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1, 1, 1, 3, 2, 1, 7, 5, 3, 1, 15, 12, 8, 4, 1, 31, 27, 20, 12, 5, 1, 63, 58, 47, 32, 17, 6, 1, 127, 121, 105, 79, 49, 23, 7, 1, 255, 248, 226, 184, 128, 72, 30, 8, 1, 511, 503, 474, 410, 312, 200, 102, 38, 9, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,4
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COMMENTS
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Riordan array ( (1-2*x+2*x^2)/((1-x)*(1-2*x)), x/(1-x) ). - Peter Bala, Mar 21 2018
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LINKS
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FORMULA
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T(n,k) = C(n, n-k) + Sum_{i = 2..n} 2^(i-1)*C(n-i, n-k-i), where C(n,k) = n!/(k!*(n-k)!) for 0 <= k <= n, otherwise 0.
Exp(x) * the e.g.f. for row n = the e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(7 + 5*x + 3*x^2/2! + x^3/3!) = 7 + 12*x + 20*x^2/2! + 32*x^3/3! + 49*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1-x) ).
(End)
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EXAMPLE
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First few rows of the triangle are:
1;
1, 1;
3, 2, 1;
7, 5, 3, 1;
15, 12, 8, 4, 1;
31, 27, 20, 12, 5, 1;
63, 58, 47, 32, 17, 6, 1;
...
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MAPLE
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C := proc (n, k) if 0 <= k and k <= n then factorial(n)/(factorial(k)*factorial(n-k)) else 0 end if;
end proc:
for n from 0 to 12 do
seq(C(n, n-k) + add(2^(i-1)*C(n-i, n-k-i), i = 2..n), k = 0..n)
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MATHEMATICA
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T[n_, k_] := Binomial[n, n-k] + Sum[2^(i-1) Binomial[n-i, n-k-i], {i, 2, n}]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 10 2019 *)
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PROG
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(Sage) # uses[riordan_array from A256893]
riordan_array((1-2*x+2*x^2)/((1-x)*(1-2*x)), x/(1-x), 8) # Peter Luschny, Mar 21 2018
(GAP) Flat(List([0..12], n->List([0..n], k->Binomial(n, k)+Sum([2..n], i->2^(i-1)*Binomial(n-i, n-k-i))))); # Muniru A Asiru, Mar 22 2018
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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