cp's OEIS Frontend

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x).

Taking p(S) = 1 - S gives the INVERT transform of s, so that p-INVERT is a generalization of the INVERT transform (e.g., A033453).

Guide to p-INVERT sequences using p(S) = 1 - S - S^2:

t(A000012) = t(1,1,1,1,1,1,1,...) = A001906

t(A000290) = t(1,4,9,16,25,36,...) = A289779

t(A000027) = t(1,2,3,4,5,6,7,8,...) = A289780

t(A000045) = t(1,2,3,5,8,13,21,...) = A289781

t(A000032) = t(2,1,3,4,7,11,14,...) = A289782

t(A000244) = t(1,3,9,27,81,243,...) = A289783

t(A000302) = t(1,4,16,64,256,...) = A289784

t(A000351) = t(1,5,25,125,625,...) = A289785

t(A005408) = t(1,3,5,7,9,11,13,...) = A289786

t(A005843) = t(2,4,6,8,10,12,14,...) = A289787

t(A016777) = t(1,4,7,10,13,16,...) = A289789

t(A016789) = t(2,5,8,11,14,17,...) = A289790

t(A008585) = t(3,6,9,12,15,18,...) = A289795

t(A000217) = t(1,3,6,10,15,21,...) = A289797

t(A000225) = t(1,3,7,15,31,63,...) = A289798

t(A000578) = t(1,8,27,64,625,...) = A289799

t(A000984) = t(1,2,6,20,70,252,...) = A289800

t(A000292) = t(1,4,10,20,35,56,...) = A289801

t(A002620) = t(1,2,4,6,9,12,16,...) = A289802

t(A001906) = t(1,3,8,21,55,144,...) = A289803

t(A001519) = t(1,1,2,5,13,34,...) = A289804

t(A103889) = t(2,1,4,3,6,5,8,7,,...) = A289805

t(A008619) = t(1,1,2,2,3,3,4,4,...) = A289806

t(A080513) = t(1,2,2,3,3,4,4,5,...) = A289807

t(A133622) = t(1,2,1,3,1,4,1,5,...) = A289809

t(A000108) = t(1,1,2,5,14,42,...) = A081696

t(A081696) = t(1,1,3,9,29,97,...) = A289810

t(A027656) = t(1,0,2,0,3,0,4,0,5...) = A289843

t(A175676) = t(1,0,0,2,0,0,3,0,...) = A289844

t(A079977) = t(1,0,1,0,2,0,3,...) = A289845

t(A059841) = t(1,0,1,0,1,0,1,...) = A289846

t(A000040) = t(2,3,5,7,11,13,...) = A289847

t(A008578) = t(1,2,3,5,7,11,13,...) = A289828

t(A000142) = t(1!, 2!, 3!, 4!, ...) = A289924

t(A000201) = t(1,3,4,6,8,9,11,...) = A289925

t(A001950) = t(2,5,7,10,13,15,...) = A289926

t(A014217) = t(1,2,4,6,11,17,29,...) = A289927

t(A000045*) = t(0,1,1,2,3,5,...) = A289975 (* indicates prepended 0's)

t(A000045*) = t(0,0,1,1,2,3,5,...) = A289976

t(A000045*) = t(0,0,0,1,1,2,3,5,...) = A289977

t(A290990*) = t(0,1,2,3,4,5,...) = A290990

t(A290990*) = t(0,0,1,2,3,4,5,...) = A290991

t(A290990*) = t(0,0,01,2,3,4,5,...) = A290992

Also called generalized decagonal numbers.

Odd triangular numbers decremented and halved.

It appears that this is zero together with the partial sums of A165998. - Omar E. Pol, Sep 10 2011 [this is correct, see the g.f., Joerg Arndt, Sep 29 2013]

Also, A033954 and positive members of A001107 interleaved. - Omar E. Pol, Aug 04 2012

Also, numbers m such that 16*m+9 is a square. After 1, therefore, there are no squares in this sequence. - Bruno Berselli, Jan 07 2016

Convolution of the sequences A047522 and A059841. - Ilya Gutkovskiy, Mar 16 2017

Numbers k such that the concatenation k5625 is a square. - Bruno Berselli, Nov 07 2018

Exponents in expansion of Product_{n >= 1} (1 + x^(8*n-7))*(1 + x^(8*n-1))*(1 - x^(8*n)) = 1 + x + x^7 + x^10 + x^22 + .... - Peter Bala, Dec 10 2020

Also the q-Stirling2 numbers at q = -1. - Peter Luschny, Mar 09 2020

Row sums give the Fibonacci sequence. So do the alternating row sums.

Triangle of coefficients of polynomials defined by p(-1,x) = p(0,x) = 1, p(n, x) = x*p(n-1, x) + p(n-2, x), for n >= 1. - Benoit Cloitre, May 08 2005 [rewritten with correct offset. - Wolfdieter Lang, Feb 18 2020]

Another version of triangle in A103631. - Philippe Deléham, Jan 01 2009

The T(n,k) coefficients appear in appendix 2 of Parks's remarkable article "A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov" if we assume that the b(n) coefficients are all equal to 1 and ignore the first column. The complete version of this triangle including the first column is A103631. - Johannes W. Meijer, Aug 11 2011

Signed ++--++..., the roots are chaotic using f(x) --> x^2 - 2 with cycle lengths shown in A003558 by n-th rows. Example: given row 3, x^3 + x^2 - 2x - 1; the roots are (a = 1.24697, ...; b = -0.445041, ...; c = -1.802937, ...). Then (say using seed b with x^2 - 2) we obtain the trajectory -0.445041, ... -> -1.80193, ... -> 1.24697, ...; matching the entry "3" in A003558(3). - Gary W. Adamson, Sep 06 2011

From Gary W. Adamson, Aug 25 2019: (Start)

Roots to the polynomials and terms in A003558 can all be obtained from the numbers below using a doubling series mod N procedure as follows: (more than one row may result). Any row ends when the trajectory produces a term already used. Then try the next higher odd term not used as the leftmost term, then repeat.

For example, for N = 11, we get: (1, 2, 4, 3, 5), showing that when confronted with two choices after the 4: (8 and -3), pick the smaller (abs) term, = 3. Then for the next row pick 7 (not used) and repeat the algorithm; succeeding only if the trajectory produces new terms. But 7 is also (-4) mod 11 and 4 was used. Therefore what I call the "r-t table" (for roots trajectory) has only one row: (1, 2, 4, 3, 5). Conjecture: The numbers of terms in the first row is equal to A003558 corresponding to N, i.e., 5 in this case with period 2.

Now for the roots to the polynomials. Pick N = 7. The polynomial is x^3 - x^2 - 2x + 1 = 0, with roots 1.8019..., -1.2469... and 0.445... corresponding to 2*cos(j*Pi/N), N = 7, and j = (1, 2, and 3). The terms (1, 2, 3) are the r-t terms for N = 7. For 11, the r-t terms are (1, 2, 4, 3, 5). This implies that given any roots of the corresponding polynomial, they are cyclic using f(x) --> x^2 - 2 with cycle lengths shown in A003558. The terms thus generated are 2*cos(j*Pi), with j = (1, 2, 4, 3, 5). Check: Begin with 2*j*Pi/N, with j = 1 (1.9189...). The other trajectory terms are: --> 1.6825..., --> 0.83083..., -1.3097...; 545...; (a 5 period and cyclic since we can begin with any of the constants). The r-t table for odd N begins as follows:

3...............1

5...............1, 2

7...............1, 2, 3

9...............1, 2, 4

...............3 (singleton terms reduce to "1") (9 has two rows)

11...............1, 2, 4, 3, 5

13...............1, 2, 4, 5, 3, 6

15...............1, 2, 4, 7

................3, 6 (dividing through by the gcd gives (1, 2))

................5. (singleton terms reduce to "1")

The result is that 15 has 3 factors (since 3 rows), and the values of those factors are the previous terms "N", corresponding to the r-t terms in each row. Thus, the first row is new, the second (1, 2), corresponds to N = 5, and the "1" in row 3 corresponds to N = 3. The factors are those values apart from 15 and 1. Note that all of the unreduced r-t terms in all rows for N form a complete set of the terms 1 through (N-1)/2 without duplication. (End)

From Gary W. Adamson, Sep 30 2019: (Start)

The 3 factors of the 7th degree polynomial for 15: (x^7 - x^6 - 6x^5 + 5x^4 + 10x^3 - 6x^2 - 4x + 1) can be determined by getting the roots for 2*cos(j*Pi/1), j = (1, 2, 4, 7) and finding the corresponding polynomial, which is x^4 + x^3 - 4x^2 - 4x + 1. This is the minimal polynomial for N = 15 as shown in Table 2, p. 46 of (Lang). The degree of this polynomial is 4, corresponding to the entry in A003558 for 15, = 4. The trajectories (3, 6) and (5) are j values for 2*cos(j*Pi/15) which are roots to x^2 - x - 1 (relating to the pentagon), and (x - 1), relating to the triangle. (End)

From Gary W. Adamson, Aug 21 2019: (Start)

Matrices M of the form: (1's in the main diagonal, -1's in the subdiagonal, and the rest zeros) are chaotic if we replace (f(x) --> x^2 - 2) with f(x) --> M^2 - 2I, where I is the Identity matrix [1, 0, 0; 0, 1, 0; 0, 0, 1]. For example, with the 3 X 3 matrix M: [0, 0, 1; 0, 1, -1; 1, -1, 0]; the f(x) trajectory is:

....M^2 - 2I: [-1, -1, 0; -1, 0, -1; 0, -1, 0], then for the latter,

....M^2 - 2I: [0, 1, 1; 1, 0, 0; 1, 0, -1]. The cycle ends with period 3 since the next matrix is (-1) * the seed matrix. As in the case with f(x) --> x^2 - 2, the eigenvalues of the 3 chaotic matrices are (abs) 1.24697, 0.44504... and 1.80193, ... Also, the characteristic equations of the 3 matrices are the same as or variants of row 4 of the triangle below: (x^3 + x - 2x - 1) with different signs. (End)

Received from Herb Conn, Jan 2004: (Start)

Let x = 2*cos(2A) (A = Angle); then

sin(A)/sin A = 1

sin(3A)/sin A = x + 1

sin(5A)/sin A = x^2 + x - 1

sin(7A)/sin A = x^3 + x - 2x - 1

sin(9A)/sin A = x^4 + x^3 - 3x^2 - 2x + 1

... (signed ++--++...). (End)

Or Pascal's triangle (A007318) with duplicated diagonals. Also triangle of coefficients of polynomials defined by P_0(x) = 1 and for n>=1, P_n(x) = F_n(x) + F_(n+1)(x), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} C(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 12 2012

The matrix inverse is given by

1;

1, 1;

0, -1, 1;

0, 1, -2, 1;

0, 0, 1, -2, 1;

0, 0, -1, 3, -3, 1;

0, 0, 0, -1, 3, -3, 1;

0, 0, 0, 1, -4, 6, -4, 1;

0, 0, 0, 0, 1, -4, 6, -4, 1;

... apart from signs the same as A124645. - R. J. Mathar, Mar 12 2013

a(0) = 1 because the empty partition vacuously has each part >= 3. - Jason Kimberley, Jan 11 2011

Number of partitions where the largest part occurs at least three times. - Joerg Arndt, Apr 17 2011

By removing a single part of size 3, an A026796 partition of n becomes an A008483 partition of n - 3.

For n >= 3 the sequence counts the isomorphism classes of authentication codes AC(2,n,n) with perfect secrecy and with largest probability 0.5 that an interceptor could deceive with a substituted message. - E. Keith Lloyd (ekl(AT)soton.ac.uk).

For n >= 1, also the number of regular graphs of degree 2. - Mitch Harris, Jun 22 2005

(1 + 0*x + 0*x^2 + x^3 + x^4 + x^5 + 2*x^6 + ...) = (1 + x + 2*x^2 + 3*x^3 + 5*x^4 + ...) * 1 / (1 + x + 2*x^2 + 2*x^3 + 3*x^4 + 3*x^5 + 4*x^6 + 4*x^7 + ...). - Gary W. Adamson, Jun 30 2009

Because the triangle A051031 is symmetric, a(n) is also the number of (n-3)-regular graphs on n vertices. Since the disconnected (n-3)-regular graph with minimum order is 2K_{n-2}, then for n > 4 there are no disconnected (n-3)-regular graphs on n vertices. Therefore for n > 4, a(n) is also the number of connected (n-3)-regular graphs on n vertices. - Jason Kimberley, Oct 05 2009

Number of partitions of n+2 such that 2*(number of parts) is a part. - Clark Kimberling, Feb 27 2014

For n >= 1, a(n) is the number of (1,1)-separable partitions of n, as defined at A239482. For example, the (1,1)-separable partitions of 11 are [10,1], [7,1,2,1], [6,1,3,1], [5,1,4,1], [4,1,2,1,2,1], [3,1,3,1,2,1], so that a(11) = 6. - Clark Kimberling, Mar 21 2014

From Peter Bala, Dec 01 2024: (Start)

Let P(3, n) denote the set of partitions of n into parts k >= 3. Then A000041(n) = (1/2) * Sum_{parts k in all partitions in P(3, n+3)} phi(k), where phi(k) is the Euler totient function (see A000010). For example, with n = 5, there are 3 partitions of n + 3 = 8 into parts greater then 3, namely, 8, 5 + 3 and 4 + 4, and (1/2)*(phi(8) + phi(5) + phi(3) + 2*phi(4)) = 7 = A000041(5). (End)

T(n,k) is the number of leaves at level k+1 in all ordered trees with n+1 edges. - Emeric Deutsch, Jan 15 2005

Riordan array ((1-2x-sqrt(1-4x))/(2x^2),(1-2x-sqrt(1-4x))/(2x)). Inverse array is A053122. - Paul Barry, Mar 17 2005

T(n,k) is the number of walks of n steps, each in direction N, S, W, or E, starting at the origin, remaining in the upper half-plane and ending at height k (see the R. K. Guy reference, p. 5). Example: T(3,2)=6 because we have ENN, WNN, NEN, NWN, NNE and NNW. - Emeric Deutsch, Apr 15 2005

Triangle T(n,k), 0<=k<=n, read by rows given by T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0) = 2*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + 2*T(n-1,k) + T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 30 2007

Number of (2n+1)-step walks from (0,0) to (2n+1,2k+1) and consisting of steps u=(1,1) and d=(1,-1) in which the path stays in the nonnegative quadrant. Examples: T(2,0)=5 because we have uuudd, uudud, uuddu, uduud, ududu; T(2,1)=4 because we have uuuud, uuudu, uuduu, uduuu; T(2,2)=1 because we have uuuuu. - Philippe Deléham, Apr 16 2007, Apr 18 2007

Triangle read by rows: T(n,k)=number of lattice paths from (0,0) to (n,k) that do not go below the line y=0 and consist of steps U=(1,1), D=(1,-1) and two types of steps H=(1,0); example: T(3,1)=14 because we have UDU, UUD, 4 HHU paths, 4 HUH paths and 4 UHH paths. - Philippe Deléham, Sep 25 2007

This triangle belongs to the family of triangles defined by T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0) = x*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + y*T(n-1,k) + T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007

With offset [1,1] this is the (ordinary) convolution triangle a(n,m) with o.g.f. of column m given by (c(x)-1)^m, where c(x) is the o.g.f. for Catalan numbers A000108. See the Riordan comment by Paul Barry.

T(n, k) is also the number of order-preserving full transformations (of an n-chain) with exactly k fixed points. - Abdullahi Umar, Oct 02 2008

T(n,k)/2^(2n+1) = coefficients of the maximally flat lowpass digital differentiator of the order N=2n+3. - Pavel Holoborodko (pavel(AT)holoborodko.com), Dec 19 2008

The signed triangle S(n,k) := (-1)^(n-k)*T(n,k) provides the transformation matrix between f(n,l) := L(2*l)*5^n*F(2*l)^(2*n+1) (F=Fibonacci numbers A000045, L=Lucas numbers A000032) and F(4*l*(k+1)), k = 0, ..., n, for each l>=0: f(n,l) = Sum_{k=0..n} S(n,k)*F(4*l*(k+1)), n>=0, l>=0. Proof: the o.g.f. of the l.h.s., G(l;x) := Sum_{n>=0} f(n,l)*x^n = F(4*l)/(1 - 5*F(2*l)^2*x) is shown to match the o.g.f. of the r.h.s.: after an interchange of the n- and k-summation, the Riordan property of S = (C(x)/x,C(x)) (compare with the above comments by Paul Barry), with C(x) := 1 - c(-x), with the o.g.f. c(x) of A000108 (Catalan numbers), is used, to obtain, after an index shift, first Sum_{k>=0} F(4*l*(k))*GS(k;x), with the o.g.f of column k of triangle S which is GS(k;x) := Sum_{n>=k} S(n,k)*x^n = C(x)^(k+1)/x. The result is GF(l;C(x))/x with the o.g.f. GF(l,x) := Sum_{k>=0} F(4*l*k)*x^k = x*F(4*l)/(1-L(4*l)*x+x^2) (see a comment on A049670, and A028412). If one uses then the identity L(4*n) - 5*F(2*n)^2 = 2 (in Koshy's book [reference under A065563] this is No. 15, p. 88, attributed to Lucas, 1876), the proof that one recovers the o.g.f. of the l.h.s. from above boils down to a trivial identity on the Catalan o.g.f., namely 1/c^2(-x) = 1 + 2*x - (x*c(-x))^2. - Wolfdieter Lang, Aug 27 2012

O.g.f. for row polynomials R(x) := Sum_{k=0..n} a(n,k)*x^k:

((1+x) - C(z))/(x - (1+x)^2*z) with C the o.g.f. of A000108 (Catalan numbers). From Riordan ((C(x)-1)/x,C(x)-1), compare with a Paul Barry comment above. This coincides with the o.g.f. given by Emeric Deutsch in the formula section. - Wolfdieter Lang, Nov 13 2012

The A-sequence for this Riordan triangle is [1,2,1] and the Z-sequence is [2,1]. See a W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 13 2012

From Wolfdieter Lang, Sep 20 2013: (Start)

T(n, k) = A053121(2*n+1, 2*k+1). T(n, k) appears in the formula for the (2*n+1)-th power of the algebraic number rho(N) := 2*cos(Pi/N) = R(N, 2) in terms of the even-indexed diagonal/side length ratios R(N, 2*(k+1)) = S(2*k+1, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310): rho(N)^(2*n+1) = Sum_{k=0..n} T(n, k)*R(N, 2*(k+1)), n >= 0, identical in N >= 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears. For the even powers of rho(n) see A039599. (End)

The tridiagonal Toeplitz production matrix P in the Example section corresponds to the unsigned Cartan matrix for the simple Lie algebra A_n as n tends to infinity (cf. Damianou ref. in A053122). - Tom Copeland, Dec 11 2015 (revised Dec 28 2015)

T(n,k) is the number of pairs of non-intersecting walks of n steps, each in direction N or E, starting at the origin, and such that the end points of the two paths are separated by a horizontal distance of k. See Shapiro 1976. - Peter Bala, Apr 12 2017

Also the convolution triangle of the Catalan numbers A000108. - Peter Luschny, Oct 07 2022

Sum of terms in row n = sigma(n) (sum of divisors of n).

Euler's derivation of A127093 in polynomial form is in his proof of the formula for Sigma(n): (let S=Sigma, then Euler proved that S(n) = S(n-1) + S(n-2) - S(n-5) - S(n-7) + S(n-12) + S(n-15) - S(n-22) - S(n-26), ...).

[Young, pp. 365-366], Euler begins, s = (1-x)*(1-x^2)*(1-x^3)*... = 1 - x - x^2 + x^5 + x^7 - x^12 ...; log s = log(1-x) + log(1-x^2) + log(1-x^3) ...; differentiating and then changing signs, Euler has t = x/(1-x) + 2x^2/(1-x^2) + 3x^3/(1-x^3) + 4x^4/(1-x^4) + 5x^5/(1-x^5) + ...

Finally, Euler expands each term of t into a geometric series, getting A127093 in polynomial form: t =

x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + ...

+ 2x^2 + 2x^4 + 2x^6 + 2x^8 + ...

+ 3x^3 + 3x^6 + ...

+ 4x^4 + 4x^8 + ...

+ 5x^5 + ...

+ 6x^6 + ...

+ 7x^7 + ...

+ 8x^8 + ...

T(n,k) is the sum of all the k-th roots of unity each raised to the n-th power. - Geoffrey Critzer, Jan 02 2016

From Davis Smith, Mar 11 2019: (Start)

For n > 1, A020639(n) is the leftmost term, other than 0 or 1, in the n-th row of this array. As mentioned in the Formula section, the k-th column is period k: repeat [k, 0, 0, ..., 0], but this also means that it's the characteristic function of the multiples of k multiplied by k. T(n,1) = A000012(n), T(n,2) = 2*A059841(n), T(n,3) = 3*A079978(n), T(n,4) = 4*A121262(n), T(n,5) = 5*A079998(n), and so on.

The terms in the n-th row, other than 0, are the factors of n. If n > 1 and for every k, 1 <= k < n, T(n,k) = 0 or 1, then n is prime. (End)

From Gary W. Adamson, Aug 07 2019: (Start)

Row terms of the triangle can be used to calculate E(n) in A002654): (1, 1, 0, 1, 2, 0, 0, 1, 1, 2, ...), and A004018, the number of points in a square lattice on the circle of radius sqrt(n), A004018: (1, 4, 4, 0, 4, 8, 0, 0, 4, ...).

As to row terms in the triangle, let E(n) of even terms = 0,

E(integers of the form 4*k - 1 = (-1), and E(integers of the form 4*k + 1 = 1.

Then E(n) is the sum of the E(n)'s of the factors of n in the triangle rows. Example: E(10) = Sum: ((E(1) + E(2) + E(5) + E(10)) = ((1 + 0 + 1 + 0) = 2, matching A002654(10).

To get A004018, multiply the result by 4, getting A004018(10) = 8.

The total numbers of lattice points = 4r^2 = E(1) + ((E(2))/2 + ((E(3))/3 + ((E(4))/4 + ((E(5))/5 + .... Since E(even integers) are zero, E(integers of the form (4*k - 1)) = (-1), and E(integers of the form (4*k + 1)) = (+1); we are left with 4r^2 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - ..., which is approximately equal to Pi(r^2). (End)

T(n,k) is also the number of parts in the partition of n into k equal parts. - Omar E. Pol, May 05 2020

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

A self-inverse permutation of the natural numbers.

Nonnegative numbers rearranged with least disturbance to maintain a(n) not equal to n. - Amarnath Murthy, Sep 13 2002

Essentially lodumo_2 of A059841. - Philippe Deléham, Apr 26 2009

a(n) = A180176(n) for n >= 20. - Reinhard Zumkeller, Aug 15 2010

Period 4: repeat [1, 0, 0, 0].

a(n) is also the number of partitions of n where each part is four (Since the empty partition has no parts, a(0) = 1). Hence a(n) is also the number of 2-regular graphs on n vertices such that each component has girth exactly four. - Jason Kimberley, Oct 01 2011

This sequence is the Euler transformation of A185014. - Jason Kimberley, Oct 01 2011

Number of permutations satisfying -k <= p(i) - i <= r and p(i)-i not in I, i = 1..n, with k = 1, r = 3, I = {0, 1, 2}. - Vladimir Baltic, Mar 07 2012

Complement of A014601. - Reinhard Zumkeller, Oct 04 2004

Let S(x) = (1, 2, 2, 2, ...). Then A042963 = ((S(x))^2 + S(x^2))/2 = ((1, 4, 8, 12, 16, 20, ...) + (1, 0, 2, 0, 2, 0, 2, ...))/2 = (1, 2, 5, 6, 9, 10, ...). - Gary W. Adamson, Jan 03 2011

(a(n)*(a(n) + 1 + 4*k))/2 is odd, for k >= 0. - Gionata Neri, Jul 19 2015

Equivalent to the following variation on Fermat's Diophantine m-tuple: 1 + the product of any two distinct terms is not a square; this sequence, which we'll call sequence S, is produced by the following algorithm. At the start, S is initially empty. At stage n, starting at n = 1, the algorithm checks whether there exists a number m already in the sequence, such that mn+1 is a perfect square. If such a number m is found, then n is not added to the sequence; otherwise, n is added. Then n is incremented to n + 1, and we repeat the procedure. Proof by Clark R. Lyons: We prove by strong induction that n is in the sequence S if and only if n == 1 (mod 4) or n == 2 (mod 4). Suppose now that this holds for all k < n. In case 1, either n == 1 (mod 4) or n == 2 (mod 4), and we wish to show that n does indeed enter the sequence S. That is, we wish to show that there does not exist m < n, already in the sequence at this point such that mn+1 is a square. By the inductive hypothesis m == 1 (mod 4) or m == 2 (mod 4). This means that both m and n are one of 1, 2, 5, or 6 mod 8. Using a multiplication table mod 8, we see that this implies mn+1 is congruent to one of 2, 3, 5, 6, or 7 mod 8. But we also see that mod 8, a perfect square is congruent to 0, 1, or 4. Thus mn+1 is not a perfect square, so n is added to the sequence. In case 2, n == 0 (mod 4) or n == 3 (mod 4), and we wish to show that n is not added to the sequence. That is, we wish to show that there exists m < n already in the sequence such that mn+1 is a perfect square. For this we let m = n - 2, which is positive since n >= 3. By the inductive hypothesis, since m == 1 (mod 4) or m == 2 (mod 4) and m < n, m is already in the sequence. And we have m*n + 1 = (n - 2)*n + 1 = n^2 - 2*n + 1 = (n - 1)^2, so mn+1 is indeed a perfect square, and so n is not added to the sequence. Thus n is added to the sequence if and only if n == 1 (mod 4) or n == 2 (mod 4). This completes the proof. - Robert C. Lyons, Jun 30 2016

Also the number of maximal cliques in the (n + 1) X (n + 1) black bishop graph. - Eric W. Weisstein, Dec 01 2017

Lexicographically earliest sequence of distinct positive integers such that the average of any two or more consecutive terms is never an integer. (For opposite property see A005408.) - Ivan Neretin, Dec 21 2017

Numbers whose binary reflected Gray code (A014550) ends with 1. - Amiram Eldar, May 17 2021

Also: append its negated last bit to n-1. - M. F. Hasler, Oct 17 2022