cp's OEIS Frontend

sigma(sigma(A046528(2,12,13,18,21,22,56,57,175,176,177,..))) is Mersenne prime.

The C program, from first link, is based on a new method, see second link. It was developed from a congruence found in the first reference "Concrete Mathematics". The function D() of this program implements the simple division algorithm found in "D. E. Knuth, The Art of Computer Programming, V.2." (second reference). Another approach can be to use Dresden's formula that can be found from the third link. One can use the function LastDigit() of the mentioned program to find the last nonzero digit of N! for very large values of N. The factorial of the 47th (known) Mersenne prime has approximately 10^12,978,195 digits.

Many other algorithms for the general problem of finding the last nonzero digit of a factorial are given in A008904. [D. S. McNeil, Dec 10 2010]

a(n) is not so far from p

All certified primes using PFGW64 from primeform group

a(34) = 1200007.

A000043(34) = 756839.

A000668(34) = 2^756839-1.

1200007*(2^756839-1)*(1200007*(2^756839-1)+1)-1 is a 455675 digits certified prime using PFGW.

The numbers k*(2^(2*A000043(n))-1)+1 may be written as k*(2^A000043(n)-1)*(2^A000043(n)+1)+1 or k*Mersenne(n)*(Mersenne(n)+2)+1 and so may be proved primes.

All the numbers a(n)*(2^(2*A000043(n))-1)+1 for n=1 to 32 have been proved to be primes.

See A065091 for comments, formulas etc. concerning only odd primes. For all information concerning prime powers, see A000961. For contributions concerning "almost primes" see A002808.

A number p is prime if (and only if) it is greater than 1 and has no positive divisors except 1 and p.

A natural number is prime if and only if it has exactly two (positive) divisors.

A prime has exactly one proper positive divisor, 1.

The paper by Kaoru Motose starts as follows: "Let q be a prime divisor of a Mersenne number 2^p-1 where p is prime. Then p is the order of 2 (mod q). Thus p is a divisor of q - 1 and q > p. This shows that there exist infinitely many prime numbers." - Pieter Moree, Oct 14 2004

1 is not a prime, for if the primes included 1, then the factorization of a natural number n into a product of primes would not be unique, since n = n*1.

Prime(n) and pi(n) are inverse functions: A000720(a(n)) = n and a(n) is the least number m such that a(A000720(m)) = a(n). a(A000720(n)) = n if (and only if) n is prime.

Second sequence ever computed by electronic computer, on EDSAC, May 09 1949 (see Renwick link). - Russ Cox, Apr 20 2006

Every prime p > 3 is a linear combination of previous primes prime(n) with nonzero coefficients c(n) and |c(n)| < prime(n). - Amarnath Murthy, Franklin T. Adams-Watters and Joshua Zucker, May 17 2006; clarified by Chayim Lowen, Jul 17 2015

The Greek transliteration of 'Prime Number' is 'Protos Arithmos'. - Daniel Forgues, May 08 2009 [Edited by Petros Hadjicostas, Nov 18 2019]

A number n is prime if and only if it is different from zero and different from a unit and each multiple of n decomposes into factors such that n divides at least one of the factors. This applies equally to the integers (where a prime has exactly four divisors (the definition of divisors is relaxed such that they can be negative)) and the positive integers (where a prime has exactly two distinct divisors). - Peter Luschny, Oct 09 2012

Motivated by his conjecture on representations of integers by alternating sums of consecutive primes, for any positive integer n, Zhi-Wei Sun conjectured that the polynomial P_n(x) = Sum_{k=0..n} a(k+1)*x^k is irreducible over the field of rational numbers with the Galois group S_n, and moreover P_n(x) is irreducible mod a(m) for some m <= n(n+1)/2. It seems that no known criterion on irreducibility of polynomials implies this conjecture. - Zhi-Wei Sun, Mar 23 2013

Questions on a(2n) and Ramanujan primes are in A233739. - Jonathan Sondow, Dec 16 2013

From Hieronymus Fischer, Apr 02 2014: (Start)

Natural numbers such that there is exactly one base b such that the base-b alternate digital sum is 0 (see A239707).

Equivalently: Numbers p > 1 such that b = p-1 is the only base >= 1 for which the base-b alternate digital sum is 0.

Equivalently: Numbers p > 1 such that the base-b alternate digital sum is <> 0 for all bases 1 <= b < p-1. (End)

An integer n > 1 is a prime if and only if it is not the sum of positive integers in arithmetic progression with common difference 2. - Jean-Christophe Hervé, Jun 01 2014

Conjecture: Numbers having prime factors <= prime(n+1) are {k|k^f(n) mod primorial(n)=1}, where f(n) = lcm(prime(i)-1, i=1..n) = A058254(n) and primorial(n) = A002110(n). For example, numbers with no prime divisor <= prime(7) = 17 are {k|k^60 mod 30030=1}. - Gary Detlefs, Jun 07 2014

Cramer conjecture prime(n+1) - prime(n) < C log^2 prime(n) is equivalent to the inequality (log prime(n+1)/log prime(n))^n < e^C, as n tend to infinity, where C is an absolute constant. - Thomas Ordowski, Oct 06 2014

I conjecture that for any positive rational number r there are finitely many primes q_1,...,q_k such that r = Sum_{j=1..k} 1/(q_j-1). For example, 2 = 1/(2-1) + 1/(3-1) + 1/(5-1) + 1/(7-1) + 1/(13-1) with 2, 3, 5, 7 and 13 all prime, 1/7 = 1/(13-1) + 1/(29-1) + 1/(43-1) with 13, 29 and 43 all prime, and 5/7 = 1/(3-1) + 1/(7-1) + 1/(31-1) + 1/(71-1) with 3, 7, 31 and 71 all prime. - Zhi-Wei Sun, Sep 09 2015

I also conjecture that for any positive rational number r there are finitely many primes p_1,...,p_k such that r = Sum_{j=1..k} 1/(p_j+1). For example, 1 = 1/(2+1) + 1/(3+1) + 1/(5+1) + 1/(7+1) + 1/(11+1) + 1/(23+1) with 2, 3, 5, 7, 11 and 23 all prime, and 10/11 = 1/(2+1) + 1/(3+1) + 1/(5+1) + 1/(7+1) + 1/(43+1) + 1/(131+1) + 1/(263+1) with 2, 3, 5, 7, 43, 131 and 263 all prime. - Zhi-Wei Sun, Sep 13 2015

Numbers k such that ((k-2)!!)^2 == +-1 (mod k). - Thomas Ordowski, Aug 27 2016

Does not satisfy Benford's law [Diaconis, 1977; Cohen-Katz, 1984; Berger-Hill, 2017]. - N. J. A. Sloane, Feb 07 2017

Prime numbers are the integer roots of 1 - sin(Pi*Gamma(s)/s)/sin(Pi/s). - Peter Luschny, Feb 23 2018

Conjecture: log log a(n+1) - log log a(n) < 1/n. - Thomas Ordowski, Feb 17 2023

D. E. Knuth writes: "Before Fibonacci wrote his work, the sequence F_{n} had already been discussed by Indian scholars, who had long been interested in rhythmic patterns that are formed from one-beat and two-beat notes. The number of such rhythms having n beats altogether is F_{n+1}; therefore both Gopāla (before 1135) and Hemachandra (c. 1150) mentioned the numbers 1, 2, 3, 5, 8, 13, 21, ... explicitly." (TAOCP Vol. 1, 2nd ed.) - Peter Luschny, Jan 11 2015

In keeping with historical accounts (see the references by P. Singh and S. Kak), the generalized Fibonacci sequence a, b, a + b, a + 2b, 2a + 3b, 3a + 5b, ... can also be described as the Gopala-Hemachandra numbers H(n) = H(n-1) + H(n-2), with F(n) = H(n) for a = b = 1, and Lucas sequence L(n) = H(n) for a = 2, b = 1. - Lekraj Beedassy, Jan 11 2015

Susantha Goonatilake writes: "[T]his sequence was well known in South Asia and used in the metrical sciences. Its development is attributed in part to Pingala (200 BC), later being associated with Virahanka (circa 700 AD), Gopala (circa 1135), and Hemachandra (circa 1150)—all of whom lived and worked prior to Fibonacci." (Toward a Global Science: Mining Civilizational Knowledge, p. 126) - Russ Cox, Sep 08 2021

Also sometimes called Hemachandra numbers.

Also sometimes called Lamé's sequence.

For a photograph of "Fibonacci"'s 1202 book, see the Leonardo of Pisa link below.

F(n+2) = number of binary sequences of length n that have no consecutive 0's.

F(n+2) = number of subsets of {1,2,...,n} that contain no consecutive integers.

F(n+1) = number of tilings of a 2 X n rectangle by 2 X 1 dominoes.

F(n+1) = number of matchings (i.e., Hosoya index) in a path graph on n vertices: F(5)=5 because the matchings of the path graph on the vertices A, B, C, D are the empty set, {AB}, {BC}, {CD} and {AB, CD}. - Emeric Deutsch, Jun 18 2001

F(n) = number of compositions of n+1 with no part equal to 1. [Cayley, Grimaldi]

Positive terms are the solutions to z = 2*x*y^4 + (x^2)*y^3 - 2*(x^3)*y^2 - y^5 - (x^4)*y + 2*y for x,y >= 0 (Ribenboim, page 193). When x=F(n), y=F(n + 1) and z > 0 then z=F(n + 1).

For Fibonacci search see Knuth, Vol. 3; Horowitz and Sahni; etc.

F(n) is the diagonal sum of the entries in Pascal's triangle at 45 degrees slope. - Amarnath Murthy, Dec 29 2001 (i.e., row sums of A030528, R. J. Mathar, Oct 28 2021)

F(n+1) is the number of perfect matchings in ladder graph L_n = P_2 X P_n. - Sharon Sela (sharonsela(AT)hotmail.com), May 19 2002

F(n+1) = number of (3412,132)-, (3412,213)- and (3412,321)-avoiding involutions in S_n.

This is also the Horadam sequence (0,1,1,1). - Ross La Haye, Aug 18 2003

An INVERT transform of A019590. INVERT([1,1,2,3,5,8,...]) gives A000129. INVERT([1,2,3,5,8,13,21,...]) gives A028859. - Antti Karttunen, Dec 12 2003

Number of meaningful differential operations of the k-th order on the space R^3. - Branko Malesevic, Mar 02 2004

F(n) = number of compositions of n-1 with no part greater than 2. Example: F(4) = 3 because we have 3 = 1+1+1 = 1+2 = 2+1.

F(n) = number of compositions of n into odd parts; e.g., F(6) counts 1+1+1+1+1+1, 1+1+1+3, 1+1+3+1, 1+3+1+1, 1+5, 3+1+1+1, 3+3, 5+1. - Clark Kimberling, Jun 22 2004

F(n) = number of binary words of length n beginning with 0 and having all runlengths odd; e.g., F(6) counts 010101, 010111, 010001, 011101, 011111, 000101, 000111, 000001. - Clark Kimberling, Jun 22 2004

The number of sequences (s(0),s(1),...,s(n)) such that 0 < s(i) < 5, |s(i)-s(i-1)|=1 and s(0)=1 is F(n+1); e.g., F(5+1) = 8 corresponds to 121212, 121232, 121234, 123212, 123232, 123234, 123432, 123434. - Clark Kimberling, Jun 22 2004 [corrected by Neven Juric, Jan 09 2009]

Likewise F(6+1) = 13 corresponds to these thirteen sequences with seven numbers: 1212121, 1212123, 1212321, 1212323, 1212343, 1232121, 1232123, 1232321, 1232323, 1232343, 1234321, 1234323, 1234343. - Neven Juric, Jan 09 2008

A relationship between F(n) and the Mandelbrot set is discussed in the link "Le nombre d'or dans l'ensemble de Mandelbrot" (in French). - Gerald McGarvey, Sep 19 2004

For n > 0, the continued fraction for F(2n-1)*phi = [F(2n); L(2n-1), L(2n-1), L(2n-1), ...] and the continued fraction for F(2n)*phi = [F(2n+1)-1; 1, L(2n)-2, 1, L(2n)-2, ...]. Also true: F(2n)*phi = [F(2n+1); -L(2n), L(2n), -L(2n), L(2n), ...] where L(i) is the i-th Lucas number (A000204). - Clark Kimberling, Nov 28 2004 [corrected by Hieronymus Fischer, Oct 20 2010]

For any nonzero number k, the continued fraction [4,4,...,4,k], which is n 4's and a single k, equals (F(3n) + k*F(3n+3))/(F(3n-3) + k*F(3n)). - Greg Dresden, Aug 07 2019

F(n+1) (for n >= 1) = number of permutations p of 1,2,3,...,n such that |k-p(k)| <= 1 for k=1,2,...,n. (For <= 2 and <= 3, see A002524 and A002526.) - Clark Kimberling, Nov 28 2004

The ratios F(n+1)/F(n) for n > 0 are the convergents to the simple continued fraction expansion of the golden section. - Jonathan Sondow, Dec 19 2004

Lengths of successive words (starting with a) under the substitution: {a -> ab, b -> a}. - Jeroen F.J. Laros, Jan 22 2005

The Fibonacci sequence, like any additive sequence, naturally tends to be geometric with common ratio not a rational power of 10; consequently, for a sufficiently large number of terms, Benford's law of first significant digit (i.e., first digit 1 <= d <= 9 occurring with probability log_10(d+1) - log_10(d)) holds. - Lekraj Beedassy, Apr 29 2005 (See Brown-Duncan, 1970. - N. J. A. Sloane, Feb 12 2017)

F(n+2) = Sum_{k=0..n} binomial(floor((n+k)/2),k), row sums of A046854. - Paul Barry, Mar 11 2003

Number of order ideals of the "zig-zag" poset. See vol. 1, ch. 3, prob. 23 of Stanley. - Mitch Harris, Dec 27 2005

F(n+1)/F(n) is also the Farey fraction sequence (see A097545 for explanation) for the golden ratio, which is the only number whose Farey fractions and continued fractions are the same. - Joshua Zucker, May 08 2006

a(n+2) is the number of paths through 2 plates of glass with n reflections (reflections occurring at plate/plate or plate/air interfaces). Cf. A006356-A006359. - Mitch Harris, Jul 06 2006

F(n+1) equals the number of downsets (i.e., decreasing subsets) of an n-element fence, i.e., an ordered set of height 1 on {1,2,...,n} with 1 > 2 < 3 > 4 < ... n and no other comparabilities. Alternatively, F(n+1) equals the number of subsets A of {1,2,...,n} with the property that, if an odd k is in A, then the adjacent elements of {1,2,...,n} belong to A, i.e., both k - 1 and k + 1 are in A (provided they are in {1,2,...,n}). - Brian Davey, Aug 25 2006

Number of Kekulé structures in polyphenanthrenes. See the paper by Lukovits and Janezic for details. - Parthasarathy Nambi, Aug 22 2006

Inverse: With phi = (sqrt(5) + 1)/2, round(log_phi(sqrt((sqrt(5) a(n) + sqrt(5 a(n)^2 - 4))(sqrt(5) a(n) + sqrt(5 a(n)^2 + 4)))/2)) = n for n >= 3, obtained by rounding the arithmetic mean of the inverses given in A001519 and A001906. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 19 2007

A result of Jacobi from 1848 states that every symmetric matrix over a p.i.d. is congruent to a triple-diagonal matrix. Consider the maximal number T(n) of summands in the determinant of an n X n triple-diagonal matrix. This is the same as the number of summands in such a determinant in which the main-, sub- and superdiagonal elements are all nonzero. By expanding on the first row we see that the sequence of T(n)'s is the Fibonacci sequence without the initial stammer on the 1's. - Larry Gerstein (gerstein(AT)math.ucsb.edu), Mar 30 2007

Suppose psi=log(phi). We get the representation F(n)=(2/sqrt(5))*sinh(n*psi) if n is even; F(n)=(2/sqrt(5))*cosh(n*psi) if n is odd. There is a similar representation for Lucas numbers (A000032). Many Fibonacci formulas now easily follow from appropriate sinh and cosh formulas. For example: the de Moivre theorem (cosh(x)+sinh(x))^m = cosh(mx)+sinh(mx) produces L(n)^2 + 5F(n)^2 = 2L(2n) and L(n)F(n) = F(2n) (setting x=n*psi and m=2). - Hieronymus Fischer, Apr 18 2007

Inverse: floor(log_phi(sqrt(5)*F(n)) + 1/2) = n, for n > 1. Also for n > 0, floor((1/2)*log_phi(5*F(n)*F(n+1))) = n. Extension valid for integer n, except n=0,-1: floor((1/2)*sign(F(n)*F(n+1))*log_phi|5*F(n)*F(n+1)|) = n (where sign(x) = sign of x). - Hieronymus Fischer, May 02 2007

F(n+2) = the number of Khalimsky-continuous functions with a two-point codomain. - Shiva Samieinia (shiva(AT)math.su.se), Oct 04 2007

This is a_1(n) in the Doroslovacki reference.

Let phi = A001622 then phi^n = (1/phi)*a(n) + a(n+1). - Gary W. Adamson, Dec 15 2007

The sequence of first differences, F(n+1)-F(n), is essentially the same sequence: 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ... - Colm Mulcahy, Mar 03 2008

Equals row sums of triangle A144152. - Gary W. Adamson, Sep 12 2008

Except for the initial term, the numerator of the convergents to the recursion x = 1/(x+1). - Cino Hilliard, Sep 15 2008

F(n) is the number of possible binary sequences of length n that obey the sequential construction rule: if last symbol is 0, add the complement (1); else add 0 or 1. Here 0,1 are metasymbols for any 2-valued symbol set. This rule has obvious similarities to JFJ Laros's rule, but is based on addition rather than substitution and creates a tree rather than a single sequence. - Ross Drewe, Oct 05 2008

F(n) = Product_{k=1..(n-1)/2} (1 + 4*cos^2 k*Pi/n), where terms = roots to the Fibonacci product polynomials, A152063. - Gary W. Adamson, Nov 22 2008

Fp == 5^((p-1)/2) mod p, p = prime [Schroeder, p. 90]. - Gary W. Adamson & Alexander R. Povolotsky, Feb 21 2009

A000032(n)^2 - 5*F(n)^2 = 4*(-1)^n. - Gary W. Adamson, Mar 11 2009

Output of Kasteleyn's formula for the number of perfect matchings of an m X n grid specializes to the Fibonacci sequence for m=2. - Sarah-Marie Belcastro, Jul 04 2009

(F(n),F(n+4)) satisfies the Diophantine equation: X^2 + Y^2 - 7XY = 9*(-1)^n. - Mohamed Bouhamida, Sep 06 2009

(F(n),F(n+2)) satisfies the Diophantine equation: X^2 + Y^2 - 3XY = (-1)^n. - Mohamed Bouhamida, Sep 08 2009

a(n+2) = A083662(A131577(n)). - Reinhard Zumkeller, Sep 26 2009

Difference between number of closed walks of length n+1 from a node on a pentagon and number of walks of length n+1 between two adjacent nodes on a pentagon. - Henry Bottomley, Feb 10 2010

F(n+1) = number of Motzkin paths of length n having exactly one weak ascent. A Motzkin path of length n is a lattice path from (0,0) to (n,0) consisting of U=(1,1), D=(1,-1) and H=(1,0) steps and never going below the x-axis. A weak ascent in a Motzkin path is a maximal sequence of consecutive U and H steps. Example: a(5)=5 because we have (HHHH), (HHU)D, (HUH)D, (UHH)D, and (UU)DD (the unique weak ascent is shown between parentheses; see A114690). - Emeric Deutsch, Mar 11 2010

(F(n-1) + F(n+1))^2 - 5*F(n-2)*F(n+2) = 9*(-1)^n. - Mohamed Bouhamida, Mar 31 2010

From the Pinter and Ziegler reference's abstract: authors "show that essentially the Fibonacci sequence is the unique binary recurrence which contains infinitely many three-term arithmetic progressions. A criterion for general linear recurrences having infinitely many three-term arithmetic progressions is also given." - Jonathan Vos Post, May 22 2010

F(n+1) = number of paths of length n starting at initial node on the path graph P_4. - Johannes W. Meijer, May 27 2010

F(k) = number of cyclotomic polynomials in denominator of generating function for number of ways to place k nonattacking queens on an n X n board. - Vaclav Kotesovec, Jun 07 2010

As n->oo, (a(n)/a(n-1) - a(n-1)/a(n)) tends to 1.0. Example: a(12)/a(11) - a(11)/a(12) = 144/89 - 89/144 = 0.99992197.... - Gary W. Adamson, Jul 16 2010

From Hieronymus Fischer, Oct 20 2010: (Start)

Fibonacci numbers are those numbers m such that m*phi is closer to an integer than k*phi for all k, 1 <= k < m. More formally: a(0)=0, a(1)=1, a(2)=1, a(n+1) = minimal m > a(n) such that m*phi is closer to an integer than a(n)*phi.

For all numbers 1 <= k < F(n), the inequality |k*phi-round(k*phi)| > |F(n)*phi-round(F(n)*phi)| holds.

F(n)*phi - round(F(n)*phi) = -((-phi)^(-n)), for n > 1.

Fract(1/2 + F(n)*phi) = 1/2 -(-phi)^(-n), for n > 1.

Fract(F(n)*phi) = (1/2)*(1 + (-1)^n) - (-phi)^(-n), n > 1.

Inverse: n = -log_phi |1/2 - fract(1/2 + F(n)*phi)|.

(End)

F(A001177(n)*k) mod n = 0, for any integer k. - Gary Detlefs, Nov 27 2010

F(n+k)^2 - F(n)^2 = F(k)*F(2n+k), for even k. - Gary Detlefs, Dec 04 2010

F(n+k)^2 + F(n)^2 = F(k)*F(2n+k), for odd k. - Gary Detlefs, Dec 04 2010

F(n) = round(phi*F(n-1)) for n > 1. - Joseph P. Shoulak, Jan 13 2012

For n > 0: a(n) = length of n-th row in Wythoff array A003603. - Reinhard Zumkeller, Jan 26 2012

From Bridget Tenner, Feb 22 2012: (Start)

The number of free permutations of [n].

The number of permutations of [n] for which s_k in supp(w) implies s_{k+-1} not in supp(w).

The number of permutations of [n] in which every decomposition into length(w) reflections is actually composed of simple reflections. (End)

The sequence F(n+1)^(1/n) is increasing. The sequence F(n+2)^(1/n) is decreasing. - Thomas Ordowski, Apr 19 2012

Two conjectures: For n > 1, F(n+2)^2 mod F(n+1)^2 = F(n)*F(n+1) - (-1)^n. For n > 0, (F(2n) + F(2n+2))^2 = F(4n+3) + Sum_{k = 2..2n} F(2k). - Alex Ratushnyak, May 06 2012

From Ravi Kumar Davala, Jan 30 2014: (Start)

Proof of Ratushnyak's first conjecture: For n > 1, F(n+2)^2 - F(n)*F(n+1) + (-1)^n = 2*F(n+1)^2.

Consider: F(n+2)^2 - F(n)*F(n+1) - 2*F(n+1)^2

= F(n+2)^2 - F(n+1)^2 - F(n+1)^2 - F(n)*F(n+1)

= (F(n+2) + F(n+1))*(F(n+2) - F(n+1)) - F(n+1)*(F(n+1) + F(n))

= F(n+3)*F(n) - F(n+1)*F(n+2) = -(-1)^n.

Proof of second conjecture: L(n) stands for Lucas number sequence from A000032.

Consider the fact that

L(2n+1)^2 = L(4n+2) - 2

(F(2n) + F(2n+2))^2 = F(4n+1) + F(4n+3) - 2

(F(2n) + F(2n+2))^2 = (Sum_{k = 2..2n} F(2k)) + F(4n+3).

(End)

The relationship: INVERT transform of (1,1,0,0,0,...) = (1, 2, 3, 5, 8, ...), while the INVERT transform of (1,0,1,0,1,0,1,...) = (1, 1, 2, 3, 5, 8, ...) is equivalent to: The numbers of compositions using parts 1 and 2 is equivalent to the numbers of compositions using parts == 1 (mod 2) (i.e., the odd integers). Generally, the numbers of compositions using parts 1 and k is equivalent to the numbers of compositions of (n+1) using parts 1 mod k. Cf. A000930 for k = 3 and A003269 for k = 4. Example: for k = 2, n = 4 we have the compositions (22; 211, 121; 112; 1111) = 5; but using parts 1 and 3 we have for n = 5: (311, 131, 113, 11111, 5) = 5. - Gary W. Adamson, Jul 05 2012

The sequence F(n) is the binomial transformation of the alternating sequence (-1)^(n-1)*F(n), whereas the sequence F(n+1) is the binomial transformation of the alternating sequence (-1)^n*F(n-1). Both of these facts follow easily from the equalities a(n;1)=F(n+1) and b(n;1)=F(n) where a(n;d) and b(n;d) are so-called "delta-Fibonacci" numbers as defined in comments to A014445 (see also the papers of Witula et al.). - Roman Witula, Jul 24 2012

F(n) is the number of different (n-1)-digit binary numbers such that all substrings of length > 1 have at least one digit equal to 1. Example: for n = 5 there are 8 binary numbers with n - 1 = 4 digits (1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111), only the F(n) = 5 numbers 1010, 1011, 1101, 1110 and 1111 have the desired property. - Hieronymus Fischer, Nov 30 2012

For positive n, F(n+1) equals the determinant of the n X n tridiagonal matrix with 1's along the main diagonal, i's along the superdiagonal and along the subdiagonal where i = sqrt(-1). Example: Det([1,i,0,0; i,1,i,0; 0,i,1,i; 0,0,i,1]) = F(4+1) = 5. - Philippe Deléham, Feb 24 2013

For n >= 1, number of compositions of n where there is a drop between every second pair of parts, starting with the first and second part; see example. Also, a(n+1) is the number of compositions where there is a drop between every second pair of parts, starting with the second and third part; see example. - Joerg Arndt, May 21 2013 [see the Hopkins/Tangboonduangjit reference for a proof, see also the Checa reference for alternative proofs and statistics]

Central terms of triangles in A162741 and A208245, n > 0. - Reinhard Zumkeller, Jul 28 2013

For n >= 4, F(n-1) is the number of simple permutations in the geometric grid class given in A226433. - Jay Pantone, Sep 08 2013

a(n) are the pentagon (not pentagonal) numbers because the algebraic degree 2 number rho(5) = 2*cos(Pi/5) = phi (golden section), the length ratio diagonal/side in a pentagon, has minimal polynomial C(5,x) = x^2 - x - 1 (see A187360, n=5), hence rho(5)^n = a(n-1)*1 + a(n)*rho(5), n >= 0, in the power basis of the algebraic number field Q(rho(5)). One needs a(-1) = 1 here. See also the P. Steinbach reference under A049310. - Wolfdieter Lang, Oct 01 2013

A010056(a(n)) = 1. - Reinhard Zumkeller, Oct 10 2013

Define F(-n) to be F(n) for n odd and -F(n) for n even. Then for all n and k, F(n+2k)^2 - F(n)^2 = F(n+k)*( F(n+3k) - F(n-k) ). - Charlie Marion, Dec 20 2013

( F(n), F(n+2k) ) satisfies the Diophantine equation: X^2 + Y^2 - L(2k)*X*Y = F(4k)^2*(-1)^n. This generalizes Bouhamida's comments dated Sep 06 2009 and Sep 08 2009. - Charlie Marion, Jan 07 2014

For any prime p there is an infinite periodic subsequence within F(n) divisible by p, that begins at index n = 0 with value 0, and its first nonzero term at n = A001602(i), and period k = A001602(i). Also see A236479. - Richard R. Forberg, Jan 26 2014

Range of row n of the circular Pascal array of order 5. - Shaun V. Ault, May 30 2014 [orig. Kicey-Klimko 2011, and observations by Glen Whitehead; more general work found in Ault-Kicey 2014]

Nonnegative range of the quintic polynomial 2*y - y^5 + 2*x*y^4 + x^2*y^3 - 2*x^3*y^2 - x^4*y with x, y >= 0, see Jones 1975. - Charles R Greathouse IV, Jun 01 2014

The expression round(1/(F(k+1)/F(n) + F(k)/F(n+1))), for n > 0, yields a Fibonacci sequence with k-1 leading zeros (with rounding 0.5 to 0). - Richard R. Forberg, Aug 04 2014

Conjecture: For n > 0, F(n) is the number of all admissible residue classes for which specific finite subsequences of the Collatz 3n + 1 function consists of n+2 terms. This has been verified for 0 < n < 51. For details see Links. - Mike Winkler, Oct 03 2014

a(4)=3 and a(6)=8 are the only Fibonacci numbers that are of the form prime+1. - Emmanuel Vantieghem, Oct 02 2014

a(1)=1=a(2), a(3)=2 are the only Fibonacci numbers that are of the form prime-1. - Emmanuel Vantieghem, Jun 07 2015

Any consecutive pair (m, k) of the Fibonacci sequence a(n) illustrates a fair equivalence between m miles and k kilometers. For instance, 8 miles ~ 13 km; 13 miles ~ 21 km. - Lekraj Beedassy, Oct 06 2014

a(n+1) counts closed walks on K_2, containing one loop on the other vertex. Equivalently the (1,1)entry of A^(n+1) where the adjacency matrix of digraph is A=(0,1; 1,1). - _David Neil McGrath, Oct 29 2014

a(n-1) counts closed walks on the graph G(1-vertex;l-loop,2-loop). - David Neil McGrath, Nov 26 2014

From Tom Copeland, Nov 02 2014: (Start)

Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1-x) = -P[-x], and C(x) = [1-sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers A000108, with inverse Cinv(x) = x * (1-x).

Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers, A000957 with inverse Fin^(-1)(x) = Cinv[Pinv(x)] = Cinv[-P(-x)].

Mot(x) = C[P(x)] = C[-Pinv(-x)] gives an o.g.f. for shifted A005043, the Motzkin or Riordan numbers with comp. inverse Mot^(-1)(x) = Pinv[Cinv(x)] = (x - x^2) / (1 - x + x^2) (cf. A057078).

BTC(x) = C[Pinv(x)] gives A007317, a binomial transform of the Catalan numbers, with BTC^(-1)(x) = P[Cinv(x)].

Fib(x) = -Fin[Cinv(Cinv(-x))] = -P[Cinv(-x)] = x + 2 x^2 + 3 x^3 + 5 x^4 + ... = (x+x^2)/[1-x-x^2] is an o.g.f. for the shifted Fibonacci sequence A000045, so the comp. inverse is Fib^(-1)(x) = -C[Pinv(-x)] = -BTC(-x) and Fib(x) = -BTC^(-1)(-x).

Generalizing to P(x,t) = x /(1 + t*x) and Pinv(x,t) = x /(1 - t*x) = -P(-x,t) gives other relations to lattice paths, such as the o.g.f. for A091867, C[P[x,1-t]], and that for A104597, Pinv[Cinv(x),t+1].

(End)

F(n+1) equals the number of binary words of length n avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015

From Russell Jay Hendel, Apr 12 2015: (Start)

We prove Conjecture 1 of Rashid listed in the Formula section.

We use the following notation: F(n)=A000045(n), the Fibonacci numbers, and L(n) = A000032(n), the Lucas numbers. The fundamental Fibonacci-Lucas recursion asserts that G(n) = G(n-1) + G(n-2), with "L" or "F" replacing "G".

We need the following prerequisites which we label (A), (B), (C), (D). The prerequisites are formulas in the Koshy book listed in the References section. (A) F(m-1) + F(m+1) = L(m) (Koshy, p. 97, #32), (B) L(2m) + 2*(-1)^m = L(m)^2 (Koshy p. 97, #41), (C) F(m+k)*F(m-k) = (-1)^n*F(k)^2 (Koshy, p. 113, #24, Tagiuri's identity), and (D) F(n)^2 + F(n+1)^2 = F(2n+1) (Koshy, p. 97, #30).

We must also prove (E), L(n+2)*F(n-1) = F(2n+1)+2*(-1)^n. To prove (E), first note that by (A), proof of (E) is equivalent to proving that F(n+1)*F(n-1) + F(n+3)*F(n-1) = F(2n+1) + 2*(-1)^n. But by (C) with k=1, we have F(n+1)*F(n-1) = F(n)^2 + (-1)^n. Applying (C) again with k=2 and m=n+1, we have F(n+3)*F(n-1) = F(n+1) + (-1)^n. Adding these two applications of (C) together and using (D) we have F(n+1)*F(n-1) + F(n+3)*F(n-1) = F(n)^2 + F(n+1)^2 + 2*(-1)^n = F(2n+1) + 2(-1)^n, completing the proof of (E).

We now prove Conjecture 1. By (A) and the Fibonacci-Lucas recursion, we have F(2n+1) + F(2n+2) + F(2n+3) + F(2n+4) = (F(2n+1) + F(2n+3)) + (F(2n+2) + F(2n+4)) = L(2n+2) +L(2n+3) = L(2n+4). But then by (B), with m=2n+4, we have sqrt(L(2n+4) + 2(-1)^n) = L(n+2). Finally by (E), we have L(n+2)*F(n-1) = F(2n+1) + 2*(-1)^n. Dividing both sides by F(n-1), we have (F(2n+1) + 2*(-1)^n)/F(n-1) = L(n+2) = sqrt(F(2n+1) + F(2n+2) + F(2n+3) + F(2n+4) + 2(-1)^n), as required.

(End)

In Fibonacci's Liber Abaci the rabbit problem appears in the translation of L. E. Sigler on pp. 404-405, and a remark [27] on p. 637. - Wolfdieter Lang, Apr 17 2015

a(n) counts partially ordered partitions of (n-1) into parts 1,2,3 where only the order of adjacent 1's and 2's are unimportant. (See example.) - David Neil McGrath, Jul 27 2015

F(n) divides F(n*k). Proved by Marjorie Bicknell and Verner E Hoggatt Jr. - Juhani Heino, Aug 24 2015

F(n) is the number of UDU-equivalence classes of ballot paths of length n. Two ballot paths of length n with steps U = (1,1), D = (1,-1) are UDU-equivalent whenever the positions of UDU are the same in both paths. - Kostas Manes, Aug 25 2015

Cassini's identity F(2n+1) * F(2n+3) = F(2n+2)^2 + 1 is the basis for a geometrical paradox (or dissection fallacy) in A262342. - Jonathan Sondow, Oct 23 2015

For n >= 4, F(n) is the number of up-down words on alphabet {1,2,3} of length n-2. - Ran Pan, Nov 23 2015

F(n+2) is the number of terms in p(n), where p(n)/q(n) is the n-th convergent of the formal infinite continued fraction [a(0),a(1),...]; e.g., p(3) = a(0)a(1)a(2)a(3) + a(0)a(1) + a(0)a(3) + a(2)a(3) + 1 has F(5) terms. Also, F(n+1) is the number of terms in q(n). - Clark Kimberling, Dec 23 2015

F(n+1) (for n >= 1) is the permanent of an n X n matrix M with M(i,j)=1 if |i-j| <= 1 and 0 otherwise. - Dmitry Efimov, Jan 08 2016

A trapezoid has three sides of lengths in order F(n), F(n+2), F(n). For increasing n a very close approximation to the maximum area will have the fourth side equal to 2*F(n+1). For a trapezoid with lengths of sides in order F(n+2), F(n), F(n+2), the fourth side will be F(n+3). - J. M. Bergot, Mar 17 2016

(1) Join two triangles with lengths of sides L(n), F(n+3), L(n+2) and F(n+2), L(n+1), L(n+2) (where L(n)=A000032(n)) along the common side of length L(n+2) to create an irregular quadrilateral. Its area is approximately 5*F(2*n-1) - (F(2*n-7) - F(2*n-13))/5. (2) Join two triangles with lengths of sides L(n), F(n+2), F(n+3) and L(n+1), F(n+1), F(n+3) along the common side F(n+3) to form an irregular quadrilateral. Its area is approximately 4*F(2*n-1) - 2*(F(2*n-7) + F(2*n-18)). - J. M. Bergot, Apr 06 2016

From Clark Kimberling, Jun 13 2016: (Start)

Let T* be the infinite tree with root 0 generated by these rules: if p is in T*, then p+1 is in T* and x*p is in T*.

Let g(n) be the set of nodes in the n-th generation, so that g(0) = {0}, g(1) = {1}, g(2) = {2, x}, g(3) = {3, 2x, x+1, x^2}, etc.

Let T(r) be the tree obtained by substituting r for x.

If a positive integer N is not a square and r = sqrt(N), then the number of (not necessarily distinct) integers in g(n) is A000045(n), for n >= 1. See A274142. (End)

Consider the partitions of n, with all summands initially listed in nonincreasing order. Freeze all the 1's in place and then allow all the other summands to change their order, without displacing any of the 1's. The resulting number of arrangements is a(n+1). - Gregory L. Simay, Jun 14 2016

Limit of the matrix power M^k shown in A163733, Sep 14 2016, as k->infinity results in a single column vector equal to the Fibonacci sequence. - Gary W. Adamson, Sep 19 2016

F(n) and Lucas numbers L(n), being related by the formulas F(n) = (F(n-1) + L(n-1))/2 and L(n) = 2 F(n+1) - F(n), are a typical pair of "autosequences" (see the link to OEIS Wiki). - Jean-François Alcover, Jun 10 2017

Also the number of independent vertex sets and vertex covers in the (n-2)-path graph. - Eric W. Weisstein, Sep 22 2017

Shifted numbers of {UD, DU, FD, DF}-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are P-equivalent iff the positions of pattern P are identical in these paths. - Sergey Kirgizov, Apr 08 2018

For n > 0, F(n) = the number of Markov equivalence classes with skeleton the path on n nodes. See Theorem 2.1 in the article by A. Radhakrishnan et al. below. - Liam Solus, Aug 23 2018

For n >= 2, also: number of terms in A032858 (every other base-3 digit is strictly smaller than its neighbors) with n-2 digits in base 3. - M. F. Hasler, Oct 05 2018

F(n+1) is the number of fixed points of the Foata transformation on S_n. - Kevin Long, Oct 17 2018

F(n+2) is the dimension of the Hecke algebra of type A_n with independent parameters (0,1,0,1,...) or (1,0,1,0,...). See Corollary 1.5 in the link "Hecke algebras with independent parameters". - Jia Huang, Jan 20 2019

The sequence is the second INVERT transform of (1, -1, 2, -3, 5, -8, 13, ...) and is the first sequence in an infinite set of successive INVERT transforms generated from (1, 0, 1, 0, 1, ...). Refer to the array shown in A073133. - Gary W. Adamson, Jul 16 2019

From Kai Wang, Dec 16 2019: (Start)

F(n*k)/F(k) = Sum_{i=0..n-1; j=0..n-1; i+2*j=n-1} (-1)^(j*(k-1))*L(k)^i*((i+j)!/(i!*j!)).

F((2*m+1)*k)/F(k) = Sum_{i=0..m-1} (-1)^(i*k)*L((2*m-2*i)*k) + (-1)^(m*k).

F(2*m*k)/F(k) = Sum_{i=0..m-1} (-1)^(i*k)*L((2*m-2*i-1)*k).

F(m+s)*F(n+r) - F(m+r)*F(n+s) = (-1)^(n+s)*F(m-n)*F(r-s).

F(m+r)*F(n+s) + F(m+s)*F(n+r) = (2*L(m+n+r+s) - (-1)^(n+s)*L(m-n)*L(r-s))/5.

L(m+r)*L(n+s) - 5*F(m+s)*F(n+r) = (-1)^(n+s)*L(m-n)*L(r-s).

L(m+r)*L(n+s) + 5*F(m+s)*F(n+r) = 2*L(m+n+r+s) + (-1)^(n+s)*5*F(m-n)*F(r-s).

L(m+r)*L(n+s) - L(m+s)*L(n+r) = (-1)^(n+s)*5*F(m-n)*F(r-s). (End)

F(n+1) is the number of permutations in S_n whose principal order ideals in the weak order are Boolean lattices. - Bridget Tenner, Jan 16 2020

F(n+1) is the number of permutations w in S_n that form Boolean intervals [s, w] in the weak order for every simple reflection s in the support of w. - Bridget Tenner, Jan 16 2020

F(n+1) is the number of subsets of {1,2,.,.,n} in which all differences between successive elements of subsets are odd. For example, for n = 6, F(7) = 13 and the 13 subsets are {6}, {1,6}, {3,6}, {5,6}, {2,3,6}, {2,5,6}, {4,5,6}, {1,2,3,6}, {1,2,5,6}, {1,4,5,6}, {3,4,5,6}, {2,3,4,5,6}, {1,2,3,4,5,6}. For even differences between elements see Comment in A016116. - Enrique Navarrete, Jul 01 2020

F(n) is the number of subsets of {1,2,...,n} in which the smallest element of the subset equals the size of the subset (this type of subset is sometimes called extraordinary). For example, F(6) = 8 and the subsets are {1}, {2,3}, {2,4}, {2,5}, {3,4,5}, {2,6}, {3,4,6}, {3,5,6}. It is easy to see that these subsets follow the Fibonacci recursion F(n) = F(n-1) + F(n-2) since we get F(n) such subsets by keeping all F(n-1) subsets from the previous stage (in the example, the F(5)=5 subsets that don't include 6), and by adding one to all elements and appending an additional element n to each subset in F(n-2) subsets (in the example, by applying this to the F(4)=3 subsets {1}, {2,3}, {2,4} we obtain {2,6}, {3,4,6}, {3,5,6}). - Enrique Navarrete, Sep 28 2020

Named "série de Fibonacci" by Lucas (1877) after the Italian mathematician Fibonacci (Leonardo Bonacci, c. 1170 - c. 1240/50). In 1876 he named the sequence "série de Lamé" after the French mathematician Gabriel Lamé (1795 - 1870). - Amiram Eldar, Apr 16 2021

F(n) is the number of edge coverings of the path with n edges. - M. Farrokhi D. G., Sep 30 2021

LCM(F(m), F(n)) is a Fibonacci number if and only if either F(m) divides F(n) or F(n) divides F(m). - M. Farrokhi D. G., Sep 30 2021

Every nonunit positive rational number has at most one representation as the quotient of two Fibonacci numbers. - M. Farrokhi D. G., Sep 30 2021

The infinite sum F(n)/10^(n-1) for all natural numbers n is equal to 100/89. More generally, the sum of F(n)/(k^(n-1)) for all natural numbers n is equal to k^2/(k^2-k-1). Jonatan Djurachkovitch, Dec 31 2023

For n >= 1, number of compositions (c(1),c(2),...,c(k)) of n where c(1), c(3), c(5), ... are 1. To obtain such compositions K(n) of length n increase all parts c(2) by one in all of K(n-1) and prepend two parts 1 in all of K(n-2). - Joerg Arndt, Jan 05 2024

Cohn (1964) proved that a(12) = 12^2 is the only square in the sequence greater than a(1) = 1. - M. F. Hasler, Dec 18 2024

Product_{i=n-2..n+2} F(i) = F(n)^5 - F(n). For example, (F(4)F(5)F(6)F(7)F(8))=(3 * 5 * 8 * 13 * 21) = 8^5 - 8. - Jules Beauchamp, Apr 28 2025

F(n) is even iff n is a multiple of 3. - Stefano Spezia, Jul 06 2025

If the canonical factorization of n into prime powers is Product p^e(p) then d(n) = Product (e(p) + 1). More generally, for k > 0, sigma_k(n) = Product_p ((p^((e(p)+1)*k))-1)/(p^k-1) is the sum of the k-th powers of the divisors of n.

Number of ways to write n as n = x*y, 1 <= x <= n, 1 <= y <= n. For number of unordered solutions to x*y=n, see A038548.

Note that d(n) is not the number of Pythagorean triangles with radius of the inscribed circle equal to n (that is A078644). For number of primitive Pythagorean triangles having inradius n, see A068068(n).

Number of factors in the factorization of the polynomial x^n-1 over the integers. - T. D. Noe, Apr 16 2003

Also equal to the number of partitions p of n such that all the parts have the same cardinality, i.e., max(p)=min(p). - Giovanni Resta, Feb 06 2006

Equals A127093 as an infinite lower triangular matrix * the harmonic series, [1/1, 1/2, 1/3, ...]. - Gary W. Adamson, May 10 2007

For odd n, this is the number of partitions of n into consecutive integers. Proof: For n = 1, clearly true. For n = 2k + 1, k >= 1, map each (necessarily odd) divisor to such a partition as follows: For 1 and n, map k + (k+1) and n, respectively. For any remaining divisor d <= sqrt(n), map (n/d - (d-1)/2) + ... + (n/d - 1) + (n/d) + (n/d + 1) + ... + (n/d + (d-1)/2) {i.e., n/d plus (d-1)/2 pairs each summing to 2n/d}. For any remaining divisor d > sqrt(n), map ((d-1)/2 - (n/d - 1)) + ... + ((d-1)/2 - 1) + (d-1)/2 + (d+1)/2 + ((d+1)/2 + 1) + ... + ((d+1)/2 + (n/d - 1)) {i.e., n/d pairs each summing to d}. As all such partitions must be of one of the above forms, the 1-to-1 correspondence and proof is complete. - Rick L. Shepherd, Apr 20 2008

Number of subgroups of the cyclic group of order n. - Benoit Jubin, Apr 29 2008

Equals row sums of triangle A143319. - Gary W. Adamson, Aug 07 2008

Equals row sums of triangle A159934, equivalent to generating a(n) by convolving A000005 prefaced with a 1; (1, 1, 2, 2, 3, 2, ...) with the INVERTi transform of A000005, (A159933): (1, 1,-1, 0, -1, 2, ...). Example: a(6) = 4 = (1, 1, 2, 2, 3, 2) dot (2, -1, 0, -1, 1, 1) = (2, -1, 0, -2, 3, 2) = 4. - Gary W. Adamson, Apr 26 2009

Number of times n appears in an n X n multiplication table. - Dominick Cancilla, Aug 02 2010

Number of k >= 0 such that (k^2 + k*n + k)/(k + 1) is an integer. - Juri-Stepan Gerasimov, Oct 25 2015

The only numbers k such that tau(k) >= k/2 are 1,2,3,4,6,8,12. - Michael De Vlieger, Dec 14 2016

a(n) is also the number of partitions of 2*n into equal parts, minus the number of partitions of 2*n into consecutive parts. - Omar E. Pol, May 03 2017

From Tomohiro Yamada, Oct 27 2020: (Start)

Let k(n) = log d(n)*log log n/(log 2 * log n), then lim sup k(n) = 1 (Hardy and Wright, Chapter 18, Theorem 317) and k(n) <= k(6983776800) = 1.537939... (the constant A280235) for every n (Nicolas and Robin, 1983).

There exist infinitely many n such that d(n) = d(n+1) (Heath-Brown, 1984). The number of such integers n <= x is at least c*x/(log log x)^3 (Hildebrand, 1987) but at most O(x/sqrt(log log x)) (Erdős, Carl Pomerance and Sárközy, 1987). (End)

Number of 2D grids of n congruent rectangles with two different side lengths, in a rectangle, modulo rotation (cf. A038548 for squares instead of rectangles). Also number of ways to arrange n identical objects in a rectangle (NOT modulo rotation, cf. A038548 for modulo rotation); cf. A007425 and A140773 for the 3D case. - Manfred Boergens, Jun 08 2021

The constant quoted above from Nicolas and Robin, 6983776800 = 2^5 * 3^3 * 5^2 * 7 * 11 * 13 * 17 * 19, appears arbitrary, but interestingly equals 2 * A095849(36). That second factor is highly composite and deeply composite. - Hal M. Switkay, Aug 08 2025

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

To test if a number is a square, see Cohen, p. 40. - N. J. A. Sloane, Jun 19 2011

Zero followed by partial sums of A005408 (odd numbers). - Jeremy Gardiner, Aug 13 2002

Begin with n, add the next number, subtract the previous number and so on ending with subtracting a 1: a(n) = n + (n+1) - (n-1) + (n+2) - (n-2) + (n+3) - (n-3) + ... + (2n-1) - 1 = n^2. - Amarnath Murthy, Mar 24 2004

Sum of two consecutive triangular numbers A000217. - Lekraj Beedassy, May 14 2004

Numbers with an odd number of divisors: {d(n^2) = A048691(n); for the first occurrence of 2n + 1 divisors, see A071571(n)}. - Lekraj Beedassy, Jun 30 2004

See also A000037.

First sequence ever computed by electronic computer, on EDSAC, May 06 1949 (see Renwick link). - Russ Cox, Apr 20 2006

Numbers k such that the imaginary quadratic field Q(sqrt(-k)) has four units. - Marc LeBrun, Apr 12 2006

For n > 0: number of divisors of (n-1)th power of any squarefree semiprime: a(n) = A000005(A006881(k)^(n-1)); a(n) = A000005(A000400(n-1)) = A000005(A011557(n-1)) = A000005(A001023(n-1)) = A000005(A001024(n-1)). - Reinhard Zumkeller, Mar 04 2007

If a 2-set Y and an (n-2)-set Z are disjoint subsets of an n-set X then a(n-2) is the number of 3-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 19 2007

Numbers a such that a^1/2 + b^1/2 = c^1/2 and a^2 + b = c. - Cino Hilliard, Feb 07 2008 (this comment needs clarification, Joerg Arndt, Sep 12 2013)

Numbers k such that the geometric mean of the divisors of k is an integer. - Ctibor O. Zizka, Jun 26 2008

Equals row sums of triangle A143470. Example: 36 = sum of row 6 terms: (23 + 7 + 3 + 1 + 1 + 1). - Gary W. Adamson, Aug 17 2008

Equals row sums of triangles A143595 and A056944. - Gary W. Adamson, Aug 26 2008

Number of divisors of 6^(n-1) for n > 0. - J. Lowell, Aug 30 2008

Denominators of Lyman spectrum of hydrogen atom. Numerators are A005563. A000290-A005563 = A000012. - Paul Curtz, Nov 06 2008

a(n) is the number of all partitions of the sum 2^2 + 2^2 + ... + 2^2, (n-1) times, into powers of 2. - Valentin Bakoev, Mar 03 2009

a(n) is the maximal number of squares that can be 'on' in an n X n board so that all the squares turn 'off' after applying the operation: in any 2 X 2 sub-board, a square turns from 'on' to 'off' if the other three are off. - Srikanth K S, Jun 25 2009

Zero together with the numbers k such that 2 is the number of perfect partitions of k. - Juri-Stepan Gerasimov, Sep 26 2009

Totally multiplicative sequence with a(p) = p^2 for prime p. - Jaroslav Krizek, Nov 01 2009

Satisfies A(x)/A(x^2), A(x) = A173277: (1, 4, 13, 32, 74, ...). - Gary W. Adamson, Feb 14 2010

Positive members are the integers with an odd number of odd divisors and an even number of even divisors. See also A120349, A120359, A181792, A181793, A181795. - Matthew Vandermast, Nov 14 2010

Besides the first term, this sequence is the denominator of Pi^2/6 = 1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36 + ... . - Mohammad K. Azarian, Nov 01 2011

Partial sums give A000330. - Omar E. Pol, Jan 12 2013

Drmota, Mauduit, and Rivat proved that the Thue-Morse sequence along the squares is normal; see A228039. - Jonathan Sondow, Sep 03 2013

a(n) can be decomposed into the sum of the four numbers [binomial(n, 1) + binomial(n, 2) + binomial(n-1, 1) + binomial(n-1, 2)] which form a "square" in Pascal's Triangle A007318, or the sum of the two numbers [binomial(n, 2) + binomial(n+1, 2)], or the difference of the two numbers [binomial(n+2, 3) - binomial(n, 3)]. - John Molokach, Sep 26 2013

In terms of triangular tiling, the number of equilateral triangles with side length 1 inside an equilateral triangle with side length n. - K. G. Stier, Oct 30 2013

Number of positive roots in the root systems of type B_n and C_n (when n > 1). - Tom Edgar, Nov 05 2013

Squares of squares (fourth powers) are also called biquadratic numbers: A000583. - M. F. Hasler, Dec 29 2013

For n > 0, a(n) is the largest integer k such that k^2 + n is a multiple of k + n. More generally, for m > 0 and n > 0, the largest integer k such that k^(2*m) + n is a multiple of k + n is given by k = n^(2*m). - Derek Orr, Sep 03 2014

For n > 0, a(n) is the number of compositions of n + 5 into n parts avoiding the part 2. - Milan Janjic, Jan 07 2016

a(n), for n >= 3, is also the number of all connected subtrees of a cycle graph, having n vertices. - Viktar Karatchenia, Mar 02 2016

On every sequence of natural continuous numbers with an even number of elements, the summatory of the second half of the sequence minus the summatory of the first half of the sequence is always a square. Example: Sequence from 61 to 70 has an even number of elements (10). Then 61 + 62 + 63 + 64 + 65 = 315; 66 + 67 + 68 + 69 + 70 = 340; 340 - 315 = 25. (n/2)^2 for n = number of elements. - César Aguilera, Jun 20 2016

On every sequence of natural continuous numbers from n^2 to (n+1)^2, the sum of the differences of pairs of elements of the two halves in every combination possible is always (n+1)^2. - César Aguilera, Jun 24 2016

Suppose two circles with radius 1 are tangent to each other as well as to a line not passing through the point of tangency. Create a third circle tangent to both circles as well as the line. If this process is continued, a(n) for n > 0 is the reciprocals of the radii of the circles, beginning with the largest circle. - Melvin Peralta, Aug 18 2016

Does not satisfy Benford's law [Ross, 2012]. - N. J. A. Sloane, Feb 08 2017

Numerators of the solution to the generalization of the Feynman triangle problem, with an offset of 2. If each vertex of a triangle is joined to the point (1/p) along the opposite side (measured say clockwise), then the area of the inner triangle formed by these lines is equal to (p - 2)^2/(p^2 - p + 1) times the area of the original triangle, p > 2. For example, when p = 3, the ratio of the areas is 1/7. The denominators of the ratio of the areas is given by A002061. [Cook & Wood, 2004] - Joe Marasco, Feb 20 2017

Equals row sums of triangle A004737, n >= 1. - Martin Michael Musatov, Nov 07 2017

Right-hand side of the binomial coefficient identity Sum_{k = 0..n} (-1)^(n+k+1)*binomial(n,k)*binomial(n + k,k)*(n - k) = n^2. - Peter Bala, Jan 12 2022

Conjecture: For n>0, min{k such that there exist subsets A,B of {0,1,2,...,a(n)-1} such that |A|=|B|=k and A+B contains {0,1,2,...,a(n)-1}} = n. - Michael Chu, Mar 09 2022

Number of 3-permutations of n elements avoiding the patterns 132, 213, 321. See Bonichon and Sun. - Michel Marcus, Aug 20 2022

Number of intercalates in cyclic Latin squares of order 2n (cyclic Latin squares of odd order do not have intercalates). - Eduard I. Vatutin, Feb 15 2024

a(n) is the number of ternary strings of length n with at most one 0, exactly one 1, and no restriction on the number of 2's. For example, a(3)=9, consisting of the 6 permutations of the string 102 and the 3 permutations of the string 122. - Enrique Navarrete, Mar 12 2025