say that A000027 is multiplicative? - R. J. Mathar, Sep 23 2011]
From Hieronymus Fischer, May 25 2007: (Start)
a(n) = (n+1)^2 - n^2.
G.f. g(x) = Sum_{k>=0} x^floor(sqrt(k)) = Sum_{k>=0} x^A000196(k). (End)
a(0) = 1, a(1) = 3, a(n) = 2*a(n-1) - a(n-2). - Jaume Oliver Lafont, May 07 2008
a(n) = A000330(A016777(n))/A000217(A016777(n)). - Pierre CAMI, Sep 27 2008
a(n) = A034856(n+1) - A000217(n) = A005843(n) + A000124(n) - A000217(n) = A005843(n) + 1. - Jaroslav Krizek, Sep 05 2009
a(n) = (n - 1) + n (sum of two sequential integers). - Dominick Cancilla, Aug 09 2010
a(n) = 4*A000217(n)+1 - 2*Sum_{i=1..n-1} a(i) for n > 1. - Bruno Berselli, Nov 17 2010
n*a(2n+1)^2+1 = (n+1)*a(2n)^2; e.g., 3*15^2+1 = 4*13^2. - Charlie Marion, Dec 31 2010
arctanh(x) = Sum_{n>=0} x^(2n+1)/a(n). - R. J. Mathar, Sep 23 2011
a(n) = det(f(i-j+1))A113311(n);%20for%20n%20%3C%200%20we%20have%20f(n)=0.%20-%20_Mircea%20Merca">{1<=i,j<=n}, where f(n) = A113311(n); for n < 0 we have f(n)=0. - _Mircea Merca, Jun 23 2012
G.f.: Q(0), where Q(k) = 1 + 2*(k+1)*x/( 1 - 1/(1 + 2*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 11 2013
a(n) = floor(sqrt(2*A000384(n+1))). - Ivan N. Ianakiev, Jun 17 2013
a(n) = 3*A000330(n)/A000217(n), n > 0. - Ivan N. Ianakiev, Jul 12 2013
a(n) = Product_{k=1..2*n} 2*sin(Pi*k/(2*n+1)) = Product_{k=1..n} (2*sin(Pi*k/(2*n+1)))^2, n >= 0 (undefined product = 1). See an Oct 09 2013 formula contribution in A000027 with a reference. - Wolfdieter Lang, Oct 10 2013
Noting that as n -> infinity, sqrt(n^2 + n) -> n + 1/2, let f(n) = n + 1/2 - sqrt(n^2 + n). Then for n > 0, a(n) = round(1/f(n))/4. - Richard R. Forberg, Feb 16 2014
a(n) = Sum_{k=0..n+1} binomial(2*n+1,2*k)*4^(k)*bernoulli(2*k). - Vladimir Kruchinin, Feb 24 2015
a(n) = Sum_{k=0..n} binomial(6*n+3, 6*k)*Bernoulli(6*k). - Michel Marcus, Jan 11 2016
a(n) = A000225(n+1) - A005803(n+1). - Miquel Cerda, Nov 25 2016
O.g.f.: Sum_{n >= 1} phi(2*n-1)*x^(n-1)/(1 - x^(2*n-1)), where phi(n) is the Euler totient function A000010. - Peter Bala, Mar 22 2019
Sum_{n>=0} 1/a(n)^2 = Pi^2/8 = A111003. - Bernard Schott, Dec 10 2020
Sum_{n >= 1} (-1)^n/(a(n)*a(n+1)) = Pi/4 - 1/2 = 1/(3 + (1*3)/(4 + (3*5)/(4 + ... + (4*n^2 - 1)/(4 + ... )))). Cf. A016754. - Peter Bala, Mar 28 2024
a(n) = A055112(n)/oblong(n) = A193218(n+1)/Hex number(n). Compare to the Sep 27 2008 comment by Pierre CAMI. - Klaus Purath, Apr 23 2024
a(k*m) = k*a(m) - (k-1). - Ya-Ping Lu, Jun 25 2024
a(n) = A000217(a(n))/n for n > 0. - Stefano Spezia, Feb 15 2025

Extensions

Incorrect comment and example removed by Joerg Arndt, Mar 11 2010
Peripheral comments deleted by N. J. A. Sloane, May 09 2022

A001700 a(n) = binomial(2*n+1, n+1): number of ways to put n+1 indistinguishable balls into n+1 distinguishable boxes = number of (n+1)-st degree monomials in n+1 variables = number of monotone maps from 1..n+1 to 1..n+1.

Original entry on oeis.org

1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220, 1052049481860, 4116715363800, 16123801841550, 63205303218876, 247959266474052
Offset: 0

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To show for example that C(2n+1, n+1) is the number of monotone maps from 1..n + 1 to 1..n + 1, notice that we can describe such a map by a nondecreasing sequence of length n + 1 with entries from 1 to n + 1. The number k of increases in this sequence is anywhere from 0 to n. We can specify these increases by throwing k balls into n+1 boxes, so the total is Sum_{k = 0..n} C((n+1) + k - 1, k) = C(2*n+1, n+1).
Also number of ordered partitions (or compositions) of n + 1 into n + 1 parts. E.g., a(2) = 10: 003, 030, 300, 012, 021, 102, 120, 210, 201, 111. - Mambetov Bektur (bektur1987(AT)mail.ru), Apr 17 2003
Also number of walks of length n on square lattice, starting at origin, staying in first and second quadrants. - David W. Wilson, May 05 2001. (E.g., for n = 2 there are 10 walks, all starting at 0, 0: 0, 1 -> 0, 0; 0, 1 -> 1, 1; 0, 1 -> 0, 2; 1, 0 -> 0, 0; 1, 0 -> 1, 1; 1, 0 -> 2, 0; 1, 0 -> 1, -1; -1, 0 -> 0, 0; -1, 0 -> -1, 1; -1, 0-> -2, 0.)
Also total number of leaves in all ordered trees with n + 1 edges.
Also number of digitally balanced numbers [A031443] from 2^(2*n+1) to 2^(2*n+2). - Naohiro Nomoto, Apr 07 2001
Also number of ordered trees with 2*n + 2 edges having root of even degree and nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002
Also number of paths of length 2*d(G) connecting two neighboring nodes in optimal chordal graph of degree 4, G(2*d(G)^2 + 2*d(G) + 1, 2d(G) + 1), where d(G) = diameter of graph G. - S. Bujnowski (slawb(AT)atr.bydgoszcz.pl), Feb 11 2002
Define an array by m(1, j) = 1, m(i, 1) = i, m(i, j) = m(i, j-1) + m(i-1, j); then a(n) = m(n, n), diagonal of A165257 - Benoit Cloitre, May 07 2002
Also the numerator of the constant term in the expansion of cos^(2*n)(x) or sin^(2*n)(x) when the denominator is 2^(2*n-1). - Robert G. Wilson v
Consider the expansion of cos^n(x) as a linear combination of cosines of multiple angles. If n is odd, then the expansion is a combination of a*cos((2*k-1)*x)/2^(n-1) for all 2*k - 1 <= n. If n is even, then the expansion is a combination of a*cos(2k*x)/2^(n-1) terms plus a constant. "The constant term, [a(n)/2^(2n-1)], is due to the fact that [cos^2n(x)] is never negative, i.e., electrical engineers would say the average or 'dc value' of [cos^(2*n)(x)] is [a(n)/2^(2*n-1)]. The dc value of [cos^(2*n-1)(x)] on the other hand, is zero because it is symmetrical about the horizontal axis, i.e., it is negative and positive equally." Nahin[62] - Robert G. Wilson v, Aug 01 2002
Also number of times a fixed Dyck word of length 2*k occurs in all Dyck words of length 2*n + 2*k. Example: if the fixed Dyck word is xyxy (k = 2), then it occurs a(1) = 3 times in the 5 Dyck words of length 6 (n = 1): (xy[xy)xy], xyxxyy, xxyyxy, x(xyxy)y, xxxyyy (placed between parentheses). - Emeric Deutsch, Jan 02 2003
a(n+1) is the determinant of the n X n matrix m(i, j) = binomial(2*n-i, j). - Benoit Cloitre, Aug 26 2003
a(n-1) = (2*n)!/(2*n!*n!), formula in [Davenport] used by Gauss for the special case prime p = 4*n + 1: x = a(n-1) mod p and y = x*(2n)! mod p are solutions of p = x^2 + y^2. - Frank Ellermann. Example: For prime 29 = 4*7 + 1 use a(7-1) = 1716 = (2*7)!/(2*7!*7!), 5 = 1716 mod 29 and 2 = 5*(2*7)! mod 29, then 29 = 5*5 + 2*2.
The number of compositions of 2*n, say c_1 + c_2 + ... + c_k = 2n, satisfy that Sum_{i = 1..j} c_i < 2*j for all j = 1..k, or equivalently, the number of subsets, say S, of [2*n-1] = {1, 2, ..., 2*n-1} with at least n elements such that if 2k is in S, then there must be at least k elements in S smaller than 2k. E.g., a(2) = 3 because we can write 4 = 1 + 1 + 1 + 1 = 1 + 1 + 2 = 1 + 2 + 1. - Ricky X. F. Chen (ricky_chen(AT)mail.nankai.edu.cn), Jul 30 2006
The number of walks of length 2*n + 1 on an infinite linear lattice that begin at the origin and end at node (1). Also the number of paths on a square lattice from the origin to (n+1, n) that use steps (1,0) and (0,1). Also number of binary numbers of length 2*n + 1 with n + 1 ones and n zeros. - Stefan Hollos (stefan(AT)exstrom.com), Dec 10 2007
If Y is a 3-subset of an 2*n-set X then, for n >= 3, a(n-1) is the number of n-subsets of X having at least two elements in common with Y. - Milan Janjic, Dec 16 2007
Also the number of rankings (preferential arrangements) of n unlabeled elements onto n levels when empty levels are allowed. - Thomas Wieder, May 24 2008
Also the Catalan transform of A000225 shifted one index, i.e., dropping A000225(0). - R. J. Mathar, Nov 11 2008
With offset 1. The number of solutions in nonnegative integers to X1 + X2 + ... + Xn = n. The number of terms in the expansion of (X1 + X2 + ... + Xn)^n. The coefficient of x^n in the expansion of (1 + x + x^2 + ...)^n. The number of distinct image sets of all functions taking [n] into [n]. - Geoffrey Critzer, Feb 22 2009
The Hankel transform of the aerated sequence 1, 0, 3, 0, 10, 0, ... is 1, 3, 3, 5, 5, 7, 7, ... (A109613(n+1)). - Paul Barry, Apr 21 2009
Also the number of distinct network topologies for a network of n items with 1 to n - 1 unidirectional connections to other objects in the network. - Anthony Bachler, May 05 2010
Equals INVERT transform of the Catalan numbers starting with offset 1. E.g.: a(3) = 35 = (1, 2, 5) dot (10, 3, 1) + 14 = 21 + 14 = 35. - Gary W. Adamson, May 15 2009
The integral of 1/(1+x^2)^(n+1) is given by a(n)/2^(2*n - 1) * (x/(1 + x^2)^n*P(x) + arctan(x)), where P(x) is a monic polynomial of degree 2*n - 2 with rational coefficients. - Christiaan van de Woestijne, Jan 25 2011
a(n) is the number of Schroder paths of semilength n in which the (2,0)-steps at level 0 come in 2 colors and there are no (2,0)-steps at a higher level. Example: a(2) = 10 because, denoting U = (1,1), H = (1,0), and D = (1,-1), we have 2^2 = 4 paths of shape HH, 2 paths of shape HUD, 2 paths of shape UDH, and 1 path of each of the shapes UDUD and UUDD. - Emeric Deutsch, May 02 2011
a(n) is the number of Motzkin paths of length n in which the (1,0)-steps at level 0 come in 3 colors and those at a higher level come in 2 colors. Example: a(3)=35 because, denoting U = (1,1), H = (1,0), and D = (1,-1), we have 3^3 = 27 paths of shape HHH, 3 paths of shape HUD, 3 paths of shape UDH, and 2 paths of shape UHD. - Emeric Deutsch, May 02 2011
Also number of digitally balanced numbers having length 2*(n + 1) in binary representation: a(n) = #{m: A070939(A031443(m)) = 2*(n + 1)}. - Reinhard Zumkeller, Jun 08 2011
a(n) equals 2^(2*n + 3) times the coefficient of Pi in 2F1([1/2, n+2]; [3/2]; -1). - John M. Campbell, Jul 17 2011
For positive n, a(n) equals 4^(n+2) times the coefficient of Pi^2 in Integral_{x = 0..Pi/2} x sin^(2*n + 2)x. - John M. Campbell, Jul 19 2011 [Apparently, the contributor means Integral_{x = 0..Pi/2} x * (sin(x))^(2*n + 2).]
a(n-1) = C(2*n, n)/2 is the number of ways to assign 2*n people into 2 (unlabeled) groups of size n. - Dennis P. Walsh, Nov 09 2011
Equals row sums of triangle A205945. - Gary W. Adamson, Feb 01 2012
a(n-1) gives the number of n-regular sequences defined by Erdős and Gallai in 1960 in connection with the degree sequences of simple graphs. - Matuszka Tamás, Mar 06 2013
a(n) is the sum of falling diagonals of squares in the comment in A085812 (equivalent to the Cloitre formula of Aug 2002). - John Molokach, Sep 26 2013
For n > 0: largest terms of Zigzag matrices as defined in A088961. - Reinhard Zumkeller, Oct 25 2013
Also the number of different possible win/loss round sequences (from the perspective of the eventual winner) in a "best of 2*n + 1" two-player game. For example, a(2) = 10 means there are 10 different win/loss sequences in a "best of 5" game (like a tennis match in which the first player to win 3 sets, out of a maximum of 5, wins the match); the 10 sequences are WWW, WWLW, WWLLW, WLWW, WLWLW, WLLWW, LWWW, LWWLW, LWLWW, LLWWW. See also A072600. - Philippe Beaudoin, May 14 2014; corrected by Jon E. Schoenfield, Nov 23 2014
When adding 1 to the beginning of the sequence: Convolving a(n)/2^n with itself equals 2^(n+1). For example, when n = 4: convolving {1, 1/1, 3/2, 10/4, 35/8, 126/16} with itself is 32 = 2^5. - Bob Selcoe, Jul 16 2014
From Tom Copeland, Nov 09 2014: (Start)
The shifted array belongs to a family of arrays associated to the Catalan A000108 (t = 1), and Riordan, or Motzkin sums A005043 (t = 0), with the o.g.f. [1 - sqrt(1 - 4x/(1 + (1 - t)x))]/2 and inverse x*(1 - x)/[1 + (t - 1)*x*(1 - x)]. See A091867 for more info on this family. Here is t = -3 (mod signs in the results).
Let C(x) = [1 - sqrt(1-4x)]/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1-x) and P(x,t) = x/(1 + t*x) with inverse P(x, -t).
O.g.f: G(x) = [-1 + sqrt(1 + 4*x/(1 - 4*x))]/2 = -C[P(-x, 4)].
Inverse o.g.f: Ginv(x) = x*(1 + x)/(1 + 4*x*(1 + x)) = -P(Cinv(-x), -4) (shifted signed A001792). A088218(x) = 1 + G(x).
Equals A001813/2 omitting the leading 1 there. (End)
Placing n distinguishable balls into n indistinguishable boxes gives A000110(n) (the number of set partitions). - N. J. A. Sloane, Jun 19 2015
The sequence is the INVERTi transform of A049027: (1, 4, 17, 74, 326, ...). - Gary W. Adamson, Jun 23 2015
a(n) is the number of compositions of 2*n + 2 such that the sum of the elements at odd positions is equal to the sum of the elements at even positions. a(2) = 10 because there are 10 such compositions of 6: (3, 3), (1, 3, 2), (2, 3, 1), (1, 1, 2, 2), (1, 2, 2, 1), (2, 2, 1, 1), (2, 1, 1, 2), (1, 2, 1, 1, 1), (1, 1, 1, 2, 1), (1, 1, 1, 1, 1, 1). - Ran Pan, Oct 08 2015
a(n-1) is also the Schur function of the partition (n) of n evaluated at x_1 = x_2 = ... = x_n = 1, i.e., the number of semistandard Young tableaux of shape (n) (weakly increasing rows with n boxes with numbers from {1, 2, ..., n}). - Wolfdieter Lang, Oct 11 2015
Also the number of ordered (rooted planar) forests with a total of n+1 edges and no trivial trees. - Nachum Dershowitz, Mar 30 2016
a(n) is the number of sets (i1,...in) of length n so that n >= i1 >= i2 >= ...>= in >= 1. For instance, n=3 as there are only 10 such sets (3,3,3) (3,3,2) (3,3,1) (3,2,2) (3,2,1) (3,1,1) (2,2,2) (2,2,1) (2,1,1) (1,1,1,) 3,2,1 is each used 10 times respectively. - Anton Zakharov, Jul 04 2016
The repeated middle term in the odd rows of Pascal's triangle, or half the central binomial coefficient in the even rows of Pascal's triangle, n >= 2. - Enrique Navarrete, Feb 12 2018
a(n) is the number of walks of length 2n+1 from the origin with steps (1,1) and (1,-1) that stay on or above the x-axis. Equivalently, a(n) is the number of walks of length 2n+1 from the origin with steps (1,0) and (0,1) that stay in the first octant. - Alexander Burstein, Dec 24 2019
Total number of nodes summed over all Dyck paths of semilength n. - Alois P. Heinz, Mar 08 2020
a(n-1) is the determinant of the n X n matrix m(i, j) = binomial(n+i-1, j). - Fabio Visonà, May 21 2022
Let X_i be iid standard Gaussian random variable N(0,1), and S_n be the partial sum S_n = X_1+...+X_n. Then P(S_1>0,S_2>0,...,S_n>0) = a(n+1)/2^(2n-1) = a(n+1) / A004171(n+1). For example, P(S_1>0) = 1/2, P(S_1>0,S_2>0) = 3/8, P(S_1>0,S_2>0,S_3>0) = 5/16, etc. This probability is also equal to the volume of the region x_1 > 0, x_2 > -x_1, x_3 > -(x_1+x_2), ..., x_n > -(x_1+x_2+...+x_(n-1)) in the hypercube [-1/2, 1/2]^n. This also holds for the Cauchy distribution and other stable distributions with mean 0, skew 0 and scale 1. - Xiaohan Zhang, Nov 01 2022
a(n) is the number of parking functions of size n+1 avoiding the patterns 132, 213, and 321. - Lara Pudwell, Apr 10 2023
Number of vectors in (Z_>=0)^(n+1) such that the sum of the components is n+1. binomial(2*n-1, n) provides this property for n. - Michael Richard, Jun 12 2023
Also number of discrete negations on the finite chain L_n={0,1,...,n-1,n}, i.e., monotone decreasing unary operators such that N(0)=n and N(n)=0. - Marc Munar, Oct 10 2023
a(n) is the number of Dyck paths of semilength n+1 having one of its peaks marked. - Juan B. Gil, Jan 03 2024
a(n) is the dimension of the (n+1)-st symmetric power of an (n+1)-dimensional vector space. - Mehmet A. Ates, Feb 15 2024
a(n) is the independence number of the twisted odd graph O^(sigma)(n+2). - _Miquel A. Fiol, Aug 26 2024
a(n) is the number of non-descending sequences with length n and the last number is less or equal to n. a(n) is also the number of integer partitions (of any positive integer) with length n and largest part is less or equal to n. - Zlatko Damijanic, Dec 06 2024
a(n) is the number of triangulations of a once-punctured (n+1)-gon [from Fontaine & Plamondon's Theorem 3.6]. - Esther Banaian, May 06 2025

Examples

			There are a(2)=10 ways to put 3 indistinguishable balls into 3 distinguishable boxes, namely, (OOO)()(), ()(OOO)(), ()()(OOO), (OO)(O)(), (OO)()(O), (O)(OO)(), ()(OO)(O), (O)()(OO), ()(O)(OO), and (O)(O)(O). - _Dennis P. Walsh_, Apr 11 2012
a(2) = 10: Semistandard Young tableaux for partition (3) of 3 (the indeterminates x_i, i = 1, 2, 3 are omitted and only their indices are given): 111, 112, 113, 122, 123, 133, 222, 223, 233, 333. - _Wolfdieter Lang_, Oct 11 2015
		

References

  • H. Davenport, The Higher Arithmetic. Cambridge Univ. Press, 7th ed., 1999, ch. V.3 (p. 122).
  • A. Frosini, R. Pinzani, and S. Rinaldi, About half the middle binomial coefficient, Pure Math. Appl., 11 (2000), 497-508.
  • Charles Jordan, Calculus of Finite Differences, Chelsea 1965, p. 449.
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • Paul J. Nahin, "An Imaginary Tale, The Story of [Sqrt(-1)]," Princeton University Press, Princeton, NJ 1998, p. 62.
  • L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.
  • 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

Equals A000984(n+1)/2.
a(n) = (2*n+1)*Catalan(n) [A000108] = A035324(n+1, 1) (first column of triangle).
Row sums of triangles A028364, A050166, A039598.
Bisections: a(2*k) = A002458(k), a(2*k+1) = A001448(k+1)/2, k >= 0.
Other versions of the same sequence: A088218, A110556, A138364.
Diagonals 1 and 2 of triangle A100257.
Second row of array A102539.
Column of array A073165.
Row sums of A103371. - Susanne Wienand, Oct 22 2011
Cf. A002054: C(2*n+1, n-1). - Bruno Berselli, Jan 20 2014

Programs

  • GAP
    List([0..30],n->Binomial(2*n+1,n+1)); # Muniru A Asiru, Feb 26 2019
  • Haskell
    a001700 n = a007318 (2*n+1) (n+1)  -- Reinhard Zumkeller, Oct 25 2013
    
  • Magma
    [Binomial(2*n, n)/2: n in [1..40]]; // Vincenzo Librandi, Nov 10 2014
    
  • Maple
    A001700 := n -> binomial(2*n+1,n+1); seq(A001700(n), n=0..20);
    A001700List := proc(m) local A, P, n; A := [1]; P := [1];
    for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), 2*P[-1]]);
    A := [op(A), P[-1]] od; A end: A001700List(27); # Peter Luschny, Mar 24 2022
  • Mathematica
    Table[ Binomial[2n + 1, n + 1], {n, 0, 23}]
    CoefficientList[ Series[2/((Sqrt[1 - 4 x] + 1)*Sqrt[1 - 4 x]), {x, 0, 22}], x] (* Robert G. Wilson v, Aug 08 2011 *)
  • Maxima
    B(n,a,x):=coeff(taylor(exp(x*t)*(t/(exp(t)-1))^a,t,0,20),t,n)*n!;
    makelist((-1)^(n)*B(n,n+1,-n-1)/n!,n,0,10); /* Vladimir Kruchinin, Apr 06 2016 */
    
  • PARI
    a(n)=binomial(2*n+1,n+1)
    
  • PARI
    z='z+O('z^50); Vec((1/sqrt(1-4*z)-1)/(2*z)) \\ Altug Alkan, Oct 11 2015
    
  • Python
    from _future_ import division
    A001700_list, b = [], 1
    for n in range(10**3):
        A001700_list.append(b)
        b = b*(4*n+6)//(n+2) # Chai Wah Wu, Jan 26 2016
    
  • Sage
    [rising_factorial(n+1,n+1)/factorial(n+1) for n in (0..22)] # Peter Luschny, Nov 07 2011
    

Formula

a(n-1) = binomial(2*n, n)/2 = A000984(n)/2 = (2*n)!/(2*n!*n!).
D-finite with recurrence: a(0) = 1, a(n) = 2*(2*n+1)*a(n-1)/(n+1) for n > 0.
G.f.: (1/sqrt(1 - 4*x) - 1)/(2*x).
L.g.f.: log((1 - sqrt(1 - 4*x))/(2*x)) = Sum_{n >= 0} a(n)*x^(n+1)/(n+1). - Vladimir Kruchinin, Aug 10 2010
G.f.: 2F1([1, 3/2]; [2]; 4*x). - Paul Barry, Jan 23 2009
G.f.: 1/(1 - 2*x - x/(1 - x/(1 - x/(1 - x/(1 - ... (continued fraction). - Paul Barry, May 06 2009
G.f.: c(x)^2/(1 - x*c(x)^2), c(x) the g.f. of A000108. - Paul Barry, Sep 07 2009
O.g.f.: c(x)/sqrt(1 - 4*x) = (2 - c(x))/(1 - 4*x), with c(x) the o.g.f. of A000108. Added second formula. - Wolfdieter Lang, Sep 02 2012
Convolution of A000108 (Catalan) and A000984 (central binomial): Sum_{k=0..n} C(k)*binomial(2*(n-k), n-k), C(k) Catalan. - Wolfdieter Lang, Dec 11 1999
a(n) = Sum_{k=0..n} C(n+k, k). - Benoit Cloitre, Aug 20 2002
a(n) = Sum_{k=0..n} C(n, k)*C(n+1, k+1). - Benoit Cloitre, Oct 19 2002
a(n) = Sum_{k = 0..n+1} binomial(2*n+2, k)*cos((n - k + 1)*Pi). - Paul Barry, Nov 02 2004
a(n) = 4^n*binomial(n+1/2, n)/(n+1). - Paul Barry, May 10 2005
E.g.f.: Sum_{n >= 0} a(n)*x^(2*n + 1)/(2*n + 1)! = BesselI(1, 2*x). - Michael Somos, Jun 22 2005
E.g.f. in Maple notation: exp(2*x)*(BesselI(0, 2*x) + BesselI(1, 2*x)). Integral representation as n-th moment of a positive function on [0, 4]: a(n) = Integral_{x = 0..4} x^n * (x/(4 - x))^(1/2)/(2*Pi) dx, n >= 0. This representation is unique. - Karol A. Penson, Oct 11 2001
Narayana transform of [1, 2, 3, ...]. Let M = the Narayana triangle of A001263 as an infinite lower triangular matrix and V = the Vector [1, 2, 3, ...]. Then A001700 = M * V. - Gary W. Adamson, Apr 25 2006
a(n) = A122366(n,n). - Reinhard Zumkeller, Aug 30 2006
a(n) = C(2*n, n) + C(2*n, n-1) = A000984(n) + A001791(n). - Zerinvary Lajos, Jan 23 2007
a(n-1) = (n+1)*(n+2)*...*(2*n-1)/(n-1)! (product of n-1 consecutive integers, divided by (n-1)!). - Jonathan Vos Post, Apr 09 2007; [Corrected and shortened by Giovanni Ciriani, Mar 26 2019]
a(n-1) = (2*n - 1)!/(n!*(n - 1)!). - William A. Tedeschi, Feb 27 2008
a(n) = (2*n + 1)*A000108(n). - Paul Barry, Aug 21 2007
Binomial transform of A005773 starting (1, 2, 5, 13, 35, 96, ...) and double binomial transform of A001405. - Gary W. Adamson, Sep 01 2007
Row sums of triangle A132813. - Gary W. Adamson, Sep 01 2007
Row sums of triangle A134285. - Gary W. Adamson, Nov 19 2007
a(n) = 2*A000984(n) - A000108(n), that is, a(n) = 2*C(2*n, n) - n-th Catalan number. - Joseph Abate, Jun 11 2010
Conjectured: 4^n GaussHypergeometric(1/2,-n; 2; 1) -- Solution for the path which stays in the first and second quadrant. - Benjamin Phillabaum, Feb 20 2011
a(n)= Sum_{k=0..n} A038231(n,k) * (-1)^k * A000108(k). - Philippe Deléham, Nov 27 2009
Let A be the Toeplitz matrix of order n defined by: A[i,i-1] = -1, A[i,j] = Catalan(j-i), (i <= j), and A[i,j] = 0, otherwise. Then, for n >= 1, a(n) = (-1)^n * charpoly(A,-2). - Milan Janjic, Jul 08 2010
a(n) is the upper left term of M^(n+1), where M is the infinite matrix in which a column of (1,2,3,...) is prepended to an infinite lower triangular matrix of all 1's and the rest zeros, as follows:
1, 1, 0, 0, 0, ...
2, 1, 1, 0, 0, ...
3, 1, 1, 1, 0, ...
4, 1, 1, 1, 1, ...
...
Alternatively, a(n) is the upper left term of M^n where M is the infinite matrix:
3, 1, 0, 0, 0, ...
1, 1, 1, 0, 0, ...
1, 1, 1, 1, 0, ...
1, 1, 1, 1, 1, ...
...
- Gary W. Adamson, Jul 14 2011
a(n) = (n + 1)*hypergeom([-n, -n], [2], 1). - Peter Luschny, Oct 24 2011
a(n) = Pochhammer(n+1, n+1)/(n+1)!. - Peter Luschny, Nov 07 2011
E.g.f.: 1 + 6*x/(U(0) - 6*x); U(k) = k^2 + (4*x + 3)*k + 6*x + 2 - 2*x*(k + 1)*(k + 2)*(2*k + 5)/U(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 18 2011
a(n) = 2*A000984(n) - A000108(n). [Abate & Whitt]
a(n) = 2^(2*n+1)*binomial(n+1/2, -1/2). - Peter Luschny, May 06 2014
For n > 1: a(n-1) = A166454(2*n, n), central terms in A166454. - Reinhard Zumkeller, Mar 04 2015
a(n) = 2*4^n*Gamma(3/2 + n)/(sqrt(Pi)*Gamma(2+n)). - Peter Luschny, Dec 14 2015
a(n) ~ 2*4^n*(1 - (5/8)/n + (73/128)/n^2 - (575/1024)/n^3 + (18459/32768)/n^4)/sqrt(n*Pi). - Peter Luschny, Dec 16 2015
a(n) = (-1)^(n)*B(n, n+1, -n-1)/n!, where B(n,a,x) is a generalized Bernoulli polynomial. - Vladimir Kruchinin, Apr 06 2016
a(n) = Gamma(2 + 2*n)/(n!*Gamma(2 + n)). Andres Cicuttin, Apr 06 2016
a(n) = (n + (n + 1))!/(Gamma(n)*Gamma(1 + n)*A002378(n)), for n > 0. Andres Cicuttin, Apr 07 2016
From Ilya Gutkovskiy, Jul 04 2016: (Start)
Sum_{n >= 0} 1/a(n) = 2*(9 + 2*sqrt(3)*Pi)/27 = A248179.
Sum_{n >= 0} (-1)^n/a(n) = 2*(5 + 4*sqrt(5)*arcsinh(1/2))/25 = 2*(5*A145433 - 1).
Sum_{n >= 0} (-1)^n*a(n)/n! = BesselI(2,2)*exp(-2) = A229020*A092553. (End)
Conjecture: a(n) = Sum_{k=2^n..2^(n+1)-1} A178244(k). - Mikhail Kurkov, Feb 20 2021
a(n-1) = 1 + (1/n)*Sum_{t=1..n/2} (2*cos((2*t-1)*Pi/(2*n)))^(2*n). - Greg Dresden, Oct 11 2022
a(n) = Product_{1 <= i <= j <= n} (i + j + 1)/(i + j - 1). Cf. A006013. - Peter Bala, Feb 21 2023
Sum_{n >= 0} a(n)*x^(n+1)/(n+1) = x + 3*x^2/2 + 10*x^3/3 + 35*x^4/4 + ... = the series reversion of exp(-x)*(1 - exp(-x)). - Peter Bala, Sep 06 2023

Extensions

Name corrected by Paul S. Coombes, Jan 11 2012
Name corrected by Robert Tanniru, Feb 01 2014

A000982 a(n) = ceiling(n^2/2).

Original entry on oeis.org

0, 1, 2, 5, 8, 13, 18, 25, 32, 41, 50, 61, 72, 85, 98, 113, 128, 145, 162, 181, 200, 221, 242, 265, 288, 313, 338, 365, 392, 421, 450, 481, 512, 545, 578, 613, 648, 685, 722, 761, 800, 841, 882, 925, 968, 1013, 1058, 1105, 1152, 1201, 1250, 1301, 1352, 1405
Offset: 0

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Comments

a(n) = number of pairs (i,j) in [1..n] X [1..n] with integral arithmetic mean. Cf. A132188, A362931. - N. J. A. Sloane, Aug 28 2023
Also, floor( (n^2+1)/2 ). - N. J. A. Sloane, Feb 08 2019
Floor(arithmetic mean of next n numbers). - Amarnath Murthy, Mar 11 2003
Pairwise sums of repeated squares (A008794).
Also, number of topologies on n+1 unlabeled elements with exactly 4 elements in the topology. a(3) gives 4 elements a,b,c,d; the valid topologies are (0,a,ab,abcd), (0,a,abc,abcd), (0,ab,abc,abcd), (0,a,bcd,abcd) and (0,ab,cd,abcd), with a count of 5. - Jon Perry, Mar 05 2004
Partition n into two parts, say, r and s, so that r^2 + s^2 is minimal, then a(n) = r^2 + s^2. Geometrical significance: folding a rod with length n units at right angles in such a way that the end points are at the least distance, which is given by a(n)^(1/2) as the hypotenuse of a right triangle with the sum of the base and height = n units. - Amarnath Murthy, Apr 18 2004
Convolution of A002061(n)-0^n and (-1)^n. Convolution of n (A001477) with {1,0,2,0,2,0,2,...}. Partial sums of repeated odd numbers {0,1,1,3,3,5,5,...}. - Paul Barry, Jul 22 2004
The ratio of the sum of terms over the total number of terms in an n X n spiral. The sum of terms of an n X n spiral is A037270, or Sum_{k=0..n^2} k = (n^4 + n^2)/2 and the total number of terms is n^2. - William A. Tedeschi, Feb 27 2008
Starting with offset 1 = row sums of triangle A158946. - Gary W. Adamson, Mar 31 2009
Partial sums of A109613. - Reinhard Zumkeller, Dec 05 2009
Also the number of compositions of even natural numbers into 2 parts < n. For example a(3)=5 are the compositions (0,0), (0,2), (2,0), (1,1), (2,2) of even natural numbers into 2 parts < 3. a(4)=8 are the compositions (0,0), (0,2), (2,0), (1,1), (2,2), (1,3), (3,1), (3,3) of even natural numbers into 2 parts < 4. - Adi Dani, Jun 05 2011
A001105 and A001844 interleaved. - Omar E. Pol, Sep 18 2011
Number of (w,x,y) having all terms in {0,...,n} and w=average(x,y). - Clark Kimberling, May 15 2012
For n > 0, minimum number of lines necessary to get through all unit cubes of an n X n X n cube (see Kantor link). - Michel Marcus, Apr 13 2013
Sum_{n > 0} 1/a(n) = Sum_{n > 0} 1/(2*n^2) + Sum_{n >= 0} 1/(2*n + 2*n^2 + 1) = (zeta(2) + (Pi* tanh(Pi/2)))/2 = 2.26312655.... - Enrique Pérez Herrero, Jun 17 2013
For n > 1, a(n) is the edge cover number of the n X n king graph. - Eric W. Weisstein, Jun 20 2017
Also the number of vertices in the n X n black bishop graph. - Eric W. Weisstein, Jun 26 2017
The same sequence arises in the triangular array of the integers >= 1, according to a simple "zig-zag" rule for selection of terms. a(n-1) lies in the (n-1)-th row of the array, and the second row of that sub-array (with apex a(n-1)) contains just two numbers, one odd, one even. The one with opposite parity to a(n-1) is a(n). - David James Sycamore, Jul 29 2018
Size of minimal ternary 1-covering code with code length n, i.e., K_n(3,1). See Kalbfleisch and Stanton. - Patrick Wienhöft, Jan 29 2019
For n > 1, a(n-1) is the maximum number of inversions in a permutation consisting of a single n-cycle on n symbols. - M. Ryan Julian Jr., Sep 10 2019
Also the number of classes of convex inscribed polyominoes in a (2,n) rectangular grid; two polyominoes are in the same class if one of them can be obtained by a reflection or 180-degree rotation of the other. - Jean-Luc Manguin, Jan 29 2020
a(n) is the number of pairs (p,q) such that 1 <= p, p+1 < q <= n+2 and q <> 2*p. - César Eliud Lozada, Oct 25 2020
a(n) is the maximum number of copies of a 12 permutation pattern in an alternating (or zig-zag) permutation of length n+1. The maximum number of copies of 123 in an alternating permutation is motivated in the Notices reference, and the argument here is analogous. - Lara Pudwell, Dec 01 2020
It appears that a(n) is the largest number of nodes of an induced path in the n X n king graph. An induced path going in a simple spiraling pattern, starting in a corner, has a(n) nodes. For even n this is optimal, because an induced path can have at most two nodes in any 2 X 2 subsquare. For odd n, I cannot see how to prove that (n^2+1)/2 is best possible. See also A357501. - Pontus von Brömssen, Oct 02 2022 [Proved by Beluhov (2023). - Pontus von Brömssen, Jan 30 2023]
a(n) = n + 2*(n-2) + 2*(n-4) + 2*(n-6) + ... number of black squares on an n X n chessboard. - R. J. Mathar, Dec 03 2022

Examples

			G.f. = x + 2*x^2 + 5*x^3 + 8*x^4 + 13*x^5 + 18*x^6 + 25*x^7 + 32*x^8 + ...
Centrosymmetric 3 X 3 matrix: [[a,b,c],[d,e,d],[c,b,a]], a(3) = 3*(3-1)/2 + (3-1)/2 + 1 = (3^2+1)/2 = 5 from a,b,c,d,e. 4 X 4 case: [[a,b,c,d],[e,f,g,h],[h,g,f,e],[d,c,b,a]], a(4) = 4*4/2 = 8. - _Wolfdieter Lang_, Oct 12 2015
a(3) = 5. The alternating permutation of length 3 + 1 = 4 with the maximum number of copies of 123 is 1324. The five copies are 12, 13, 14, 23, and 24. - _Lara Pudwell_, Dec 01 2020
		

References

  • 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

Formula

a(2*n) = 2*n^2, a(2*n+1) = 2*n^2 + 2*n + 1.
G.f.: -x*(1+x^2) / ( (1+x)*(x-1)^3 ). - Simon Plouffe in his 1992 dissertation
From Benoit Cloitre, Nov 06 2002: (Start)
a(n) = (2*n^2 + 1 - (-1)^n) / 4.
a(0)=0, a(1)=1; for n>1, a(n+1) = n + 1 + max(2*floor(a(n)/2), 3*floor(a(n)/3)). (End)
G.f.: (x + x^2 + x^3 + x^4)/((1 - x)*(1 - x^2)^2), not reduced. - Len Smiley
a(n) = a(n-2) + 2n - 2. - Paul Barry, Jul 17 2004
From Paul Barry, Jul 22 2004: (Start)
G.f.: x*(1+x^2)/((1-x^2)*(1-x)^2) = x*(1+x^2)/((1+x)*(1-x)^3);
a(n) = Sum_{k=0..n} (k^2 - k + 1 - 0^k)*(-1)^(n-k);
a(n) = Sum_{k=0..n} (1 + (-1)^(n-k) - 0^(n-k))*k. (End)
From Reinhard Zumkeller, Feb 27 2006: (Start)
a(0) = 0, a(n+1) = a(n) + 2*floor(n/2) + 1.
a(n) = A116940(n) - A005843(n). (End)
Starting with offset 1, = row sums of triangle A134444. Also, with offset 1, = binomial transform of [1, 1, 2, -2, 4, -8, 16, -32, ...]. - Gary W. Adamson, Oct 25 2007
a(n) = floor((n^2+1)/2). - William A. Tedeschi, Feb 27 2008
a(n) = A004526(n+1) + A000217(n-1). - Yosu Yurramendi, Sep 12 2008, corrected by Klaus Purath, Jun 15 2021
From Jaume Oliver Lafont, Dec 05 2008: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3) + 2.
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). (End)
a(n) = A004526(n)^2 + A110654(n)^2. - Philippe Deléham, Mar 12 2009
a(n) = n^2 - floor(n^2/2). - Wesley Ivan Hurt, Jun 14 2013
Euler transform is length 4 sequence [2, 2, 0, -1].
a(n) = a(-n) for all n in Z. - Michael Somos, May 05 2015
a(n) is also the number of independent entries in a centrosymmetric n X n matrix: M(i, j) = M(n-i+1, n-j+1). - Wolfdieter Lang, Oct 12 2015
For n > 1, a(n+1)/a(n) = 3 - A081352(n-2)/a(n). - Miko Labalan, Mar 26 2016
E.g.f.: (1/2)*(x*(1 + x)*cosh(x) + (1 + x + x^2)*sinh(x)). - Stefano Spezia, Feb 03 2020
a(n) = binomial(n+1,2) - floor(n/2). - César Eliud Lozada, Oct 25 2020
From Klaus Purath, Jun 15 2021: (Start)
a(n-1) + a(n) = A002061(n).
a(n) = (a(n-1)^2 + 1) / a(n-2), n >= 3 odd.
a(n) = (a(n-1)^2 - (n-1)^2) / a(n-2), n >= 4 even. (End)

A052928 The even numbers repeated.

Original entry on oeis.org

0, 0, 2, 2, 4, 4, 6, 6, 8, 8, 10, 10, 12, 12, 14, 14, 16, 16, 18, 18, 20, 20, 22, 22, 24, 24, 26, 26, 28, 28, 30, 30, 32, 32, 34, 34, 36, 36, 38, 38, 40, 40, 42, 42, 44, 44, 46, 46, 48, 48, 50, 50, 52, 52, 54, 54, 56, 56, 58, 58, 60, 60, 62, 62, 64, 64, 66, 66, 68, 68, 70, 70, 72, 72
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

a(n) is also the binary rank of the complete graph K(n). - Alessandro Cosentino (cosenal(AT)gmail.com), Feb 07 2009
Let I=I_n be the n X n identity matrix and P=P_n be the incidence matrix of the cycle (1,2,3,...,n). Then, for n >= 6, a(n) is the number of (0,1) n X n matrices A <= P^(-1)+I+P having exactly two 1's in every row and column with perA=2. - Vladimir Shevelev, Apr 12 2010
a(n+2) is the number of symmetry allowed, linearly independent terms at n-th order in the series expansion of the (E+A)xe vibronic perturbation matrix, H(Q) (cf. Eisfeld & Viel). - Bradley Klee, Jul 21 2015
The arithmetic function v_2(n,1) as defined in A289187. - Robert Price, Aug 22 2017
For n > 1, also the chromatic number of the n X n white bishop graph. - Eric W. Weisstein, Nov 17 2017
For n > 2, also the maximum vertex degree of the n-polygon diagonal intersection graph. - Eric W. Weisstein, Mar 23 2018
For n >= 2, a(n+2) gives the minimum weight of a Boolean function of algebraic degree at most n-2 whose support contains n linearly independent elements. - Christof Beierle, Nov 25 2019

References

  • C. D. Godsil and G. Royle, Algebraic Graph Theory, Springer, 2001, page 181. - Alessandro Cosentino (cosenal(AT)gmail.com), Feb 07 2009
  • V. S. Shevelyov (Shevelev), Extension of the Moser class of four-line Latin rectangles, DAN Ukrainy, 3(1992),15-19.

Crossrefs

First differences: A010673; partial sums: A007590; partial sums of partial sums: A212964(n+1).
Complement of A109613 with respect to universe A004526. - Guenther Schrack, Dec 07 2017
Is first differences of A099392. Fixed point sequence: A005843. - Guenther Schrack, May 30 2019
For n >= 3, A329822(n) gives the minimum weight of a Boolean function of algebraic degree at most n-3 whose support contains n linearly independent elements. - Christof Beierle, Nov 25 2019

Programs

  • Haskell
    a052928 = (* 2) . flip div 2
    a052928_list = 0 : 0 : map (+ 2) a052928_list
    -- Reinhard Zumkeller, Jun 20 2015
  • Magma
    [2*Floor(n/2) : n in [0..50]]; // Wesley Ivan Hurt, Sep 13 2014
    
  • Maple
    spec := [S,{S=Union(Sequence(Prod(Z,Z)),Prod(Sequence(Z),Sequence(Z)))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    Flatten[Table[{2n, 2n}, {n, 0, 39}]] (* Alonso del Arte, Jun 24 2012 *)
    With[{ev=2Range[0,40]},Riffle[ev,ev]] (* Harvey P. Dale, May 08 2021 *)
    Table[Round[n + 1/2], {n, -1, 72}] (* Ed Pegg Jr, Jul 28 2025 *)
  • PARI
    a(n)=n\2*2 \\ Charles R Greathouse IV, Nov 20 2011
    

Formula

a(n) = 2*floor(n/2).
G.f.: 2*x^2/((-1+x)^2*(1+x)).
a(n) + a(n+1) + 2 - 2*n = 0.
a(n) = n - 1/2 + (-1)^n/2.
a(n) = n + Sum_{k=1..n} (-1)^k. - William A. Tedeschi, Mar 20 2008
a(n) = a(n-1) + a(n-2) - a(n-3). - R. J. Mathar, Feb 19 2010
a(n) = |A123684(n) - A064455(n)| = A032766(n) - A008619(n-1). - Jaroslav Krizek, Mar 22 2011
For n > 0, a(n) = floor(sqrt(n^2+(-1)^n)). - Francesco Daddi, Aug 02 2011
a(n) = Sum_{k>=0} A030308(n,k)*b(k) with b(0)=0 and b(k)=2^k for k>0. - Philippe Deléham, Oct 19 2011
a(n) = A109613(n) - 1. - M. F. Hasler, Oct 22 2012
a(n) = n - (n mod 2). - Wesley Ivan Hurt, Jun 29 2013
a(n) = a(a(n-1)) + a(n-a(n-1)) for n>2. - Nathan Fox, Jul 24 2016
a(n) = 2*A004526(n). - Filip Zaludek, Oct 28 2016
E.g.f.: x*exp(x) - sinh(x). - Ilya Gutkovskiy, Oct 28 2016
a(-n) = -a(n+1); a(n) = A005843(A004526(n)). - Guenther Schrack, Sep 11 2018
From Guenther Schrack, May 29 2019: (Start)
a(b(n)) = b(n) + ((-1)^b(n) - 1)/2 for any sequence b(n) of offset 0.
a(a(n)) = a(n), idempotent.
a(A086970(n)) = A124356(n-1) for n > 1.
a(A000124(n)) = A192447(n+1).
a(n)*a(n+1)/2 = A007590(n), also equals partial sums of a(n).
A007590(a(n)) = 2*A008794(n). (End)

Extensions

More terms from James Sellers, Jun 05 2000
Removed duplicate of recurrence; corrected original recurrence and g.f. against offset - R. J. Mathar, Feb 19 2010
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