A002476 -id:A002476 - cp's OEIS frontend

A000040 The prime numbers.

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271

Offset: 1

N. J. A. Sloane

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

			From _David A. Corneth_, Oct 22 2024: (Start)
7 is a prime number as it has exactly two divisors, 1 and 7.
8 is not a prime number as it does not have exactly two divisors (it has 1, 2, 4 and 8 as divisors though it is sufficient to find one other divisor than 1 and 8)
55 is not a prime number as it does not have exactly two divisors. One other divisor than 1 and 55 is 5.
59 is a prime number as it has exactly two divisors; 1 and 59. (End)

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M. Agrawal, A Short History of "PRIMES is in P"
P. Alfeld, Notes and Literature on Prime Numbers
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Anonymous, Prime Number Master Index (for primes up to 2*10^7)
Anonymous, prime number
Juan Arias de Reyna and Jeremy Toulisse, The n-th prime asymptotically, arxiv:1203.5413 [math.NT], 2012.
Christian Axler, New estimates for the n-th prime number, arXiv:1706.03651 [math.NT], 2017.
P. T. Bateman & H. G. Diamond, A Hundred Years of Prime Numbers, Amer. Math. Month., Vol. 103 (9), Nov. 1996, pp. 729-741.
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Index entries for "core" sequences
Index entries for sequences related to Benford's law
Additional (less important) items -- comments, formulas, references, links, programs, etc. -- related to the prime numbers, A000040. [HTML version] - [Plain text (TXT) version].

For is_prime and next_prime, see A010051 and A151800.
Cf. A000720 ("pi"), A001223 (differences between primes), A002476, A002808, A003627, A006879, A006880, A008578, A080339, A233588.
Cf. primes in lexicographic order: A210757, A210758, A210759, A210760, A210761.
Cf. A003558, A179480 (relating to the Quasi-order theorem of Hilton and Pedersen).
Boustrophedon transforms: A000747, A000732, A230953.
a(2n) = A104272(n) - A233739(n).
Related sequences:
Primes (p) and composites (c): A002808, A000720, A065855.
Primes between p(n) and 2*p(n): A063124, A070046; between c(n) and 2*c(n): A376761; between n and 2*n: A035250, A060715, A077463, A108954.
Composites between p(n) and 2*p(n): A246514; between c(n) and 2*c(n): A376760; between n and 2*n: A075084, A307912, A307989, A376759.

A000040:=Filtered([1..10^5],IsPrime); # Muniru A Asiru, Sep 04 2017

-- See also Haskell Wiki Link.
import Data.List (genericIndex)
a000040 n = genericIndex a000040_list (n - 1)
a000040_list = base ++ larger where
base = [2,3,5,7,11,13,17]
larger = p : filter prime more
prime n = all ((> 0) . mod n) $ takeWhile (\x -> x*x <= n) larger
_ : p : more = roll $ makeWheels base
roll (Wheel n rs) = [n * k + r | k <- [0..], r <- rs]
makeWheels = foldl nextSize (Wheel 1 [1])
nextSize (Wheel size bs) p = Wheel (size * p)
[r | k <- [0..p-1], b <- bs, let r = size*k+b, mod r p > 0]
data Wheel = Wheel Integer [Integer]
-- Reinhard Zumkeller, Apr 07 2014

Magma
```
[n : n in [2..500] | IsPrime(n)];
```
Magma
```
a := func< n | NthPrime(n) >;
```

A000040 := n->ithprime(n); [ seq(ithprime(i),i=1..100) ];
# For illustration purposes only:
isPrime := s -> is(1 = sin(Pi*GAMMA(s)/s)/sin(Pi/s)):
select(isPrime, [$2..100]); # Peter Luschny, Feb 23 2018

Mathematica
```
Prime[Range[60]]
```

A000040(n) := block(
if n = 1 then return(2),
return( next_prime(A000040(n-1)))
)$ /* recursive, to be replaced if possible - R. J. Mathar, Feb 27 2012 */

PARI
```
{a(n) = if( n<1, 0, prime(n))};
```

/* The following functions provide asymptotic approximations, one based on the asymptotic formula cited above (slight overestimate for n > 10^8), the other one based on pi(x) ~ li(x) = Ei(log(x)) (slight underestimate): */
prime1(n)=n*(log(n)+log(log(n))-1+(log(log(n))-2)/log(n)-((log(log(n))-6)*log(log(n))+11)/log(n)^2/2)
prime2(n)=solve(X=n*log(n)/2,2*n*log(n),real(eint1(-log(X)))+n)
\\ M. F. Hasler, Oct 21 2013

forprime(p=2, 10^3, print1(p, ", ")) \\ Felix Fröhlich, Jun 30 2014

primes(10^5) \\ Altug Alkan, Mar 26 2018

from sympy import primerange
print(list(primerange(2, 272))) # Michael S. Branicky, Apr 30 2022

a = sloane.A000040
a.list(58)  # Jaap Spies, 2007

prime_range(1, 300)  # Zerinvary Lajos, May 27 2009

The prime number theorem is the statement that a(n) ~ n * log n as n -> infinity (Hardy and Wright, page 10).
For n >= 2, n*(log n + log log n - 3/2) < a(n); for n >= 20, a(n) < n*(log n + log log n - 1/2). [Rosser and Schoenfeld]
For all n, a(n) > n log n. [Rosser]
n log(n) + n (log log n - 1) < a(n) < n log n + n log log n for n >= 6. [Dusart, quoted in the Wikipedia article]
a(n) = n log n + n log log n + (n/log n)*(log log n - log n - 2) + O( n (log log n)^2/ (log n)^2). [Cipolla, see also Cesàro or the "Prime number theorem" Wikipedia article for more terms in the expansion]
a(n) = 2 + Sum_{k = 2..floor(2n*log(n)+2)} (1-floor(pi(k)/n)), for n > 1, where the formula for pi(k) is given in A000720 (Ruiz and Sondow 2002). - Jonathan Sondow, Mar 06 2004
I conjecture that Sum_{i>=1} (1/(prime(i)*log(prime(i)))) = Pi/2 = 1.570796327...; Sum_{i=1..100000} (1/(prime(i)*log(prime(i)))) = 1.565585514... It converges very slowly. - Miklos Kristof, Feb 12 2007
The last conjecture has been discussed by the math.research newsgroup recently. The sum, which is greater than Pi/2, is shown in sequence A137245. - T. D. Noe, Jan 13 2009
A000005(a(n)) = 2; A002033(a(n+1)) = 1. - Juri-Stepan Gerasimov, Oct 17 2009
A001222(a(n)) = 1. - Juri-Stepan Gerasimov, Nov 10 2009
From Gary Detlefs, Sep 10 2010: (Start)
Conjecture:
a(n) = {n| n! mod n^2 = n(n-1)}, n <> 4.
a(n) = {n| n!*h(n) mod n = n-1}, n <> 4, where h(n) = Sum_{k=1..n} 1/k. (End)
For n = 1..15, a(n) = p + abs(p-3/2) + 1/2, where p = m + int((m-3)/2), and m = n + int((n-2)/8) + int((n-4)/8). - Timothy Hopper, Oct 23 2010
a(2n) <= A104272(n) - 2 for n > 1, and a(2n) ~ A104272(n) as n -> infinity. - Jonathan Sondow, Dec 16 2013
Conjecture: Sequence = {5 and n <> 5| ( Fibonacci(n) mod n = 1 or Fibonacci(n) mod n = n - 1) and 2^(n-1) mod n = 1}. - Gary Detlefs, May 25 2014
Conjecture: Sequence = {5 and n <> 5| ( Fibonacci(n) mod n = 1 or Fibonacci(n) mod n = n - 1) and 2^(3*n) mod 3*n = 8}. - Gary Detlefs, May 28 2014
Satisfies a(n) = 2*n + Sum_{k=1..(a(n)-1)} cot(k*Pi/a(n))*sin(2*k*n^a(n)*Pi/a(n)). - Ilya Gutkovskiy, Jun 29 2016
Sum_{n>=1} 1/a(n)^s = P(s), where P(s) is the prime zeta function. - Eric W. Weisstein, Nov 08 2016
a(n) = floor(1 - log(-1/2 + Sum_{ d | A002110(n-1) } mu(d)/(2^d-1))/log(2)) where mu(d) = A008683(d) [Ghandi, 1971] (see Ribenboim). Golomb gave a proof in 1974: Give each positive integer a probability of W(n) = 1/2^n, then the probability M(d) of the integer multiple of number d equals 1/(2^d-1). Suppose Q = a(1)*a(2)*...*a(n-1) = A002110(n-1), then the probability of random integers that are mutually prime with Q is Sum_{ d | Q } mu(d)*M(d) = Sum_{ d | Q } mu(d)/(2^d-1) = Sum_{ gcd(m, Q) = 1 } W(m) = 1/2 + 1/2^a(n) + 1/2^a(n+1) + 1/2^a(n+2) + ... So ((Sum_{ d | Q } mu(d)/(2^d-1)) - 1/2)*2^a(n) = 1 + x(n), which means that a(n) is the only integer so that 1 < ((Sum_{ d | Q } mu(d)/(2^d-1)) - 1/2)*2^a(n) < 2. - Jinyuan Wang, Apr 08 2019
Conjecture: n * (log(n)+log(log(n))-1+((log(log(n))-A)/log(n))) is asymptotic to a(n) if and only if A=2. - Alain Rocchelli, Feb 12 2025
From Stefano Spezia, Apr 13 2025: (Start)
a(n) = 1 + Sum_{m=1..2^n} floor(floor(n/Sum_{j=1..m} A080339(j))^(1/n)) [Willans, 1964].
a(n) = 1 + Sum_{m=1..2^n} floor(floor(n/(1 + A000720(m)))^(1/n)) [Willans, 1964]. (End)

A001006 Motzkin numbers: number of ways of drawing any number of nonintersecting chords joining n (labeled) points on a circle.

Original entry on oeis.org

1, 1, 2, 4, 9, 21, 51, 127, 323, 835, 2188, 5798, 15511, 41835, 113634, 310572, 853467, 2356779, 6536382, 18199284, 50852019, 142547559, 400763223, 1129760415, 3192727797, 9043402501, 25669818476, 73007772802, 208023278209, 593742784829, 1697385471211

Offset: 0

N. J. A. Sloane

Number of 4321-, (3412,2413)-, (3412,3142)- and 3412-avoiding involutions in S_n.
Number of sequences of length n-1 consisting of positive integers such that the first and last elements are 1 or 2 and the absolute difference between any 2 consecutive elements is 0 or 1. - Jon Perry, Sep 04 2003
From David Callan, Jul 15 2004: (Start)
Also number of Motzkin n-paths: paths from (0,0) to (n,0) in an n X n grid using only steps U = (1,1), F = (1,0) and D = (1,-1).
Number of Dyck n-paths with no UUU. (Given such a Dyck n-path, change each UUD to U, then change each remaining UD to F. This is a bijection to Motzkin n-paths. Example with n=5: U U D U D U U D D D -> U F U D D.)
Number of Dyck (n+1)-paths with no UDU. (Given such a Dyck (n+1)-path, mark each U that is followed by a D and each D that is not followed by a U. Then change each unmarked U whose matching D is marked to an F. Lastly, delete all the marked steps. This is a bijection to Motzkin n-paths. Example with n=6 and marked steps in small type: U U u d D U U u d d d D u d -> U U u d D F F u d d d D u d -> U U D F F D.) (End)
a(n) is the number of strings of length 2n+2 from the following recursively defined set: L contains the empty string and, for any strings a and b in L, we also find (ab) in L. The first few elements of L are e, (), (()), ((())), (()()), (((()))), ((()())), ((())()), (()(())) and so on. This proves that a(n) is less than or equal to C(n), the n-th Catalan number. See Orrick link (2024). - Saul Schleimer (saulsch(AT)math.rutgers.edu), Feb 23 2006 (Additional linked comment added by William P. Orrick, Jun 13 2024.)
a(n) = number of Dyck n-paths all of whose valleys have even x-coordinate (when path starts at origin). For example, T(4,2)=3 counts UDUDUUDD, UDUUDDUD, UUDDUDUD. Given such a path, split it into n subpaths of length 2 and transform UU->U, DD->D, UD->F (there will be no DUs for that would entail a valley with odd x-coordinate). This is a bijection to Motzkin n-paths. - David Callan, Jun 07 2006
Also the number of standard Young tableaux of height <= 3. - Mike Zabrocki, Mar 24 2007
a(n) is the number of RNA shapes of size 2n+2. RNA Shapes are essentially Dyck words without "directly nested" motifs of the form A[[B]]C, for A, B and C Dyck words. The first RNA Shapes are []; [][]; [][][], [[][]]; [][][][], [][[][]], [[][][]], [[][]][]; ... - Yann Ponty (ponty(AT)lri.fr), May 30 2007
The sequence is self-generated from top row A going to the left starting (1,1) and bottom row = B, the same sequence but starting (0,1) and going to the right. Take dot product of A and B and add the result to n-th term of A to get the (n+1)-th term of A. Example: a(5) = 21 as follows: Take dot product of A = (9, 4, 2, 1, 1) and (0, 1, 1, 2, 4) = (0, + 4 + 2 + 2 + 4) = 12; which is added to 9 = 21. - Gary W. Adamson, Oct 27 2008
Equals A005773 / A005773 shifted (i.e., (1,2,5,13,35,96,...) / (1,1,2,5,13,35,96,...)). - Gary W. Adamson, Dec 21 2008
Starting with offset 1 = iterates of M * [1,1,0,0,0,...], where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 07 2009
a(n) is the number of involutions of {1,2,...,n} having genus 0. The genus g(p) of a permutation p of {1,2,...,n} is defined by g(p)=(1/2)[n+1-z(p)-z(cp')], where p' is the inverse permutation of p, c = 234...n1 = (1,2,...,n), and z(q) is the number of cycles of the permutation q. Example: a(4)=9; indeed, p=3412=(13)(24) is the only involution of {1,2,3,4} with genus > 0. This follows easily from the fact that a permutation p of {1,2,...,n} has genus 0 if and only if the cycle decomposition of p gives a noncrossing partition of {1,2,...,n} and each cycle of p is increasing (see Lemma 2.1 of the Dulucq-Simion reference). [Also, redundantly, for p=3412=(13)(24) we have cp'=2341*3412=4123=(1432) and so g(p)=(1/2)(4+1-2-1)=1.] - Emeric Deutsch, May 29 2010
Let w(i,j,n) denote walks in N^2 which satisfy the multivariate recurrence w(i,j,n) = w(i, j + 1, n - 1) + w(i - 1, j, n - 1) + w(i + 1, j - 1, n - 1) with boundary conditions w(0,0,0) = 1 and w(i,j,n) = 0 if i or j or n is < 0. Then a(n) = Sum_{i = 0..n, j = 0..n} w(i,j,n) is the number of such walks of length n. - Peter Luschny, May 21 2011
a(n)/a(n-1) tends to 3.0 as N->infinity: (1+2*cos(2*Pi/N)) relating to longest odd N regular polygon diagonals, by way of example, N=7: Using the tridiagonal generator [cf. comment of Jan 07 2009], for polygon N=7, we extract an (N-1)/2 = 3 X 3 matrix, [0,1,0; 1,1,1; 0,1,1] with an e-val of 2.24697...; the longest Heptagon diagonal with edge = 1. As N tends to infinity, the diagonal lengths tend to 3.0, the convergent of the sequence. - Gary W. Adamson, Jun 08 2011
Number of (n+1)-length permutations avoiding the pattern 132 and the dotted pattern 23\dot{1}. - Jean-Luc Baril, Mar 07 2012
Number of n-length words w over alphabet {a,b,c} such that for every prefix z of w we have #(z,a) >= #(z,b) >= #(z,c), where #(z,x) counts the letters x in word z. The a(4) = 9 words are: aaaa, aaab, aaba, abaa, aabb, abab, aabc, abac, abca. - Alois P. Heinz, May 26 2012
Number of length-n restricted growth strings (RGS) [r(1), r(2), ..., r(n)] such that r(1)=1, r(k)<=k, and r(k)!=r(k-1); for example, the 9 RGS for n=4 are 1010, 1012, 1201, 1210, 1212, 1230, 1231, 1232, 1234. - Joerg Arndt, Apr 16 2013
Number of length-n restricted growth strings (RGS) [r(1), r(2), ..., r(n)] such that r(1)=0, r(k)<=k and r(k)-r(k-1) != 1; for example, the 9 RGS for n=4 are 0000, 0002, 0003, 0004, 0022, 0024, 0033, 0222, 0224. - Joerg Arndt, Apr 17 2013
Number of (4231,5276143)-avoiding involutions in S_n. - Alexander Burstein, Mar 05 2014
a(n) is the number of increasing unary-binary trees with n nodes that have an associated permutation that avoids 132. For more information about unary-binary trees with associated permutations, see A245888. - Manda Riehl, Aug 07 2014
a(n) is the number of involutions on [n] avoiding the single pattern p, where p is any one of the 8 (classical) patterns 1234, 1243, 1432, 2134, 2143, 3214, 3412, 4321. Also, number of (3412,2413)-, (3412,3142)-, (3412,2413,3142)-avoiding involutions on [n] because each of these 3 sets actually coincides with the 3412-avoiding involutions on [n]. This is a complete list of the 8 singles, 2 pairs, and 1 triple of 4-letter classical patterns whose involution avoiders are counted by the Motzkin numbers. (See Barnabei et al. 2011 reference.) - David Callan, Aug 27 2014
From Tony Foster III, Jul 28 2016: (Start)
A series created using 2*a(n) + a(n+1) has Hankel transform of F(2n), offset 3, F being the Fibonacci bisection, A001906 (empirical observation).
A series created using 2*a(n) + 3*a(n+1) + a(n+2) gives the Hankel transform of Sum_{k=0..n} k*Fibonacci(2*k), offset 3, A197649 (empirical observation). (End)
Conjecture: (2/n)*Sum_{k=1..n} (2k+1)*a(k)^2 is an integer for each positive integer n. - Zhi-Wei Sun, Nov 16 2017
The Rubey and Stump reference proves a refinement of a conjecture of René Marczinzik, which they state as: "The number of 2-Gorenstein algebras which are Nakayama algebras with n simple modules and have an oriented line as associated quiver equals the number of Motzkin paths of length n." - Eric M. Schmidt, Dec 16 2017
Number of U_{k}-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
If tau_1 and tau_2 are two distinct permutation patterns chosen from the set {132,231,312}, then a(n) is the number of valid hook configurations of permutations of [n+1] that avoid the patterns tau_1 and tau_2. - Colin Defant, Apr 28 2019
Number of permutations of length n that are sorted to the identity by a consecutive-321-avoiding stack followed by a classical-21-avoiding stack. - Colin Defant, Aug 29 2020
From Helmut Prodinger, Dec 13 2020: (Start)
a(n) is the number of paths in the first quadrant starting at (0,0) and consisting of n steps from the infinite set {(1,1), (1,-1), (1,-2), (1,-3), ...}.
For example, denoting U=(1,1), D=(1,-1), D_ j=(1,-j) for j >= 2, a(4) counts UUUU, UUUD, UUUD_2, UUUD_3, UUDU, UUDD, UUD_2U, UDUU, UDUD.
This step set is inspired by {(1,1), (1,-1), (1,-3), (1,-5), ...}, suggested by Emeric Deutsch around 2000.
See Prodinger link that contains a bijection to Motzkin paths. (End)
Named by Donaghey (1977) after the Israeli-American mathematician Theodore Motzkin (1908-1970). In Sloane's "A Handbook of Integer Sequences" (1973) they were called "generalized ballot numbers". - Amiram Eldar, Apr 15 2021
Number of Motzkin n-paths a(n) is split into A107587(n), number of even Motzkin n-paths, and A343386(n), number of odd Motzkin n-paths. The value A107587(n) - A343386(n) can be called the "shadow" of a(n) (see A343773). - Gennady Eremin, May 17 2021
Conjecture: If p is a prime of the form 6m+1 (A002476), then a(p-2) is divisible by p. Currently, no counterexample exists for p < 10^7. Personal communication from Robert Gerbicz: mod such p this is equivalent to A066796 with comment: "Every A066796(n) from A066796((p-1)/2) to A066796(p-1) is divisible by prime p of form 6m+1". - Serge Batalov, Feb 08 2022
From Rob Burns, Nov 11 2024: (Start)
The conjecture is proved in the 2017 paper by Rob Burns in the Links below. The result is contained in Tables 4 and 5 of the paper, which show that a(p-2) == 0 (mod p) when p == 1 (mod 6) and a(p-2) == -1 (mod p) when p == -1 (mod 6).
In fact, the 2017 paper by Burns establishes more general congruences for a(p^k - 2) where k >= 1.
If p == 1 (mod 6) then a(p^k - 2) == 0 (mod p) for k >= 1.
If p == -1 (mod 6) then a(p^k - 2) == -1 (mod p) when k is odd and a(p^k - 2) == 0 (mod p) when k is even.
These are consequences of the transitions provided in Tables 4, 5 and 6 of the paper.
The 2024 paper by Nadav Kohen also proves the conjecture. Proposition 6 of the paper states that a prime p divides a(p-2) if and only if p = (1 mod 3). (End)
From Peter Bala, Feb 10 2022: (Start)
Conjectures:
(1) For prime p == 1 (mod 6) and n, r >= 1, a(n*p^r - 2) == -A005717(n-1) (mod p), where we take A005717(0) = 0 to match Batalov's conjecture above.
(2) For prime p == 5 (mod 6) and n >= 1, a(n*p - 2) == -A005773(n) (mod p).
(3) For prime p >= 3 and k >= 1, a(n + p^k) == a(n) (mod p) for 0 <= n <= (p^k - 3).
(4) For prime p >= 5 and k >= 2, a(n + p^k) == a(n) (mod p^2) for 0 <= n <= (p^(k-1) - 3). (End)
The Hankel transform of this sequence with a(0) omitted gives the period-6 sequence [1, 0, -1, -1, 0, 1, ...] which is A010892 with its first term omitted, while the Hankel transform of the current sequence is the all-ones sequence A000012, and also it is the unique sequence with this property which is similar to the unique Hankel transform property of the Catalan numbers. - Michael Somos, Apr 17 2022
The number of terms in which the exponent of any variable x_i is not greater than 2 in the expansion of Product_{j=1..n} Sum_{i=1..j} x_i. E.g.: a(4) = 9: 3*x1^2*x2^2, 4*x1^2*x2*x3, 2*x1^2*x2*x4, x1^2*x3^2, x1^2*x3*x4, 2*x1*x2^2*x3, x1*x2^2*x4, x1*x2*x3^2, x1*x2*x3*x4. - Elif Baser, Dec 20 2024

			G.f.: 1 + x + 2*x^2 + 4*x^3 + 9*x^4 + 21*x^5 + 51*x^6 + 127*x^7 + 323*x^8 + ...
.
The 21 Motzkin-paths of length 5: UUDDF, UUDFD, UUFDD, UDUDF, UDUFD, UDFUD, UDFFF, UFUDD, UFDUD, UFDFF, UFFDF, UFFFD, FUUDD, FUDUD, FUDFF, FUFDF, FUFFD, FFUDF, FFUFD, FFFUD, FFFFF.

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Index entries for "core" sequences

Cf. A026300, A005717, A020474, A001850, A004148. First column of A064191, A064189, A000108, A088615, A007971, A001405, A005817, A049401, A007579, A007578, A097862, A005773, A178515, A217275. First row of A064645.
Bisections: A026945, A099250.
Sequences related to chords in a circle: A001006, A054726, A006533, A006561, A006600, A007569, A007678. See also entries for chord diagrams in Index file.
a(n) = A005043(n)+A005043(n+1).
A086246 is another version, although this is the main entry. Column k=3 of A182172.
Motzkin numbers A001006 read mod 2,3,4,5,6,7,8,11: A039963, A039964, A299919, A258712, A299920, A258711, A299918, A258710.
Cf. A004148, A004149, A023421, A023422, A023423, A290277 (inv. Euler Transf.).

a001006 n = a001006_list !! n
a001006_list = zipWith (+) a005043_list $ tail a005043_list
-- Reinhard Zumkeller, Jan 31 2012

# Three different Maple scripts for this sequence:
A001006 := proc(n)
    add(binomial(n,2*k)*A000108(k),k=0..floor(n/2)) ;
end proc:
A001006 := proc(n) option remember; local k; if n <= 1 then 1 else procname(n-1) + add(procname(k)*procname(n-k-2),k=0..n-2); end if; end proc:
# n -> [a(0),a(1),..,a(n)]
A001006_list := proc(n) local w, m, j, i; w := proc(i,j,n) option remember;
if min(i,j,n) < 0 or max(i,j) > n then 0
elif n = 0 then if i = 0 and j = 0 then 1 else 0 fi else
w(i, j + 1, n - 1) + w(i - 1, j, n - 1) + w(i + 1, j - 1, n - 1) fi end:
[seq( add( add( w(i, j, m), i = 0..m), j = 0..m), m = 0..n)] end:
A001006_list(29); # Peter Luschny, May 21 2011

a[0] = 1; a[n_Integer] := a[n] = a[n - 1] + Sum[a[k] * a[n - 2 - k], {k, 0, n - 2}]; Array[a, 30]
(* Second program: *)
CoefficientList[Series[(1 - x - (1 - 2x - 3x^2)^(1/2))/(2x^2), {x, 0, 29}], x] (* Jean-François Alcover, Nov 29 2011 *)
Table[Hypergeometric2F1[(1-n)/2, -n/2, 2, 4], {n,0,29}] (* Peter Luschny, May 15 2016 *)
Table[GegenbauerC[n,-n-1,-1/2]/(n+1),{n,0,100}] (* Emanuele Munarini, Oct 20 2016 *)
MotzkinNumber = DifferenceRoot[Function[{y, n}, {(-3n-3)*y[n] + (-2n-5)*y[n+1] + (n+4)*y[n+2] == 0, y[0] == 1, y[1] == 1}]];
Table[MotzkinNumber[n], {n, 0, 29}] (* Jean-François Alcover, Oct 27 2021 *)

a[0]:1$
a[1]:1$
a[n]:=((2*n+1)*a[n-1]+(3*n-3)*a[n-2])/(n+2)$
makelist(a[n],n,0,12); /* Emanuele Munarini, Mar 02 2011 */

M(n) := coeff(expand((1+x+x^2)^(n+1)),x^n)/(n+1);
makelist(M(n),n,0,60); /* Emanuele Munarini, Apr 04 2012 */

makelist(ultraspherical(n,-n-1,-1/2)/(n+1),n,0,12); /* Emanuele Munarini, Oct 20 2016 */

{a(n) = polcoeff( ( 1 - x - sqrt((1 - x)^2 - 4 * x^2 + x^3 * O(x^n))) / (2 * x^2), n)}; /* Michael Somos, Sep 25 2003 */

{a(n) = if( n<0, 0, n++; polcoeff( serreverse( x / (1 + x + x^2) + x * O(x^n)), n))}; /* Michael Somos, Sep 25 2003 */

{a(n) = if( n<0, 0, n! * polcoeff( exp(x + x * O(x^n)) * besseli(1, 2 * x + x * O(x^n)), n))}; /* Michael Somos, Sep 25 2003 */

from gmpy2 import divexact
A001006 = [1, 1]
for n in range(2, 10**3):
    A001006.append(divexact(A001006[-1]*(2*n+1)+(3*n-3)*A001006[-2],n+2))
# Chai Wah Wu, Sep 01 2014

def mot():
    a, b, n = 0, 1, 1
    while True:
        yield b//n
        n += 1
        a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)//((n+1)*(n-1))
A001006 = mot()
print([next(A001006) for n in range(30)]) # Peter Luschny, May 16 2016

# A simple generator of Motzkin-paths (see the first comment of David Callan).
C = str.count
def aGen(n: int):
    a = [""]
    for w in a:
        if len(w) == n:
            if C(w, "U") == C(w, "D"): yield w
        else:
            for j in "UDF":
                u = w + j
                if C(u, "U") >= C(u, "D"): a += [u]
    return a
for n in range(6):
    MP = [w for w in aGen(n)];
    print(len(MP), ":", MP)  # Peter Luschny, Dec 03 2024

G.f.: A(x) = ( 1 - x - (1-2*x-3*x^2)^(1/2) ) / (2*x^2).
G.f. A(x) satisfies A(x) = 1 + x*A(x) + x^2*A(x)^2.
G.f.: F(x)/x where F(x) is the reversion of x/(1+x+x^2). - Joerg Arndt, Oct 23 2012
a(n) = (-1/2) Sum_{i+j = n+2, i >= 0, j >= 0} (-3)^i*C(1/2, i)*C(1/2, j).
a(n) = (3/2)^(n+2) * Sum_{k >= 1} 3^(-k) * Catalan(k-1) * binomial(k, n+2-k). [Doslic et al.]
a(n) ~ 3^(n+1)*sqrt(3)*(1 + 1/(16*n))/((2*n+3)*sqrt((n+2)*Pi)). [Barcucci, Pinzani and Sprugnoli]
Limit_{n->infinity} a(n)/a(n-1) = 3. [Aigner]
a(n+2) - a(n+1) = a(0)*a(n) + a(1)*a(n-1) + ... + a(n)*a(0). [Bernhart]
a(n) = (1/(n+1)) * Sum_{i} (n+1)!/(i!*(i+1)!*(n-2*i)!). [Bernhart]
From Len Smiley: (Start)
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*A000108(k+1), inv. Binomial Transform of A000108.
a(n) = (1/(n+1))*Sum_{k=0..ceiling((n+1)/2)} binomial(n+1, k)*binomial(n+1-k, k-1);
D-finite with recurrence: (n+2)*a(n) = (2*n+1)*a(n-1) + (3*n-3)*a(n-2). (End)
a(n) = Sum_{k=0..n} C(n, 2k)*A000108(k). - Paul Barry, Jul 18 2003
E.g.f.: exp(x)*BesselI(1, 2*x)/x. - Vladeta Jovovic, Aug 20 2003
a(n) = A005043(n) + A005043(n+1).
The Hankel transform of this sequence gives A000012 = [1, 1, 1, 1, 1, 1, ...]. E.g., Det([1, 1, 2, 4; 1, 2, 4, 9; 2, 4, 9, 21; 4, 9, 21, 51]) = 1. - Philippe Deléham, Feb 23 2004
a(m+n) = Sum_{k>=0} A064189(m, k)*A064189(n, k). - Philippe Deléham, Mar 05 2004
a(n) = (1/(n+1))*Sum_{j=0..floor(n/3)} (-1)^j*binomial(n+1, j)*binomial(2*n-3*j, n). - Emeric Deutsch, Mar 13 2004
a(n) = A086615(n) - A086615(n-1) (n >= 1). - Emeric Deutsch, Jul 12 2004
G.f.: A(x)=(1-y+y^2)/(1-y)^2 where (1+x)*(y^2-y)+x=0; A(x)=4*(1+x)/(1+x+sqrt(1-2*x-3*x^2))^2; a(n)=(3/4)*(1/2)^n*Sum_(k=0..2*n, 3^(n-k)*C(k)*C(k+1, n+1-k) ) + 0^n/4 [after Doslic et al.]. - Paul Barry, Feb 22 2005
G.f.: c(x^2/(1-x)^2)/(1-x), c(x) the g.f. of A000108. - Paul Barry, May 31 2006
Asymptotic formula: a(n) ~ sqrt(3/4/Pi)*3^(n+1)/n^(3/2). - Benoit Cloitre, Jan 25 2007
a(n) = A007971(n+2)/2. - Zerinvary Lajos, Feb 28 2007
a(n) = (1/(2*Pi))*Integral_{x=-1..3} x^n*sqrt((3-x)*(1+x)) is the moment representation. - Paul Barry, Sep 10 2007
Given an integer t >= 1 and initial values u = [a_0, a_1, ..., a_{t-1}], we may define an infinite sequence Phi(u) by setting a_n = a_{n-1} + a_0*a_{n-1} + a_1*a_{n-2} + ... + a_{n-2}*a_1 for n >= t. For example, Phi([1]) is the Catalan numbers A000108. The present sequence is Phi([0,1,1]), see the 6th formula. - Gary W. Adamson, Oct 27 2008
G.f.: 1/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/.... (continued fraction). - Paul Barry, Dec 06 2008
G.f.: 1/(1-(x+x^2)/(1-x^2/(1-(x+x^2)/(1-x^2/(1-(x+x^2)/(1-x^2/(1-.... (continued fraction). - Paul Barry, Feb 08 2009
a(n) = (-3)^(1/2)/(6*(n+2)) * (-1)^n*(3*hypergeom([1/2, n+1],[1],4/3) - hypergeom([1/2, n+2],[1],4/3)). - Mark van Hoeij, Nov 12 2009
G.f.: 1/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-... (continued fraction). - Paul Barry, Mar 02 2010
G.f.: 1/(1-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-... (continued fraction). - Paul Barry, Jan 26 2011 [Adds apparently a third '1' in front. - R. J. Mathar, Jan 29 2011]
Let A(x) be the g.f., then B(x)=1+x*A(x) = 1 + 1*x + 1*x^2 + 2*x^3 + 4*x^4 + 9*x^5 + ... = 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1+x) (continued fraction); more generally B(x)=C(x/(1+x)) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011
a(n) = (2/Pi)*Integral_{x=-1..1} (1+2*x)^n*sqrt(1-x^2). - Peter Luschny, Sep 11 2011
G.f.: (1-x-sqrt(1-2*x-3*(x^2)))/(2*(x^2)) = 1/2/(x^2)-1/2/x-1/2/(x^2)*G(0); G(k) = 1+(4*k-1)*x*(2+3*x)/(4*k+2-x*(2+3*x)*(4*k+1)*(4*k+2) /(x*(2+3*x)*(4*k+1)+(4*k+4)/G(k+1))), if -1 < x < 1/3; (continued fraction). - Sergei N. Gladkovskii, Dec 01 2011
G.f.: (1-x-sqrt(1-2*x-3*(x^2)))/(2*(x^2)) = (-1 + 1/G(0))/(2*x); G(k) = 1-2*x/(1+x/(1+x/(1-2*x/(1-x/(2-x/G(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, Dec 11 2011
0 = a(n) * (9*a(n+1) + 15*a(n+2) - 12*a(n+3)) + a(n+1) * ( -3*a(n+1) + 10*a(n+2) - 5*a(n+3)) + a(n+2) * (a(n+2) + a(n+3)) unless n=-2. - Michael Somos, Mar 23 2012
a(n) = (-1)^n*hypergeometric([-n,3/2],[3],4). - Peter Luschny, Aug 15 2012
Representation in terms of special values of Jacobi polynomials P(n,alpha,beta,x), in Maple notation: a(n)= 2*(-1)^n*n!*JacobiP(n,2,-3/2-n,-7)/(n+2)!, n>=0. - Karol A. Penson, Jun 24 2013
G.f.: Q(0)/x - 1/x, where Q(k) = 1 + (4*k+1)*x/((1+x)*(k+1) - x*(1+x)*(2*k+2)*(4*k+3)/(x*(8*k+6)+(2*k+3)*(1+x)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 14 2013
Catalan(n+1) = Sum_{k=0..n} binomial(n,k)*a(k). E.g.: 42 = 1*1 + 4*1 + 6*2 + 4*4 + 1*9. - Doron Zeilberger, Mar 12 2015
G.f. A(x) with offset 1 satisfies: A(x)^2 = A( x^2/(1-2*x) ). - Paul D. Hanna, Nov 08 2015
a(n) = GegenbauerPoly(n,-n-1,-1/2)/(n+1). - Emanuele Munarini, Oct 20 2016
a(n) = a(n-1) + A002026(n-1). Number of Motzkin paths that start with an F step plus number of Motzkin paths that start with an U step. - R. J. Mathar, Jul 25 2017
G.f. A(x) satisfies A(x)*A(-x) = F(x^2), where F(x) is the g.f. of A168592. - Alexander Burstein, Oct 04 2017
G.f.: A(x) = exp(int((E(x)-1)/x dx)), where E(x) is the g.f. of A002426. Equivalently, E(x) = 1 + x*A'(x)/A(x). - Alexander Burstein, Oct 05 2017
G.f. A(x) satisfies: A(x) = Sum_{j>=0} x^j * Sum_{k=0..j} binomial(j,k)*x^k*A(x)^k. - Ilya Gutkovskiy, Apr 11 2019
From Gennady Eremin, May 08 2021: (Start)
G.f.: 2/(1 - x + sqrt(1-2*x-3*x^2)).
a(n) = A107587(n) + A343386(n) = 2*A107587(n) - A343773(n) = 2*A343386(n) + A343773(n). (End)
Revert transform of A049347 (after Michael Somos). - Gennady Eremin, Jun 11 2021
Sum_{n>=0} 1/a(n) = 2.941237337631025604300320152921013604885956025483079699366681494505960039781389... - Vaclav Kotesovec, Jun 17 2021
Let a(-1) = (1 - sqrt(-3))/2 and a(n) = a(-3-n)*(-3)^(n+3/2) for all n in Z. Then a(n) satisfies my previous formula relation from Mar 23 2012 now for all n in Z. - Michael Somos, Apr 17 2022
Let b(n) = 1 for n <= 1, otherwise b(n) = Sum_{k=2..n} b(k-1) * b(n-k), then a(n) = b(n+1) (conjecture). - Joerg Arndt, Jan 16 2023
From Peter Bala, Feb 03 2024: (Start)
G.f.: A(x) = 1/(1 + x)*c(x/(1 + x))^2 = 1 + x/(1 + x)*c(x/(1 + x))^3, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108.
A(x) = 1/(1 - 3*x)*c(-x/(1 -3*x))^2.
a(n+1) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n, k)*A000245(k+1).
a(n) = 3^n * Sum_{k = 0..n} (-3)^(-k)*binomial(n, k)*Catalan(k+1).
a(n) = 3^n * hypergeom([3/2, -n], [3], 4/3). (End)
G.f. A(x) satisfies A(x) = exp( x*A(x) + Integral x*A(x)/(1 - x^2*A(x)) dx ). - Paul D. Hanna, Mar 04 2024
a(n) = hypergeom([-n/2,1/2-n/2],[2],4). - Karol A. Penson, May 18 2025

{b(n)} is an analog of A037019 and of A340388: all prime factors of b(n) are all congruent to 1 modulo 6 and b(n) has exactly n divisors, so A002324(b(n)) = n. By definition we have A343771(n) <= b(n), and it seems that the equality holds for most n. This sequence lists the exceptions.
Since {b(n)} agrees with A343771(n) for most n, it cannot have its own entry.
Let q be a prime, then q^e is here if and only if e >= N+1, where N is the number of primes congruent to 1 modulo 6 below 7^q (N = 6, 32, 958, ... for q = 2, 3, 5, ...).
Proof: p_1 < p_2 < ... be the primes congruent to 1 modulo 6. Suppose that A343771(q^e) = (p_1)^(q^(m_1)-1) * (p_2)^(q^(m_2)-1) * ... * (p_r)^(q^(m_r)-1) with r <= e, m_1 >= m_2 >= ... >= m_r. If m_1 >= 2, then r < e, so we can substitute (p_1)^(q^(m_1)-1) with (p_1)^(q^(m_1-1)-1) * (p_{r+1})^(q-1), which a smaller number with exactly q^e divisors, a contradiction. So we have m_1 = 1, namely A343771(q^e) = b(q^e). On the other hand, if e >= N+1, then A343771(q^e) <= (p_1)^(q^2-1) * (p_2)^(q-1) * ... * (p_{e-1})^(q-1) < b(q^e).
It seems that q^(N+1) is the smallest q-rough term in this sequence.

Each term of this sequence is a multiple of 6.
Since no term of A002476 ends in 5, the longest run of 6's in this sequence will have length 3 (e.g., 61, 67, 73, 79 in A002476), the longest run of 12's will have length 3 (e.g., 397, 409, 421, 433 in A002476), the longest run of 18's will have length 3 (e.g., 673, 691, 709, 727 in A002476), and the longest run of 24's will have length 3 (e.g., 14149, 14173, 14197, 14221 in A002476). This run limit of length 3 also extends to other multiples of 6 that are not divisible by 5.
For multiples of 6 that are divisible by 5, the length of the longest run does not appear to be bounded.
Of course there cannot be 6 or more consecutive 30s in this sequence because then one of the primes must be divisible by 7, but there could be up to 10 consecutive 210s. The first run of four 30s corresponds to the primes 320149, 320179, 320209, 320239, 320269 and the first run of five 30s corresponds to the primes 28204591, 28204621, 28204651, 28204681, 28204711, 28204741. - Charles R Greathouse IV, Jan 27 2023

Discriminant=-7. Binary quadratic forms ax^2 + bxy + cy^2 have discriminant d = b^2 - 4ac.

Consider sequences of primes produced by forms with -100

The Mathematica function QuadPrimes2 is useful for finding the primes less than "lim" represented by the positive definite quadratic form ax^2 + bxy + cy^2 for any a, b and c satisfying a>0, c>0, and discriminant d<0. It does this by examining all x>=0 and y>=0 in the ellipse ax^2 + bxy + cy^2 <= lim. To find the primes generated by positive and negative x and y, compute the union of QuadPrimes2[a,b,c,lim] and QuadPrimes2[a,-b,c,lim]. - T. D. Noe, Sep 01 2009

For other programs see the "Binary Quadratic Forms and OEIS" link.

David A. Cox, Primes of the Form x^2 + n y^2, Wiley, 1989.
L. E. Dickson, History of the Theory of Numbers, Vol. 3, Chelsea, 1923.

Zak Seidov and N. J. A. Sloane, Table of n, a(n) for n = 1..10000 (The first 1225 terms were found by Zak Seidov)
N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references)

Discriminants in the range -3 to -100: A007645 (d=-3), A002313 (d=-4), A045373, A106856 (d=-7), A033203 (d=-8), A056874, A106857 (d=-11), A002476 (d=-12), A033212, A106858-A106861 (d=-15), A002144, A002313 (d=-16), A106862-A106863 (d=-19), A033205, A106864-A106865 (d=-20), A106866-A106869 (d=-23), A033199, A084865 (d=-24), A002476, A106870 (d=-27), A033207 (d=-28), A033221, A106871-A106874 (d=-31), A007519, A007520, A106875-A106876 (d=-32), A106877-A106881 (d=-35), A040117, A068228, A106882 (d=-36), A033227, A106883-A106888 (d=-39), A033201, A106889 (d=-40), A106890-A106891 (d=-43), A033209, A106282, A106892-A106893 (d=-44), A033232, A106894-A106900 (d=-47), A068229 (d=-48), A106901-A106904 (d=-51), A033210, A106905-A106906 (d=-52), A033235, A106907-A106913 (d=-55), A033211, A106914-A106917 (d=-56), A106918-A106922 (d=-59), A033212, A106859 (d=-60), A106923-A106930 (d=-63), A007521, A106931 (d=-64), A106932-A106933 (d=-67), A033213, A106934-A106938 (d=-68), A033246, A106939-A106948 (d=-71), A106949-A106950 (d=-72), A033212, A106951-A106952 (d=-75), A033214, A106953-A106955 (d=-76), A033251, A106956-A106962 (d=-79), A047650, A106963-A106965 (d=-80), A106966-A106970 (d=-83), A033215, A102271, A102273, A106971-A106974 (d=-84), A033256, A106975-A106983 (d=-87), A033216, A106984 (d=-88), A106985-A106989 (d=-91), A033217 (d=-92), A033206, A106990-A107001 (d=-95), A107002-A107008 (d=-96), A107009-A107013 (d=-99).
Other collections of quadratic forms: A139643, A139827.
For a more comprehensive list of sequences giving numbers and/or primes represented by binary quadratic forms, see the "Binary Quadratic Forms and OEIS" link.
Cf. also A242660.

QuadPrimes2[a_, b_, c_, lmt_] := Module[{p, d, lst = {}, xMax, yMax}, d = b^2 - 4a*c; If[a > 0 && c > 0 && d < 0, xMax = Sqrt[lmt/a]*(1+Abs[b]/Floor[Sqrt[-d]])]; Do[ If[ 4c*lmt + d*x^2 >= 0, yMax = ((-b)*x + Sqrt[4c*lmt + d*x^2])/(2c), yMax = 0 ]; Do[p = a*x^2 + b*x*y + c*y^2; If[ PrimeQ[ p]  && p <= lmt && !MemberQ[ lst, p], AppendTo[ lst, p]], {y, 0, yMax}], {x, 0, xMax}]; Sort[ lst]];
QuadPrimes2[1, 1, 2, 1000]
(This is a corrected version of the old, incorrect, program QuadPrimes. - N. J. A. Sloane, Jun 15 2014)
max = 1000; Table[yy = {y, 1, Floor[Sqrt[8 max - 7 x^2]/4 - x/4]}; Table[ x^2 + x y + 2 y^2, yy // Evaluate], {x, 0, Floor[Sqrt[max]]}] // Flatten // Union // Select[#, PrimeQ]& (* Jean-François Alcover, Oct 04 2018 *)

list(lim)=my(q=Qfb(1,1,2), v=List([2])); forprime(p=2, lim, if(vecmin(qfbsolve(q, p))>0, listput(v,p))); Vec(v) \\ Charles R Greathouse IV, Aug 05 2016

Removed old Mathematica programs - T. D. Noe, Sep 09 2009

Edited (pointed out error in QuadPrimes, added new version of program, checked and extended b-file). - N. J. A. Sloane, Jun 06 2014

cp's OEIS Frontend

A349772 Consider primitive 120-degree integer triangles with sides A < B < C = A002476(n). This sequence gives the values of A.

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A350347 Consider primitive 120-degree integer triangles with sides A < B < C = A002476(n). This sequence gives the values of B.

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A354543 Convolution of A007528 and A002476.

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A171836 a(n) = A002476(A002476(n)).

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A355919 Let n = p_1*p_2*...*p_k be the prime factorization of n, with the primes sorted in descending order; let b(n) = 7^(p_1 - 1)*13^(p_2 - 1)*19^(p_3 - 1)*...*A002476(k)^(p_k - 1). Sequence lists m such that b(m) > A343771(m).

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A359257 First differences of A002476.

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A144921 Duplicate of A002476.

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A000040 The prime numbers.

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A355919 Let n = p_1p_2...p_k be the prime factorization of n, with the primes sorted in descending order; let b(n) = 7^(p_1 - 1)13^(p_2 - 1)19^(p_3 - 1)...*A002476(k)^(p_k - 1). Sequence lists m such that b(m) > A343771(m).