cp's OEIS Frontend

These are Hogben's central polygonal numbers denoted by the symbol

...2....

....P...

...2.n..

(P with three attachments).

Also the maximal number of 1's that an n X n invertible {0,1} matrix can have. (See Halmos for proof.) - Felix Goldberg (felixg(AT)tx.technion.ac.il), Jul 07 2001

Maximal number of interior regions formed by n intersecting circles, for n >= 1. - Amarnath Murthy, Jul 07 2001

The terms are the smallest of n consecutive odd numbers whose sum is n^3: 1, 3 + 5 = 8 = 2^3, 7 + 9 + 11 = 27 = 3^3, etc. - Amarnath Murthy, May 19 2001

(n*a(n+1)+1)/(n^2+1) is the smallest integer of the form (n*k+1)/(n^2+1). - Benoit Cloitre, May 02 2002

For n >= 3, a(n) is also the number of cycles in the wheel graph W(n) of order n. - Sharon Sela (sharonsela(AT)hotmail.com), May 17 2002

Let b(k) be defined as follows: b(1) = 1 and b(k+1) > b(k) is the smallest integer such that Sum_{i=b(k)..b(k+1)} 1/sqrt(i) > 2; then b(n) = a(n) for n > 0. - Benoit Cloitre, Aug 23 2002

Drop the first three terms. Then n*a(n) + 1 = (n+1)^3. E.g., 7*1 + 1 = 8 = 2^3, 13*2 + 1 = 27 = 3^3, 21*3 + 1 = 64 = 4^3, etc. - Amarnath Murthy, Oct 20 2002

Arithmetic mean of next 2n - 1 numbers. - Amarnath Murthy, Feb 16 2004

The n-th term of an arithmetic progression with first term 1 and common difference n: a(1) = 1 -> 1, 2, 3, 4, 5, ...; a(2) = 3 -> 1, 3, ...; a(3) = 7 -> 1, 4, 7, ...; a(4) = 13 -> 1, 5, 9, 13, ... - Amarnath Murthy, Mar 25 2004

Number of walks of length 3 between any two distinct vertices of the complete graph K_{n+1} (n >= 1). Example: a(2) = 3 because in the complete graph ABC we have the following walks of length 3 between A and B: ABAB, ACAB and ABCB. - Emeric Deutsch, Apr 01 2004

Narayana transform of [1, 2, 0, 0, 0, ...] = [1, 3, 7, 13, 21, ...]. Let M = the infinite lower triangular matrix of A001263 and let V = the Vector [1, 2, 0, 0, 0, ...]. Then A002061 starting (1, 3, 7, ...) = M * V. - Gary W. Adamson, Apr 25 2006

The sequence 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, ... is the trajectory of 3 under repeated application of the map n -> n + 2 * square excess of n, cf. A094765.

Also n^3 mod (n^2+1). - Zak Seidov, Aug 31 2006

Also, omitting the first 1, the main diagonal of A081344. - Zak Seidov, Oct 05 2006

Ignoring the first ones, these are rectangular parallelepipeds with integer dimensions that have integer interior diagonals. Using Pythagoras: sqrt(a^2 + b^2 + c^2) = d, an integer; then this sequence: sqrt(n^2 + (n+1)^2 + (n(n+1))^2) = 2T_n + 1 is the first and most simple example. Problem: Are there any integer diagonals which do not satisfy the following general formula? sqrt((k*n)^2 + (k*(n+(2*m+1)))^2 + (k*(n*(n+(2*m+1)) + 4*T_m))^2) = k*d where m >= 0, k >= 1, and T is a triangular number. - Marco Matosic, Nov 10 2006

Numbers n such that a(n) is prime are listed in A055494. Prime a(n) are listed in A002383. All terms are odd. Prime factors of a(n) are listed in A007645. 3 divides a(3*k-1), 7 divides a(7*k-4) and a(7*k-2), 7^2 divides a(7^2*k-18) and a(7^2*k+19), 7^3 divides a(7^3*k-18) and a(7^3*k+19), 7^4 divides a(7^4*k+1048) and a(7^4*k-1047), 7^5 divides a(7^5*k+1354) and a(7^5*k-1353), 13 divides a(13*k-9) and a(13*k-3), 13^2 divides a(13^2*k+23) and a(13^2*k-22), 13^3 divides a(13^3*k+1037) and a(13^3*k-1036). - Alexander Adamchuk, Jan 25 2007

Complement of A135668. - Kieren MacMillan, Dec 16 2007

From William A. Tedeschi, Feb 29 2008: (Start)

Numbers (sorted) on the main diagonal of a 2n X 2n spiral. For example, when n=2:

.

7---8---9--10

| |

6 1---2 11

| | |

5---4---3 12

|

16--15--14--13

.

Cf. A137928. (End)

a(n) = AlexanderPolynomial[n] defined as Det[Transpose[S]-n S] where S is Seifert matrix {{-1, 1}, {0, -1}}. - Artur Jasinski, Mar 31 2008

Starting (1, 3, 7, 13, 21, ...) = binomial transform of [1, 2, 2, 0, 0, 0]; example: a(4) = 13 = (1, 3, 3, 1) dot (1, 2, 2, 0) = (1 + 6 + 6 + 0). - Gary W. Adamson, May 10 2008

Starting (1, 3, 7, 13, ...) = triangle A158821 * [1, 2, 3, ...]. - Gary W. Adamson, Mar 28 2009

Starting with offset 1 = triangle A128229 * [1,2,3,...]. - Gary W. Adamson, Mar 26 2009

a(n) = k such that floor((1/2)*(1 + sqrt(4*k-3))) + k = (n^2+1), that is A000037(a(n)) = A002522(n) = n^2 + 1, for n >= 1. - Jaroslav Krizek, Jun 21 2009

For n > 0: a(n) = A170950(A002522(n-1)), A170950(a(n)) = A174114(n), A170949(a(n)) = A002522(n-1). - Reinhard Zumkeller, Mar 08 2010

From Emeric Deutsch, Sep 23 2010: (Start)

a(n) is also the Wiener index of the fan graph F(n). The fan graph F(n) is defined as the graph obtained by joining each node of an n-node path graph with an additional node. The Wiener index of a connected graph is the sum of the distances between all unordered pairs of vertices in the graph. The Wiener polynomial of the graph F(n) is (1/2)t[(n-1)(n-2)t + 2(2n-1)]. Example: a(2)=3 because the corresponding fan graph is a cycle on 3 nodes (a triangle), having distances 1, 1, and 1.

(End)

For all elements k = n^2 - n + 1 of the sequence, sqrt(4*(k-1)+1) is an integer because 4*(k-1) + 1 = (2*n-1)^2 is a perfect square. Building the intersection of this sequence with A000225, k may in addition be of the form k = 2^x - 1, which happens only for k = 1, 3, 7, 31, and 8191. [Proof: Still 4*(k-1)+1 = 2^(x+2) - 7 must be a perfect square, which has the finite number of solutions provided by A060728: x = 1, 2, 3, 5, or 13.] In other words, the sequence A038198 defines all elements of the form 2^x - 1 in this sequence. For example k = 31 = 6*6 - 6 + 1; sqrt((31-1)*4+1) = sqrt(121) = 11 = A038198(4). - Alzhekeyev Ascar M, Jun 01 2011

a(n) such that A002522(n-1) * A002522(n) = A002522(a(n)) where A002522(n) = n^2 + 1. - Michel Lagneau, Feb 10 2012

Left edge of the triangle in A214661: a(n) = A214661(n, 1), for n > 0. - Reinhard Zumkeller, Jul 25 2012

a(n) = A215630(n, 1), for n > 0; a(n) = A215631(n-1, 1), for n > 1. - Reinhard Zumkeller, Nov 11 2012

Sum_{n > 0} arccot(a(n)) = Pi/2. - Franz Vrabec, Dec 02 2012

If you draw a triangle with one side of unit length and one side of length n, with an angle of Pi/3 radians between them, then the length of the third side of the triangle will be the square root of a(n). - Elliott Line, Jan 24 2013

a(n+1) is the number j such that j^2 = j + m + sqrt(j*m), with corresponding number m given by A100019(n). Also: sqrt(j*m) = A027444(n) = n * a(n+1). - Richard R. Forberg, Sep 03 2013

Let p(x) the interpolating polynomial of degree n-1 passing through the n points (n,n) and (1,1), (2,1), ..., (n-1,1). Then p(n+1) = a(n). - Giovanni Resta, Feb 09 2014

The number of square roots >= sqrt(n) and < n+1 (n >= 0) gives essentially the same sequence, 1, 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, 133, 157, 183, 211, ... . - Michael G. Kaarhus, May 21 2014

For n > 1: a(n) is the maximum total number of queens that can coexist without attacking each other on an [n+1] X [n+1] chessboard. Specifically, this will be a lone queen of one color placed in any position on the perimeter of the board, facing an opponent's "army" of size a(n)-1 == A002378(n-1). - Bob Selcoe, Feb 07 2015

a(n+1) is, for n >= 1, the number of points as well as the number of lines of a finite projective plane of order n (cf. Hughes and Piper, 1973, Theorem 3.5., pp. 79-80). For n = 3, a(4) = 13, see the 'Finite example' in the Wikipedia link, section 2.3, for the point-line matrix. - Wolfdieter Lang, Nov 20 2015

Denominators of the solution to the generalization of the Feynman triangle problem. 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 numerators of the ratio of the areas is given by A000290 with an offset of 2. [Cook & Wood, 2004.] - Joe Marasco, Feb 20 2017

n^2 equal triangular tiles with side lengths 1 X 1 X 1 may be put together to form an n X n X n triangle. For n>=2 a(n-1) is the number of different 2 X 2 X 2 triangles being contained. - Heinrich Ludwig, Mar 13 2017

For n >= 0, the continued fraction [n, n+1, n+2] = (n^3 + 3n^2 + 4n + 2)/(n^2 + 3n + 3) = A034262(n+1)/a(n+2) = n + (n+2)/a(n+2); e.g., [2, 3, 4] = A034262(3)/a(4) = 30/13 = 2 + 4/13. - Rick L. Shepherd, Apr 06 2017

Starting with b(1) = 1 and not allowing the digit 0, let b(n) = smallest nonnegative integer not yet in the sequence such that the last digit of b(n-1) plus the first digit of b(n) is equal to k for k = 1, ..., 9. This defines 9 finite sequences, each of length equal to a(k), k = 1, ..., 9. (See A289283-A289287 for the cases k = 5..9.) For k = 10, the sequence is infinite (A289288). For example, for k = 4, b(n) = 1,3,11,31,32,2,21,33,12,22,23,13,14. These terms can be ordered in the following array of size k*(k-1)+1:

1 2 3

21 22 23

31 32 33

11 12 13 14

.

The sequence ends with the term 1k, which lies outside the rectangular array and gives the term +1 (see link).- Enrique Navarrete, Jul 02 2017

The central polygonal numbers are the delimiters (in parenthesis below) when you write the natural numbers in groups of odd size 2*n+1 starting with the group {2} of size 1: (1) 2 (3) 4,5,6 (7) 8,9,10,11,12 (13) 14,15,16,17,18,19,20 (21) 22,23,24,25,26,27,28,29,30 (31) 32,33,34,35,36,37,38,39,40,41,42 (43) ... - Enrique Navarrete, Jul 11 2017

Also the number of (non-null) connected induced subgraphs in the n-cycle graph. - Eric W. Weisstein, Aug 09 2017

Since (n+1)^2 - (n+1) + 1 = n^2 + n + 1 then from 7 onwards these are also exactly the numbers that are represented as 111 in all number bases: 111(2)=7, 111(3)=13, ... - Ron Knott, Nov 14 2017

Number of binary 2 X (n-1) matrices such that each row and column has at most one 1. - Dmitry Kamenetsky, Jan 20 2018

Observed to be the squares visited by bishop moves on a spirally numbered board and moving to the lowest available unvisited square at each step, beginning at the second term (cf. A316667). It should be noted that the bishop will only travel to squares along the first diagonal of the spiral. - Benjamin Knight, Jan 30 2019

From Ed Pegg Jr, May 16 2019: (Start)

Bound for n-subset coverings. Values in A138077 covered by difference sets.

C(7,3,2), {1,2,4}

C(13,4,2), {0,1,3,9}

C(21,5,2), {3,6,7,12,14}

C(31,6,2), {1,5,11,24,25,27}

C(43,7,2), existence unresolved

C(57,8,2), {0,1,6,15,22,26,45,55}

Next unresolved cases are C(111,11,2) and C(157,13,2). (End)

"In the range we explored carefully, the optimal packings were substantially irregular only for n of the form n = k(k+1)+1, k = 3, 4, 5, 6, 7, i.e., for n = 13, 21, 31, 43, and 57." (cited from Lubachevsky, Graham link, Introduction). - Rainer Rosenthal, May 27 2020

From Bernard Schott, Dec 31 2020: (Start)

For n >= 1, a(n) is the number of solutions x in the interval 1 <= x <= n of the equation x^2 - [x^2] = (x - [x])^2, where [x] = floor(x). For n = 3, the a(3) = 7 solutions in the interval [1, 3] are 1, 3/2, 2, 9/4, 5/2, 11/4 and 3.

This sequence is the answer to the 4th problem proposed during the 20th British Mathematical Olympiad in 1984 (see link B.M.O 1984. and Gardiner reference). (End)

Called "Hogben numbers" after the British zoologist, statistician and writer Lancelot Thomas Hogben (1895-1975). - Amiram Eldar, Jun 24 2021

Minimum Wiener index of 2-degenerate graphs with n+1 vertices (n>0). A maximal 2-degenerate graph can be constructed from a 2-clique by iteratively adding a new 2-leaf (vertex of degree 2) adjacent to two existing vertices. The extremal graphs are maximal 2-degenerate graphs with diameter at most 2. - Allan Bickle, Oct 14 2022

a(n) is the number of parking functions of size n avoiding the patterns 123, 213, and 312. - Lara Pudwell, Apr 10 2023

Repeated iteration of a(k) starting with k=2 produces Sylvester's sequence, i.e., A000058(n) = a^n(2), where a^n is the n-th iterate of a(k). - Curtis Bechtel, Apr 04 2024

a(n) is the maximum number of triangles that can be traversed by starting from a triangle and moving to adjacent triangles via an edge, without revisiting any triangle, in an n X n X n equilateral triangular grid made up of n^2 unit equilateral triangles. - Kiran Ananthpur Bacche, Jan 16 2025

Inert rational primes in the field Q(sqrt(-3)). - N. J. A. Sloane, Dec 25 2017

Primes p such that 1+x+x^2 is irreducible over GF(p). - Joerg Arndt, Aug 10 2011

Primes p dividing sum(k=0,p,C(2k,k)) -1 = A006134(p)-1. - Benoit Cloitre, Feb 08 2003

A039701(A049084(a(n))) = 2; A134323(A049084(a(n))) = -1. - Reinhard Zumkeller, Oct 21 2007

The set of primes of the form 3n - 1 is a superset of the set of lesser of twin primes larger than three (A001359). - Paul Muljadi, Jun 05 2008

Primes of this form do not occur in or as divisors of {n^2+n+1}. See A002383 (n^2+n+1 = prime), A162471 (prime divisors of n^2+n+1 not in A002383), and A002061 (numbers of the form n^2-n+1). - Daniel Tisdale, Jul 04 2009

Or, primes not in A007645. A003627 UNION A007645 = A000040. Also, primes of the form 6*k-5/2-+3/2. - Juri-Stepan Gerasimov, Jan 28 2010

Except for first term "2", all these prime numbers are of the form: 6*n-1. - Vladimir Joseph Stephan Orlovsky, Jul 13 2011

A088534(a(n)) = 0. - Reinhard Zumkeller, Oct 30 2011

For n>1: Numbers k such that (k-4)! mod k =(-1)^(floor(k/3)+1)*floor((k+1)/6), k>4. - Gary Detlefs, Jan 02 2012

Binomial(a(n),3)/a(n)= (3*A024893(n)^2+A024893(n))/2, n>1. - Gary Detlefs, May 06 2012

For every prime p in this sequence, 3 is a 9th power mod p. See Williams link. - Michel Marcus, Nov 12 2017

2 adjoined to A007528. - David A. Corneth, Nov 12 2017

For n >= 2 there exists a polygonal number P_s(3) = 3s - 3 = a(n) + 1. These are the only primes p with P_s(k) = p + 1, s >= 3, k >= 3, since P_s(k) - 1 is composite for k > 3. - Ralf Steiner, May 17 2018

Also, odd primes p such that -3 is a square mod p. - N. J. A. Sloane, Dec 25 2017

Equivalently, primes of the form p = (x^3 - y^3)/(x - y). If x=y+1 we get the cuban primes A002407, which is therefore a subsequence.

These are not to be confused with the Eisenstein primes, which are the primes in the ring of integers Z[w], where w = (-1+sqrt(-3))/2. The present sequence gives the rational primes which are also Eisenstein primes. - N. J. A. Sloane, Feb 06 2008

Also primes of the form x^2+3y^2 and, except for 3, x^2+xy+7y^2. See A140633. - T. D. Noe, May 19 2008

Conjecture: this sequence is Union(A002383,A162471). - Daniel Tisdale, Jul 04 2009

Primes p such that antiharmonic mean B(p) of the numbers k < p such that gcd(k, p) = 1 is not integer, where B(p) = A053818(p) / A023896(p) = A175505(p) / A175506(p) = (2p - 1) / 3. Primes p such that A175506(p) > 1. Subsequence of A179872. Union a(n) + A179891 = A179872. Example: a(6) = 37 because B(37) = A053818(37) / A023896(37) = A175505(37) / A175506(37) = 16206 / 666 = 73 / 3 (not integer). Cf. A179871, A179872, A179873, A179874, A179875, A179876, A179877, A179878, A179879, A179880, A179882, A179883, A179884, A179885, A179886, A179887, A179890, A179891, A003627, A034934. - Jaroslav Krizek, Aug 01 2010

Subsequence of Loeschian numbers, cf. A003136 and A024614; A088534(a(n)) > 0. - Reinhard Zumkeller, Oct 30 2011

Primes such that there exist a unique x, y, with 1 < x <= y < p, x + y == 1 (mod p) and x * y == 1 (mod p). - Jon Perry, Feb 02 2014

The prime factors of A002061. - Richard R. Forberg, Dec 10 2014

This sequence gives the primes p which solve s^2 == -3 (mod 4*p) (see Buell, Proposition 4.1., p. 50, for Delta = -3). p = 2 is not a solution. x^2 == -3 (mod 4) has solutions for all odd x. x^2 == -3 (mod p) has for odd primes p, not 3, the solutions of Legendre(-3|p) = +1 which are p == {1, 7} (mod 12). For p = 3 the representative solution is x = 0. Hence the solution of s^2 == -3 (mod 4*p) are the odd primes p = 3 and p == {1, 7} (mod 12) (or the primes p = 0, 1 (mod 3)). - Wolfdieter Lang, May 22 2021

A002383 lists the corresponding primes. - Bernard Schott, Dec 22 2012

This is also the list of bases where 111 represents a prime number. - Christian N. K. Anderson, Mar 28 2013

If d>1 divides m^2 + m + 1, then m + k*d is not in the sequence, for all k>=1. - Gionata Neri, Mar 04 2017

Primes that are base-b repunits with three or more digits for at least one b >= 2: Primes in A053696. Subsequence of A000668 U A076481 U A086122 U A165210 U A102170 U A004022 U ... (for each possible b). - Rick L. Shepherd, Sep 07 2009

From Bernard Schott, Dec 18 2012: (Start)

Also known as Brazilian primes. The primes that are not Brazilian primes are in A220627.

The number of terms k+1 is always an odd prime, but this is not enough to guarantee a prime, for example 111 = 1 + 10 + 100 = 3*37.

The inverses of the Brazilian primes form a convergent series; the sum is slightly larger than 0.33 (see Theorem 4 of Quadrature article in the Links). (End)

It is not known whether there are infinitely many Brazilian primes. See A002383. - Bernard Schott, Jan 11 2013

Primes of the form (n^p - 1)/(n - 1), where p is odd prime and n > 1. - Thomas Ordowski, Apr 25 2013

Number of terms less than 10^n: 1, 5, 14, 34, 83, 205, 542, 1445, 3880, 10831, 30699, 88285, ..., . - Robert G. Wilson v, Mar 31 2014

From Bernard Schott, Apr 08 2017: (Start)

Brazilian primes fall into two classes:

1) when n is prime, we get sequence A023195 except 3 which is not Brazilian,

2) when n is composite, we get sequence A285017. (End)

The conjecture proposed in Quadrature "No Sophie Germain prime is Brazilian (prime)" (see link Bernard Schott, Quadrature, Conjecture 1, page 36) is false. Thanks to Giovanni Resta, who found that a(856) = 28792661 = 1 + 73 + 73^2 + 73^3 + 73^4 = (11111)73 is the 141385th Sophie Germain prime. - _Bernard Schott, Mar 08 2019

Also nonnegative integers such that a(n) + w is an Eisenstein prime, where w is the primitive cube root of unity. - Frank M Jackson, Jul 15 2025

Terms are listed in the order of their appearance in sequence b.

This is a transformed version of the polynomial P(x) = 47*x^2 + 9*x - 5209 whose absolute value gives 43 distinct primes for -24 <= x <= 18, found by G. W. Fung in 1988. - Hugo Pfoertner, Dec 13 2019

Record values appear to match the terms of A002383 for n>1. - Bill McEachen, Oct 18 2023

Also primes in A001001. - Philippe Deléham, Feb 21 2004

This sequence is a subsequence of A002383. These numbers are repunit primes 111_n, so they are Brazilian primes belonging to A085104. - Bernard Schott, Dec 21 2012

Also, primes in A060800. - Zak Seidov, Mar 21 2014

Also subsequence of A002061, A193574. - Hartmut F. W. Hoft, May 05 2017

As p^2 + p + 1 is the sum of divisors of p^2 for any prime p, this is a subsequence of A023195. - Peter Munn, Feb 16 2018