cp's OEIS Frontend

a(n-3) is the number of aperiodic necklaces (Lyndon words) with 3 black beads and n-3 white beads.

Number of triangular partitions (see Almkvist).

Consists of arithmetic progression quadruples of common difference n+1 starting at A045943(n). Refers to the least number of coins needed to be rearranged in order to invert the pattern of a (n+1)-rowed triangular array. For instance, a 5-rowed triangular array requires a minimum of a(4)=5 rearrangements (shown bracketed here) for it to be turned upside down.

.....{*}..................{*}*.*{*}{*}

.....*.*....................*.*.*.{*}

....*.*.*....---------\......*.*.*

..{*}*.*.*...---------/.......*.*

{*}{*}*.*{*}..................{*}

- Lekraj Beedassy, Oct 13 2003

Partial sums of 1,1,1,2,2,2,3,3,3,4,4,4,... - Jon Perry, Mar 01 2004

Sum of three successive terms is a triangular number in natural order starting with 3: a(n)+a(n+1)+a(n+2) = T(n+2) = (n+2)*(n+3)/2. - Amarnath Murthy, Apr 25 2004

Apply Riordan array (1/(1-x^3),x) to n. - Paul Barry, Apr 16 2005

Absolute values of numbers that appear in A145919. - Matthew Vandermast, Oct 28 2008

In the Moree definition, (-1)^n*a(n) is the 3rd Witt transform of A033999 and (-1)^n*A004524(n) with 2 leading zeros dropped is the 2nd Witt transform of A033999. - R. J. Mathar, Nov 08 2008

Column sums of:

1 2 3 4 5 6 7 8 9.....

1 2 3 4 5 6.....

1 2 3.....

........................

----------------------

1 2 3 5 7 9 12 15 18 - Jon Perry, Nov 16 2010

a(n) is the sum of the positive integers <= n that have the same residue modulo 3 as n. They are the additive counterpart of the triple factorial numbers. - Peter Luschny, Jul 06 2011

a(n+1) is the number of 3-tuples (w,x,y) with all terms in {0,...,n} and w=3*x+y. - Clark Kimberling, Jun 04 2012

a(n+1) is the number of pairs (x,y) with x and y in {0,...,n}, x-y = (1 mod 3), and x+y < n. - Clark Kimberling, Jul 02 2012

a(n+1) is the number of partitions of n into two sorts of part(s) 1 and one sort of (part) 3. - Joerg Arndt, Jun 10 2013

Arrange A004523 in rows successively shifted to the right two spaces and sum the columns:

1 2 2 3 4 4 5 6 6...

1 2 2 3 4 4 5...

1 2 2 3 4...

1 2 2...

1...

------------------------------

1 2 3 5 7 9 12 15 18... - L. Edson Jeffery, Jul 30 2014

a(n) = A258708(n+1,1) for n > 0. - Reinhard Zumkeller, Jun 23 2015

Also the number of triples of positive integers summing to n + 4, the first less than each of the other two. Also the number of triples of positive integers summing to n + 2, the first less than or equal to each of the other two. - Gus Wiseman, Oct 11 2020

Also the lower matching number of the (n+1)-triangular honeycomb king graph = n-triangular grid graph (West convention). - Eric W. Weisstein, Dec 14 2024

Inverse binomial transform of A057083. Binomial transform of A061347. The sums of consecutive pairs of elements give A084103. - Paul Barry, May 15 2003

Hexaperiodic sequence identical to its third differences. - Paul Curtz, Dec 13 2007

a(n+1) is the Hankel transform of A001700(n+1)-A001700(n). - Paul Barry, Apr 21 2009

Non-simple continued fraction expansion of 1 = 1+1/(2+1/(1+1/(-1+...))). - R. J. Mathar, Mar 08 2012

Pisano period lengths: 1, 3, 2, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, ... - R. J. Mathar, Aug 10 2012

Alternating row sums of Riordan triangle A111125. - Wolfdieter Lang, Oct 18 2012

Periodic sequences of this type can be also calculated by a(n) = c + floor(q/(p^m-1)*p^n) mod p, where c is a constant, q is the number representing the periodic digit pattern and m is the period length. c, p and q can be calculated as follows: Let D be the array representing the number pattern to be repeated, m = size of D, max = maximum value of elements in D, min = minimum value of elements in D. Than c := min, p := max - min + 1 and q := p^m*sum_{i=1..m} (D(i)-min)/p^i. Example: D = (1, 2, 1, -1, -2, -1), c = -2, m = 6, p = 5 and q = 12276 for this sequence. - Hieronymus Fischer, Jan 04 2013

T(n,k) is the number of compatible k-sets of cluster variables in Fomin and Zelevinsky's Cluster algebra of finite type B_n. Take a row of this triangle regarded as a polynomial in x and rewrite as a polynomial in y := x+1. The coefficients of the polynomial in y give a row of triangle A008459 (squares of binomial coefficients). For example, x^2+6*x+6 = y^2+4*y+1. - Paul Boddington, Mar 07 2003

T(n,k) is the number of lattice paths from (0,0) to (n,n) using steps E=(1,0), N=(0,1) and D=(1,1) (i.e., bilateral Schroeder paths), having k N=(0,1) steps. E.g. T(2,0)=1 because we have DD; T(2,1) = 6 because we have NED, NDE, EDN, END, DEN and DNE; T(2,2)=6 because we have NNEE, NENE, NEEN, EENN, ENEN and ENNE. - Emeric Deutsch, Apr 20 2004

Another version of [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] DELTA [0, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...] = 1; 1, 0; 1, 2, 0; 1, 6, 6, 0; 1, 12, 30, 20, 0; ..., where DELTA is the operator defined in A084938. - Philippe Deléham Apr 15 2005

Terms in row n are the coefficients of the Legendre polynomial P(n,2x+1) with increasing powers of x.

From Peter Bala, Oct 28 2008: (Start)

Row n of this triangle is the f-vector of the simplicial complex dual to an associahedron of type B_n (a cyclohedron) [Fomin & Reading, p.60]. See A008459 for the corresponding h-vectors for associahedra of type B_n and A001263 and A033282 respectively for the h-vectors and f-vectors for associahedra of type A_n.

An alternative description of this triangle in terms of f-vectors is as follows. Let A_n be the root lattice generated as a monoid by {e_i - e_j: 0 <= i,j <= n+1}. Let P(A_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the f-vectors of a unimodular triangulation of P(A_n) [Ardila et al.]. A008459 is the corresponding array of h-vectors for these type A_n polytopes. See A127674 (without the signs) for the array of f-vectors for type C_n polytopes and A108556 for the array of f-vectors associated with type D_n polytopes.

The S-transform on the ring of polynomials is the linear transformation of polynomials that is defined on the basis monomials x^k by S(x^k) = binomial(x,k) = x(x-1)...(x-k+1)/k!. Let P_n(x) denote the S-transform of the n-th row polynomial of this array. In the notation of [Hetyei] these are the Stirling polynomials of the type B associahedra. The first few values are P_1(x) = 2*x + 1, P_2(x) = 3*x^2 + 3*x + 1 and P_3(x) = (10*x^3 + 15*x^2 + 11*x + 3)/3. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane, that is, the polynomials P_n(-x) satisfy a Riemann hypothesis. See A142995 for further details. The sequence of values P_n(k) for k = 0,1,2,3, ... produces the n-th row of A108625. (End)

This is the row reversed version of triangle A104684. - Wolfdieter Lang, Sep 12 2016

T(n, k) is also the number of (n-k)-dimensional faces of a convex n-dimensional Lipschitz polytope of real functions f defined on the set X = {1, 2, ..., n+1} which satisfy the condition f(n+1) = 0 (see Gordon and Petrov). - Stefano Spezia, Sep 25 2021

The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial ((x+1)*(x+2)*(x+3)*...*(x+n) / n!)^2 in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 09 2022

Chapoton's observation above is correct: the precise expansion is ((x+1)*(x+2)*(x+3)*...*(x+n)/ n!)^2 = Sum_{k = 0..n} (-1)^k*T(n,n-k)*binomial(x+2*n-k, 2*n-k), as can be verified using the WZ algorithm. For example, n = 3 gives ((x+1)*(x+2)*(x+3)/3!)^2 = 20*binomial(x+6,6) - 30*binomial(x+5,5) + 12*binomial(x+4,4) - binomial(x+3,3). - Peter Bala, Jun 24 2023

According to rook theory, John Riordan considered a(1) to be -1. - Vladimir Shevelev, Apr 02 2010

This is also the value that the formulas of Touchard and of Wyman and Moser give and is compatible with many recurrences. - William P. Orrick, Aug 31 2020

Or, for n >= 3, the number of 3 X n Latin rectangles the second row of which is full cycle with a fixed order of its elements, e.g., the cycle (x_2,x_3,...,x_n,x_1) with x_1 < x_2 < ... < x_n. - Vladimir Shevelev, Mar 22 2010

Muir (p. 112) gives essentially this recurrence (although without specifying any initial conditions). Compare A186638. - N. J. A. Sloane, Feb 24 2011

Sequence discovered by Touchard in 1934. - L. Edson Jeffery, Nov 13 2013

Although these are also known as Touchard numbers, the problem was formulated by Lucas in 1891, who gave the recurrence formula shown below. See Cerasoli et al., 1988. - Stanislav Sykora, Mar 14 2014

An equivalent problem was formulated by Tait; solutions to Tait's problem were given by Muir (1878) and Cayley (1878). - William P. Orrick, Aug 31 2020

From Vladimir Shevelev, Jun 25 2015: (Start)

According to the ménage problem, 2*n!*a(n) is the number of ways of seating n married couples at 2*n chairs around a circular table, men and women in alternate positions, so that no husband is next to his wife.

It is known [Riordan, ch. 7] that a(n) is the number of arrangements of n non-attacking rooks on the positions of the 1's in an n X n (0,1)-matrix A_n with 0's in positions (i,i), i = 1,...,n, (i,i+1), i = 1,...,n-1, and (n,1). This statement could be written as a(n) = per(A_n). For example, A_5 has the form

001*11

1*0011

11001* (1)

11*100

0111*0,

where 5 non-attacking rooks are denoted by {1*}.

We can indicate a one-to-one correspondence between arrangements of n non-attacking rooks on the 1's of a matrix A_n and arrangements of n married couples around a circular table by the rules of the ménage problem, after the ladies w_1, w_2, ..., w_n have taken the chairs numbered

2*n, 2, 4, ..., 2*n-2 (2)

respectively. Suppose we consider an arrangement of rooks: (1,j_1), (2,j_2), ..., (n,j_n). Then the men m_1, m_2, ..., m_n took chairs with numbers

2*j_i - 3 (mod 2*n), (3)

where the residues are chosen from the interval[1,2*n]. Indeed {j_i} is a permutation of 1,...,n. So {2*j_i-3}(mod 2*n) is a permutation of odd positive integers <= 2*n-1. Besides, the distance between m_i and w_i cannot be 1. Indeed, the equality |2*(j_i-i)-1| = 1 (mod 2*n) is possible if and only if either j_i=i or j_i=i+1 (mod n) that correspond to positions of 0's in matrix A_n.

For example, in the case of positions of {1*} in(1) we have j_1=3, j_2=1, j_3=5, j_4=2, j_5=4. So, by(2) and (3) the chairs 1,2,...,10 are taken by m_4, w_2, m_1, w_3, m_5, w_4, m_3, w_5, m_2, w_1, respectively. (End)

The first 20 terms of this sequence were calculated in 1891 by E. Lucas (see [Lucas, p. 495]). - Peter J. C. Moses, Jun 26 2015

From Ira M. Gessel, Nov 27 2018: (Start)

If we invert the formula

Sum_{ n>=0 } u_n z^n = ((1-z)/(1+z)) F(z/(1+z)^2)

that Don Knuth mentions (see link) (i.e., set x=z/(1+z)^2 and solve for z in terms of x), we get a formula for F(z) = Sum_{n >= 0} n! z^n as a sum with all positive coefficients of (almost) powers of the Catalan number generating function.

The exact formula is (5) of the Yiting Li article.

This article also gives a combinatorial proof of this formula (though it is not as simple as one might want). (End)

Also gives the number of points of period n in the subshift of finite type corresponding to the square matrix A=[1,2;1,0] (this is then given by trace(A^n)). - Thomas Ward, Mar 07 2001

Sequence is identical to its signed inverse binomial transform (autosequence of the second kind). - Paul Curtz, Jul 11 2008

a(n) can be expressed in terms of values of the Fibonacci polynomials F_n(x), computed at x=1/sqrt(2). - Tewodros Amdeberhan (tewodros(AT)math.mit.edu), Dec 15 2008

Pisano period lengths: 1, 1, 2, 2, 4, 2, 6, 2, 6, 4, 10, 2, 12, 6, 4, 2, 8, 6, 18, 4, ... - R. J. Mathar, Aug 10 2012

Let F(x) = Product_{n >= 0} (1 - x^(3*n+1))/(1 - x^(3*n+2)). This sequence is the simple continued fraction expansion of the real number 1 + F(-1/2) = 2.83717 78068 73232 99799 ... = 2 + 1/(1 + 1/(5 + 1/(7 + 1/(17 + ...)))). See A111317. - Peter Bala, Dec 26 2012

With different signs, 2, -1, 5, -7, 17, -31, 65, -127, 257, -511, 1025, -2047, ... is the Lucas V(-1,-2) sequence. - R. J. Mathar, Jan 08 2013

The identity 2 = 2/2 + 2^2/(2*1) - 2^3/(2*1*5) - 2^4/(2*1*5*7) + 2^5/(2*1*5*7*17) + 2^6/(2*1*5*7*17*31) - - + + can be viewed as a generalized Engel-type expansion of the number 2 to the base 2. Compare with A062510. - Peter Bala, Nov 13 2013

For n >= 2, a(n) is the number of ways to tile a 2 X n strip, where the first two columns have an extra cell at the top, with 1 X 2 dominoes and 2 X 2 squares. Shown here is one of the a(7)=127 ways for the n=7 case:

._.

|_|_________.

| | | |_| |

||__|_|_|_|. - Greg Dresden, Sep 26 2021

Named by Horadam (1988) after the German mathematician Ernst Jacobsthal (1882-1965) and the French mathematician Édouard Lucas (1842-1891). - Amiram Eldar, Oct 02 2023

Matrix inverse of A124645.

Let L(n,x) = Sum_{k=0..n} T(n,k)*x^(n-k) and Pi=3.14...:

L(n,x) = Product_{k=1..n} (x - 2*cos((2*k-1)*Pi/(2*n+1)));

Sum_{k=0..n} T(n,k) = L(n,1) = A010892(n+1);

Sum_{k=0..n} abs(T(n,k)) = A000045(n+2);

abs(T(n,k)) = A065941(n,k), T(n,k) = A065941(n,k)*A087960(k);

T(2*n,k) + T(2*n+1,k+1) = 0 for 0 <= k <= 2*n;

T(n,0) = A000012(n) = 1; T(n,1) = -1 for n > 0;

T(n,2) = -(n-1) for n > 1; T(n,3) = A000027(n)=n for n > 2;

T(n,4) = A000217(n-3) for n > 3; T(n,5) = -A000217(n-4) for n > 4;

T(n,6) = -A000292(n-5) for n > 5; T(n,7) = A000292(n-6) for n > 6;

T(n,n-3) = A058187(n-3)*(-1)^floor(n/2) for n > 2;

T(n,n-2) = A008805(n-2)*(-1)^floor((n+1)/2) for n > 1;

T(n,n-1) = A008619(n-1)*(-1)^floor(n/2) for n > 0;

T(n,n) = L(n,0) = (-1)^floor((n+1)/2);

L(n,1) = A010892(n+1); L(n,-1) = A061347(n+2);

L(n,2) = 1; L(n,-2) = A005408(n)*(-1)^n;

L(n,3) = A001519(n); L(n,-3) = A002878(n)*(-1)^n;

L(n,4) = A001835(n+1); L(n,-4) = A001834(n)*(-1)^n;

L(n,5) = A004253(n); L(n,-5) = A030221(n)*(-1)^n;

L(n,6) = A001653(n); L(n,-6) = A002315(n)*(-1)^n;

L(n,7) = A049685(n); L(n,-7) = A033890(n)*(-1)^n;

L(n,8) = A070997(n); L(n,-8) = A057080(n)*(-1)^n;

L(n,9) = A070998(n); L(n,-9) = A057081(n)*(-1)^n;

L(n,10) = A072256(n+1); L(n,-10) = A054320(n)*(-1)^n;

L(n,11) = A078922(n+1); L(n,-11) = A097783(n)*(-1)^n;

L(n,12) = A077417(n); L(n,-12) = A077416(n)*(-1)^n;

L(n,13) = A085260(n);

L(n,14) = A001570(n); L(n,-14) = A028230(n)*(-1)^n;

L(n,n) = A108366(n); L(n,-n) = A108367(n).

Row n of the matrix inverse (A124645) has g.f.: x^floor(n/2)*(1-x)^(n-floor(n/2)). - Paul D. Hanna, Jun 12 2005

From L. Edson Jeffery, Mar 12 2011: (Start)

Conjecture: Let N=2*n+1, with n > 2. Then T(n,k) (0 <= k <= n) gives the k-th coefficient in the characteristic function p_N(x)=0, of degree n in x, for the n X n tridiagonal unit-primitive matrix G_N (see [Jeffery]) of the form

G_N=A_{N,1}=

(0 1 0 ... 0)

(1 0 1 0 ... 0)

(0 1 0 1 0 ... 0)

...

(0 ... 0 1 0 1)

(0 ... 0 1 1),

with solutions phi_j = 2*cos((2*j-1)*Pi/N), j=1,2,...,n. For example, for n=3,

G_7=A_{7,1}=

(0 1 0)

(1 0 1)

(0 1 1).

We have {T(3,k)}=(1,-1,-2,1), while the characteristic function of G_7 is p(x) = x^3-x^2-2*x+1 = 0, with solutions phi_j = 2*cos((2*j-1)*Pi/7), j=1,2,3. (End)

The triangle sums, see A180662 for their definitions, link A108299 with several sequences, see the crossrefs. - Johannes W. Meijer, Aug 08 2011

The roots to the polynomials are chaotic using iterates of the operation (x^2 - 2), with cycle lengths L and initial seeds returning to the same term or (-1)* the seed. Periodic cycle lengths L are shown in A003558 such that for the polynomial represented by row r, the cycle length L is A003558(r-1). The matrices corresponding to the rows as characteristic polynomials are likewise chaotic [cf. Kappraff et al., 2005] with the same cycle lengths but substituting 2*I for the "2" in (x^2 - 2), where I = the Identity matrix. For example, the roots to x^3 - x^2 - 2x + 1 = 0 are 1.801937..., -1.246979..., and 0.445041... With 1.801937... as the initial seed and using (x^2 - 2), we obtain the 3-period trajectory of 8.801937... -> 1.246979... -> -0.445041... (returning to -1.801937...). We note that A003558(2) = 3. The corresponding matrix M is: [0,1,0; 1,0,1; 0,1,1,]. Using seed M with (x^2 - 2*I), we obtain the 3-period with the cycle completed at (-1)*M. - Gary W. Adamson, Feb 07 2012

Other names are the Rudin-Shapiro or Golay-Rudin-Shapiro infinite word.

The Shapiro polynomials are defined by P_0 = Q_0 = 1; for n>=0, P_{n+1} = P_n + x^(2^n)*Q_n, Q_{n+1} = P_n - x^(2^n)*Q_n. Then P_n = Sum_{m=0..2^n-1} a(m)*x^m, where the a(m) (the present sequence) do not depend on n. - N. J. A. Sloane, Aug 12 2016

Related to paper-folding sequences - see the Mendès France and Tenenbaum article.

a(A022155(n)) = -1; a(A203463(n)) = 1. - Reinhard Zumkeller, Jan 02 2012

a(n) = 1 if and only if the numbers of 1's and runs of 1's in binary representation of n have the same parity: A010060(n) = A268411(n); otherwise, when A010060(n) = 1 - A268411(n), a(n) = -1. - Vladimir Shevelev, Feb 10 2016. Typo corrected and comment edited by Antti Karttunen, Jul 11 2017

A word that is uniform primitive morphic, but not pure morphic. - N. J. A. Sloane, Jul 14 2018

Named after the Austrian-American mathematician Walter Rudin (1921-2010), the mathematician Harold S. Shapiro (1928-2021) and the Swiss mathematician and physicist Marcel Jules Edouard Golay (1902-1989). - Amiram Eldar, Jun 13 2021

Rows equal columns and are periodic. No zero elements are found (conjecture). The recurrence is related to the recurrence for the Mahonian numbers. The main diagonal is the Dirichlet inverse of the Euler totient function A023900 (conjecture). [The 2nd and 3rd formulas state that the conjecture is correct. R. J. Mathar, Sep 16 2017]

The sums from n=1 to infinity of T(n,k)/n converge to the Mangoldt function for column k (conjecture).

If gcd(n,k)=1 then T(n,k)=1 and if T(n,k)=1 then gcd(n,k)=1 (conjecture).

The Dirichlet generating functions for s > 1 for the columns appear to be (see A054535):

Zeta(s)*(1 + (Sum over all possible combinations of products of negative distinct prime factors of k, up to rearrangement, 1/((-1* first distinct prime factor)*(-1*second distinct prime factor)*(-1*third distinct prime factor * ...))^(s-1))).

Examples:

k=1: Zeta(s)

k=2: Zeta(s)*(1 - 1/2^(s-1))

k=3: Zeta(s)*(1 - 1/3^(s-1))

k=4: Zeta(s)*(1 - 1/2^(s-1))

k=5: Zeta(s)*(1 - 1/5^(s-1))

k=6: Zeta(s)*(1 - 1/2^(s-1) - 1/3^(s-1) + 1/6^(s-1))

k=7: Zeta(s)*(1 - 1/7^(s-1))

...

k=30: Zeta(s)*(1 - 1/2^(s-1) - 1/3^(s-1) - 1/5^(s-1) + 1/6^(s-1) + 1/10^(s-1) + 1/15^(s-1) - 1/30^(s-1))

...

(conjecture)

See triangle A142971 for negative distinct prime factors.

This could probably be checked by matrix multiplication.

The signs of the eigenvalues of this matrix are a rearrangement of the Mobius function A008683 (conjecture). The first few eigenvalues are:

{1.0000}

{-1.4142, 1.4142}

{-2.6554, 1.8662, -1.2108}

{-3.4393, 2.1004, -1.6611, 0}

{-4.7711, -3.3867, 2.5910, -1.4332, 0}

{-5.2439, -3.4641, 3.4641, 2.5169, -2.2730, 0}

The relation to Dirichlet characters for the entries in this matrix appears appears to be formulated in terms of the sequence A089026 which is equal to n if n is a prime, otherwise equal to 1. See Mathematica program below. [Mats Granvik, Nov 23 2013]

From Mats Granvik, Jun 19 2016: (Start)

Remark about the Dirichlet generating function for the whole matrix: Subtracting the first column (in the form of zeta(c)) of the matrix gives us the limit: lim_{c->1} zeta(s)*zeta(c)/zeta(c+s-1)-zeta(c) = -zeta'(s)/zeta(s) which is the classical Dirichlet generating function for the von Mangoldt function.

For n >= k, see A231425, this matrix has row sums equal to zero except for the first row:

1=1

1-1=0

1+1-2=0

1-1+1-1=0

1+1+1+1-4=0

...

log(A014963(n)) = Sum_{k>=1} A191898(n,k)/k, for n>1.

log(A014963(k)) = Sum_{n>=1} A191898(n,k)/n, for k>1.

log(A014963(n)) = limit of zeta(s)*(Sum_{d divides n} A008683(d)/d^(s-1)) as s->1, for n>1.

A008683(n) = Sum_{k=1..n} A191898(n,k)*exp(-i*2*Pi*k/n)/n.

A008683(n) = Sum_{n=1..k} A191898(n,k)*exp(-i*2*Pi*n/k)/k.

(End)

From Mats Granvik, Aug 09 2016: (Start)

The Dirichlet generating function for the matrix is zero for any pair c and s = 2 - c and any pair s and c = 2 - s except at the pole c = 1 and s = 1 where it is indeterminate.

In the Mathematica program section, in the expression for the matrix as Dirichlets characters, the variables s and c can apparently be any pair of positive integers.

Limits related to the Dirichlet generating function for the matrix: Let s = ZetaZero(n), then lim_{c->1} zeta(s*c)/zeta(c+s-1) = ZetaZero(n). Let s = ZetaZero(n), then lim_{c->1} zeta(s*c)/zeta(c+s*c-1) = ZetaZero(n)/(1+ZetaZero(n)).

(End)