Search: seq:1,0,1,1,1,1,0,2,2,1
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A059260
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Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-y-x*y-x^2) = 1/((1+x)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...
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+30
24
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1, 0, 1, 1, 1, 1, 0, 2, 2, 1, 1, 2, 4, 3, 1, 0, 3, 6, 7, 4, 1, 1, 3, 9, 13, 11, 5, 1, 0, 4, 12, 22, 24, 16, 6, 1, 1, 4, 16, 34, 46, 40, 22, 7, 1, 0, 5, 20, 50, 80, 86, 62, 29, 8, 1, 1, 5, 25, 70, 130, 166, 148, 91, 37, 9, 1, 0, 6, 30, 95, 200, 296, 314, 239, 128, 46, 10, 1
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OFFSET
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0,8
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COMMENTS
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Coefficients of the (left, normalized) shifted cyclotomic polynomial. Or, coefficients of the basic n-th q-series for q=-2. Indeed, let Y_n(x) = Sum_{k=0..n} x^k, having as roots all the n-th roots of unity except for 0; then coefficients in x of (-1)^n Y_n(-x-1) give exactly the n-th row of A059260 and a practical way to compute it. - Olivier Gérard, Jul 30 2002
The maximum in the (2n)-th row is T(n,n), which is A026641; also T(n,n) ~ (2/3)*binomial(2n,n). The maximum in the (2n-1)-th row is T(n-1,n), which is A014300 (but T does not have the same definition as in A026637); also T(n-1,n) ~ (1/3)*binomial(2n,n). Here is a generalization of the formula given in A026641: T(i,j) = Sum_{k=0..j} binomial(i+k-x,j-k)*binomial(j-k+x,k) for all x real (the proof is easy by induction on i+j using T(i,j) = T(i-1,j) + T(i,j-1)). - Claude Morin, May 21 2002
The second greatest term in the (2n)-th row is T(n-1,n+1), which is A014301; the second greatest term in the (2n+1)-th row is T(n+1,n) = 2*T(n-1,n+1), which is 2*A014301. - Claude Morin
Riordan array (1/(1-x^2), x/(1-x)). As a product of Riordan arrays, factors into the product of (1/(1+x),x) and (1/(1-x),1/(1-x)) (binomial matrix). - Paul Barry, Oct 25 2004
Signed version is A239473 with relations to partial sums of sequences. - Tom Copeland, Mar 24 2014
Columns of the triangle (cf. Example below) give alternate partial sums along nw-se diagonals of the Pascal triangle, i.e., sequences A000035, A004526, A002620 (or A087811), A002623 (or A173196), A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808, etc.
The dimension of the space of closed currents (distributional forms) of degree p on Gr(n), the Grassmann algebra with n generators, equivalently, the dimension of the space of Gr(n)-valued symmetric multilinear forms with vanishing graded divergence, is V(n,p) = 2^n T(p,n-1) - (-1)^p.
If p is odd V(n,p) is also the dimension of the cyclic cohomology group of order p of the Z2 graded algebra Gr(n).
If p is even the dimension of this cohomology group is V(n,p)+1.
The following remarks assume the row indexing starts at n = 1.
The sequence of row polynomials R(n,x), beginning R(1,x) = 1, R(2,x) = x, R(3,x) = 1 + x + x^2 , ..., is a strong divisibility sequence of polynomials in the ring Z[x]; that is, for all positive integers n and m, poly_gcd( R(n,x), R(m,x)) = R(gcd(n, m), x) - apply Norfleet (2005), Theorem 3. Consequently, the polynomial sequence {R(n,x): n >= 1} is a divisibility sequence; that is, if n divides m then R(n,x) divides R(m,x) in Z[x]. (End)
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LINKS
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FORMULA
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G.f.: 1/(1-y-x*y-x^2) = 1 + y + x^2 + xy + y^2 + 2x^2y + 2xy^2 + y^3 + ...
E.g.f: (exp(-t)+(x+1)*exp((x+1)*t))/(x+2). - Tom Copeland, Mar 19 2014
O.g.f. (n-th row): ((-1)^n+(x+1)^(n+1))/(x+2). - Tom Copeland, Mar 19 2014
T(i, 0) = 1 if i is even or 0 if i is odd, T(0, i) = 1 and otherwise T(i, j) = T(i-1, j) + T(i, j-1); also T(i, j) = Sum_{m=j..i+j} (-1)^(i+j+m)*binomial(m, j). - Robert FERREOL, May 17 2002
T(i, j) ~ (i+j)/(2*i+j)*binomial(i+j, j); more precisely, abs(T(i, j)/binomial(i+j, j) - (i+j)/(2*i+j) )<=1/(4*(i+j)-2); the proof is by induction on i+j using the formula 2*T(i, j) = binomial(i+j, j)+T(i, j-1). - Claude Morin, May 21 2002
T(n, k) = Sum_{j=0..n} (-1)^(n-j)binomial(j, k). - Paul Barry, Aug 25 2004
T(n, k) = Sum_{j=0..n-k} binomial(n-j, j)*binomial(j, n-k-j). - Paul Barry, Jul 25 2005
T(i, j) = binomial(i+j, j)-T(i-1, j). - Laszlo Major, Apr 11 2017
Recurrence for row polynomials (with row indexing starting at n = 1): R(n,x) = x*R(n-1,x) + (x + 1)*R(n-2,x) with R(1,x) = 1 and R(2,x) = x. - Peter Bala, Feb 07 2024
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EXAMPLE
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Triangle begins
1;
0, 1;
1, 1, 1;
0, 2, 2, 1;
1, 2, 4, 3, 1;
0, 3, 6, 7, 4, 1;
1, 3, 9, 13, 11, 5, 1;
0, 4, 12, 22, 24, 16, 6, 1;
1, 4, 16, 34, 46, 40, 22, 7, 1;
0, 5, 20, 50, 80, 86, 62, 29, 8, 1;
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MAPLE
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read transforms; 1/(1-y-x*y-x^2); SERIES2(%, x, y, 12); SERIES2TOLIST(%, x, y, 12);
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MATHEMATICA
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t[n_, k_] := Sum[ (-1)^(n-j)*Binomial[j, k], {j, 0, n}]; Flatten[ Table[t[n, k], {n, 0, 12}, {k, 0, n}]] (* Jean-François Alcover, Oct 20 2011, after Paul Barry *)
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PROG
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(Sage)
@cached_function
def prec(n, k):
if k==n: return 1
if k==0: return 0
return -prec(n-1, k-1)-sum(prec(n, k+i-1) for i in (2..n-k+1))
return [(-1)^(n-k+1)*prec(n+1, n-k+1) for k in (1..n)]
(PARI) T(n, k) = sum(j=0, n, (-1)^(n - j)*binomial(j, k));
for(n=0, 12, for(k=0, n, print1(T(n, k), ", "); ); print(); ) \\ Indranil Ghosh, Apr 11 2017
(Python)
from sympy import binomial
def T(n, k): return sum((-1)**(n - j)*binomial(j, k) for j in range(n + 1))
for n in range(13): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 11 2017
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CROSSREFS
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Seen as a square array read by antidiagonals this is the coefficient of x^k in expansion of 1/((1-x^2)*(1-x)^n) with rows A002620, A002623, A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808 etc. (allowing for signs). A058393 would then effectively provide the table for nonpositive n. - Henry Bottomley, Jun 25 2001
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A247495
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Generalized Motzkin numbers: Square array read by descending antidiagonals, T(n, k) = k!*[x^k](exp(n*x)* BesselI_{1}(2*x)/x), n>=0, k>=0.
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+30
5
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1, 0, 1, 1, 1, 1, 0, 2, 2, 1, 2, 4, 5, 3, 1, 0, 9, 14, 10, 4, 1, 5, 21, 42, 36, 17, 5, 1, 0, 51, 132, 137, 76, 26, 6, 1, 14, 127, 429, 543, 354, 140, 37, 7, 1, 0, 323, 1430, 2219, 1704, 777, 234, 50, 8, 1, 42, 835, 4862, 9285, 8421, 4425, 1514, 364, 65, 9, 1
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OFFSET
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0,8
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COMMENTS
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This two-dimensional array of numbers can be seen as a generalization of the Motzkin numbers A001006 for two reasons: The case n=1 reduces to the Motzkin numbers and the columns are the values of the Motzkin polynomials M_{k}(x) = sum_{j=0..k} A097610(k,j)*x^j evaluated at the nonnegative integers.
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LINKS
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FORMULA
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T(n,k) = (n*(2*k+1)*T(n,k-1)-(n-2)*(n+2)*(k-1)*T(n,k-2))/(k+2) for k>=2.
T(n,k) = sum_{j=0..floor(k/2)}(n^(k-2*j)*binomial(k,2*j)* binomial(2*j,j)/(j+1).
T(n,k) = n^k*hypergeom([(1-k)/2,-k/2], [2], 4/n^2) for n>0.
O.g.f. for row n: (1-n*x-sqrt(((n-2)*x-1)*((n+2)*x-1)))/(2*x^2).
O.g.f. for row n: R(x)/x where R(x) is series reversion of x/(1+n*x+x^2).
E.g.f. for row n: exp(n*x)*hypergeom([],[2],x^2).
O.g.f. for column k: the k-th column consists of the values of the k-th Motzkin polynomial M_{k}(x) evaluated at x = 0,1,2,...; M_{k}(x) = sum_{j=0..k} A097610(k,j)*x^j = sum_{j=0..k} (-1)^j*binomial(k,j)*A001006(j)*(x+1)^(k-j).
O.g.f. for row n: 1/(1 - n*x - x^2/(1 - n*x - x^2/(1 - n*x - x^2/(1 - n*x - x^2/(1 - ...))))), a continued fraction. - Ilya Gutkovskiy, Sep 21 2017
T(n,k) is the coefficient of x^k in the expansion of 1/(k+1) * (1 + n*x + x^2)^(k+1). - Seiichi Manyama, May 07 2019
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EXAMPLE
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Square array starts:
[n\k][0][1] [2] [3] [4] [5] [6] [7] [8]
[0] 1, 0, 1, 0, 2, 0, 5, 0, 14, ... A126120
[1] 1, 1, 2, 4, 9, 21, 51, 127, 323, ... A001006
[2] 1, 2, 5, 14, 42, 132, 429, 1430, 4862, ... A000108
[3] 1, 3, 10, 36, 137, 543, 2219, 9285, 39587, ... A002212
[4] 1, 4, 17, 76, 354, 1704, 8421, 42508, 218318, ... A005572
[5] 1, 5, 26, 140, 777, 4425, 25755, 152675, 919139, ... A182401
[6] 1, 6, 37, 234, 1514, 9996, 67181, 458562, 3172478, ... A025230
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Triangular array starts:
1,
0, 1,
1, 1, 1,
0, 2, 2, 1,
2, 4, 5, 3, 1,
0, 9, 14, 10, 4, 1,
5, 21, 42, 36, 17, 5, 1,
0, 51, 132, 137, 76, 26, 6, 1.
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MAPLE
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# RECURRENCE
T := proc(n, k) option remember; if k=0 then 1 elif k=1 then n else
(n*(2*k+1)*T(n, k-1)-(n-2)*(n+2)*(k-1)*T(n, k-2))/(k+2) fi end:
seq(print(seq(T(n, k), k=0..9)), n=0..6);
# OGF (row)
ogf := n -> (1-n*x-sqrt(((n-2)*x-1)*((n+2)*x-1)))/(2*x^2):
seq(print(seq(coeff(series(ogf(n), x, 12), x, k), k=0..9)), n=0..6);
# EGF (row)
egf := n -> exp(n*x)*hypergeom([], [2], x^2):
seq(print(seq(k!*coeff(series(egf(n), x, k+2), x, k), k=0..9)), n=0..6);
# MOTZKIN polynomial (column)
A097610 := proc(n, k) if type(n-k, odd) then 0 else n!/(k!*((n-k)/2)!^2* ((n-k)/2+1)) fi end: M := (k, x) -> add(A097610(k, j)*x^j, j=0..k):
seq(print(seq(M(k, n), n=0..9)), k=0..6);
# OGF (column)
col := proc(n, len) local G; G := A247497_row(n); (-1)^(n+1)* add(G[k+1]/(x-1)^(k+1), k=0..n); seq(coeff(series(%, x, len+1), x, j), j=0..len) end: seq(print(col(n, 8)), n=0..6); # Peter Luschny, Dec 14 2014
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MATHEMATICA
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T[0, k_] := If[EvenQ[k], CatalanNumber[k/2], 0];
T[n_, k_] := n^k*Hypergeometric2F1[(1 - k)/2, -k/2, 2, 4/n^2];
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PROG
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(Sage)
if n==0: return(k//2+1)*factorial(k)/factorial(k//2+1)^2 if is_even(k) else 0
return n^k*hypergeometric([(1-k)/2, -k/2], [2], 4/n^2).simplify()
for n in (0..7): print([A247495(n, k) for k in range(11)])
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CROSSREFS
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Cf. A126120, A001006, A000108, A002212, A005572, A182401, A025230, A002522, A079908, A055151, A097610, A247497, A306684.
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KEYWORD
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AUTHOR
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STATUS
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approved
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A340453
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G.f.: Product_{n>=0} (1 - x^(5*n+5))^2 / ( (1 - x^(5*n+1))*(1 - x^(5*n+4)) ).
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+30
3
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1, 1, 1, 1, 2, 0, 1, 1, 2, 1, 1, 0, 2, 1, 0, 2, 2, 0, 1, 1, 2, 1, 1, -1, 2, 1, 2, 1, 2, 1, 0, 1, 1, 1, 1, 0, 2, 2, 1, 1, 2, -1, 1, 2, 2, 1, 0, 0, 3, 0, 1, 1, 2, 0, 1, 1, 2, 2, 1, 0, 2, 1, 0, 1, 2, 0, 1, 1, 0, 1, 2, 2, 2, 1, 2, 0, 2, -1, 0, 1, 2
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OFFSET
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0,5
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LINKS
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FORMULA
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G.f.: Sum_{n>=0} x^(1*n)/(1 - x^(5*n+2)) - x^2 * Sum_{n>=0} x^(3*n)/(1 - x^(5*n+4)).
G.f.: Sum_{n>=0} x^(2*n)/(1 - x^(5*n+1)) - x^2 * Sum_{n>=0} x^(4*n)/(1 - x^(5*n+3)).
G.f.: Sum_{n>=0} x^(1*n)/(1 - x^(5*n+2)) - x^2 * Sum_{n>=0} x^(4*n)/(1 - x^(5*n+3)).
G.f.: Sum_{n>=0} x^(2*n)/(1 - x^(5*n+1)) - x^2 * Sum_{n>=0} x^(3*n)/(1 - x^(5*n+4)).
G.f.: [ Sum_{n>=0} x^n/(1 - x^(5*n+1)) - x^3 * Sum_{n>=0} x^(4*n)/(1 - x^(5*n+4)) ] * R(x), where R(q) is the expansion of Ramanujan's continued fraction (A007325).
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EXAMPLE
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G.f.: P(q) = 1 + q + q^2 + q^3 + 2*q^4 + q^6 + q^7 + 2*q^8 + q^9 + q^10 + 2*q^12 + q^13 + 2*q^15 + 2*q^16 + q^18 + q^19 + 2*q^20 + ...
Given the g.f. of this sequence,
P(q) = Product_{n>=0} (1 - q^(5*n+5))^2 / ( (1 - q^(5*n+1))*(1 - q^(5*n+4)) ),
Q(q) = Product_{n>=0} (1 - q^(5*n+5))^2 / ( (1 - q^(5*n+2))*(1 - q^(5*n+3)) ),
then R(q) = P(q)/Q(q) where
Q(q) = 1 + q^2 + q^3 + q^4 - q^5 + 2*q^6 + q^8 + q^9 + q^10 + q^12 - q^13 + q^14 + 2*q^15 + q^16 + 2*q^18 - q^19 + q^20 + ...
and
R(q) = 1 + q - q^3 + q^5 + q^6 - q^7 - 2*q^8 + 2*q^10 + 2*q^11 - q^12 - 3*q^13 - q^14 + 3*q^15 + 3*q^16 - 2*q^17 - 5*q^18 - q^19 + 6*q^20 + ...;
here, R(q) is the expansion of Ramanujan's continued fraction (A007325).
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PROG
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(PARI) {a(n) = my(A = prod(m=0, n, (1 - x^(5*m+5))^2 / ( (1 - x^(5*m+1))*(1 - x^(5*m+4)) +x*O(x^n)) )); polcoeff(A, n)}
for(n=0, 80, print1(a(n), ", "))
(PARI) {a(n) = my(A = sum(m=0, n, x^(1*m)/(1 - x^(5*m+2) +x*O(x^n)) ) - x^2 * sum(m=0, n, x^(3*m)/(1 - x^(5*m+4) +x*O(x^n)) )); polcoeff(A, n)}
for(n=0, 80, print1(a(n), ", "))
(PARI) {a(n) = my(A = sum(m=0, n, x^(2*m)/(1 - x^(5*m+1) +x*O(x^n)) ) - x^2 * sum(m=0, n, x^(4*m)/(1 - x^(5*m+3) +x*O(x^n)) )); polcoeff(A, n)}
for(n=0, 80, print1(a(n), ", "))
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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A123682
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First in an infinite series of triangular arrays which, when taken together, sum to 1,1,3,5,11,21,43,85,... cf. A001045.
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+30
2
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1, 0, 1, 1, 1, 1, 0, 2, 2, 1, 1, 1, 3, 3, 1, 0, 2, 2, 4, 4, 1, 1, 1, 3, 3, 5, 5, 1, 0, 2, 2, 4, 4, 6, 6, 1, 1, 1, 3, 3, 5, 5, 7, 7, 1
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OFFSET
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1,8
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COMMENTS
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The starting row for each triangle is determined by the applicable source integer partition; thus for partitions 1,22,33,222,44,332,2222,333,... the starting rows are 1,4,6,6,8,8,8,9,...
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LINKS
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EXAMPLE
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Summing the row sums of the triangular arrays we have
1 1
1 1
3 3
5 5 0
11 9 2
21 13 7 1 0
43 19 18 4 2
85 25 36 12 10 2
...
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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A365573
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Number of primes between prime(n) and prime(n)+log(prime(n)), exclusive.
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+30
1
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0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 2, 2, 1, 0, 0, 1, 0, 1
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OFFSET
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1,88
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COMMENTS
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LINKS
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FORMULA
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Conjecture: Limit_{N->oo} (Ratio_{n=1..N} a(n)=0) = 1/e (A068985).
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PROG
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(PARI) a(n) = primepi(prime(n)+log(prime(n))) - primepi(prime(n))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A330608
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T(n, k) = P(n-k, k) where P(n, x) = Sum_{k=0..n} A053121(n, k)*x^k. Triangle read by rows, for 0 <= k <= n.
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+30
0
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1, 0, 1, 1, 1, 1, 0, 2, 2, 1, 2, 3, 5, 3, 1, 0, 6, 12, 10, 4, 1, 5, 10, 30, 33, 7, 5, 1, 0, 20, 74, 110, 72, 26, 6, 1, 14, 35, 185, 366, 306, 135, 37, 7, 1, 0, 70, 460, 1220, 1300, 702, 228, 50, 8, 1, 42, 126, 1150, 4065, 5525, 3650, 1406, 357, 65, 9, 1
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OFFSET
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0,8
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LINKS
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EXAMPLE
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Triangle starts:
[0] [ 1]
[1] [ 0, 1]
[2] [ 1, 1, 1]
[3] [ 0, 2, 2, 1]
[4] [ 2, 3, 5, 3, 1]
[5] [ 0, 6, 12, 10, 4, 1]
[6] [ 5, 10, 30, 33, 17, 5, 1]
[7] [ 0, 20, 74, 110, 72, 26, 6, 1]
[8] [14, 35, 185, 366, 306, 135, 37, 7, 1]
[9] [ 0, 70, 460, 1220, 1300, 702, 228, 50, 8, 1]
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MAPLE
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A053121 := (n, k, x) -> irem(n+k+1, 2)*x^k*(k+1)*binomial(n+1, (n-k)/2)/(n+1):
P := (n, x) -> add(A053121(n, k, x), k=0..n):
seq(seq(P(n-k, k), k=0..n), n=0..10);
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A239473
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Triangle read by rows: signed version of A059260: coefficients for expansion of partial sums of sequences a(n,x) in terms of their binomial transforms (1+a(.,x))^n ; Laguerre polynomial expansion of the truncated exponential.
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+20
9
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1, 0, 1, 1, -1, 1, 0, 2, -2, 1, 1, -2, 4, -3, 1, 0, 3, -6, 7, -4, 1, 1, -3, 9, -13, 11, -5, 1, 0, 4, -12, 22, -24, 16, -6, 1, 1, -4, 16, -34, 46, -40, 22, -7, 1, 0, 5, -20, 50, -80, 86, -62, 29, -8, 1, 1, -5, 25, -70, 130, -166, 148, -91, 37, -9, 1, 0, 6, -30, 95, -200, 296, -314, 239, -128, 46, -10, 1
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OFFSET
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0,8
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COMMENTS
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With T the lower triangular array above and the Laguerre polynomials L(k,x) = Sum_{j=0..k} (-1)^j binomial(k, j) x^j/j!, the following identities hold:
(A) Sum_{k=0..n} (-1)^k L(k,x) = Sum_{k=0..n} T(n,k) x^k/k!;
(B) Sum_{k=0..n} x^k/k! = Sum_{k=0..n} T(n,k) L(k,-x);
(C) Sum_{k=0..n} x^k = Sum_{k=0..n} T(n,k) (1+x)^k = (1-x^(n+1))/(1-x).
More generally, for polynomial sequences,
(D) Sum_{k=0..n} P(k,x) = Sum_{k=0..n} T(n,k) (1+P(.,x))^k,
where, e.g., for an Appell sequence, such as the Bernoulli polynomials, umbrally, (1+ Ber(.,x))^k = Ber(k,x+1).
Identity B follows from A through umbral substitution of j!L(j,-x) for x^j in A. Identity C, related to the cyclotomic polynomials for prime index, follows from B through the Laplace transform.
Integrating C gives Sum_{k=0..n} T(n,k) (2^(k+1)-1)/(k+1) = H(n+1), the harmonic numbers.
Identity A >= 0 for x >= 0 (see MathOverflow link for evaluation in terms of Hermite polynomials).
From identity C, W(m,n) = (-1)^n Sum_{k=0..n} T(n,k) (2-m)^k = number of walks of length n+1 between any two distinct vertices of the complete graph K_m for m > 2.
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LINKS
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FORMULA
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T(n, k) = Sum_{j=0..n} (-1)^(j+k) * binomial(j, k).
E.g.f: (exp(t) - (x-1)*exp((x-1)*t))/(2-x).
O.g.f. (n-th row): (1-(x-1)^(n+1))/(2-x).
Associated operator identities:
With D=d/dx, :xD:^n=x^n*D^n, and :Dx:^n=D^n*x^n, then bin(xD,n)= binomial(xD,n)=:xD:^n/n! and L(n,-:xD:)=:Dx:^n/n!=bin(xD+n,n)=(-1)^n bin(-xD-1,n),
A-o) Sum_{k=0..n} (-1)^k L(k,-:xD:) = Sum_{k=0..n} :-Dx:^k/k!
= Sum_{k=0..n} T(n,k) :-xD:^k/k! = Sum_{k=0..n} (-1)^k T(n,k)bin(xD,k)
B-o) Sum_{k=0..n} :xD:^k/k! = Sum_{k=0..n}, T(n,k) L(k,-:xD:)
= Sum_{k=0..n} T(n,k) :Dx:^k/k! = Sum_{k=0..n}, bin(xD,k).
Associated binomial identities:
A-b) Sum_{k=0..n} (-1)^k bin(s+k,k) = Sum_{k=0..n} (-1)^k T(n,k) bin(s,k)
= Sum_{k=0..n} bin(-s-1,k) = Sum{k=0..n} T(n,k) bin(-s-1+k,k)
B-b) Sum_{k=0..n} bin(s,k) = Sum_{k=0..n} T(n,k) bin(s+k,k)
= Sum_{k=0..n} (-1)^k bin(-s-1+k,k)
= Sum_{k=0..n} (-1)^k T(n,k) bin(-s-1,k).
In particular, from B-b with s=n, Sum_{k=0..n} T(n,k) bin(n+k,k) = 2^n. From B-b with s=0, row sums are all 1.
From identity C with x=-2, the unsigned row sums are the Jacobsthal sequence, i.e., Sum_{k=0..n} T(n,k) (1+(-2))^k = (-1)^n A001045(n+1); for x=2, the Mersenne numbers A000225; for x=-3, A014983 or signed A015518; for x=3, A003462; for x=-4, A014985 or signed A015521; for x=4, A002450; for x=-5, A014986 or signed A015531; and for x=5, A003463; for x=-6, A014987 or signed A015540; and for x=6, A003464.
With -s-1 = m = 0,1,2,..., B-b gives finite differences (recursions):
Sum_{k=0..n} (-1)^k T(n,k) bin(m,k) = Sum_{k=0..n} (-1)^k bin(m+k,k) = T(n+m,m), i.e., finite differences of the columns of T generate shifted columns of T. The columns of T are signed, shifted versions of sequences listed in the cross-references. Since the finite difference is an involution, T(n,k) = Sum_{j=0..k} (-1)^j T(n+j,j) bin(k,j)}. Gauss-Newton interpolation can be applied to give a generalized T(n,s) for s noninteger.
From identity C, S(n,m) = Sum_{k=0..n} T(n,k) bin(k,m) = 1 for m < n+1 and 0 otherwise, i.e., S = T*P, where S = A000012, as a lower triangular matrix and P = Pascal = A007318, so T = S*P^(-1), where P^(-1) = A130595, the signed Pascal array (see A132440), the inverse of P, and T^(-1) = P*S^(-1) = P*A167374 = A156644.
U(n,cos(x)) = e^(-n*i*x)*Sum_{k=0..n} T(n,k)*(1+e^(2*i*x))^k = sin((n+1)x)/sin(x), where U is the Chebyschev polynomial of the second kind A053117 and i^2 = -1. - Tom Copeland, Oct 18 2014
With a(n,x) = e^(nx), the partial sums are 1+e^x+...+e^(nx) = Sum_{k=0..n} T(n,k) (1+e^x)^k = [ x / (e^x-1) ] [ e^((n+1)x) -1 ] / x = [ (x / (e^x-1)) e^((n+1)x) - (x / (e^x-1)) ] / x = Sum_{k>=0} [ (Ber(k+1,n+1) - Ber(k+1,0)) / (k+1) ] * x^k/k!, where Ber(n,x) are the Bernoulli polynomials (cf. Adams p. 140). Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of the m-th powers of the integers, their binomial transforms, and the Bernoulli polynomials.
With a(n,x) = (-1)^n e^(nx), the partial sums are 1-e^x+...+(-1)^n e^(nx) = Sum_{k=0..n} T(n,k) (1-e^x)^k = [ (-1)^n e^((n+1)x) + 1 ] / (e^x+1) = [ (-1)^n (2 / (e^x+1)) e^((n+1)x) + (2 / (e^x+1)) ] / 2 = (1/2) Sum_{k>=0} [ (-1)^n Eul(k,n+1) + Eul(k,0) ] * x^k/k!, where Eul(n,x) are the Euler polynomials. Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of signed m-th powers of the integers; their binomial transforms, related to the Stirling numbers of the second kind and face numbers of the permutahedra; and the Euler polynomials.
(End)
As in A059260, a generator in terms of bivariate polynomials with the coefficients of this entry is given by (1/(1-y))*1/(1 + (y/(1-y))*x - (1/(1-y))*x^2) = 1 + y + (x^2 - x*y + y^2) + (2*x^2*y - 2*x*y^2 + y^3) + (x^4 - 2*x^3*y + 4*x^2*y^2 - 3*x*y^3 + y^4) + ... . This is of the form -h2 * 1 / (1 + h1*x + h2*x^2), related to the bivariate generator of A049310 with h1 = y/(1-y) and h2 = -1/(1-y) = -(1+h1). - Tom Copeland, Feb 16 2016
Letting P(k,x) = x in D gives Sum_{k=0..n} T(n,k)*Sum_{j=0..k} binomial(k,j) = Sum_{k=0..n} T(n,k) 2^k = n + 1.
The quantum integers [n+1]_q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)) = q^(-n) (1 - q^(2(n+1)) / (1 - q^2) = q^(-n)*Sum_{k=0..n} q^(2k) = q^(-n)*Sum_{k=0..n} T(n,k)*(1 + q^2)^k. (End)
T(n, k) = [x^k] Sum_{j=0..n} (x-1)^j. - Peter Luschny, Jul 09 2019
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EXAMPLE
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1
0 1
1 -1 1
0 2 -2 1
1 -2 4 -3 1
0 3 -6 7 -4 1
1 -3 9 -13 11 -5 1
0 4 -12 22 -24 16 -6 1
1 -4 16 -34 46 -40 22 -7 1
0 5 -20 50 -80 86 -62 29 -8 1
1 -5 25 -70 130 -166 148 -91 37 -9 1
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MAPLE
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add(binomial(j, k)*(-1)^(j+k), j=k..n) ;
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MATHEMATICA
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Table[Sum[(-1)^(j+k)*Binomial[j, k], {j, 0, n}], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Feb 06 2018 *)
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PROG
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(PARI) for(n=0, 10, for(k=0, n, print1(sum(j=0, n, (-1)^(j+k)*binomial(j, k)), ", "))) \\ G. C. Greubel, Feb 06 2018
(Magma) [[(&+[(-1)^(j+k)*Binomial(j, k): j in [0..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Feb 06 2018
(Sage)
Trow = lambda n: sum((x-1)^j for j in (0..n)).list()
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CROSSREFS
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Cf. Bottomley's cross-references in A059260.
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KEYWORD
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AUTHOR
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EXTENSIONS
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Formula re Euler polynomials corrected by Tom Copeland, Mar 08 2024
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STATUS
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approved
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A015504
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Inverse of 1495th cyclotomic polynomial.
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+20
1
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1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 1, -1, 1, -1, 0, 1, -1, 1, -1, 0, 2, -2, 1, 0, -1, 2, -2, 1, 0, -1, 2, -2, 1, 1, -2, 2, -1, 0, 1, -2, 2, -1, 0, 2, -3, 2, 0, -1, 2, -2, 1, 0, -1, 2, -2, 1, 1, -2, 2, -1, 0, 1, -2, 2, -1, 0, 2, -3, 2, 0, -1, 2, -2, 1, 0, -1, 2, -2
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OFFSET
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0,24
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COMMENTS
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Periodic with period length 1495. - Ray Chandler, Apr 07 2017
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LINKS
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MAPLE
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with(numtheory, cyclotomic); c := n->series(1/cyclotomic(n, x), x, 80);
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MATHEMATICA
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CoefficientList[Series[1/Cyclotomic[1495, x], {x, 0, 120}], x] (* Harvey P. Dale, Jan 05 2021 *)
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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1, 0, 1, -1, -1, 1, 0, -2, -2, 1, 0, 0, -3, -3, 1, 0, 0, 0, -4, -4, 1, 0, 0, 0, 0, -5, -5, 1, 0, 0, 0, 0, 0, -6, -6, 1, 0, 0, 0, 0, 0, 0, -7, -7, 1, 0, 0, 0, 0, 0, 0, 0, -8, -8, 1, 0, 0, 0, 0, 0, 0, 0, 0, -9, -9, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -10, -10, 1
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OFFSET
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0,8
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LINKS
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FORMULA
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T(n,n) = 1 for n >= 0, and T(n,n-1) = 1 - n for n > 0, and T(n,n-2) = 1 - n for n > 1, and T(n,k) = 0 if n < 0 or k < 0 or n < k or n > k+2.
G.f.: Sum_{n>=0, k=0..n} T(n,k) * x^k * t^n = (1 + t) * (1 - (1 + x) * t) / (1 - x * t)^2.
Alt. row sums equal (-1)^n for n >= 0.
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EXAMPLE
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The triangle T(n,k) for 0 <= k <= n starts:
n\k : 0 1 2 3 4 5 6 7 8 9
======================================================
0 : 1
1 : 0 1
2 : -1 -1 1
3 : 0 -2 -2 1
4 : 0 0 -3 -3 1
5 : 0 0 0 -4 -4 1
6 : 0 0 0 0 -5 -5 1
7 : 0 0 0 0 0 -6 -6 1
8 : 0 0 0 0 0 0 -7 -7 1
9 : 0 0 0 0 0 0 0 -8 -8 1
etc.
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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