cp's OEIS Frontend

Moebius inversion: f(n) = Sum_{d|n} g(d) for all n <=> g(n) = Sum_{d|n} mu(d)*f(n/d) for all n.

a(n) depends only on prime signature of n (cf. A025487). So a(24) = a(375) since 24 = 2^3 * 3 and 375 = 3 * 5^3 both have prime signature (3, 1).

A008683 = A140579^(-1) * A140664. - Gary W. Adamson, May 20 2008

Coons & Borwein prove that Sum_{n>=1} mu(n) z^n is transcendental. - Jonathan Vos Post, Jun 11 2008; edited by Charles R Greathouse IV, Sep 06 2017

Equals row sums of triangle A144735 (the square of triangle A054533). - Gary W. Adamson, Sep 20 2008

Conjecture: a(n) is the determinant of Redheffer matrix A143104 where T(n, n) = 0. Verified for the first 50 terms. - Mats Granvik, Jul 25 2008

From Mats Granvik, Dec 06 2008: (Start)

The Editorial Office of the Journal of Number Theory kindly provided (via B. Conrey) the following proof of the conjecture: Let A be A143104 and B be A143104 where T(n, n) = 0.

"Suppose you expand det(B_n) along the bottom row. There is only a 1 in the first position and so the answer is (-1)^n times det(C_{n-1}) say, where C_{n-1} is the (n-1) by (n-1) matrix obtained from B_n by deleting the first column and the last row. Now the determinant of the Redheffer matrix is det(A_n) = M(n) where M(n) is the sum of mu(m) for 1 <= m <= n. Expanding det(A_n) along the bottom row, we see that det(A_n) = (-1)^n * det(C_{n-1}) + M(n-1). So we have det(B_n) = (-1)^n * det(C_{n-1}) = det(A_n) - M(n-1) = M(n) - M(n-1) = mu(n)." (End)

Conjecture: Consider the table A051731 and treat 1 as a divisor. Move the value in the lower right corner vertically to a divisor position in the transpose of the table and you will find that the determinant is the Moebius function. The number of permutation matrices that contribute to the Moebius function appears to be A074206. - Mats Granvik, Dec 08 2008

Convolved with A152902 = A000027, the natural numbers. - Gary W. Adamson, Dec 14 2008

[Pickover, p. 226]: "The probability that a number falls in the -1 mailbox turns out to be 3/Pi^2 - the same probability as for falling in the +1 mailbox". - Gary W. Adamson, Aug 13 2009

Let A = A176890 and B = A * A * ... * A, then the leftmost column in matrix B converges to the Moebius function. - Mats Granvik, Gary W. Adamson, Apr 28 2010 and May 28 2020

Equals row sums of triangle A176918. - Gary W. Adamson, Apr 29 2010

Calculate matrix powers: A175992^0 - A175992^1 + A175992^2 - A175992^3 + A175992^4 - ... Then the Mobius function is found in the first column. Compare this to the binomial series for (1+x)^-1 = 1 - x + x^2 - x^3 + x^4 - ... . - Mats Granvik, Gary W. Adamson, Dec 06 2010

From Richard L. Ollerton, May 08 2021: (Start)

Formulas for the numerous OEIS entries involving the Möbius transform (Dirichlet convolution of a(n) and some sequence h(n)) can be derived using the following (n >= 1):

Sum_{d|n} mu(d)*h(n/d) = Sum_{k=1..n} h(gcd(n,k))*mu(n/gcd(n,k))/phi(n/gcd(n,k)) = Sum_{k=1..n} h(n/gcd(n,k))*mu(gcd(n,k))/phi(n/gcd(n,k)), where phi = A000010.

Use of gcd(n,k)*lcm(n,k) = n*k provides further variations. (End)

Formulas for products corresponding to the sums above are also available for sequences f(n) > 0: Product_{d|n} f(n/d)^mu(d) = Product_{k=1..n} f(gcd(n,k))^(mu(n/gcd(n,k))/phi(n/gcd(n,k))) = Product_{k=1..n} f(n/gcd(n,k))^(mu(gcd(n,k))/phi(n/gcd(n,k))). - Richard L. Ollerton, Nov 08 2021

When convolved with the partition numbers A000041 gives 1, 0, 0, 0, 0, ...

Also, number of different partitions of n into parts of -1 different kinds (based upon formal analogy). - Michele Dondi (blazar(AT)lcm.mi.infn.it), Jun 29 2004

The comment that "when convolved with the partition numbers gives [1, 0, 0, 0, ...]" is equivalent to row sums of triangle A145975 = [1, 0, 0, 0, ...]; where A145975 is a partition number convolution triangle. - Gary W. Adamson, Oct 25 2008

When convolved with n-th partial sums of A000041 = the binomial sequence starting (1, n, ...). Example: A010815 convolved with A014160 (partial sum operation applied thrice to the partition numbers) = (1, 3, 6, 10, ...). - Gary W. Adamson, Nov 11 2008

(A000012^(-n) * A000041) convolved with A010815 = n-th row of the inverse of Pascal's triangle, (as a vector, followed by zeros); where A000012^(-1) = the pairwise difference operator. Example: (A000012^(-4) * A000041) convolved with A010815 = (1, -4, 6, -4, 1, 0, 0, 0, ...). - Gary W. Adamson, Nov 11 2008

Also sum of [product of (1-2/(hook lengths)^2)] over all partitions of n. - Wouter Meeussen, Sep 16 2010

Cayley (1895) begins article 387 with "Write for shortness sqrt(2k'K / pi) / [1-q^{2m-1}]^2 = G, ..." which is a convoluted way of writing G = [1-q^{2m}] = (1-q^2)(1-q^4)... - Michael Somos, Aug 01 2011

This is an example of the quintuple product identity in the form f(a*b^4, a^2/b) - (a/b) * f(a^4*b, b^2/a) = f(-a*b, -a^2*b^2) * f(-a/b, -b^2) / f(a, b) where a = x^3, b = x. - Michael Somos, Jan 21 2012

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

Number 1 of the 14 primitive eta-products which are holomorphic modular forms of weight 1/2 listed by D. Zagier on page 30 of "The 1-2-3 of Modular Forms". - Michael Somos, May 04 2016

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

This is an example of the quintuple product identity in the form f(a*b^4, a^2/b) - (a/b) * f(a^4*b, b^2/a) = f(-a*b, -a^2*b^2) * f(-a/b, -b^2) / f(a, b) where a = -x^3, b = -x. - Michael Somos, Jul 11 2012

Number 5 of the 14 primitive eta-products which are holomorphic modular forms of weight 1/2 listed by D. Zagier on page 30 of "The 1-2-3 of Modular Forms". - Michael Somos, May 04 2016

G.f. for row polynomials S(n,x) (signed triangle): 1/(1-x*z+z^2). Unsigned triangle |a(n,m)| has Fibonacci polynomials F(n+1,x) as row polynomials with g.f. 1/(1-x*z-z^2). |a(n,m)| triangle has rows of Pascal's triangle A007318 in the even-numbered diagonals (odd-numbered ones have only 0's).

Row sums (unsigned triangle) A000045(n+1) (Fibonacci). Row sums (signed triangle) S(n,1) sequence = periodic(1,1,0,-1,-1,0) = A010892.

Alternating row sums A049347(n) = S(n,-1) = periodic(1,-1,0). - Wolfdieter Lang, Nov 04 2011

S(n,x) is the characteristic polynomial of the adjacency matrix of the n-path. - Michael Somos, Jun 24 2002

S(n,x) is also the matching polynomial of the n-path. - Eric W. Weisstein, Apr 10 2017

|T(n,k)| = number of compositions of n+1 into k+1 odd parts. Example: |T(7,3)| = 10 because we have (1,1,3,3), (1,3,1,3), (1,3,3,1), (3,1,1,3), (3,1,3,1), (3,3,1,1), (1,1,1,5), (1,1,5,1), (1,5,1,1) and (5,1,1,1). - Emeric Deutsch, Apr 09 2005

S(n,x)= R(n,x) + S(n-2,x), n >= 2, S(-1,x)=0, S(0,x)=1, R(n,x):=2*T(n,x/2) = Sum_{m=0..n} A127672(n,m)*x^m (monic integer Chebyshev T-Polynomials). This is the rewritten so-called trace of the transfer matrix formula for the T-polynomials. - Wolfdieter Lang, Dec 02 2010

In a regular N-gon inscribed in a unit circle, the side length is d(N,1) = 2*sin(Pi/N). The length ratio R(N,k):=d(N,k)/d(N,1) for the (k-1)-th diagonal, with k from {2,3,...,floor(N/2)}, N >= 4, equals S(k-1,x) = sin(k*Pi/N)/sin(Pi/N) with x=rho(N):=R(N,2) = 2*cos(Pi/N). Example: N=7 (heptagon), rho=R(7,2), sigma:=R(N,3) = S(2,rho) = rho^2 - 1. Motivated by the quoted paper by P. Steinbach. - Wolfdieter Lang, Dec 02 2010

From Wolfdieter Lang, Jul 12 2011: (Start)

In q- or basic analysis, q-numbers are [n]_q := S(n-1,q+1/q) = (q^n-(1/q)^n)/(q-1/q), with the row polynomials S(n,x), n >= 0.

The zeros of the row polynomials S(n-1,x) are (from those of Chebyshev U-polynomials):

x(n-1;k) = +- t(k,rho(n)), k = 1..ceiling((n-1)/2), n >= 2, with t(n,x) the row polynomials of A127672 and rho(n):= 2*cos(Pi/n). The simple vanishing zero for even n appears here as +0 and -0.

Factorization of the row polynomials S(n-1,x), x >= 1, in terms of the minimal polynomials of cos(2 Pi/2), called Psi(n,x), with coefficients given by A181875/A181876:

S(n-1,x) = (2^(n-1))*Product_{n>=1}(Psi(d,x/2), 2 < d | 2n).

(From the rewritten eq. (3) of the Watkins and Zeitlin reference, given under A181872.) [See the W. Lang ArXiv link, Proposition 9, eq. (62). - Wolfdieter Lang, Apr 14 2018]

(End)

The discriminants of the S(n,x) polynomials are found in A127670. - Wolfdieter Lang, Aug 03 2011

This is an example for a subclass of Riordan convolution arrays (lower triangular matrices) called Bell arrays. See the L. W. Shapiro et al. reference under A007318. If a Riordan array is named (G(z),F(z)) with F(z)=z*Fhat(z), the o.g.f. for the row polynomials is G(z)/(1-x*z*Fhat(z)), and it becomes a Bell array if G(z)=Fhat(z). For the present Bell type triangle G(z)=1/(1+z^2) (see the o.g.f. comment above). This leads to the o.g.f. for the column no. k, k >= 0, x^k/(1+x^2)^(k+1) (see the formula section), the one for the row sums and for the alternating row sums (see comments above). The Riordan (Bell) A- and Z-sequences (defined in a W. Lang link under A006232, with references) have o.g.f.s 1-x*c(x^2) and -x*c(x^2), with the o.g.f. of the Catalan numbers A000108. Together they lead to a recurrence given in the formula section. - Wolfdieter Lang, Nov 04 2011

The determinant of the N x N matrix S(N,[x[1], ..., x[N]]) with elements S(m-1,x[n]), for n, m = 1, 2, ..., N, and for any x[n], is identical with the determinant of V(N,[x[1], ..., x[N]]) with elements x[n]^(m-1) (a Vandermondian, which equals Product_{1 <= i < j<= N} (x[j] - x[i])). This is a special instance of a theorem valid for any N >= 1 and any monic polynomial system p(m,x), m>=0, with p(0,x) = 1. For this theorem see the Vein-Dale reference, p. 59. Thanks to L. Edson Jeffery for an email asking for a proof of the non-singularity of the matrix S(N,[x[1], ...., x[N]]) if and only if the x[j], j = 1..N, are pairwise distinct. - Wolfdieter Lang, Aug 26 2013

These S polynomials also appear in the context of modular forms. The rescaled Hecke operator T*n = n^((1-k)/2)*T_n acting on modular forms of weight k satisfies T*(p^n) = S(n, T*p), for each prime p and positive integer n. See the Koecher-Krieg reference, p. 223. - _Wolfdieter Lang, Jan 22 2016

For a shifted o.g.f. (mod signs), its compositional inverse, and connections to Motzkin and Fibonacci polynomials, non-crossing partitions and other combinatorial structures, see A097610. - Tom Copeland, Jan 23 2016

From M. Sinan Kul, Jan 30 2016; edited by Wolfdieter Lang, Jan 31 2016 and Feb 01 2016: (Start)

Solutions of the Diophantine equation u^2 + v^2 - k*u*v = 1 for integer k given by (u(k,n), v(k,n)) = (S(n,k), S(n-1,k)) because of the Cassini-Simson identity: S(n,x)^2 - S(n+1,x)*S(n-1, x) = 1, after use of the S-recurrence. Note that S(-n, x) = -S(-n-2, x), n >= 1, and the periodicity of some S(n, k) sequences.

Hence another way to obtain the row polynomials would be to take powers of the matrix [x, -1; 1,0]: S(n, x) = (([x, -1; 1, 0])^n)[1,1], n >= 0.

See also a Feb 01 2016 comment on A115139 for a well-known S(n, x) sum formula.

Then we have with the present T triangle

A039834(n) = -i^(n+1)*T(n-1, k) where i is the imaginary unit and n >= 0.

A051286(n) = Sum_{i=0..n} T(n,i)^2 (see the Philippe Deléham, Nov 21 2005 formula),

A181545(n) = Sum_{i=0..n+1} abs(T(n,i)^3),

A181546(n) = Sum_{i=0..n+1} T(n,i)^4,

A181547(n) = Sum_{i=0..n+1} abs(T(n,i)^5).

S(n, 0) = A056594(n), and for k = 1..10 the sequences S(n-1, k) with offset n = 0 are A128834, A001477, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189.

(End)

For more on the Diophantine equation presented by Kul, see the Ismail paper. - Tom Copeland, Jan 31 2016

The o.g.f. for the Legendre polynomials L(n,x) is 1 / sqrt(1- 2x*z + z^2), and squaring it gives the o.g.f. of U(n,x), A053117, so Sum_{k=0..n} L(k,x/2) L(n-k,x/2) = S(n,x). This gives S(n,x) = L(n/2,x/2)^2 + 2*Sum_{k=0..n/2-1} L(k,x/2) L(n-k,x/2) for n even and S(n,x) = 2*Sum_{k=0..(n-1)/2} L(k,x/2) L(n-k,x/2) for odd n. For a connection to elliptic curves and modular forms, see A053117. For the normalized Legendre polynomials, see A100258. For other properties and relations to other polynomials, see Allouche et al. - Tom Copeland, Feb 04 2016

LG(x,h1,h2) = -log(1 - h1*x + h2*x^2) = Sum_{n>0} F(n,-h1,h2,0,..,0) x^n/n is a log series generator of the bivariate row polynomials of A127672 with A127672(0,0) = 0 and where F(n,b1,b2,..,bn) are the Faber polynomials of A263916. Exp(LG(x,h1,h2)) = 1 / (1 - h1*x + h2*x^2 ) is the o.g.f. of the bivariate row polynomials of this entry. - Tom Copeland, Feb 15 2016 (Instances of the bivariate o.g.f. for this entry are on pp. 5 and 18 of Sunada. - Tom Copeland, Jan 18 2021)

For distinct odd primes p and q the Legendre symbol can be written as Legendre(q,p) = Product_{k=1..P} S(q-1, 2*cos(2*Pi*k/p)), with P = (p-1)/2. See the Lemmermeyer reference, eq. (8.1) on p. 236. Using the zeros of S(q-1, x) (see above) one has S(q-1, x) = Product_{l=1..Q} (x^2 - (2*cos(Pi*l/q))^2), with Q = (q-1)/2. Thus S(q-1, 2*cos(2*Pi*k/p)) = ((-4)^Q)*Product_{l=1..Q} (sin^2(2*Pi*k/p) - sin^2(Pi*l/q)) = ((-4)^Q)*Product_{m=1..Q} (sin^2(2*Pi*k/p) - sin^2(2*Pi*m/q)). For the proof of the last equality see a W. Lang comment on the triangle A057059 for n = Q and an obvious function f. This leads to Eisenstein's proof of the quadratic reciprocity law Legendre(q,p) = ((-1)^(P*Q)) * Legendre(p,q), See the Lemmermeyer reference, pp. 236-237. - Wolfdieter Lang, Aug 28 2016

For connections to generalized Fibonacci polynomials, compare their generating function on p. 5 of the Amdeberhan et al. link with the o.g.f. given above for the bivariate row polynomials of this entry. - Tom Copeland, Jan 08 2017

The formula for Ramanujan's tau function (see A000594) for prime powers is tau(p^k) = p^(11*k/2)*S(k, p^(-11/2)*tau(p)) for k >= 1, and p = A000040(n), n >= 1. See the Hardy reference, p. 164, eqs. (10.3.4) and (10.3.6) rewritten in terms of S. - Wolfdieter Lang, Jan 27 2017

From Wolfdieter Lang, May 08 2017: (Start)

The number of zeros Z(n) of the S(n, x) polynomials in the open interval (-1,+1) is 2*b(n) for even n >= 0 and 1 + 2*b(n) for odd n >= 1, where b(n) = floor(n/2) - floor((n+1)/3). This b(n) is the number of integers k in the interval (n+1)/3 < k <= floor(n/2). See a comment on the zeros of S(n, x) above, and b(n) = A008615(n-2), n >= 0. The numbers Z(n) have been proposed (with a conjecture related to A008611) by Michel Lagneau, as the number of zeros of Fibonacci polynomials on the imaginary axis (-I,+I), with I=sqrt(-1). They are Z(n) = A008611(n-1), n >= 0, with A008611(-1) = 0. Also Z(n) = A194960(n-4), n >= 0. Proof using the A008611 version. A194960 follows from this.

In general the number of zeros Z(a;n) of S(n, x) for n >= 0 in the open interval (-a,+a) for a from the interval (0,2) (x >= 2 never has zeros, and a=0 is trivial: Z(0;n) = 0) is with b(a;n) = floor(n//2) - floor((n+1)*arccos(a/2)/Pi), as above Z(a;n) = 2*b(a;n) for even n >= 0 and 1 + 2*b(a;n) for odd n >= 1. For the closed interval [-a,+a] Z(0;n) = 1 and for a from (0,1) one uses for Z(a;n) the values b(a;n) = floor(n/2) - ceiling((n+1)*arccos(a/2)/Pi) + 1. (End)

The Riordan row polynomials S(n, x) (Chebyshev S) belong to the Boas-Buck class (see a comment and references in A046521), hence they satisfy the Boas-Buck identity: (E_x - n*1)*S(n, x) = (E_x + 1)*Sum_{p=0..n-1} (1 - (-1)^p)*(-1)^((p+1)/2)*S(n-1-p, x), for n >= 0, where E_x = x*d/dx (Euler operator). For the triangle T(n, k) this entails a recurrence for the sequence of column k, given in the formula section. - Wolfdieter Lang, Aug 11 2017

The e.g.f. E(x,t) := Sum_{n>=0} (t^n/n!)*S(n,x) for the row polynomials is obtained via inverse Laplace transformation from the above given o.g.f. as E(x,t) = ((1/xm)*exp(t/xm) - (1/xp)*exp(t/xp) )/(xp - xm) with xp = (x + sqrt(x^2-4))/2 and xm = (x - sqrt(x^2-4))/2. - Wolfdieter Lang, Nov 08 2017

From Wolfdieter Lang, Apr 12 2018: (Start)

Factorization of row polynomials S(n, x), for n >= 1, in terms of C polynomials (not Chebyshev C) with coefficients given in A187360. This is obtained from the factorization into Psi polynomials (see the Jul 12 2011 comment above) but written in terms of minimal polynomials of 2*cos(2*Pi/n) with coefficients in A232624:

S(2*k, x) = Product_{2 <= d | (2*k+1)} C(d, x)*(-1)^deg(d)*C(d, -x), with deg(d) = A055034(d) the degree of C(d, x).

S(2*k+1, x) = Product_{2 <= d | 2*(k+1)} C(d, x) * Product_{3 <= 2*d + 1 | (k+1)} (-1)^(deg(2*d+1))*C(2*d+1, -x).

Note that (-1)^(deg(2*d+1))*C(2*d+1, -x)*C(2*d+1, x) pairs always appear.

The number of C factors of S(2*k, x), for k >= 0, is 2*(tau(2*k+1) - 1) = 2*(A099774(k+1) - 1) = 2*A095374(k), and for S(2*k+1, x), for k >= 0, it is tau(2*(k+1)) + tau_{odd}(k+1) - 2 = A302707(k), with tau(2*k+1) = A099774(k+1), tau(n) = A000005 and tau(2*(k+1)) = A099777(k+1).

For the reverse problem, the factorization of C polynomials into S polynomials, see A255237. (End)

The S polynomials with general initial conditions S(a,b;n,x) = x*S(a,b;n-1,x) - S(a,b;n-2,x), for n >= 1, with S(a,b;-1,x) = a and S(a,b;0,x) = b are S(a,b;n,x) = b*S(n, x) - a*S(n-1, x), for n >= -1. Recall that S(-2, x) = -1 and S(-1, x) = 0. The o.g.f. is G(a,b;z,x) = (b - a*z)/(1 - x*z + z^2). - Wolfdieter Lang, Oct 18 2019

Also the convolution triangle of A101455. - Peter Luschny, Oct 06 2022

From Wolfdieter Lang, Apr 26 2023: (Start)

Multi-section of S-polynomials: S(m*n+k, x) = S(m+k, x)*S(n-1, R(m, x)) - S(k, x)*S(n-2, R(m, x)), with R(n, x) = S(n, x) - S(n-2, x) (see A127672), S(-2, x) = -1, and S(-1, x) = 0, for n >= 0, m >= 1, and k = 0, 1, ..., m-1.

O.g.f. of {S(m*n+k, y)}_{n>=0}: G(m,k,y,x) = (S(k, y) - (S(k, y)*R(m, y) - S(m+k, y))*x)/(1 - R(m,y)*x + x^2).

See eqs. (40) and (49), with r = x or y and s =-1, of the G. Detlefs and W. Lang link at A034807. (End)

S(n, x) for complex n and complex x: S(n, x) = ((-i/2)/sqrt(1 - (x/2)^2))*(q(x/2)*exp(+n*log(q(x/2))) - (1/q(x/2))*exp(-n*log(q(x/2)))), with q(x) = x + sqrt(1 - x^2)*i. Here log(z) = |z| + Arg(z)*i, with Arg(z) from [-Pi,+Pi) (principal branch). This satisfies the recurrence relation for S because it is derived from the Binet - de Moivre formula for S. Examples: S(n/m, 0) = cos((n/m)*Pi/4), for n >= 0 and m >= 1. S(n*i, 0) = (1/2)*(1 + exp(n*Pi))*exp(-(n/2)*Pi), for n >= 0. S(1+i, 2+i) = 0.6397424847... + 1.0355669490...*i. Thanks to Roberto Alfano for asking a question leading to this formula. - Wolfdieter Lang, Jun 05 2023

Lim_{n->oo} S(n, x)/S(n-1, x) = r(x) = (x - sqrt(x^2 -4))/2, for |x| >= 2. For x = +-2, this limit is +-1. - Wolfdieter Lang, Nov 15 2023

The unsigned numbers are also called Stirling cycle numbers: |s(n,k)| = number of permutations of n objects with exactly k cycles.

The unsigned numbers (read from right to left) also give the number of permutations of 1..n with complexity k, where the complexity of a permutation is defined to be the sum of the lengths of the cycles minus the number of cycles. In other words, the complexity equals the sum of (length of cycle)-1 over all cycles. For n=5, the numbers of permutations with complexity 0,1,2,3,4 are 1, 10, 35, 50, 24. - N. J. A. Sloane, Feb 08 2019

The unsigned numbers are also the number of permutations of 1..n with k left to right maxima (see Khovanova and Lewis, Smith).

With P(n) = the number of integer partitions of n, T(i,n) = the number of parts of the i-th partition of n, D(i,n) = the number of different parts of the i-th partition of n, p(j,i,n) = the j-th part of the i-th partition of n, m(j,i,n) = multiplicity of the j-th part of the i-th partition of n, Sum_[T(i,n)=k]{i=1}^{P(n)} = sum running from i=1 to i=p(n) but taking only partitions with T(i,n)=k parts into account, Product{j=1..T(i,n)} = product running from j=1 to j=T(i,n), Product_{j=1..D(i,n)} = product running from j=1 to j=D(i,n) one has S1(n,k) = Sum_[T(i,n)=k]{i=1}^{P(n)} (n!/Product{j=1..T(i,n)} p(j,i,n))* (1/Product_{j=1..D(i,n)} m(j,i,n)!). For example, S1(6,3) = 225 because n=6 has the following partitions with k=3 parts: (114), (123), (222). Their complexions are: (114): (6!/1*1*4)*(1/2!*1!) = 90, (123): (6!/1*2*3)*(1/1!*1!*1!) = 120, (222): (6!/2*2*2)*(1/3!) = 15. The sum of the complexions is 90+120+15 = 225 = S1(6,3). - Thomas Wieder, Aug 04 2005

Row sums equal 0. - Jon Perry, Nov 14 2005

|s(n,k)| enumerates unordered n-vertex forests composed of k increasing non-plane (unordered) trees. Proof from the e.g.f. of the first column and the F. Bergeron et al. reference, especially Table 1, last row (non-plane "recursive"), given in A049029. - Wolfdieter Lang, Oct 12 2007

|s(n,k)| enumerates unordered increasing n-vertex k-forests composed of k unary trees (out-degree r from {0,1}) whose vertices of depth (distance from the root) j >= 0 come in j+1 colors (j=0 for the k roots). - Wolfdieter Lang, Oct 12 2007, Feb 22 2008

A refinement of the unsigned array is A036039. For an association to forests of "naturally grown" rooted non-planar trees, dispositions of flags on flagpoles, and colorings of the vertices of the complete graphs K_n, see A130534. - Tom Copeland, Mar 30 and Apr 05 2014

The Stirling numbers of the first kind were related to the falling factorial and the convolved, or generalized, Bernoulli numbers B_n by Norlund in 1924 through Sum_{k=1..n+1} T(n+1,k) * x^(k-1) = (x-1)!/(x-1-n)! = (x + B.(0))^n = B_n(x), umbrally evaluated with (B.(0))^k = B_k(0) and the associated Appell polynomial B_n(x) defined by the e.g.f. (t/(exp(t) - 1))^(n+1) * exp(x*t) = exp(B.(x)t). - Tom Copeland, Sep 29 2015

With x = e^z, D_x = d/dx, D_z = d/dz, and p_n(x) the row polynomials of this entry, x^n (D_x)^n = p_n(D_z) = (D_z)! / (D_z - n)! = (xD_x)! / (xD_x - n)!. - Tom Copeland, Nov 27 2015

From the operator relation z + Psi(1) + sum_{n > 0} (-1)^n (-1/n) binomial(D,n) = z + Psi(1+D) with D = d/dz and Psi the digamma function, Zeta(n+1) = Sum_{k > n-1} (1/k) |S(k,n)| / k! for n > 0 and Zeta the Riemann zeta function. - Tom Copeland, Aug 12 2016

Let X_1,...,X_n be i.i.d. random variables with exponential distribution having mean = 1. Let Y = max{X_1,...,X_n}. Then (-1)^n*n!/( Sum_{k=1..n+1} a(n+1,k) t^(k-1) ) is the moment generating function of Y. The expectation of Y is the n-th harmonic number. - Geoffrey Critzer, Dec 25 2018

In the Ewens sampling theory describing the multivariate probability distribution of the sizes of the allelic classes in a sample of size n under the Infinite Alleles Model, |s(n,k)| gives the coefficient in the formula for the probability that a sample of n alleles has exactly k distinct types. - Noah A Rosenberg, Feb 10 2019

Named by Nielson (1906) after the Scottish mathematician James Stirling (1692-1770). - Amiram Eldar, Jun 11 2021 and Oct 02 2023

The first few row polynomials along with a recursion formula are found in a manuscript by Newton written in 1664 or 1665 (p. 169 of Turnbull) giving a geometric presentation of the binomial theorem for rational powers. - Tom Copeland, Dec 10 2022

a(n)/A027642(n) (Bernoulli numbers) provide the a-sequence for the Sheffer matrix A094816 (coefficients of orthogonal Poisson-Charlier polynomials). See the W. Lang link under A006232 for a- and z-sequences for Sheffer matrices. The corresponding z-sequence is given by the rationals A130189(n)/A130190(n).

Harvey (2008) describes a new algorithm for computing Bernoulli numbers. His method is to compute B(k) modulo p for many small primes p and then reconstruct B(k) via the Chinese Remainder Theorem. The time complexity is O(k^2 log(k)^(2+eps)). The algorithm is especially well-suited to parallelization. - Jonathan Vos Post, Jul 09 2008

Regard the Bernoulli numbers as forming a vector = B_n, and the variant starting (1, 1/2, 1/6, 0, -1/30, ...), (i.e., the first 1/2 has sign +) as forming a vector Bv_n. The relationship between the Pascal triangle matrix, B_n, and Bv_n is as follows: The binomial transform of B_n = Bv_n. B_n is unchanged when multiplied by the Pascal matrix with rows signed (+-+-, ...), i.e., (1; -1,-1; 1,2,1; ...). Bv_n is unchanged when multiplied by the Pascal matrix with columns signed (+-+-, ...), i.e., (1; 1,-1; 1,-2,1; 1,-3,3,-1; ...). - Gary W. Adamson, Jun 29 2012

The sequence of the Bernoulli numbers B_n = a(n)/A027642(n) is the inverse binomial transform of the sequence {A164555(n)/A027642(n)}, illustrated by the fact that they appear as top row and left column in A190339. - Paul Curtz, May 13 2016

Named by de Moivre (1773; "the numbers of Mr. James Bernoulli") after the Swiss mathematician Jacob Bernoulli (1655-1705). - Amiram Eldar, Oct 02 2023

The unsigned numbers are also called Stirling cycle numbers: |s(n,k)| = number of permutations of n objects with exactly k cycles.

Mirror image of the triangle A054654. - Philippe Deléham, Dec 30 2006

Also the triangle gives coefficients T(n,k) of x^k in the expansion of C(x,n) = (a(k)*x^k)/n!. - Mokhtar Mohamed, Dec 04 2012

From Wolfdieter Lang, Nov 14 2018: (Start)

This is the Sheffer triangle of Jabotinsky type (1, log(1 + x)). See the e.g.f. of the triangle below.

This is the inverse Sheffer triangle of the Stirling2 Sheffer triangle A008275.

The a-sequence of this Sheffer triangle (see a W. Lang link in A006232)

is from the e.g.f. A(x) = x/(exp(x) -1) a(n) = Bernoulli(n) = A027641(n)/A027642(n), for n >= 0. The z-sequence vanishes.

The Boas-Buck sequence for the recurrences of columns has o.g.f. B(x) = Sum_{n>=0} b(n)*x^n = 1/((1 + x)*log(1 + x)) - 1/x. b(n) = (-1)^(n+1)*A002208(n+1)/A002209(n+1), b = {-1/2, 5/12, -3/8, 251/720, -95/288, 19087/60480,...}. For the Boas-Buck recurrence of Riordan and Sheffer triangles see the Aug 10 2017 remark in A046521, adapted to the Sheffer case, also for two references. See the recurrence and example below. (End)

Let G(n,m,k) be the number of simple labeled graphs on [n] with m edges and k components. Then T(n,k) = Sum (-1)^m*G(n,m,k). See the Read link below. Equivalently, T(n,k) = Sum mu(0,p) where the sum is over all set partitions p of [n] containing k blocks and mu is the Moebius function in the incidence algebra associated to the set partition lattice on [n]. - Geoffrey Critzer, May 11 2024

Unsigned sequence is expansion of theta series of hexagonal net with respect to a node.

Cubic AGM theta functions: a(q) (see A004016), b(q) (this: A005928), c(q) (A005882).

Denoted by a_3(n) in Kassel and Reutenauer 2015. - Michael Somos, Jun 04 2015

Row sums (signed triangle): A000012 (powers of 1). Row sums (unsigned triangle): A001333(n).

From Wolfdieter Lang, Oct 21 2013: (Start)

The row polynomials T(n,x) equal (S(n,2*x) - S(n-2,2*x))/2, n >= 0, with the row polynomials S from A049310, with S(-1,x) = 0, and S(-2,x) = -1.

The zeros of T(n,x) are x(n,k) = cos((2*k+1)*Pi/(2*n)), k = 0, 1, ..., n-1, n >= 1. (End)

From Wolfdieter Lang, Jan 03 2020 and Paul Weisenhorn: (Start)

The (sub)diagonal sequences {D_{2*k}(m)}{m >= 0}, for k >= 0, have o.g.f. GD{2*k}(x) = (-1)^k*(1-x)/(1-2*x)^(k+1), for k >= 0, and GD_{2*k+1}(x) = 0, for k >= 0. This follows from their o.g.f. GGD(z, x) := Sum_{k>=0} GD_k(x)*z^n which is obtained from the o.g.f. of the T-triangle GT(z, x) = (1-x*z)/(1 - 2*x + z^2) (see the formula section) by GGD(z, x) = GT(z, x/z).

The explicit form is then D_{2*k}(m) = (-1)^k, for m = 0, and

(-1)^k*(2*k+m)*2^(m-1)*risefac(k+1, m-1)/m!, for m >= 1, with the rising factorial risefac(x, n). (End)

Coefficients of the cusp form of weight 12 for the full modular group.

It is conjectured that tau(n) is never zero (this has been verified for n < 816212624008487344127999, see the Derickx, van Hoeij, Zeng reference).

M. J. Hopkins mentions that the only known primes p for which tau(p) == 1 (mod p) are 11, 23 and 691, that it is an open problem to decide if there are infinitely many such p and that no others are known below 35000. Simon Plouffe has now searched up to tau(314747) and found no other examples. - N. J. A. Sloane, Mar 25 2007

Number 1 of the 74 eta-quotients listed in Table I of Martin (1996).

With Dedekind's eta function and the discriminant Delta one has eta(z)^24 = Delta(z)/(2*Pi)^12 = Sum_{m >= 1} tau(m)*q^m, with q = exp(2*Pi*i*z), and z in the complex upper half plane, where i is the imaginary unit. Delta is the eigenfunction of the Hecke operator T_n (n >= 1) with eigenvalue tau(n): T_n Delta = tau(n) Delta. From this the formula for tau(m)*tau(n) given below in the formula section follows. See, e.g., the Koecher-Krieg reference, Lemma and Satz, p. 212. Or the Apostol reference, eq. (3) on p. 114 and the first part of section 6.13 on p. 131. - Wolfdieter Lang, Jan 26 2016

For the functional equation satisfied by the Dirichlet series F(s), Re(s) > 7, of a(n) see the Hardy reference, p. 173, (10.9.4). It is (2*Pi)^(-s) * Gamma(s) * F(s) = (2*Pi)^(s-12) * Gamma(12-s) * F(12-s). This is attributed to J. R. Wilton, 1929, on p. 185. - Wolfdieter Lang, Feb 08 2017

Conjecture: |a(n)| with n > 1 can never be a perfect power. This has been verified for n up to 10^6. - Zhi-Wei Sun, Dec 18 2024

Conjecture: The numbers |a(n)| (n = 1,2,3,...) are distinct. This has been verified for the first 10^6 terms. - Zhi-Wei Sun, Dec 21 2024

Conjecture: |a(n)| > 2*n^4 for all n > 2. This has been verified for n = 3..10^6. - Zhi-Wei Sun, Dec 25 2024

Conjecture: a(m)^2 + a(n)^2 can never be a perfect power. This implies Lehmer's conjecture that a(n) is never zero. We have verified that there is no perfect power among a(m)^2 + a(n)^2 with m,n <= 1000 . - Zhi-Wei Sun, Dec 28 2024

Conjecture: The equation |a(m)a(n)| = x^k with m < n, k > 1 and x >= 0 has no solution. This has been verified for m < n <= 5000. - Zhi-Wei Sun, Dec 29 2024

For some conjectures motivated by additive combinatorics, one may consult the link to Question 485138 at MathOverflow. - Zhi-Wei Sun, Jan 25 2025