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A105438 Triangle, row sums = (Fibonacci numbers - 2). 4
1, 2, 1, 3, 2, 1, 4, 4, 2, 1, 5, 6, 5, 2, 1, 6, 9, 8, 6, 2, 1, 7, 12, 14, 10, 7, 2, 1, 8, 16, 20, 20, 12, 8, 2, 1, 9, 20, 30, 30, 27, 14, 9, 2, 1, 10, 25, 40, 50, 42, 35, 16, 10, 2, 1, 11, 30, 55, 70, 77, 56, 44, 18, 11, 2, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
Row sums = 1, 3, 6, 11, 19, 32, 53...(Fibonacci numbers - 2; starting with F(4)) The first few rows of the triangle are:
The first few rows of the triangle are:
1;
2, 1;
3, 2, 1;
4, 4, 2, 1;
5, 6, 5, 2, 1;
6, 9, 8, 6, 2, 1;
7, 12, 14, 10, 7, 2, 1;
8, 16, 20, 20, 12, 8, 2, 1;
9, 20, 30, 30, 27, 14, 9, 2, 1;
10, 25, 40, 50, 42, 35, 16, 10, 2, 1;
...
Row sums = (Fibonacci numbers - 2; starting 1, 3, 6...).
Column 1 = A002620; Column 2 = A006918; Column 3 = A096338.
Inverse array is A105522. - Paul Barry, Apr 11 2005
Diagonal sums are A027383(n). - Philippe Deléham, Jan 16 2014
LINKS
T. Mansour, A. O. Munagi, Alternating subsets modulo m, Rocky Mt. J. Math. 42, No. 4, 1313-1325 (2012), eq. (2)
FORMULA
By columns (k = 0, 1, 2...); use partial sum operator on (bin(n, k) numbers repeated).
T(n, k)=sum{j=0..n-k, C((j+2k)/2, k)(1+(-1)^j)+C((j-1+2k)/2, k)(1-(-1)^j)/2; Riordan array (1/(1-x)^2, x/(1-x^2)). - Paul Barry, Apr 11 2005
T(n,k)=T(n-1,k)+T(n-1,k-1)+T(n-2,k)-T(n-2,k-1)-T(n-3,k), T(0,0)=1, T(1,0)=2, T(1,1)= 1, T(n,k)= 0 if k<0 or if k>n. - Philippe Deléham, Jan 16 2014
EXAMPLE
Column 2: 1, 2, 5, 8, 14, 20, 30...is generated by using the partial sum operator on 1, 1, 3, 3, 6, 6, 10, 10...
CROSSREFS
Sequence in context: A355474 A137679 A152072 * A062001 A361043 A181847
KEYWORD
nonn,tabl,easy
AUTHOR
Gary W. Adamson, Apr 09 2005
STATUS
approved

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Last modified March 18 22:34 EDT 2024. Contains 370951 sequences. (Running on oeis4.)