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Search: id:A059260
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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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11
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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
(list; table; graph; listen)
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OFFSET
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0,8
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COMMENT
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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( x^k, k=0..n), having as roots all the n-th root of unity except 0; then coefficients in x of (-1)^n Y_n(-x-1) gives exactly the n-th row of A059260 and a practical way to compute it. - Olivier Gerard (olivier.gerard(AT)gmail.com), Jul 30 2002
The maximum in the 2n-row is T(n,n) which is A026641; also T(n,n)~2/3*binomial(2n,n). The maximum in the (2n-1)-row is T(n-1,n) which is A014300 (but T has not 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(binomial(i+k-x,j-k)*binomial(j-k+x,k),k=0..j) 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 (claude.morin4(AT)libertysurf.fr), May 21 2002
The second greatest term in the 2n-row is T(n-1,n+1) which is A014301; the second greatest term in the (2n+1)-row is T(n+1,n)=2*T(n-1,n+1) which is 2*A014301. - Claude MORIN.
Diagonal sums give A008346. - Paul Barry (pbarry(AT)wit.ie), Sep 23 2004
Riordan array (1/(1-x^2), 1/(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 (pbarry(AT)wit.ie), Oct 25 2004
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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 + ...
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((-1)^(i+j+m)*binomial(m, j), m=j..i+j). - Robert FERREOL (rferreol(AT)noos.fr), 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
T(n, k)=sum{j=0..n, (-1)^(n-j)binomial(j, k)}. - Paul Barry (pbarry(AT)wit.ie), Aug 25 2004
T(n, k)=sum{j=0..n-k, C(n-j, j)*C(j, n-k-j)} - Paul Barry (pbarry(AT)wit.ie), Jul 25 2005
Equals A097807 * A007318. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Feb 21 2007
Equals A128173 * A007318 as infinite lower triangular matrices. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Feb 17 2007
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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; ...
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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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CROSSREFS
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Cf. A059259. Row sums give A001045.
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 (se16(AT)btinternet.com), Jun 25 2001
Cf. A026641, A014300.
Sequence in context: A034851 A122085 A066287 this_sequence A135229 A081372 A101489
Adjacent sequences: A059257 A059258 A059259 this_sequence A059261 A059262 A059263
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KEYWORD
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nonn,tabl,nice
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AUTHOR
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N. J. A. Sloane (njas(AT)research.att.com), Jan 23 2001
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