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Triangle read by rows: given the left border = the Lucas numbers, (1, 3, 4, 7, ...), T(n,k) = (n-1,k) + (n-1,k-1).
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%I #14 Nov 19 2024 07:27:09

%S 1,3,1,4,4,1,7,8,5,1,11,15,13,6,1,18,26,28,19,7,1,29,44,54,47,26,8,1,

%T 47,73,98,101,73,34,9,1,76,120,171,199,174,107,43,10,1,123,196,291,

%U 370,373,281,150,53,11,1,199,319,487,661,743,654,431,203,64,12,1,322,518,806,1148,1404,1397,1085,634,267,76,13,1

%N Triangle read by rows: given the left border = the Lucas numbers, (1, 3, 4, 7, ...), T(n,k) = (n-1,k) + (n-1,k-1).

%C Row sums = A027973: (1, 4, 9, 21, 46, 99, 209, ...).

%e First few rows of the triangle:

%e 1;

%e 3, 1;

%e 4, 4, 1;

%e 7, 8, 5, 1;

%e 11, 15, 13, 6, 1;

%e 18, 26, 28, 19, 7, 1;

%e ...

%e (6,3) = 28 = 13 + 15 = (5,3) + (5,2).

%p L[1]:=1: L[2]:=3: for n from 3 to 12 do L[n]:=L[n-1]+L[n-2] od: T:=proc(n,k) if k=1 then L[n] elif n=1 then 0 else T(n-1,k)+T(n-1,k-1) fi end: for n from 1 to 12 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form - _Emeric Deutsch_, Jan 01 2007

%p A000204 := proc(n) if n =1 then RETURN(1) ; elif n = 2 then RETURN(3) ; else RETURN( A000204(n-1)+A000204(n-2)) ; fi ; end ; A125608 := proc(nmax) local a,row,col,anext ; a := [1] ; row := 1 ; while nops(a) < nmax do row := row+1 ; a := [op(a),A000204(row)] ; for col from 2 to row-1 do anext := op(-row,a)+op(-row+1,a) ; a := [op(a),anext] ; od ; a := [op(a),1] ; od ; RETURN(a) ; end ; A125608(80) ; # _R. J. Mathar_, Jan 07 2007

%t T[n_, 1] := LucasL[n];

%t T[n_, k_] /; 2 <= k <= n := T[n, k] = T[n - 1, k] + T[n - 1, k - 1];

%t T[_, _] = 0;

%t Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Nov 19 2024 *)

%Y Cf. A027973.

%K nonn,tabl

%O 1,2

%A _Gary W. Adamson_, Nov 27 2006

%E More terms from _Emeric Deutsch_, Jan 01 2007

%E More terms from _R. J. Mathar_, Jan 07 2007