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A101447
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Triangle read by rows: T(n,k) = (2*k+1)*(n+1-k), 0<=k<n.
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0
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1, 2, 3, 3, 6, 5, 4, 9, 10, 7, 5, 12, 15, 14, 9, 6, 15, 20, 21, 18, 11, 7, 18, 25, 28, 27, 22, 13, 8, 21, 30, 35, 36, 33, 26, 15, 9, 24, 35, 42, 45, 44, 39, 30, 17, 10, 27, 40, 49, 54, 55, 52, 45, 34, 19, 11, 30, 45, 56, 63, 66, 65, 60, 51, 38, 21, 12, 33, 50, 63, 72, 77, 78, 75, 68
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| The triangle is generated from the product of matrix A and matrix B, I.e. A * B where A = the infinite lower triangular matrix:
1 0 0 0 0 ...
1 1 0 0 0 ...
1 1 1 0 0 ...
1 1 1 1 0 ...
1 1 1 1 1 ...
... and B = the infinite lower triangular matrix:
1 0 0 0 0...
1 3 0 0 0...
1 3 5 0 0...
1 3 5 7 0...
1 3 5 7 9...
...
Row sums give the square pyramidal numbers A000330.
T(n+0,0)=1*n=A000027(n+1); T(n+1,1)=3*n=A008585(n); T(n+2,2)=5*n=A008587(n); T(n+3,3)=7*n=A008589(n); etc. So T(n,0)*T(n,1)=3*n*(n+1)=A028896(n) (6 times triangular numbers.) T(n,1)*T(n,2)/10=3*n*(n+1)/2=A045943(n) for n>0 T(n,2)*T(n,3)/10=7/2*n*(n+1)=A024966(n) for n>1 (7 times triangular numbers.) etc.
Contribution from Gary W. Adamson (qntmpkt(AT)yahoo.com), Apr 25 2010: (Start)
Consider the following array, signed as shown:
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1,...3,...5,...7,...9,...11,...
2,..-6,..10,.-14,..18,..-22,...
3,...9,..15,..21,..27,...33,...
4,.-12,..20,.-28,..36,..-44,...
5,..15,..25,..35,..45,...55,...
6,.-18,..30,.-42,..54,..-66,...
7,..21,..35,..49,..63,...77,...
...
Let each term (+, -)k = (+, -) phi^(-k).
Consider the inverse terms of the Lucas series (1/1, 1/3, 1/4, 1/7,...).
By way of example, let q = phi = 1.6180339,...; then
...
1/1. = q^(-1) + q^(-3) + q^(-5) + q^(-7) + q^-9) + ...
1/3. = q^(^2) - q^(-6) + q^(-10) - q^(-14) + q^(-18) + ...
1/4. = q^(-3) + q^(-9) + q^(-15) + q^(-21) + q^(-27) +...
1/7. = q^(-4) - q^-12) + q^(-20) - q^(-28) + q^(-36) + ...
1/11 = q^(-5) + q^-15) + q^(-25) + q^(-35) + q^(-45) + ...
...
Relating to the Pell series, the corresponding "Lucas"-like series is:
(2, 6, 14, 34, 82, 198,...) such that herein, q = 2.414213,...= (1 + sqrt(2))
Then analogous to the previous set,
...
1/2 = q^(-1) + q^(-3) + q^(-5) + q^(-7) + ...
1/6 = q^(-2) - q^(-6) + q^(-10) - q^(-14) + q^(-18) + ...
... (End)
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MATHEMATICA
| t[n_, k_] := If[n < k, 0, (2*k + 1)*(n - k + 1)]; Flatten[ Table[ t[n, k], {n, 0, 11}, {k, 0, n}]] (from Robert G. Wilson v Jan 20 2005)
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PROG
| (PARI) T(n, k)=if(n<k, 0, (2*k+1)*(n-k+1)) for(i=0, 15, for(j=0, i, print1(T(i, j), ", ")); print())
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CROSSREFS
| Cf. A094728 (triangle generated by B*A), A000330.
Sequence in context: A160791 A115973 A057047 * A119322 A014498 A186286
Adjacent sequences: A101444 A101445 A101446 * A101448 A101449 A101450
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KEYWORD
| nonn,tabl
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AUTHOR
| Lambert Klasen (lambert.klasen(AT)gmx.de) and Gary W. Adamson (qntmpkt(AT)yahoo.com), Jan 19 2005
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