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A097805
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Riordan array (1,1/(1-x)) read by rows.
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48
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1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 4, 1, 0, 1, 5, 10, 10, 5, 1, 0, 1, 6, 15, 20, 15, 6, 1, 0, 1, 7, 21, 35, 35, 21, 7, 1, 0, 1, 8, 28, 56, 70, 56, 28, 8, 1, 0, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 0, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 0, 1, 11, 55, 165, 330, 462
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,9
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COMMENTS
| Columns have g.f. (x/(1-x))^k. Reverse of A071919. Row sums are A011782. Diagonal sums are Fib(n-1). Inverse as Riordan array is (1,1/(1+x)). A097805=B*A059260*B^(-1), where B is the binomial matrix.
(0,1)-Pascal triangle . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Nov 21 2006
(n+1) * each term of row n generates triangle A127952: (1; 0, 2; 0, 3, 3; 0, 4, 8, 4;...). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Feb 09 2007
Triangle T(n,k), 0<=k<=n, read by rows, given by [0,1,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938 . [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Dec 12 2008]
Contribution from Paul Weisenhorn, Feb 09 2011: (Start)
Triangle read by rows: T(r,c) = number of unordered partitions of the number n=r*(r+1)/2+c with (r+1) summands < (r+1) and at most pairs of equal summands and summands in neighbouring pairs have difference 2.
Triangle read by rows: T(r,c) = number of unordered partitions of the number n=r*(r+1)/2+(c-1) with r summands < (r+1) and at most pairs of equal summands and summands in neighbouring pairs have difference 2. (End)
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FORMULA
| Number triangle T(n, k) defined by T(n, k)=sum{j=0..n, binomial(n, j)*if(k<=j, (-1)^(j-k), 0)}
G.f.: 1 +x(x +x^3(1+x) +x^6(1+x)^2 +x^10(1+x)^3 +...) . - Michael Somos Aug 20 2006
T(r,c)=binomial(r-1,c-1), 0<=c<=r. Paul Weisenhorn, Feb 09 2011
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EXAMPLE
| Rows begin {1}, {0,1}, {0,1,1}, {0,1,2,1}, ....
From Paul Weisenhorn, Feb 09 2011 (Start)
T(r=5,c=3) = binomial(4,2) = 6 unordered partitions of the number n = r*(r+1)/2+c = 18 with (r+1)=6 summands : (5+5+4+2+1+1), (5+5+3+3+1+1), (5+4+4+3+1+1), (5+5+3+2+2+1), (5+4+4+2+2+1), (5+4+3+3+2+1).
T(r=5,c=3) = binomial(4,2) = 6 unordered partitions of the number n = r*(r+1)/2+(c-1) = 17 with r=5 summands : (5+5+4+2+1), (5+5+3+3+1), (5+5+3+2+2), (5+4+4+3+1), (5+4+4+2+2), (5+4+3+3+2). (End)
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PROG
| (PARI) {a(n)=local(m); if(n<2, n==0, n--; m=(sqrtint(8*n+1)-1)\2; binomial(m-1, n-m*(m+1)/2))} /* Michael Somos Aug 20 2006 */
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CROSSREFS
| Cf. A127952.
Sequence in context: A119337 A110555 A071919 * A167763 A127839 A017827
Adjacent sequences: A097802 A097803 A097804 * A097806 A097807 A097808
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KEYWORD
| easy,nonn,tabl
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AUTHOR
| Paul Barry (pbarry(AT)wit.ie), Aug 25 2004
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EXTENSIONS
| Corrected by Philippe DELEHAM, Oct 05 2005
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