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A063394
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Border sum triangle, read by rows: Let T(n,0)=T(n,n)=1. In general T(n,m) is the sum of the elements (apart from T(n,m) itself) in the border of the rectangle with vertices T(0,0), T(n-m,0), T(n,m) and T(m,m).
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4
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1, 1, 1, 1, 3, 1, 1, 7, 7, 1, 1, 15, 19, 15, 1, 1, 31, 47, 47, 31, 1, 1, 63, 111, 131, 111, 63, 1, 1, 127, 255, 343, 343, 255, 127, 1, 1, 255, 575, 863, 979, 863, 575, 255, 1, 1, 511, 1279, 2111, 2655, 2655, 211, 1279, 511, 1, 1, 1023, 2815, 5055, 6943, 7683, 6943
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,5
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FORMULA
| If m < 0 or m > n then T(n, m) = 0; if m = 0 or m = n then T(n, m) = 1; otherwise T(n, m) = Sum( T(n-i, m-i), i=1..m) + Sum( T(n-i, m), i=1..n-m) + Sum( T(n-m-i, 0), i=1..n-m) + Sum( T(i, i), i=1..m-1).
The U-coordinates are nicer. Label the elements U(0, 0), U(1, 0), U(0, 1), U(2, 0), U(1, 1), U(0, 2), ...
Then U(n, 0) = U(0, m) = 1; for n>=1, m>=1, U(n, m) = Sum_{i=0..n-1} U(i, 0) + Sum_{j=0..m-1} U(0, j) - U(0, 0) + Sum_{j=0..m-1} U(n, j) + Sum_{i=0..n-1} U(i, m). Hence U(z, w) = Sum U(n, m) z^n w^m = (1-2*z-2*w+5*z*w-2*z^2*w^2)/((1-z)*(1-w)*(1-2*z-2*w+3*z*w)). - N. J. A. Sloane (njas(AT)research.att.com), Jun 16 2005
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EXAMPLE
| The triangle begins:
..........1
........1...1
......1...3...1
....1...7...7...1
..1..15..19...15..1
E.g. 19 = 7 + 1 + 1 + 1 + 1 + 1 + 7.
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MAPLE
| T:=proc(n, m) option remember; local i, j, k, t1, t2, t3; if m < 0 or m > n then RETURN(0); fi; if m = 0 or m = n then RETURN(1); fi; add( T(n-i, m-i), i=1..m) + add( T(n-i, m), i=1..n-m) + add( T(n-m-i, 0), i=1..n-m) + add( T(i, i), i=1..m-1); end;
U:=(1-2*z-2*w+5*z*w-2*z^2*w^2)/(1-z)/(1-w)/(1-2*z-2*w+3*z*w);
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CROSSREFS
| T(1, n) gives A000225(n+1), T(2, n) for n>0 gives A006589.
Other diagonals: A063396, A063397, A063398, A063395.
Sequence in context: A176331 A157836 A205497 * A193871 A108470 A157152
Adjacent sequences: A063391 A063392 A063393 * A063395 A063396 A063397
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KEYWORD
| easy,nonn,tabl
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AUTHOR
| Floor van Lamoen (fvlamoen(AT)hotmail.com), Jul 16 2001
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EXTENSIONS
| Entry revised by N. J. A. Sloane (njas(AT)research.att.com), Jun 15 2005
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