cp's OEIS Frontend

Leibniz's series: Pi/4 = Sum_{n>=0} (-1)^n/(2n+1) (cf. A072172).

Beginning of the ordering of the natural numbers used in Sharkovski's theorem - see the Cielsielski-Pogoda paper.

The Sharkovski ordering begins with the odd numbers >= 3, then twice these numbers, then 4 times them, then 8 times them, etc., ending with the powers of 2 in decreasing order, ending with 2^0 = 1.

Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0(6).

Also continued fraction for coth(1) (A073747 is decimal expansion). - Rick L. Shepherd, Aug 07 2002

a(1) = 1; a(n) is the smallest number such that a(n) + a(i) is composite for all i = 1 to n-1. - Amarnath Murthy, Jul 14 2003

Smallest number greater than n, not a multiple of n, but containing it in binary representation. - Reinhard Zumkeller, Oct 06 2003

Numbers n such that phi(2n) = phi(n), where phi is Euler's totient (A000010). - Lekraj Beedassy, Aug 27 2004

Pi*sqrt(2)/4 = Sum_{n>=0} (-1)^floor(n/2)/(2n+1) = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 ... [since periodic f(x)=x over -Pi < x < Pi = 2(sin(x)/1 - sin(2x)/2 + sin(3x)/3 - ...) using x = Pi/4 (Maor)]. - Gerald McGarvey, Feb 04 2005

For n > 1, numbers having 2 as an anti-divisor. - Alexandre Wajnberg, Oct 02 2005

a(n) = shortest side a of all integer-sided triangles with sides a <= b <= c and inradius n >= 1.

First differences of squares (A000290). - Lekraj Beedassy, Jul 15 2006

The odd numbers are the solution to the simplest recursion arising when assuming that the algorithm "merge sort" could merge in constant unit time, i.e., T(1):= 1, T(n):= T(floor(n/2)) + T(ceiling(n/2)) + 1. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 14 2006

2n-5 counts the permutations in S_n which have zero occurrences of the pattern 312 and one occurrence of the pattern 123. - David Hoek (david.hok(AT)telia.com), Feb 28 2007

For n > 0: number of divisors of (n-1)th power of any squarefree semiprime: a(n) = A000005(A001248(k)^(n-1)); a(n) = A000005(A000302(n-1)) = A000005(A001019(n-1)) = A000005(A009969(n-1)) = A000005(A087752(n-1)). - Reinhard Zumkeller, Mar 04 2007

For n > 2, a(n-1) is the least integer not the sum of < n n-gonal numbers (0 allowed). - Jonathan Sondow, Jul 01 2007

A134451(a(n)) = abs(A134452(a(n))) = 1; union of A134453 and A134454. - Reinhard Zumkeller, Oct 27 2007

Numbers n such that sigma(2n) = 3*sigma(n). - Farideh Firoozbakht, Feb 26 2008

a(n) = A139391(A016825(n)) = A006370(A016825(n)). - Reinhard Zumkeller, Apr 17 2008

Number of divisors of 4^(n-1) for n > 0. - J. Lowell, Aug 30 2008

Equals INVERT transform of A078050 (signed - cf. comments); and row sums of triangle A144106. - Gary W. Adamson, Sep 11 2008

Odd numbers(n) = 2*n+1 = square pyramidal number(3*n+1) / triangular number(3*n+1). - Pierre CAMI, Sep 27 2008

A000035(a(n))=1, A059841(a(n))=0. - Reinhard Zumkeller, Sep 29 2008

Multiplicative closure of A065091. - Reinhard Zumkeller, Oct 14 2008

a(n) is also the maximum number of triangles that n+2 points in the same plane can determine. 3 points determine max 1 triangle; 4 points can give 3 triangles; 5 points can give 5; 6 points can give 7 etc. - Carmine Suriano, Jun 08 2009

Binomial transform of A130706, inverse binomial transform of A001787(without the initial 0). - Philippe Deléham, Sep 17 2009

Also the 3-rough numbers: positive integers that have no prime factors less than 3. - Michael B. Porter, Oct 08 2009

Or n without 2 as prime factor. - Juri-Stepan Gerasimov, Nov 19 2009

Given an L(2,1) labeling l of a graph G, let k be the maximum label assigned by l. The minimum k possible over all L(2,1) labelings of G is denoted by lambda(G). For n > 0, this sequence gives lambda(K_{n+1}) where K_{n+1} is the complete graph on n+1 vertices. - K.V.Iyer, Dec 19 2009

A176271 = odd numbers seen as a triangle read by rows: a(n) = A176271(A002024(n+1), A002260(n+1)). - Reinhard Zumkeller, Apr 13 2010

For n >= 1, a(n-1) = numbers k such that arithmetic mean of the first k positive integers is an integer. A040001(a(n-1)) = 1. See A145051 and A040001. - Jaroslav Krizek, May 28 2010

Union of A179084 and A179085. - Reinhard Zumkeller, Jun 28 2010

For n>0, continued fraction [1,1,n] = (n+1)/a(n); e.g., [1,1,7] = 8/15. - Gary W. Adamson, Jul 15 2010

Numbers that are the sum of two sequential integers. - Dominick Cancilla, Aug 09 2010

Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n + (h-4)*(-1)^n - h)/4 (h and n in A000027), therefore ((2*h*n + (h-4)*(-1)^n - h)/4)^2 - 1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 4). Also a(n)^2 - 1 == 0 (mod 8). - Bruno Berselli, Nov 17 2010

A004767 = a(a(n)). - Reinhard Zumkeller, Jun 27 2011

A001227(a(n)) = A000005(a(n)); A048272(a(n)) < 0. - Reinhard Zumkeller, Jan 21 2012

a(n) is the minimum number of tosses of a fair coin needed so that the probability of more than n heads is at least 1/2. In fact, Sum_{k=n+1..2n+1} Pr(k heads|2n+1 tosses) = 1/2. - Dennis P. Walsh, Apr 04 2012

A007814(a(n)) = 0; A037227(a(n)) = 1. - Reinhard Zumkeller, Jun 30 2012

1/N (i.e., 1/1, 1/2, 1/3, ...) = Sum_{j=1,3,5,...,infinity} k^j, where k is the infinite set of constants 1/exp.ArcSinh(N/2) = convergents to barover(N). The convergent to barover(1) or [1,1,1,...] = 1/phi = 0.6180339..., whereas c.f. barover(2) converges to 0.414213..., and so on. Thus, with k = 1/phi we obtain 1 = k^1 + k^3 + k^5 + ..., and with k = 0.414213... = (sqrt(2) - 1) we get 1/2 = k^1 + k^3 + k^5 + .... Likewise, with the convergent to barover(3) = 0.302775... = k, we get 1/3 = k^1 + k^3 + k^5 + ..., etc. - Gary W. Adamson, Jul 01 2012

Conjecture on primes with one coach (A216371) relating to the odd integers: iff an integer is in A216371 (primes with one coach either of the form 4q-1 or 4q+1, (q > 0)); the top row of its coach is composed of a permutation of the first q odd integers. Example: prime 19 (q = 5), has 5 terms in each row of its coach: 19: [1, 9, 5, 7, 3] ... [1, 1, 1, 2, 4]. This is interpreted: (19 - 1) = (2^1 * 9), (19 - 9) = (2^1 * 5), (19 - 5) = (2^1 - 7), (19 - 7) = (2^2 * 3), (19 - 3) = (2^4 * 1). - Gary W. Adamson, Sep 09 2012

A005408 is the numerator 2n-1 of the term (1/m^2 - 1/n^2) = (2n-1)/(mn)^2, n = m+1, m > 0 in the Rydberg formula, while A035287 is the denominator (mn)^2. So the quotient a(A005408)/a(A035287) simulates the Hydrogen spectral series of all hydrogen-like elements. - Freimut Marschner, Aug 10 2013

This sequence has unique factorization. The primitive elements are the odd primes (A065091). (Each term of the sequence can be expressed as a product of terms of the sequence. Primitive elements have only the trivial factorization. If the products of terms of the sequence are always in the sequence, and there is a unique factorization of each element into primitive elements, we say that the sequence has unique factorization. So, e.g., the composite numbers do not have unique factorization, because for example 36 = 4*9 = 6*6 has two distinct factorizations.) - Franklin T. Adams-Watters, Sep 28 2013

These are also numbers k such that (k^k+1)/(k+1) is an integer. - Derek Orr, May 22 2014

a(n-1) gives the number of distinct sums in the direct sum {1,2,3,..,n} + {1,2,3,..,n}. For example, {1} + {1} has only one possible sum so a(0) = 1. {1,2} + {1,2} has three distinct possible sums {2,3,4} so a(1) = 3. {1,2,3} + {1,2,3} has 5 distinct possible sums {2,3,4,5,6} so a(2) = 5. - Derek Orr, Nov 22 2014

The number of partitions of 4*n into at most 2 parts. - Colin Barker, Mar 31 2015

a(n) is representable as a sum of two but no fewer consecutive nonnegative integers, e.g., 1 = 0 + 1, 3 = 1 + 2, 5 = 2 + 3, etc. (see A138591). - Martin Renner, Mar 14 2016

Unique solution a( ) of the complementary equation a(n) = a(n-1)^2 - a(n-2)*b(n-1), where a(0) = 1, a(1) = 3, and a( ) and b( ) are increasing complementary sequences. - Clark Kimberling, Nov 21 2017

Also the number of maximal and maximum cliques in the n-centipede graph. - Eric W. Weisstein, Dec 01 2017

Lexicographically earliest sequence of distinct positive integers such that the average of any number of consecutive terms is always an integer. (For opposite property see A042963.) - Ivan Neretin, Dec 21 2017

Maximum number of non-intersecting line segments between vertices of a convex (n+2)-gon. - Christoph B. Kassir, Oct 21 2022

a(n) is the number of parking functions of size n+1 avoiding the patterns 123, 132, and 231. - Lara Pudwell, Apr 10 2023

Maximal number of regions in the plane that can be formed with n hyperbolas.

Also the number of different 2 X 2 determinants with integer entries from 0 to n.

Number of lattice points in an n-dimensional ball of radius sqrt(2). - David W. Wilson, May 03 2001

Equals A112295(unsigned) * [1, 2, 3, ...]. - Gary W. Adamson, Oct 07 2007

Binomial transform of A166926. - Gary W. Adamson, May 03 2008

a(n) = longest side a of all integer-sided triangles with sides a <= b <= c and inradius n >= 1. Triangle has sides (2n^2 + 1, 2n^2 + 2, 4n^2 + 1).

{a(k): 0 <= k < 3} = divisors of 9. - Reinhard Zumkeller, Jun 17 2009

Number of ways to partition a 3*n X 2 grid into 3 connected equal-area regions. - R. H. Hardin, Oct 31 2009

Let A be the Hessenberg matrix of order n defined by: A[1, j] = 1, A[i, i] := 2, (i > 1), A[i, i - 1] = -1, and A[i, j] = 0 otherwise. Then, for n >= 3, a(n - 1) = coeff(charpoly(A, x), x^(n - 2)). - Milan Janjic, Jan 26 2010

Except for the first term of [A002522] and [A058331] if X = [A058331], Y = [A087113], A = [A002522], we have, for all other terms, Pell's equation: [A058331]^2 - [A002522]*[A087113]^2 = 1; (X^2 - A*Y^2 = 1); e.g., 3^2 -2*2^2 = 1; 9^2 - 5*4^2 = 1; 129^2 - 65*16^2 = 1, and so on. - Vincenzo Librandi, Aug 07 2010

Niven (1961) gives this formula as an example of a formula that does not contain all odd integers, in contrast to 2n + 1 and 2n - 1. - Alonso del Arte, Dec 05 2012

Numbers m such that 2*m-2 is a square. - Vincenzo Librandi, Apr 10 2015

Number of n-tuples from the set {1,0,-1} where at most two elements are nonzero. - Michael Somos, Oct 19 2022

a(n) gives the x-value of the integral solution (x,y) of the Pellian equation x^2 - (n^2 + 1)*y^2 = 1. The y-value is given by 2*n (see Tattersall). - Stefano Spezia, Jul 23 2025

a(n) = longest side b of all integer-sided triangles with sides a <= b <= c and inradius n >= 1. Triangle has sides (n^2+2, n^4+2*n^2+1, n^4+3*n^2+1).

Longest possible side c of a triangle with integer sides a <= b < c and inradius n. Triangle has sides (n^2+2, n^4+2n^2+1, n^4+3n^2+1). Proved by Joseph Myers, Jun 11 2006.

a(n) is divisible by 24, and the positive squares A000290(n) are included in the sequence a(n)/24 = {1, 4, 5, 7, 9, 10, 14, 16, 18, 20, 25, 26, 28, 30, 32, 35, 36, 40, 45, 49, 55, 56, 63, 64, 65, ...}.

The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2. The inradius r is given by r = A/s and the circumradius is given by R = abc/4A.

a(n) = #{k | A070083(k) = n and A070200(k) = exact inradius};

a(n) = A070203(n) + A070204(n);

a(n) = A070205(n) + A070206(n) + A024155(n);

a(odd) = 0.

The radius rho of the inscribed circle of a triangle (a,b,c) is rho = sqrt((s-a)*(s-b)*(s-c)/s), with s=(a+b+c)/2. For given integer values of a <= b and a rational target value r2 of the squared incircle radius, c is given by the two positive real roots of the polynomial P(a,b,x,r2) = x^3 - x^2 * (a+b) + x * (4*r2-(b-a)^2) + (a+b)^3 + 4*(a+b)*(r2-a*b). P(a,b,x,r2) = 0 may have 0, 1 or 2 positive integer solutions.

The potential ranges of the side lengths of the triangles can be determined in analogy to the ranges for the case of integer radii of the incircles, see A120062 for the relevant formulas and sequences.

a(n) == 0 (mod 6).

Empirically, 3*sqrt(3) < a(n)/n^2 <= 6. The lower bound is provably tight, the upper bound seems to be achieved infinitely often, e.g., for prime n >= 5.

From Michel Lagneau, Mar 02 2012: (Start)

Subset of A188158.

The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s(s-a)(s-b)(s-c)), where s = (a+b+c)/2. The radius of the incircle or inscribed circle (also known as the inradius, r) is given by r = A/s.

From n = 17, it is possible to find couples of triangles with the property: r1 > r2 and A1 < A2 where A1, A2 are the consecutive areas corresponding to the inradius r1, r2. For example, a(17) = 1734 with (a,b,c) = (51, 68, 85) and a(18) = 1710 with (a,b,c) = (57, 65, 68).

Another interesting property of this sequence is that a(n) is divisible by 6 and, except n = 3, a(n)/6 = n^2 if n prime, hence the proposition:

Among the set of triangles whose area and sides are integers and whose inradius r is a prime number other than 3, the smallest area A is given by A = 6r^2.

Example: if r = 5, the areas of the triangles are {150, 210, 270, 330, 390, 510, ...} and the smallest area of them is A = 6*5^2 = 150 because 5 is prime.

Proof: Let r be a number such that the sides of a triangle are a = 3r, b = 4r, c = 5r. Then s = (a+b+c)/2 = 6r and A = sqrt(s(s-a)(s-b)(s-c)) = sqrt(36r^4) = 6r^2 is a possible area. Is 6r^2 the smallest area? The response is no in the general case for the composite numbers.

Writing a = (m+n)/2, b = (n+l)/2, c = (l+m)/2, and using rs = A and Heron's formula for A, we find lmn = 4r^2(l+m+n). Since m, n and l have to be of the same parity for a, b and c to be integral, they must therefore be even. Setting l = 2u, m = 2v, and n = 2w, we have a = v+w, b = w+u, c = u+v, and uvw = r^2(u+v+w). Then r^2 = uvw/(u+v+w).

First case: If r = p is prime, we prove that A = 6p^2 is the smallest area of all the triangles whose inradius is p. Suppose A' < A with inradius(A') = p. The area A is the corresponding value of the triangle (u,v,w) = (1*p, 2*p, 3*p) because 6p^3/6p = p^2. However, inradius(A') = p => u'v'w'/(u'+v'+w') = p^2 => (u',v',w') = (u,v,w) and A is the smallest area.

Second case: If r = q is composite, the triangle (u,v,w) = (1*q, 2*q, 3*q) gives an area A with inradius(A) = q, but it is possible to find A' < A with inradius(A') = q; for example, if q = 10, the triangle (u,v,w) = (30, 20, 10) whose area is A = 600 gives sqrt{(30*20*10)/(30+20+10)} = sqrt(6000/60) = 10 and the triangle(u',v',w') = (24,15,15) whose area is A' = 540 gives sqrt{(24*15*15)/(24+15+15)} = sqrt(5400/54) = 10.

(End)

A120062(n) = sum_{k:k|n} a(k)