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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A276645 Numbers n such that A002088(n) is a triangular number.

Original entry on oeis.org

0, 1, 4, 5, 9, 19, 148, 209, 404, 652, 763, 1739, 2790, 9551, 11117, 13173, 25641, 170003, 181927, 519900, 3253342, 4816257, 7103608, 8269588, 64261168, 177847445, 294878941, 819667577, 844965349, 2944185248
Offset: 1

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Author

Altug Alkan, Sep 09 2016

Keywords

Comments

Corresponding triangular numbers are 0, 1, 6, 10, 28, 120, 6670, 13366, 49770, 129286, 177310, ...

Examples

			9 is a term because A002088(9) = 1+1+2+2+4+2+6+4+6 = 28 = 1+2+3+4+5+6+7.
		

Crossrefs

Programs

  • PARI
    lista(nn) = {print1(st = 0, ", "); for(n=1, nn, st += eulerphi(n); if (ispolygonal(st, 3), print1(n, ", ")));} \\ Michel Marcus, Sep 09 2016

Extensions

More terms from Michel Marcus, Sep 09 2016
a(25)-a(30) from Lars Blomberg, Oct 15 2016

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

A000010 Euler totient function phi(n): count numbers <= n and prime to n.

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4, 12, 6, 8, 8, 16, 6, 18, 8, 12, 10, 22, 8, 20, 12, 18, 12, 28, 8, 30, 16, 20, 16, 24, 12, 36, 18, 24, 16, 40, 12, 42, 20, 24, 22, 46, 16, 42, 20, 32, 24, 52, 18, 40, 24, 36, 28, 58, 16, 60, 30, 36, 32, 48, 20, 66, 32, 44
Offset: 1

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Author

Keywords

Comments

Number of elements in a reduced residue system modulo n.
Degree of the n-th cyclotomic polynomial (cf. A013595). - Benoit Cloitre, Oct 12 2002
Number of distinct generators of a cyclic group of order n. Number of primitive n-th roots of unity. (A primitive n-th root x is such that x^k is not equal to 1 for k = 1, 2, ..., n - 1, but x^n = 1.) - Lekraj Beedassy, Mar 31 2005
Also number of complex Dirichlet characters modulo n; Sum_{k=1..n} a(k) is asymptotic to (3/Pi^2)*n^2. - Steven Finch, Feb 16 2006
a(n) is the highest degree of irreducible polynomial dividing 1 + x + x^2 + ... + x^(n-1) = (x^n - 1)/(x - 1). - Alexander Adamchuk, Sep 02 2006, corrected Sep 27 2006
a(p) = p - 1 for prime p. a(n) is even for n > 2. For n > 2, a(n)/2 = A023022(n) = number of partitions of n into 2 ordered relatively prime parts. - Alexander Adamchuk, Jan 25 2007
Number of automorphisms of the cyclic group of order n. - Benoit Jubin, Aug 09 2008
a(n+2) equals the number of palindromic Sturmian words of length n which are "bispecial", prefix or suffix of two Sturmian words of length n + 1. - Fred Lunnon, Sep 05 2010
Suppose that a and n are coprime positive integers, then by Euler's totient theorem, any factor of n divides a^phi(n) - 1. - Lei Zhou, Feb 28 2012
If m has k prime factors, (p_1, p_2, ..., p_k), then phi(m*n) = (Product_{i=1..k} phi (p_i*n))/phi(n)^(k-1). For example, phi(42*n) = phi(2*n)*phi(3*n)*phi(7*n)/phi(n)^2. - Gary Detlefs, Apr 21 2012
Sum_{n>=1} a(n)/n! = 1.954085357876006213144... This sum is referenced in Plouffe's inverter. - Alexander R. Povolotsky, Feb 02 2013 (see A336334. - Hugo Pfoertner, Jul 22 2020)
The order of the multiplicative group of units modulo n. - Michael Somos, Aug 27 2013
A strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n, m)) for all positive integers n and m. - Michael Somos, Dec 30 2016
From Eric Desbiaux, Jan 01 2017: (Start)
a(n) equals the Ramanujan sum c_n(n) (last term on n-th row of triangle A054533).
a(n) equals the Jordan function J_1(n) (cf. A007434, A059376, A059377, which are the Jordan functions J_2, J_3, J_4, respectively). (End)
For n > 1, a(n) appears to be equal to the number of semi-meander solutions for n with top arches containing exactly 2 mountain ranges and exactly 2 arches of length 1. - Roger Ford, Oct 11 2017
a(n) is the minimum dimension of a lattice able to generate, via cut-and-project, the quasilattice whose diffraction pattern features n-fold rotational symmetry. The case n=15 is the first n > 1 in which the following simpler definition fails: "a(n) is the minimum dimension of a lattice with n-fold rotational symmetry". - Felix Flicker, Nov 08 2017
Number of cyclic Latin squares of order n with the first row in ascending order. - Eduard I. Vatutin, Nov 01 2020
a(n) is the number of rational numbers p/q >= 0 (in lowest terms) such that p + q = n. - Rémy Sigrist, Jan 17 2021
From Richard L. Ollerton, May 08 2021: (Start)
Formulas for the numerous OEIS entries involving Dirichlet convolution of a(n) and some sequence h(n) can be derived using the following (n >= 1):
Sum_{d|n} phi(d)*h(n/d) = Sum_{k=1..n} h(gcd(n,k)) [see P. H. van der Kamp link] = Sum_{d|n} h(d)*phi(n/d) = Sum_{k=1..n} h(n/gcd(n,k))*phi(gcd(n,k))/phi(n/gcd(n,k)). Similarly,
Sum_{d|n} phi(d)*h(d) = Sum_{k=1..n} h(n/gcd(n,k)) = Sum_{k=1..n} h(gcd(n,k))*phi(gcd(n,k))/phi(n/gcd(n,k)).
More generally,
Sum_{d|n} h(d) = Sum_{k=1..n} h(gcd(n,k))/phi(n/gcd(n,k)) = Sum_{k=1..n} h(n/gcd(n,k))/phi(n/gcd(n,k)).
In particular, for sequences involving the Möbius transform:
Sum_{d|n} mu(d)*h(n/d) = Sum_{k=1..n} h(gcd(n,k))*mu(n/gcd(n,k))/phi(n/gcd(n,k)) = Sum_{k=1..n} h(n/gcd(n,k))*mu(gcd(n,k))/phi(n/gcd(n,k)), where mu = A008683.
Use of gcd(n,k)*lcm(n,k) = n*k and phi(gcd(n,k))*phi(lcm(n,k)) = phi(n)*phi(k) provide further variations. (End)
From Richard L. Ollerton, Nov 07 2021: (Start)
Formulas for products corresponding to the sums above may found using the substitution h(n) = log(f(n)) where f(n) > 0 (for example, cf. formulas for the sum A018804 and product A067911 of gcd(n,k)):
Product_{d|n} f(n/d)^phi(d) = Product_{k=1..n} f(gcd(n,k)) = Product_{d|n} f(d)^phi(n/d) = Product_{k=1..n} f(n/gcd(n,k))^(phi(gcd(n,k))/phi(n/gcd(n,k))),
Product_{d|n} f(d)^phi(d) = Product_{k=1..n} f(n/gcd(n,k)) = Product_{k=1..n} f(gcd(n,k))^(phi(gcd(n,k))/phi(n/gcd(n,k))),
Product_{d|n} f(d) = Product_{k=1..n} f(gcd(n,k))^(1/phi(n/gcd(n,k))) = Product_{k=1..n} f(n/gcd(n,k))^(1/phi(n/gcd(n,k))),
Product_{d|n} f(n/d)^mu(d) = Product_{k=1..n} f(gcd(n,k))^(mu(n/gcd(n,k))/phi(n/gcd(n,k))) = Product_{k=1..n} f(n/gcd(n,k))^(mu(gcd(n,k))/phi(n/gcd(n,k))), where mu = A008683. (End)
a(n+1) is the number of binary words with exactly n distinct subsequences (when n > 0). - Radoslaw Zak, Nov 29 2021

Examples

			G.f. = x + x^2 + 2*x^3 + 2*x^4 + 4*x^5 + 2*x^6 + 6*x^7 + 4*x^8 + 6*x^9 + 4*x^10 + ...
a(8) = 4 with {1, 3, 5, 7} units modulo 8. a(10) = 4 with {1, 3, 7, 9} units modulo 10. - _Michael Somos_, Aug 27 2013
From _Eduard I. Vatutin_, Nov 01 2020: (Start)
The a(5)=4 cyclic Latin squares with the first row in ascending order are:
  0 1 2 3 4   0 1 2 3 4   0 1 2 3 4   0 1 2 3 4
  1 2 3 4 0   2 3 4 0 1   3 4 0 1 2   4 0 1 2 3
  2 3 4 0 1   4 0 1 2 3   1 2 3 4 0   3 4 0 1 2
  3 4 0 1 2   1 2 3 4 0   4 0 1 2 3   2 3 4 0 1
  4 0 1 2 3   3 4 0 1 2   2 3 4 0 1   1 2 3 4 0
(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. 840.
  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 24.
  • M. Baake and U. Grimm, Aperiodic Order Vol. 1: A Mathematical Invitation, Encyclopedia of Mathematics and its Applications 149, Cambridge University Press, 2013: see Tables 3.1 and 3.2.
  • Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 409.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 193.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 154-156.
  • C. W. Curtis, Pioneers of Representation Theory ..., Amer. Math. Soc., 1999; see p. 3.
  • J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Ellipses, Paris, 2004, Problème 529, pp. 71-257.
  • 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. 1, Chapter V.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 115-119.
  • Carl Friedrich Gauss, "Disquisitiones Arithmeticae", Yale University Press, 1965; see p. 21.
  • Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Math., 2n-d ed.; Addison-Wesley, 1994, p. 137.
  • R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section B36.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford Univ. Press, 1979, th. 60, 62, 63, 288, 323, 328, 330.
  • Peter Hilton and Jean Pedersen, A Mathematical Tapestry, Demonstrating the Beautiful Unity of Mathematics, Cambridge University Press, pages 261-264, the Coach theorem.
  • Jean-Marie Monier, Analyse, Exercices corrigés, 2ème année MP, Dunod, 1997, Exercice 3.2.21 pp. 281-294.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, Springer-Verlag, New York, Heidelberg, Berlin, 2 vols., 1976, Vol. II, problem 71, p. 126.
  • Paulo Ribenboim, The New Book of Prime Number Records.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 28-33.
  • 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 162-167.

Crossrefs

Cf. A002088 (partial sums), A008683, A003434 (steps to reach 1), A007755, A049108, A002202 (values), A011755 (Sum k*phi(k)).
Cf. also A005277 (nontotient numbers). For inverse see A002181, A006511, A058277.
Jordan function J_k(n) is a generalization - see A059379 and A059380 (triangle of values of J_k(n)), this sequence (J_1), A007434 (J_2), A059376 (J_3), A059377 (J_4), A059378 (J_5).
Row sums of triangles A134540, A127448, A143239, A143353 and A143276.
Equals right and left borders of triangle A159937. - Gary W. Adamson, Apr 26 2009
Values for prime powers p^e: A006093 (e=1), A036689 (e=2), A135177 (e=3), A138403 (e=4), A138407 (e=5), A138412 (e=6).
Values for perfect powers n^e: A002618 (e=2), A053191 (e=3), A189393 (e=4), A238533 (e=5), A306411 (e=6), A239442 (e=7), A306412 (e=8), A239443 (e=9).
Cf. A076479.
Cf. A023900 (Dirichlet inverse of phi), A306633 (Dgf at s=3).

Programs

  • Axiom
    [eulerPhi(n) for n in 1..100]
    
  • Haskell
    a n = length (filter (==1) (map (gcd n) [1..n])) -- Allan C. Wechsler, Dec 29 2014
    
  • Julia
    # Computes the first N terms of the sequence.
    function A000010List(N)
        phi = [i for i in 1:N + 1]
        for i in 2:N + 1
            if phi[i] == i
                for j in i:i:N + 1
                    phi[j] -= div(phi[j], i)
        end end end
    return phi end
    println(A000010List(68))  # Peter Luschny, Sep 03 2023
  • Magma
    [ EulerPhi(n) : n in [1..100] ]; // Sergei Haller (sergei(AT)sergei-haller.de), Dec 21 2006
    
  • Maple
    with(numtheory): A000010 := phi; [ seq(phi(n), n=1..100) ]; # version 1
    with(numtheory): phi := proc(n) local i,t1,t2; t1 := ifactors(n)[2]; t2 := n*mul((1-1/t1[i][1]),i=1..nops(t1)); end; # version 2
    # Alternative without library function:
    A000010List := proc(N) local i, j, phi;
        phi := Array([seq(i, i = 1 .. N+1)]);
        for i from 2 to N + 1 do
            if phi[i] = i then
                for j from i by i to N + 1 do
                    phi[j] := phi[j] - iquo(phi[j], i) od
            fi od;
    return phi end:
    A000010List(68);  # Peter Luschny, Sep 03 2023
  • Mathematica
    Array[EulerPhi, 70]
  • Maxima
    makelist(totient(n),n,0,1000); /* Emanuele Munarini, Mar 26 2011 */
    
  • PARI
    {a(n) = if( n==0, 0, eulerphi(n))}; /* Michael Somos, Feb 05 2011 */
    
  • Python
    from sympy.ntheory import totient
    print([totient(i) for i in range(1, 70)])  # Indranil Ghosh, Mar 17 2017
    
  • Python
    # Note also the implementation in A365339.
    
  • Sage
    def A000010(n): return euler_phi(n) # Jaap Spies, Jan 07 2007
    
  • Sage
    [euler_phi(n) for n in range(1, 70)]  # Zerinvary Lajos, Jun 06 2009
    

Formula

phi(n) = n*Product_{distinct primes p dividing n} (1 - 1/p).
Sum_{d divides n} phi(d) = n.
phi(n) = Sum_{d divides n} mu(d)*n/d, i.e., the Moebius transform of the natural numbers; mu() = Moebius function A008683().
Dirichlet generating function Sum_{n>=1} phi(n)/n^s = zeta(s-1)/zeta(s). Also Sum_{n >= 1} phi(n)*x^n/(1 - x^n) = x/(1 - x)^2.
Multiplicative with a(p^e) = (p - 1)*p^(e-1). - David W. Wilson, Aug 01 2001
Sum_{n>=1} (phi(n)*log(1 - x^n)/n) = -x/(1 - x) for -1 < x < 1 (cf. A002088) - Henry Bottomley, Nov 16 2001
a(n) = binomial(n+1, 2) - Sum_{i=1..n-1} a(i)*floor(n/i) (see A000217 for inverse). - Jon Perry, Mar 02 2004
It is a classical result (certainly known to Landau, 1909) that lim inf n/phi(n) = 1 (taking n to be primes), lim sup n/(phi(n)*log(log(n))) = e^gamma, with gamma = Euler's constant (taking n to be products of consecutive primes starting from 2 and applying Mertens' theorem). See e.g. Ribenboim, pp. 319-320. - Pieter Moree, Sep 10 2004
a(n) = Sum_{i=1..n} |k(n, i)| where k(n, i) is the Kronecker symbol. Also a(n) = n - #{1 <= i <= n : k(n, i) = 0}. - Benoit Cloitre, Aug 06 2004 [Corrected by Jianing Song, Sep 25 2018]
Conjecture: Sum_{i>=2} (-1)^i/(i*phi(i)) exists and is approximately 0.558 (A335319). - Orges Leka (oleka(AT)students.uni-mainz.de), Dec 23 2004
From Enrique Pérez Herrero, Sep 07 2010: (Start)
a(n) = Sum_{i=1..n} floor(sigma_k(i*n)/sigma_k(i)*sigma_k(n)), where sigma_2 is A001157.
a(n) = Sum_{i=1..n} floor(tau_k(i*n)/tau_k(i)*tau_k(n)), where tau_3 is A007425.
a(n) = Sum_{i=1..n} floor(rad(i*n)/rad(i)*rad(n)), where rad is A007947. (End)
a(n) = A173557(n)*A003557(n). - R. J. Mathar, Mar 30 2011
a(n) = A096396(n) + A096397(n). - Reinhard Zumkeller, Mar 24 2012
phi(p*n) = phi(n)*(floor(((n + p - 1) mod p)/(p - 1)) + p - 1), for primes p. - Gary Detlefs, Apr 21 2012
For odd n, a(n) = 2*A135303((n-1)/2)*A003558((n-1)/2) or phi(n) = 2*c*k; the Coach theorem of Pedersen et al. Cf. A135303. - Gary W. Adamson, Aug 15 2012
G.f.: Sum_{n>=1} mu(n)*x^n/(1 - x^n)^2, where mu(n) = A008683(n). - Mamuka Jibladze, Apr 05 2015
a(n) = n - cototient(n) = n - A051953(n). - Omar E. Pol, May 14 2016
a(n) = lim_{s->1} n*zeta(s)*(Sum_{d divides n} A008683(d)/(e^(1/d))^(s-1)), for n > 1. - Mats Granvik, Jan 26 2017
Conjecture: a(n) = Sum_{a=1..n} Sum_{b=1..n} Sum_{c=1..n} 1 for n > 1. The sum is over a,b,c such that n*c - a*b = 1. - Benedict W. J. Irwin, Apr 03 2017
a(n) = Sum_{j=1..n} gcd(j, n) cos(2*Pi*j/n) = Sum_{j=1..n} gcd(j, n) exp(2*Pi*i*j/n) where i is the imaginary unit. Notice that the Ramanujan's sum c_n(k) := Sum_{j=1..n, gcd(j, n) = 1} exp(2*Pi*i*j*k/n) gives a(n) = Sum_{k|n} k*c_(n/k)(1) = Sum_{k|n} k*mu(n/k). - Michael Somos, May 13 2018
G.f.: x*d/dx(x*d/dx(log(Product_{k>=1} (1 - x^k)^(-mu(k)/k^2)))), where mu(n) = A008683(n). - Mamuka Jibladze, Sep 20 2018
a(n) = Sum_{d|n} A007431(d). - Steven Foster Clark, May 29 2019
G.f. A(x) satisfies: A(x) = x/(1 - x)^2 - Sum_{k>=2} A(x^k). - Ilya Gutkovskiy, Sep 06 2019
a(n) >= sqrt(n/2) (Nicolas). - Hugo Pfoertner, Jun 01 2020
a(n) > n/(exp(gamma)*log(log(n)) + 5/(2*log(log(n)))), except for n=223092870 (Rosser, Schoenfeld). - Hugo Pfoertner, Jun 02 2020
From Bernard Schott, Nov 28 2020: (Start)
Sum_{m=1..n} 1/a(m) = A028415(n)/A048049(n) -> oo when n->oo.
Sum_{n >= 1} 1/a(n)^2 = A109695.
Sum_{n >= 1} 1/a(n)^3 = A335818.
Sum_{n >= 1} 1/a(n)^k is convergent iff k > 1.
a(2n) = a(n) iff n is odd, and, a(2n) > a(n) iff n is even. (End) [Actually, a(2n) = 2*a(n) for even n. - Jianing Song, Sep 18 2022]
a(n) = 2*A023896(n)/n, n > 1. - Richard R. Forberg, Feb 03 2021
From Richard L. Ollerton, May 09 2021: (Start)
For n > 1, Sum_{k=1..n} phi^{(-1)}(n/gcd(n,k))*a(gcd(n,k))/a(n/gcd(n,k)) = 0, where phi^{(-1)} = A023900.
For n > 1, Sum_{k=1..n} a(gcd(n,k))*mu(rad(gcd(n,k)))*rad(gcd(n,k))/gcd(n,k) = 0.
For n > 1, Sum_{k=1..n} a(gcd(n,k))*mu(rad(n/gcd(n,k)))*rad(n/gcd(n,k))*gcd(n,k) = 0.
Sum_{k=1..n} a(gcd(n,k))/a(n/gcd(n,k)) = n. (End)
a(n) = Sum_{d|n, e|n} gcd(d, e)*mobius(n/d)*mobius(n/e) (the sum is a multiplicative function of n by Tóth, and takes the value p^e - p^(e-1) for n = p^e, a prime power). - Peter Bala, Jan 22 2024
Sum_{n >= 1} phi(n)*x^n/(1 + x^n) = x + 3*x^3 + 5*x^5 + 7*x^7 + ... = Sum_{n >= 1} phi(2*n-1)*x^(2*n-1)/(1 - x^(4*n-2)). For the first equality see Pólya and Szegő, problem 71, p. 126. - Peter Bala, Feb 29 2024
Conjecture: a(n) = lim_{k->oo} (n^(k + 1))/A000203(n^k). - Velin Yanev, Dec 04 2024 [A000010(p) = p-1, A000203(p^k) = (p^(k+1)-1)/(p-1), so the conjecture is true if n is prime. - Vaclav Kotesovec, Dec 19 2024]

A003418 Least common multiple (or LCM) of {1, 2, ..., n} for n >= 1, a(0) = 1.

Original entry on oeis.org

1, 1, 2, 6, 12, 60, 60, 420, 840, 2520, 2520, 27720, 27720, 360360, 360360, 360360, 720720, 12252240, 12252240, 232792560, 232792560, 232792560, 232792560, 5354228880, 5354228880, 26771144400, 26771144400, 80313433200, 80313433200, 2329089562800, 2329089562800
Offset: 0

Views

Author

Roland Anderson (roland.anderson(AT)swipnet.se)

Keywords

Comments

The minimal exponent of the symmetric group S_n, i.e., the least positive integer for which x^a(n)=1 for all x in S_n. - Franz Vrabec, Dec 28 2008
Product over all primes of highest power of prime less than or equal to n. a(0) = 1 by convention.
Also smallest number whose set of divisors contains an n-term arithmetic progression. - Reinhard Zumkeller, Dec 09 2002
An assertion equivalent to the Riemann hypothesis is: | log(a(n)) - n | < sqrt(n) * log(n)^2. - Lekraj Beedassy, Aug 27 2006. (This is wrong for n = 1 and n = 2. Should "for n large enough" be added? - Georgi Guninski, Oct 22 2011)
Corollary 3 of Farhi gives a proof that a(n) >= 2^(n-1). - Jonathan Vos Post, Jun 15 2009
Appears to be row products of the triangle T(n,k) = b(A010766) where b = A130087/A130086. - Mats Granvik, Jul 08 2009
Greg Martin (see link) proved that "the product of the Gamma function sampled over the set of all rational numbers in the open interval (0,1) whose denominator in lowest terms is at most n" equals (2*Pi)^(1/2)*a(n)^(-1/2). - Jonathan Vos Post, Jul 28 2009
a(n) = lcm(A188666(n), A188666(n)+1, ..., n). - Reinhard Zumkeller, Apr 25 2011
a(n+1) is the smallest integer such that all polynomials a(n+1)*(1^i + 2^i + ... + m^i) in m, for i=0,1,...,n, are polynomials with integer coefficients. - Vladimir Shevelev, Dec 23 2011
It appears that A020500(n) = a(n)/a(n-1). - Asher Auel, corrected by Bill McEachen, Apr 05 2024
n-th distinct value = A051451(n). - Matthew Vandermast, Nov 27 2009
a(n+1) = least common multiple of n-th row in A213999. - Reinhard Zumkeller, Jul 03 2012
For n > 2, (n-1) = Sum_{k=2..n} exp(a(n)*2*i*Pi/k). - Eric Desbiaux, Sep 13 2012
First column minus second column of A027446. - Eric Desbiaux, Mar 29 2013
For n > 0, a(n) is the smallest number k such that n is the n-th divisor of k. - Michel Lagneau, Apr 24 2014
Slowest growing integer > 0 in Z converging to 0 in Z^ when considered as profinite integer. - Herbert Eberle, May 01 2016
What is the largest number of consecutive terms that are all equal? I found 112 equal terms from a(370261) to a(370372). - Dmitry Kamenetsky, May 05 2019
Answer: there exist arbitrarily long sequences of consecutive terms with the same value; also, the maximal run of consecutive terms with different values is 5 from a(1) to a(5) (see link Roger B. Eggleton). - Bernard Schott, Aug 07 2019
Related to the inequality (54) in Ramanujan's paper about highly composite numbers A002182, also used in A199337: a(A329570(m))^2 is a (not minimal) bound above which all highly composite numbers are divisible by m, according to the right part of that inequality. - M. F. Hasler, Jan 04 2020
For n > 2, a(n) is of the form 2^e_1 * p_2^e_2 * ... * p_m^e_m, where e_m = 1 and e = floor(log_2(p_m)) <= e_1. Therefore, 2^e * p_m^e_m is a primitive Zumkeler number (A180332). Therefore, 2^e_1 * p_m^e_m is a Zumkeller number (A083207). Therefore, for n > 2, a(n) = 2^e_1 * p_m^e_m * r, where r is relatively prime to 2*p_m, is a Zumkeller number (see my proof at A002182 for details). - Ivan N. Ianakiev, May 10 2020
For n > 1, 2|(a(n)+2) ... n|(a(n)+n), so a(n)+2 .. a(n)+n are all composite and (part of) a prime gap of at least n. (Compare n!+2 .. n!+n). - Stephen E. Witham, Oct 09 2021

Examples

			LCM of {1,2,3,4,5,6} = 60. The primes up to 6 are 2, 3 and 5. floor(log(6)/log(2)) = 2 so the exponent of 2 is 2.
floor(log(6)/log(3)) = 1 so the exponent of 3 is 1.
floor(log(6)/log(5)) = 1 so the exponent of 5 is 1. Therefore, a(6) = 2^2 * 3^1 * 5^1 = 60. - _David A. Corneth_, Jun 02 2017
		

References

  • J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 365.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row products of A133233.
Cf. A025528 (number of prime factors of a(n) with multiplicity).
Cf. A275120 (lengths of runs of consecutive equal terms), A276781 (ordinal transform from term a(1)=1 onward).

Programs

  • Haskell
    a003418 = foldl lcm 1 . enumFromTo 2
    -- Reinhard Zumkeller, Apr 04 2012, Apr 25 2011
    
  • Magma
    [1] cat [Exponent(SymmetricGroup(n)) : n in [1..28]]; // Arkadiusz Wesolowski, Sep 10 2013
    
  • Magma
    [Lcm([1..n]): n in [0..30]]; // Bruno Berselli, Feb 06 2015
    
  • Maple
    A003418 := n-> lcm(seq(i,i=1..n));
    HalfFarey := proc(n) local a,b,c,d,k,s; a := 0; b := 1; c := 1; d := n; s := NULL; do k := iquo(n + b, d); a, b, c, d := c, d, k*c - a, k*d - b; if 2*a > b then break fi; s := s,(a/b); od: [s] end: LCM := proc(n) local i; (1/2)*mul(2*sin(Pi*i),i=HalfFarey(n))^2 end: # Peter Luschny
    # next Maple program:
    a:= proc(n) option remember; `if`(n=0, 1, ilcm(n, a(n-1))) end:
    seq(a(n), n=0..33);  # Alois P. Heinz, Jun 10 2021
  • Mathematica
    Table[LCM @@ Range[n], {n, 1, 40}] (* Stefan Steinerberger, Apr 01 2006 *)
    FoldList[ LCM, 1, Range@ 28]
    A003418[0] := 1; A003418[1] := 1; A003418[n_] := A003418[n] = LCM[n,A003418[n-1]]; (* Enrique Pérez Herrero, Jan 08 2011 *)
    Table[Product[Prime[i]^Floor[Log[Prime[i], n]], {i, PrimePi[n]}], {n, 0, 28}] (* Wei Zhou, Jun 25 2011 *)
    Table[Product[Cyclotomic[n, 1], {n, 2, m}], {m, 0, 28}] (* Fred Daniel Kline, May 22 2014 *)
    a1[n_] := 1/12 (Pi^2+3(-1)^n (PolyGamma[1,1+n/2] - PolyGamma[1,(1+n)/2])) // Simplify
    a[n_] := Denominator[Sqrt[a1[n]]];
    Table[If[IntegerQ[a[n]], a[n], a[n]*(a[n])[[2]]], {n, 0, 28}] (* Gerry Martens, Apr 07 2018 [Corrected by Vaclav Kotesovec, Jul 16 2021] *)
  • PARI
    a(n)=local(t); t=n>=0; forprime(p=2,n,t*=p^(log(n)\log(p))); t
    
  • PARI
    a(n)=if(n<1,n==0,1/content(vector(n,k,1/k)))
    
  • PARI
    a(n)=my(v=primes(primepi(n)),k=sqrtint(n),L=log(n+.5));prod(i=1,#v,if(v[i]>k,v[i],v[i]^(L\log(v[i])))) \\ Charles R Greathouse IV, Dec 21 2011
    
  • PARI
    a(n)=lcm(vector(n,i,i)) \\ Bill Allombert, Apr 18 2012 [via Charles R Greathouse IV]
    
  • PARI
    n=1; lim=100; i=1; j=1; until(n==lim, a=lcm(j,i+1); i++; j=a; n++; print(n" "a);); \\ Mike Winkler, Sep 07 2013
    
  • Python
    from functools import reduce
    from operator import mul
    from sympy import sieve
    def integerlog(n,b): # find largest integer k>=0 such that b^k <= n
        kmin, kmax = 0,1
        while b**kmax <= n:
            kmax *= 2
        while True:
            kmid = (kmax+kmin)//2
            if b**kmid > n:
                kmax = kmid
            else:
                kmin = kmid
            if kmax-kmin <= 1:
                break
        return kmin
    def A003418(n):
        return reduce(mul,(p**integerlog(n,p) for p in sieve.primerange(1,n+1)),1) # Chai Wah Wu, Mar 13 2021
    
  • Python
    # generates initial segment of sequence
    from math import gcd
    from itertools import accumulate
    def lcm(a, b): return a * b // gcd(a, b)
    def aupton(nn): return [1] + list(accumulate(range(1, nn+1), lcm))
    print(aupton(30)) # Michael S. Branicky, Jun 10 2021
  • Sage
    [lcm(range(1,n)) for n in range(1, 30)] # Zerinvary Lajos, Jun 06 2009
    
  • Scheme
    (define (A003418 n) (let loop ((n n) (m 1)) (if (zero? n) m (loop (- n 1) (lcm m n))))) ;; Antti Karttunen, Jan 03 2018
    

Formula

The prime number theorem implies that lcm(1,2,...,n) = exp(n(1+o(1))) as n -> infinity. In other words, log(lcm(1,2,...,n))/n -> 1 as n -> infinity. - Jonathan Sondow, Jan 17 2005
a(n) = Product (p^(floor(log n/log p))), where p runs through primes not exceeding n (i.e., primes 2 through A007917(n)). - Lekraj Beedassy, Jul 27 2004
Greg Martin showed that a(n) = lcm(1,2,3,...,n) = Product_{i = Farey(n), 0 < i < 1} 2*Pi/Gamma(i)^2. This can be rewritten (for n > 1) as a(n) = (1/2)*(Product_{i = Farey(n), 0 < i <= 1/2} 2*sin(i*Pi))^2. - Peter Luschny, Aug 08 2009
Recursive formula useful for computations: a(0)=1; a(1)=1; a(n)=lcm(n,a(n-1)). - Enrique Pérez Herrero, Jan 08 2011
From Enrique Pérez Herrero, Jun 01 2011: (Start)
a(n)/a(n-1) = A014963(n).
if n is a prime power p^k then a(n)=a(p^k)=p*a(n-1), otherwise a(n)=a(n-1).
a(n) = Product_{k=2..n} (1 + (A007947(k)-1)*floor(1/A001221(k))), for n > 1. (End)
a(n) = A079542(n+1, 2) for n > 1.
a(n) = exp(Sum_{k=1..n} Sum_{d|k} moebius(d)*log(k/d)). - Peter Luschny, Sep 01 2012
a(n) = A025529(n) - A027457(n). - Eric Desbiaux, Mar 14 2013
a(n) = exp(Psi(n)) = 2 * Product_{k=2..A002088(n)} (1 - exp(2*Pi*i * A038566(k+1) / A038567(k))), where i is the imaginary unit, and Psi the second Chebyshev's function. - Eric Desbiaux, Aug 13 2014
a(n) = A064446(n)*A038610(n). - Anthony Browne, Jun 16 2016
a(n) = A000142(n) / A025527(n) = A000793(n) * A225558(n). - Antti Karttunen, Jun 02 2017
log(a(n)) = Sum_{k>=1} (A309229(n, k)/k - 1/k). - Mats Granvik, Aug 10 2019
From Petros Hadjicostas, Jul 24 2020: (Start)
Nair (1982) proved that 2^n <= a(n) <= 4^n for n >= 9. See also Farhi (2009). Nair also proved that
a(n) = lcm(m*binomial(n,m): 1 <= m <= n) and
a(n) = gcd(a(m)*binomial(n,m): n/2 <= m <= n). (End)
Sum_{n>=1} 1/a(n) = A064859. - Bernard Schott, Aug 24 2020

A002321 Mertens's function: Sum_{k=1..n} mu(k), where mu is the Moebius function A008683.

Original entry on oeis.org

1, 0, -1, -1, -2, -1, -2, -2, -2, -1, -2, -2, -3, -2, -1, -1, -2, -2, -3, -3, -2, -1, -2, -2, -2, -1, -1, -1, -2, -3, -4, -4, -3, -2, -1, -1, -2, -1, 0, 0, -1, -2, -3, -3, -3, -2, -3, -3, -3, -3, -2, -2, -3, -3, -2, -2, -1, 0, -1, -1, -2, -1, -1, -1, 0, -1, -2, -2, -1, -2, -3, -3, -4, -3, -3, -3, -2, -3, -4, -4, -4
Offset: 1

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Keywords

Comments

Partial sums of the Moebius function A008683.
Also determinant of n X n (0,1) matrix defined by A(i,j)=1 if j=1 or i divides j.
The first positive value of Mertens's function for n > 1 is for n = 94. The graph seems to show a negative bias for the Mertens function which is eerily similar to the Chebyshev bias (described in A156749 and A156709). The purported bias seems to be empirically approximated to - (6 / Pi^2) * (sqrt(n) / 4) (by looking at the graph) (see MathOverflow link, May 28 2012) where 6 / Pi^2 = 1 / zeta(2) is the asymptotic density of squarefree numbers (the squareful numbers having Moebius mu of 0). This would be a growth pattern akin to the Chebyshev bias. - Daniel Forgues, Jan 23 2011
All integers appear infinitely often in this sequence. - Charles R Greathouse IV, Aug 06 2012
Soundararajan proves that, on the Riemann Hypothesis, a(n) << sqrt(n) exp(sqrt(log n)*(log log n)^14), sharpening the well-known equivalence. - Charles R Greathouse IV, Jul 17 2015
Balazard & De Roton improve this (on the Riemann Hypothesis) to a(n) << sqrt(n) exp(sqrt(log n)*(log log n)^k) for any k > 5/2, where the implied constant in the Vinogradov symbol depends on k. Saha & Sankaranarayanan reduce the exponent to 5/4 on additional hypotheses. - Charles R Greathouse IV, Feb 02 2023

Examples

			G.f. = x - x^3 - x^4 - 2*x^5 - x^6 - 2*x^7 - 2*x^8 - 2*x^9 - x^10 - 2*x^11 - 2*x^12 - ...
		

References

  • E. Landau, Vorlesungen über Zahlentheorie, Chelsea, NY, Vol. 2, p. 157.
  • D. H. Lehmer, Guide to Tables in the Theory of Numbers. Bulletin No. 105, National Research Council, Washington, DC, 1941, pp. 7-10.
  • F. Mertens, "Über eine zahlentheoretische Funktion", Akademie Wissenschaftlicher Wien Mathematik-Naturlich Kleine Sitzungsber, IIa 106, (1897), p. 761-830.
  • D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Section VI.1.
  • Biswajyoti Saha and Ayyadurai Sankaranarayanan, On estimates of the Mertens function, International Journal of Number Theory, Vol. 15, No. 02 (2019), pp. 327-337.
  • 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).
  • J. von zur Gathen and J. Gerhard, Modern Computer Algebra, Cambridge, 1999, see p. 482.

Crossrefs

First column of A134541.
First column of A179287.

Programs

  • Haskell
    import Data.List (genericIndex)
    a002321 n = genericIndex a002321_list (n-1)
    a002321_list = scanl1 (+) a008683_list
    -- Reinhard Zumkeller, Jul 14 2014, Dec 26 2012
    
  • Magma
    [&+[MoebiusMu(k): k in [1..n]]: n in [1..81]]; // Bruno Berselli, Jul 12 2021
  • Maple
    with(numtheory); A002321 := n->add(mobius(k),k=1..n);
  • Mathematica
    Rest[ FoldList[ #1+#2&, 0, Array[ MoebiusMu, 100 ] ] ]
    Accumulate[Array[MoebiusMu,100]] (* Harvey P. Dale, May 11 2011 *)
  • PARI
    a(n) = sum( k=1, n, moebius(k))
    
  • PARI
    a(n) = if( n<1, 0, matdet( matrix(n, n, i, j, j==1 || 0==j%i)))
    
  • PARI
    a(n)=my(s); forsquarefree(k=1,n, s+=moebius(k)); s \\ Charles R Greathouse IV, Jan 08 2018
    
  • Python
    from sympy import mobius
    def M(n): return sum(mobius(k) for k in range(1,n + 1))
    print([M(n) for n in range(1, 151)]) # Indranil Ghosh, Mar 18 2017
    
  • Python
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def A002321(n):
        if n == 0:
            return 0
        c, j = n, 2
        k1 = n//j
        while k1 > 1:
            j2 = n//k1 + 1
            c += (j2-j)*A002321(k1)
            j, k1 = j2, n//j2
        return j-c # Chai Wah Wu, Mar 30 2021
    

Formula

Assuming the Riemann hypothesis, a(n) = O(x^(1/2 + eps)) for every eps > 0 (Littlewood - see Landau p. 161).
Lambert series: Sum_{n >= 1} a(n)*(x^n/(1-x^n)-x^(n+1)/(1-x^(n+1))) = x and -1/x. - Mats Granvik, Sep 09 2010 and Sep 23 2010
a(n)+2 = A192763(n,1) for n>1, and A192763(1,k) for k>1 (conjecture). - Mats Granvik, Jul 10 2011
Sum_{k = 1..n} a(floor(n/k)) = 1. - David W. Wilson, Feb 27 2012
a(n) = Sum_{k = 1..n} tau_{-2}(k) * floor(n/k), where tau_{-2} is A007427. - Enrique Pérez Herrero, Jan 23 2013
a(n) = Sum_{k=1..A002088(n)} exp(2*Pi*i*A038566(k)/A038567(k-1)) where i is the imaginary unit. - Eric Desbiaux, Jul 31 2014
Schoenfeld proves that |a(n)| < 5.3*n/(log n)^(10/9) for n > 1. - Charles R Greathouse IV, Jan 17 2018
G.f. A(x) satisfies: A(x) = (1/(1 - x)) * (x - Sum_{k>=2} (1 - x^k) * A(x^k)). - Ilya Gutkovskiy, Aug 11 2021

A048691 a(n) = d(n^2), where d(k) = A000005(k) is the number of divisors of k.

Original entry on oeis.org

1, 3, 3, 5, 3, 9, 3, 7, 5, 9, 3, 15, 3, 9, 9, 9, 3, 15, 3, 15, 9, 9, 3, 21, 5, 9, 7, 15, 3, 27, 3, 11, 9, 9, 9, 25, 3, 9, 9, 21, 3, 27, 3, 15, 15, 9, 3, 27, 5, 15, 9, 15, 3, 21, 9, 21, 9, 9, 3, 45, 3, 9, 15, 13, 9, 27, 3, 15, 9, 27, 3, 35, 3, 9, 15, 15, 9, 27, 3, 27
Offset: 1

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Comments

Inverse Moebius transform of A034444: Sum_{d|n} 2^omega(d), where omega(n) = A001221(n) is the number of distinct primes dividing n.
Number of elements in the set {(x,y): x|n, y|n, gcd(x,y)=1}.
Number of elements in the set {(x,y): lcm(x,y)=n}.
Also gives total number of positive integral solutions (x,y), order being taken into account, to the optical or parallel resistor equation 1/x + 1/y = 1/n. Indeed, writing the latter as X*Y=N, with X=x-n, Y=y-n, N=n^2, the one-to-one correspondence between solutions (X, Y) and (x, y) is obvious, so that clearly, the solution pairs (x, y) are tau(N)=tau(n^2) in number. - Lekraj Beedassy, May 31 2002
Number of ordered pairs of positive integers (a,c) such that n^2 - ac = 0. Therefore number of quadratic equations of the form ax^2 + 2nx + c = 0 where a,n,c are positive integers and each equation has two equal (rational) roots, -n/a. (If a and c are positive integers, but, instead, the coefficient of x is odd, it is impossible for the equation to have equal roots.) - Rick L. Shepherd, Jun 19 2005
Problem A1 on the 21st Putnam competition in 1960 (see John Scholes link) asked for the number of pairs of positive integers (x,y) such that xy/(x+y) = n: the answer is a(n); for n = 4, the a(4) = 5 solutions (x,y) are (5,20), (6,12), (8,8), (12,6), (20,5). - Bernard Schott, Feb 12 2023
Numbers k such that a(k)/d(k) is an integer are in A217584 and the corresponding quotients are in A339055. - Bernard Schott, Feb 15 2023

References

  • A. M. Gleason et al., The William Lowell Putnam Mathematical Competitions, Problems & Solutions:1938-1960 Soln. to Prob. 1 1960, p. 516, MAA, 1980.
  • Ross Honsberger, More Mathematical Morsels, Morsel 43, pp. 232-3, DMA No. 10 MAA, 1991.
  • Loren C. Larson, Problem-Solving Through Problems, Prob. 3.3.7, p. 102, Springer 1983.
  • Alfred S. Posamentier and Charles T. Salkind, Challenging Problems in Algebra, Prob. 9-9 pp. 143 Dover NY, 1988.
  • D. O. Shklarsky et al., The USSR Olympiad Problem Book, Soln. to Prob. 123, pp. 28, 217-8, Dover NY.
  • Wacław Sierpiński, Elementary Theory of Numbers, pp. 71-2, Elsevier, North Holland, 1988.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 91.
  • Charles W. Trigg, Mathematical Quickies, Question 194, pp. 53, 168, Dover, 1985.

Crossrefs

Partial sums give A061503.
For similar LCM sequences, see A070919, A070920, A070921.
For the earliest occurrence of 2n-1 see A016017.

Programs

Formula

a(n) = A000005(A000290(n)).
tau(n^2) = Sum_{d|n} mu(n/d)*tau(d)^2, where mu(n) = A008683(n), cf. A061391.
Multiplicative with a(p^e) = 2e+1. - Vladeta Jovovic, Jul 23 2001
Also a(n) = Sum_{d|n} (tau(d)*moebius(n/d)^2), Dirichlet convolution of A000005 and A008966. - Benoit Cloitre, Sep 08 2002
a(n) = A055205(n) + A000005(n). - Reinhard Zumkeller, Dec 08 2009
Dirichlet g.f.: (zeta(s))^3/zeta(2s). - R. J. Mathar, Feb 11 2011
a(n) = Sum_{d|n} 2^omega(d). Inverse Mobius transform of A034444. - Enrique Pérez Herrero, Apr 14 2012
G.f.: Sum_{k>=1} 2^omega(k)*x^k/(1 - x^k). - Ilya Gutkovskiy, Mar 10 2018
Sum_{k=1..n} a(k) ~ n*(6/Pi^2)*(log(n)^2/2 + log(n)*(3*gamma - 1) + 1 - 3*gamma + 3*gamma^2 - 3*gamma_1 + (2 - 6*gamma - 2*log(n))*zeta'(2)/zeta(2) + (2*zeta'(2)/zeta(2))^2 - 2*zeta''(2)/zeta(2)), where gamma is Euler's constant (A001620) and gamma_1 is the first Stieltjes constant (A082633). - Amiram Eldar, Jan 26 2023

Extensions

Additional comments from Vladeta Jovovic, Apr 29 2001

A005728 Number of fractions in Farey series of order n.

Original entry on oeis.org

1, 2, 3, 5, 7, 11, 13, 19, 23, 29, 33, 43, 47, 59, 65, 73, 81, 97, 103, 121, 129, 141, 151, 173, 181, 201, 213, 231, 243, 271, 279, 309, 325, 345, 361, 385, 397, 433, 451, 475, 491, 531, 543, 585, 605, 629, 651, 697, 713, 755, 775, 807, 831, 883, 901, 941, 965
Offset: 0

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Keywords

Comments

Sometimes called Phi(n).
Leo Moser found an interesting way to generate this sequence, see Gardner.
a(n) is a prime number for nine consecutive values of n: n = 1, 2, 3, 4, 5, 6, 7, 8, 9. - Altug Alkan, Sep 26 2015
Named after the English geologist and writer John Farey, Sr. (1766-1826). - Amiram Eldar, Jun 17 2021

Examples

			a(5)=11 because the fractions are 0/1, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1/1.
		

References

  • Martin Gardner, The Last Recreations, 1997, chapter 12.
  • Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, a foundation for computer science, Chapter 4.5 - Relative Primality, pages 118 - 120 and Chapter 9 - Asymptotics, Problem 6, pages 448 - 449, Addison-Wesley Publishing Co., Reading, Mass., 1989.
  • William Judson LeVeque, Topics in Number Theory, Addison-Wesley, Reading, MA, 2 vols., 1956, Vol. 1, p. 154.
  • Andrey O. Matveev, Farey Sequences, De Gruyter, 2017, Table 1.7.
  • Leo Moser, Solution to Problem P42, Canadian Mathematical Bulletin, Vol. 5, No. 3 (1962), pp. 312-313.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

For the Farey series see A006842/A006843.
Essentially the same as A049643.

Programs

  • GAP
    List([0..60],n->Sum([1..n],i->Phi(i)))+1; # Muniru A Asiru, Jul 31 2018
    
  • Haskell
    a005728 n = a005728_list
    a005728_list = scanl (+) 1 a000010_list
    -- Reinhard Zumkeller, Aug 04 2012
    
  • Magma
    [1] cat [n le 1 select 2 else Self(n-1)+EulerPhi(n): n in [1..60]]; // Vincenzo Librandi, Sep 27 2015
    
  • Maple
    A005728 := proc(n)
        1+add(numtheory[phi](i),i=1..n) ;
    end proc:
    seq(A005728(n),n=0..80) ; # R. J. Mathar, Nov 29 2017
  • Mathematica
    Accumulate@ Array[ EulerPhi, 54, 0] + 1
    f[n_] := 1 + Sum[ EulerPhi[m], {m, n}]; Array[f, 55, 0] (* or *)
    f[n_] := (Sum[ MoebiusMu[m] Floor[n/m]^2, {m, n}] + 3)/2; f[0] = 1; Array[f, 55, 0] (* or *)
    f[n_] := n (n + 3)/2 - Sum[f[Floor[n/m]], {m, 2, n}]; f[0] = 1; Array[f, 55, 0] (* Robert G. Wilson v, Sep 26 2015 *)
    a[n_] := If[n == 0, 1, FareySequence[n] // Length];
    Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Jul 16 2022 *)
  • PARI
    a(n)=1+sum(k=1,n,eulerphi(k)) \\ Charles R Greathouse IV, Jun 03 2013
    
  • Python
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def A005728(n): # based on second formula in A018805
        if n == 0:
            return 1
        c, j = -2, 2
        k1 = n//j
        while k1 > 1:
            j2 = n//k1 + 1
            c += (j2-j)*(2*A005728(k1)-3)
            j, k1 = j2, n//j2
        return (n*(n-1)-c+j)//2 # Chai Wah Wu, Mar 24 2021

Formula

a(n) = 1 + Sum_{i=1..n} phi(i).
a(n) = n*(n+3)/2 - Sum_{k=2..n} a(floor(n/k)). - David W. Wilson, May 25 2002
a(n) = a(n-1) + phi(n) with a(0) = 1. - Arkadiusz Wesolowski, Oct 13 2012
a(n) = 1 + A002088(n). - Robert G. Wilson v, Sep 26 2015

A020492 Balanced numbers: numbers k such that phi(k) (A000010) divides sigma(k) (A000203).

Original entry on oeis.org

1, 2, 3, 6, 12, 14, 15, 30, 35, 42, 56, 70, 78, 105, 140, 168, 190, 210, 248, 264, 270, 357, 418, 420, 570, 594, 616, 630, 714, 744, 812, 840, 910, 1045, 1240, 1254, 1485, 1672, 1848, 2090, 2214, 2376, 2436, 2580, 2730, 2970, 3080, 3135, 3339, 3596, 3720, 3828
Offset: 1

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Comments

The quotient A020492(n)/A002088(n) = SummatorySigma/SummatoryTotient as n increases seems to approach Pi^4/36 or zeta(2)^2 [~2.705808084277845]. - Labos Elemer, Sep 20 2004, corrected by Charles R Greathouse IV, Jun 20 2012
If 2^p-1 is prime (a Mersenne prime) then m = 2^(p-2)*(2^p-1) is in the sequence because when p = 2 we get m = 3 and phi(3) divides sigma(3) and for p > 2, phi(m) = 2^(p-2)*(2^(p-1)-1); sigma(m) = (2^(p-1)-1)*2^p hence sigma(m)/phi(m) = 4 is an integer. So for each n, A133028(n) = 2^(A000043(n)-2)*(2^A000043(n)-1) is in the sequence. - Farideh Firoozbakht, Nov 28 2005
Phi and sigma are both multiplicative functions and for this reason if m and n are coprime and included in this sequence then m*n is also in this sequence. - Enrique Pérez Herrero, Sep 05 2010
The quotients sigma(n)/phi(n) are in A023897. - Bernard Schott, Jun 06 2017
There are 544768 balanced numbers < 10^14. - Jud McCranie, Sep 10 2017
a(975807) = 419998185095132. - Jud McCranie, Nov 28 2017

Examples

			sigma(35) = 1+5+7+35 = 48, phi(35) = 24, hence 35 is a term.
		

References

  • D. Chiang, "N's for which phi(N) divides sigma(N)", Mathematical Buds, Chap. VI pp. 53-70 Vol. 3 Ed. H. D. Ruderman, Mu Alpha Theta 1984.

Crossrefs

Positions of 0's in A063514.

Programs

  • Magma
    [ n: n in [1..3900] | SumOfDivisors(n) mod EulerPhi(n) eq 0 ]; // Klaus Brockhaus, Nov 09 2008
    
  • Mathematica
    Select[ Range[ 4000 ], IntegerQ[ DivisorSigma[ 1, # ]/EulerPhi[ # ] ]& ]
    (* Second program: *)
    Select[Range@ 4000, Divisible[DivisorSigma[1, #], EulerPhi@ #] &] (* Michael De Vlieger, Nov 28 2017 *)
  • PARI
    select(n->sigma(n)%eulerphi(n)==0,vector(10^4,i,i)) \\ Charles R Greathouse IV, Jun 20 2012
    
  • Python
    from sympy import totient, divisor_sigma
    print([n for n in range(1, 4001) if divisor_sigma(n)%totient(n)==0]) # Indranil Ghosh, Jul 06 2017
    
  • Python
    from math import prod
    from itertools import count, islice
    from sympy import factorint
    def A020492_gen(startvalue=1): # generator of terms >= startvalue
        for m in count(max(startvalue,1)):
            f = factorint(m)
            if not prod(p**(e+2)-p for p,e in f.items())%(m*prod((p-1)**2 for p in f)):
                yield m
    A020492_list = list(islice(A020492_gen(),20)) # Chai Wah Wu, Aug 12 2024

Extensions

More terms from Farideh Firoozbakht, Nov 28 2005

A038566 Numerators in canonical bijection from positive integers to positive rationals <= 1: arrange fractions by increasing denominator then by increasing numerator.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 1, 2, 3, 4, 1, 5, 1, 2, 3, 4, 5, 6, 1, 3, 5, 7, 1, 2, 4, 5, 7, 8, 1, 3, 7, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 5, 7, 11, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 3, 5, 9, 11, 13, 1, 2, 4, 7, 8, 11, 13, 14, 1, 3, 5, 7, 9, 11, 13, 15, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16
Offset: 1

Views

Author

Keywords

Comments

For denominators see A038567.
Row n has length A000010(n).
Also numerators in canonical bijection from positive integers to all positive rational numbers: arrange fractions in triangle in which in the n-th row the phi(n) numbers are the fractions i/j with gcd(i,j) = 1, i+j=n, i=1..n-1, j=n-1..1. n>=2. Denominators (A020653) are obtained by reversing each row.
Also triangle in which n-th row gives phi(n) numbers between 1 and n that are relatively prime to n.
A038610(n) = least common multiple of n-th row. - Reinhard Zumkeller, Sep 21 2013
Row n has sum A023896(n). - Jamie Morken, Dec 17 2019
This irregular triangle gives in row n the smallest positive reduced residue system modulo n, for n >= 1. If one takes 0 for n = 1 it becomes the smallest nonnegative residue system modulo n. - Wolfdieter Lang, Feb 29 2020

Examples

			The beginning of the list of positive rationals <= 1: 1/1, 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, .... This is A038566/A038567.
The beginning of the triangle giving all positive rationals: 1/1; 1/2, 2/1; 1/3, 3/1; 1/4, 2/3, 3/2, 4/1; 1/5, 5/1; 1/6, 2/5, 3/4, 4/3, 5/2, 6/1; .... This is A020652/A020653, with A020652(n) = A038566(n+1). [Corrected by _M. F. Hasler_, Mar 06 2020]
The beginning of the triangle in which n-th row gives numbers between 1 and n that are relatively prime to n:
n\k 1 2 3  4  5  6  7  8 9 10 11 12 13 14 15 16 17 18
1:  1
2:  1
3:  1 2
4:  1 3
5:  1 2 3  4
6:  1 5
7:  1 2 3  4  5  6
8:  1 3 5  7
9:  1 2 4  5  7  8
10: 1 3 7  9
11: 1 2 3  4  5  6  7  8 9 10
12: 1 5 7 11
13: 1 2 3  4  5  6  7  8 9 10 11 12
14: 1 3 5  9 11 13
15: 1 2 4  7  8 11 13 14
16: 1 3 5  7  9 11 13 15
17: 1 2 3  4  5  6  7  8 9 10 11 12 13 14 15 16
18: 1 5 7 11 13 17
19: 1 2 3  4  5  6  7  8 9 10 11 12 13 14 15 16 17 18
20: 1 3 7  9 11 13 17 19
... Reformatted. - _Wolfdieter Lang_, Jan 18 2017
------------------------------------------------------
		

References

  • Richard Courant and Herbert Robbins. What Is Mathematics?, Oxford, 1941, pp. 79-80.
  • H. Lauwerier, Fractals, Princeton Univ. Press, p. 23.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 163.

Crossrefs

A054424 gives mapping to Stern-Brocot tree.
Row sums give rationals A111992(n)/A069220(n), n>=1.
A112484 (primes, rows n >=3).

Programs

  • Haskell
    a038566 n k = a038566_tabf !! (n-1) !! (k-1)
    a038566_row n = a038566_tabf !! (n-1)
    a038566_tabf=
       zipWith (\v ws -> filter ((== 1) . (gcd v)) ws) [1..] a002260_tabl
    a038566_list = concat a038566_tabf
    -- Reinhard Zumkeller, Sep 21 2013, Feb 23 2012
    
  • Maple
    s := proc(n) local i,j,k,ans; i := 0; ans := [ ]; for j while i
    				
  • Mathematica
    Flatten[Table[Flatten[Position[GCD[Table[Mod[j, w], {j, 1, w-1}], w], 1]], {w, 1, 100}], 2]
    row[n_]:=Select[Range[n],GCD[n,#]==1 &]; Array[row,17]//Flatten (* Stefano Spezia, Jul 20 2025 *)
  • PARI
    first(n)=my(v=List(),i,j);while(iCharles R Greathouse IV, Feb 07 2013
    
  • PARI
    row(n) = select(x->gcd(n, x)==1, [1..n]); \\ Michel Marcus, May 05 2020
    
  • SageMath
    def aRow(n):
        if n == 1: return 1
        return [k for k in ZZ(n).coprime_integers(n+1)]
    print(flatten([aRow(n) for n in range(1, 18)])) # Peter Luschny, Aug 17 2020

Formula

The n-th "clump" consists of the phi(n) integers <= n and prime to n.
a(n) = A002260(A169581(n)). - Reinhard Zumkeller, Dec 02 2009
a(n+1) = A020652(n) for n > 1. - Georg Fischer, Oct 27 2020

Extensions

More terms from Erich Friedman
Offset corrected by Max Alekseyev, Apr 26 2010

A018805 Number of elements in the set {(x,y): 1 <= x,y <= n, gcd(x,y)=1}.

Original entry on oeis.org

1, 3, 7, 11, 19, 23, 35, 43, 55, 63, 83, 91, 115, 127, 143, 159, 191, 203, 239, 255, 279, 299, 343, 359, 399, 423, 459, 483, 539, 555, 615, 647, 687, 719, 767, 791, 863, 899, 947, 979, 1059, 1083, 1167, 1207, 1255, 1299, 1391, 1423, 1507, 1547, 1611, 1659, 1763
Offset: 1

Views

Author

Keywords

Comments

Number of positive rational numbers of height at most n, where the height of p/q is max(p, q) when p and q are relatively prime positive integers. - Charles R Greathouse IV, Jul 05 2012
The number of ordered pairs (i,j) with 1<=i<=n, 1<=j<=n, gcd(i,j)=d is a(floor(n/d)). - N. J. A. Sloane, Jul 29 2012
Equals partial sums of A140434 (1, 2, 4, 4, 8, 4, 12, 8, ...) and row sums of triangle A143469. - Gary W. Adamson, Aug 17 2008
Number of distinct solutions to k*x+h=0, where 1 <= k,h <= n. - Giovanni Resta, Jan 08 2013
a(n) is the number of rational numbers which can be constructed from the set of integers between 1 and n, without combination of multiplication and division. a(3) = 7 because {1, 2, 3} can only create {1/3, 1/2, 2/3, 1, 3/2, 2, 3}. - Bernard Schott, Jul 07 2019

References

  • S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 110-112.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954. See Theorem 332.

Crossrefs

Cf. A177853 (partial sums).
The main diagonal of A331781, also of A333295.

Programs

  • Haskell
    a018805 n = length [()| x <- [1..n], y <- [1..n], gcd x y == 1]
    -- Reinhard Zumkeller, Jan 21 2013
    
  • Magma
    /* based on the first formula */ A018805:=func< n | 2*&+[ EulerPhi(k): k in [1..n] ]-1 >; [ A018805(n): n in [1..60] ]; // Klaus Brockhaus, Jan 27 2011
    
  • Magma
    /* based on the second formula */ A018805:=func< n | n eq 1 select 1 else n^2-&+[ $$(n div j): j in [2..n] ] >; [ A018805(n): n in [1..60] ]; // Klaus Brockhaus, Feb 07 2011
    
  • Maple
    N:= 1000; # to get the first N entries
    P:= Array(1..N,numtheory:-phi);
    A:= map(t -> 2*round(t)-1, Statistics:-CumulativeSum(P));
    convert(A,list); # Robert Israel, Jul 16 2014
  • Mathematica
    FoldList[ Plus, 1, 2 Array[ EulerPhi, 60, 2 ] ] (* Olivier Gérard, Aug 15 1997 *)
    Accumulate[2*EulerPhi[Range[60]]]-1 (* Harvey P. Dale, Oct 21 2013 *)
  • PARI
    a(n)=sum(k=1,n,moebius(k)*(n\k)^2)
    
  • PARI
    A018805(n)=2 *sum(j=1, n, eulerphi(j)) - 1;
    for(n=1, 99, print1(A018805(n), ", ")); /* show terms */
    
  • PARI
    a(n)=my(s); forsquarefree(k=1,n, s+=moebius(k)*(n\k[1])^2); s \\ Charles R Greathouse IV, Jan 08 2018
    
  • Python
    from sympy import sieve
    def A018805(n): return 2*sum(t for t in sieve.totientrange(1,n+1)) - 1 # Chai Wah Wu, Mar 23 2021
    
  • Python
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def A018805(n): # based on second formula
        if n == 0:
            return 0
        c, j = 1, 2
        k1 = n//j
        while k1 > 1:
            j2 = n//k1 + 1
            c += (j2-j)*A018805(k1)
            j, k1 = j2, n//j2
        return n*(n-1)-c+j # Chai Wah Wu, Mar 24 2021

Formula

a(n) = 2*(Sum_{j=1..n} phi(j)) - 1.
a(n) = n^2 - Sum_{j=2..n} a(floor(n/j)).
a(n) = 2*A015614(n) + 1. - Reinhard Zumkeller, Apr 08 2006
a(n) = 2*A002088(n) - 1. - Hugo van der Sanden, Nov 22 2008
a(n) ~ (1/zeta(2)) * n^2 = (6/Pi^2) * n^2 as n goes to infinity (zeta is the Riemann zeta function, A013661, and the constant 6/Pi^2 is 0.607927..., A059956). - Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 18 2001
a(n) ~ 6*n^2/Pi^2 + O(n*log n). - N. J. A. Sloane, May 31 2020
a(n) = Sum_{k=1..n} mu(k)*floor(n/k)^2. - Benoit Cloitre, May 11 2003
a(n) = A000290(n) - A100613(n) = A015614(n) + A002088(n). - Reinhard Zumkeller, Jan 21 2013
a(n) = A242114(floor(n/k),1), 1<=k<=n; particularly a(n) = A242114(n,1). - Reinhard Zumkeller, May 04 2014
a(n) = 2 * A005728(n) - 3. - David H Post, Dec 20 2016
a(n) ~ 6*n^2/Pi^2, cf. A059956. [Hardy and Wright] - M. F. Hasler, Jan 20 2017
G.f.: (1/(1 - x)) * (-x + 2 * Sum_{k>=1} mu(k) * x^k / (1 - x^k)^2). - Ilya Gutkovskiy, Feb 14 2020

Extensions

More terms from Reinhard Zumkeller, Apr 08 2006
Link to Moree's paper corrected by Peter Luschny, Aug 08 2009
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