cp's OEIS Frontend

Apparently a(n)/3 mod 2 = A010060(n-1), the Thue-Morse sequence.

a(n+1) is the smallest number with exactly n+1 partitions into distinct powers of 2 or of 3: A131996(a(n+1)) = n+1 and A131996(m) < n+1 for m < a(n+1). - Reinhard Zumkeller, Aug 06 2007

a(n) is always an integer.

Primes p such that p^2 divides A006134((p-1)/2) - Legendre(p, 3) are p = 103, ... What's the next?

a(n) is always an integer.

a(6) has 215 digits and a(7) has 654 digits.

For primes p we have A006134((p-1)/2) == Legendre(p, 3) (mod p). For composite n equal to a power of 3, n is also divisible by A006134((n-1)/2). Odd composite n not a power of 3 such that n divides A006134((n-1)/2) are n = 99, 1539, ... What's the next? Are there similar values of n not divisible by 3?

Conjecture: for n > 0, a(n) == 1 (mod 3) for odd n, a(n) == 2 (mod 3) for even n.

a(n) is always an integer.

Are there any primes p such that p^3 divides A006134(p-1) - Legendre(p, 3)?

a(n) is always an integer. k = A038754(n) - 1 is the smallest index that A006134(k) is divisible by 3^n.

The next term has 140 digits.

For primes p we have A006134(p-1) == Legendre(p, 3) (mod p^2). For composite n that is a power of 3, n^2 is also divisible by A006134(n-1). Are there any other such n?

Conjecture: for n > 1, a(n) == 1 (mod 27) for even n, a(n) == 13 (mod 27) for odd n.

The congruence A006134(k) == A102283(k) (mod k) holds for all values of k that are primes or squares of primes.

This is also the result of applying the transformation on generating functions A -> 1/((1 - x)*(1 - x*A)) to the g.f. for the Catalan numbers.

p divides a(p) - 3 for prime p = 3 and p = {7, 13, 19, 31, 37, 43, ...} = A002476 (Primes of the form 6*n + 1). p^2 divides a(p^2) - 3 for prime p > 3. - Alexander Adamchuk, Jul 11 2006

Prime p divides a(p) for p = {2, 3, 5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89, 101, ...} = A045309 (Primes congruent to {0, 2} mod 3); and A045309 (Primes p such that x^3 = n (integer) has only one solution mod p). Nonprime numbers n such that n divides a(n) are listed in A128287 = {1, 8, 133, ...}. - Alexander Adamchuk, Feb 23 2007

For p prime >= 5, a(p-1) = 1 or -2 (mod p) according as p = 1 or -1 (mod 3) (see Pan and Sun link). For example, with p=5, a(p-1) = 23 = -2 (mod p). - David Callan, Nov 29 2007

Hankel transform is A010892(n+1). - Paul Barry, Apr 24 2009

Equals INVERTi transform of A000245: (1, 3, 9, 28, ...). - Gary W. Adamson, May 15 2009

The subsequence of prime partial sums of Catalan numbers begins: a(1) = 2, a(4) = 23, a(6) = 197, a(16) = 48760367; see A121852. - Jonathan Vos Post, Feb 10 2010

Number of lattice paths from (0,0) to (n,n) which do not go above the diagonal x=y using steps (1,k), (k,1) with k >= 1 including two kinds of (1,1). - Alois P. Heinz, Oct 14 2015

Binomial transform of A086246(n+1) = [1, 1, 1, 2, 4, 9, ...], or, equivalently, of A001006 (Motzkin numbers) with 1 prepended.

Equivalently, primes of the form 3m + 1.

Rational primes that decompose in the field Q(sqrt(-3)). - N. J. A. Sloane, Dec 25 2017

Primes p dividing Sum_{k=0..p} binomial(2k, k) - 3 = A006134(p) - 3. - Benoit Cloitre, Feb 08 2003

Primes p such that tau(p) == 2 (mod 3) where tau(x) is the Ramanujan tau function (cf. A000594). - Benoit Cloitre, May 04 2003

Primes of the form x^2 + xy - 2y^2 = (x+2y)(x-y). - N. J. A. Sloane, May 31 2014

Primes of the form x^2 - xy + 7y^2 with x and y nonnegative. - T. D. Noe, May 07 2005

Primes p such that p^2 divides Sum_{m=1..2(p-1)} Sum_{k=1..m} (2k)!/(k!)^2. - Alexander Adamchuk, Jul 04 2006

A006512 larger than 5 (Greater of twin primes) is a subsequence of this. - Jonathan Vos Post, Sep 03 2006

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

Also primes p such that the arithmetic mean of divisors of p^2 is an integer: sigma_1(p^2)/sigma_0(p^2) = C. (A000203(p^2)/A000005(p^2) = C). - Ctibor O. Zizka, Sep 15 2008

Fermat knew that these numbers can also be expressed as x^2 + 3y^2 and are therefore not prime in Z[omega], where omega is a complex cubic root of unity. - Alonso del Arte, Dec 07 2012

Primes of the form x^2 + xy + y^2 with x < y and nonnegative. Also see A007645 which also applies when x=y, adding an initial 3. - Richard R. Forberg, Apr 11 2016

For any term p in this sequence, let k = (p^2 - 1)/6; then A016921(k) = p^2. - Sergey Pavlov, Dec 16 2016; corrected Dec 18 2016

For the decomposition p=x^2+3*y^2, x(n) = A001479(n+1) and y(n) = A001480(n+1). - R. J. Mathar, Apr 16 2024

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

Triangle of coefficients of the polynomial (x+1)(x+2)...(x+n), expanded in decreasing powers of x. - T. D. Noe, Feb 22 2008

Row n also gives the number of permutation of 1..n with complexity 0,1,...,n-1. See the comments in A008275. - N. J. A. Sloane, Feb 08 2019

T(n,k) is the number of deco polyominoes of height n and having k columns. A deco polyomino is a directed column-convex polyomino in which the height, measured along the diagonal, is attained only in the last column. Example: T(2,1)=1 and T(2,2)=1 because the deco polyominoes of height 2 are the vertical and horizontal dominoes, having, respectively, 1 and 2 columns. - Emeric Deutsch, Aug 14 2006

Sum_{k=1..n} k*T(n,k) = A121586. - Emeric Deutsch, Aug 14 2006

Let the triangle U(n,k), 0 <= k <= n, read by rows, be given by [1,0,1,0,1,0,1,0,1,0,1,...] DELTA [1,1,2,2,3,3,4,4,5,5,6,...] where DELTA is the operator defined in A084938; then T(n,k) = U(n-1,k-1). - Philippe Deléham, Jan 06 2007

From Tom Copeland, Dec 15 2007: (Start)

Consider c(t) = column vector(1, t, t^2, t^3, t^4, t^5, ...).

Starting at 1 and sampling every integer to the right, we obtain (1,2,3,4,5,...). And T * c(1) = (1, 1*2, 1*2*3, 1*2*3*4,...), giving n! for n > 0. Call this sequence the right factorial (n+)!.

Starting at 1 and sampling every integer to the left, we obtain (1,0,-1,-2,-3,-4,-5,...). And T * c(-1) = (1, 1*0, 1*0*-1, 1*0*-1*-2,...) = (1, 0, 0, 0, ...), the left factorial (n-)!.

Sampling every other integer to the right, we obtain (1,3,5,7,9,...). T * c(2) = (1, 1*3, 1*3*5, ...) = (1,3,15,105,945,...), giving A001147 for n > 0, the right double factorial, (n+)!!.

Sampling every other integer to the left, we obtain (1,-1,-3,-5,-7,...). T * c(-2) = (1, 1*-1, 1*-1*-3, 1*-1*-3*-5,...) = (1,-1,3,-15,105,-945,...) = signed A001147, the left double factorial, (n-)!!.

Sampling every 3 steps to the right, we obtain (1,4,7,10,...). T * c(3) = (1, 1*4, 1*4*7,...) = (1,4,28,280,...), giving A007559 for n > 0, the right triple factorial, (n+)!!!.

Sampling every 3 steps to the left, we obtain (1,-2,-5,-8,-11,...), giving T * c(-3) = (1, 1*-2, 1*-2*-5, 1*-2*-5*-8,...) = (1,-2,10,-80,880,...) = signed A008544, the left triple factorial, (n-)!!!.

The list partition transform A133314 of [1,T * c(t)] gives [1,T * c(-t)] with all odd terms negated; e.g., LPT[1,T*c(2)] = (1,-1,-1,-3,-15,-105,-945,...) = (1,-A001147). And e.g.f. for [1,T * c(t)] = (1-xt)^(-1/t).

The above results hold for t any real or complex number. (End)

Let R_n(x) be the real and I_n(x) the imaginary part of Product_{k=0..n} (x + I*k). Then, for n=1,2,..., we have R_n(x) = Sum_{k=0..floor((n+1)/2)}(-1)^k*Stirling1(n+1,n+1-2*k)*x^(n+1-2*k), I_n(x) = Sum_{k=0..floor(n/2)}(-1)^(k+1)*Stirling1(n+1,n-2*k)*x^(n-2*k). - Milan Janjic, May 11 2008

T(n,k) is also the number of permutations of n with "reflection length" k (i.e., obtained from 12..n by k not necessarily adjacent transpositions). For example, when n=3, 132, 213, 321 are obtained by one transposition, while 231 and 312 require two transpositions. - Kyle Petersen, Oct 15 2008

From Tom Copeland, Nov 02 2010: (Start)

[x^(y+1) D]^n = x^(n*y) [T(n,1)(xD)^n + T(n,2)y (xD)^(n-1) + ... + T(n,n)y^(n-1)(xD)], with D the derivative w.r.t. x.

E.g., [x^(y+1) D]^4 = x^(4*y) [(xD)^4 + 6 y(xD)^3 + 11 y^2(xD)^2 + 6 y^3(xD)].

(xD)^m can be further expanded in terms of the Stirling numbers of the second kind and operators of the form x^j D^j. (End)

With offset 0, 0 <= k <= n: T(n,k) is the sum of products of each size k subset of {1,2,...,n}. For example, T(3,2) = 11 because there are three subsets of size two: {1,2},{1,3},{2,3}. 1*2 + 1*3 + 2*3 = 11. - Geoffrey Critzer, Feb 04 2011

The Kn11, Fi1 and Fi2 triangle sums link this triangle with two sequences, see the crossrefs. For the definitions of these triangle sums see A180662. The mirror image of this triangle is A130534. - Johannes W. Meijer, Apr 20 2011

T(n+1,k+1) is the elementary symmetric function a_k(1,2,...,n), n >= 0, k >= 0, (a_0(0):=1). See the T. D. Noe and Geoffrey Critzer comments given above. For a proof see the Stanley reference, p. 19, Second Proof. - Wolfdieter Lang, Oct 24 2011

Let g(t) = 1/d(log(P(j+1,-t)))/dt (see Tom Copeland's 2007 formulas). The Mellin transform (t to s) of t*Dirac[g(t)] gives Sum_{n=1..j} n^(-s), which as j tends to infinity gives the Riemann zeta function for Re(s) > 1. Dirac(x) is the Dirac delta function. The complex contour integral along a circle of radius 1 centered at z=1 of z^s/g(z) gives the same result. - Tom Copeland, Dec 02 2011

Rows are coefficients of the polynomial expansions of the Pochhammer symbol, or rising factorial, Pch(n,x) = (x+n-1)!/(x-1)!. Expansion of Pch(n,xD) = Pch(n,Bell(.,:xD:)) in a polynomial with terms :xD:^k=x^k*D^k gives the Lah numbers A008297. Bell(n,x) are the unsigned Bell polynomials or Stirling polynomials of the second kind A008277. - Tom Copeland, Mar 01 2014

From Tom Copeland, Dec 09 2016: (Start)

The Betti numbers, or dimension, of the pure braid group cohomology. See pp. 12 and 13 of the Hyde and Lagarias link.

Row polynomials and their products appear in presentation of the Jack symmetric functions of R. Stanley. See Copeland link on the Witt differential generator.

(End)

From Tom Copeland, Dec 16 2019: (Start)

The e.g.f. given by Copeland in the formula section appears in a combinatorial Dyson-Schwinger equation of quantum field theory in Yeats in Thm. 2 on p. 62 related to a Hopf algebra of rooted trees. See also the Green function on p. 70.

Per comments above, this array contains the coefficients in the expansion in polynomials of the Euler, or state number, operator xD of the rising factorials Pch(n,xD) = (xD+n-1)!/(xD-1)! = x [:Dx:^n/n!]x^{-1} = L_n^{-1}(-:xD:), where :Dx:^n = D^n x^n and :xD:^n = x^n D^n. The polynomials L_n^{-1} are the Laguerre polynomials of order -1, i.e., normalized Lah polynomials.

The Witt differential operators L_n = x^(n+1) D and the row e.g.f.s appear in Hopf and dual Hopf algebra relations presented by Foissy. The Witt operators satisfy L_n L_k - L_k L_n = (k-n) L_(n+k), as for the dual Hopf algebra. (End)