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A164844
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Generalized Pascal Triangle - satisfying the same recurrence as Pascal's triangle, but with a(n,0)=1 and a(n,n)=10^n (instead of both being 1).
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6
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1, 1, 10, 1, 11, 100, 1, 12, 111, 1000, 1, 13, 123, 1111, 10000, 1, 14, 136, 1234, 11111, 100000, 1, 15, 150, 1370, 12345, 111111, 1000000, 1, 16, 165, 1520, 13715, 123456, 1111111, 10000000, 1, 17, 181, 1685, 15235, 137171, 1234567, 11111111, 100000000, 1, 18, 198, 1866, 16920, 152406, 1371738, 12345678, 111111111, 1000000000, 1, 19, 216, 2064, 18786, 169326, 1524144, 13717416, 123456789, 1111111111, 10000000000
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| Like with Pascal's triangle, the columns grown polynomially. For example, a(n,1)=10+n, a(n,2)=(1/2)*(180+19n+n^2), a(n,3)=(1/6)*(5400 + 569n + 30n^2 + n^3). Likewise, diagonals grow exponentially: a(n,n)=10^n, a(n,n-1) = (10^n-1) / 9. (Kellen Myers Jan 24 2010)
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FORMULA
| Defined for 0<=k<=n.
a(n,k) = sum( 10^i * ( r-i-1 choose r-k-1 ) , i = 0..k ).
a(0,n)=1, a(n,n)=10^n, a(n,k)=a(n-1,k-1)+a(n-1,k).
[Kellen Myers Jan 24 2010]
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EXAMPLE
| Triangle begins:
1
1,10
1,11,100
1,12,111,1000
1,13,123,1111,10000
1,14,136,1234,11111,100000
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MATHEMATICA
| f[r_, k_] := Sum[10^i*Binomial[r - i - 1, r - k - 1], {i, 0, k}];
TableForm[Table[f[n, k], {n, 0, 15}, {k, 0, n}]]
(*Alex Meiburg, 8/21/2010. timeroot.alex(AT)gmail.com*)
a[n_, k_] := a[n, k] =
Piecewise[{{0, k > n || k < 0}, {1, k == 0}, {10^n, k == n}},
a[n - 1, k - 1] + a[n - 1, k]]
TableForm[Table[a[n, k], {n, 0, 10}, {k, 0, n}]]
(*Kellen Myers Jan 24 2010*)
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CROSSREFS
| A007318, A093645, A011557
Sequence in context: A107353 A172171 A164899 * A130311 A063672 A070606
Adjacent sequences: A164841 A164842 A164843 * A164845 A164846 A164847
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KEYWORD
| nonn
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AUTHOR
| Mark Dols (markdols99(AT)yahoo.com), Aug 28 2009
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EXTENSIONS
| Definition clarified, more terms, and overhaul of Meiburg's Mathematica code by Kellen Myers Jan 24 2010.
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