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Search: id:A128813
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| A128813 |
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Triangle of coefficients of (x+1)(x+3)(x+6)...(x+n(n+1)/2). |
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1
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| 1, 1, 1, 1, 4, 3, 1, 10, 27, 18, 1, 20, 127, 288, 180, 1, 35, 427, 2193, 4500, 2700, 1, 56, 1162, 11160, 50553, 97200, 56700, 1, 84, 2730, 43696, 363033, 1512684, 2778300, 1587600, 1, 120, 5754, 141976, 1936089, 14581872, 57234924, 101606400, 57153600
(list; graph; listen)
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OFFSET
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0,5
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COMMENT
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The triangle begins: 1 1,1 1,4,3 1,10,27,18 1, 20, 127, 288, 180 1, 35, 427, 2193, 4500, 2700 1, 56, 1162, 11160, 50553, 97200, 56700
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FORMULA
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a(0,0)=1, a(1,0)=1, a(1,1)=1, a(i,j)=i*(i+1)/2*a(i-1,j-1)+a(i-1,j), j=0..i-1, a(i,i)= i*(i+1)/2*a(i-1,i-1). a(n,n)=Product(k*(k+1)/2, k=1..n) = 1, 1, 3, 18, 180, 2700, 56700,.. Sum(a(n,m), m=0..n)= Product(k*(k+1)/2+1, k=1..n) = 1, 2, 8, 56, 616, 9856,..
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EXAMPLE
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(x+1)(x+3)(x+6)=x^3+10x^2+27x+18, so a(3,j)=1, 10, 27, 18
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MAPLE
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for n from 1 to 9 do b[n]:=n*(n+1)/2 od: a[0, 0]:=1:a[1, 0]:=1:a[1, 1]:=1:for i from 2 to 9 do a[i, 0]:=1:for j from 1 to i-1 do a[i, j]:=b[i]*a[i-1, j-1]+ a[i-1, j] od:a[i, i]:=b[i]*a[i-1, i-1] od: seq(seq(a[i, j], j=0..i), i=0..9);
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CROSSREFS
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Cf. A007318, A094638, A000142.
Adjacent sequences: A128810 A128811 A128812 this_sequence A128814 A128815 A128816
Sequence in context: A123160 A039758 A109692 this_sequence A109062 A112493 A010305
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KEYWORD
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easy,nonn
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AUTHOR
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Miklos Kristof (kristmikl(AT)freemail.hu), Apr 10 2007
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