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A094435
Triangular array read by rows: T(n,k) = Fibonacci(k)*C(n,k), k = 1...n; n>=1.
11
1, 2, 1, 3, 3, 2, 4, 6, 8, 3, 5, 10, 20, 15, 5, 6, 15, 40, 45, 30, 8, 7, 21, 70, 105, 105, 56, 13, 8, 28, 112, 210, 280, 224, 104, 21, 9, 36, 168, 378, 630, 672, 468, 189, 34, 10, 45, 240, 630, 1260, 1680, 1560, 945, 340, 55, 11, 55, 330, 990, 2310, 3696, 4290, 3465, 1870, 605, 89
OFFSET
1,2
COMMENTS
Let F(n) denote the n-th Fibonacci number (A000045). Then n-th row sum of T is F(2n) and n-th alternating row sum is F(n).
FORMULA
From G. C. Greubel, Oct 30 2019: (Start)
T(n, k) = binomial(n, k)*Fibonacci(k).
Sum_{k=1..n} binomial(n,k)*Fibonacci(k) = Fibonacci(2*n).
Sum_{k=1..n} (-1)^(k-1)*binomial(n,k)*Fibonacci(k) = Fibonacci(n). (End)
EXAMPLE
First few rows:
1;
2 1;
3 3 2;
4 6 8 3;
5, 10, 20, 15, 5;
6, 15, 40, 45, 30, 8;
MAPLE
with(combinat); seq(seq(binomial(n, k)*fibonacci(k), k=1..n), n=1..12); # G. C. Greubel, Oct 30 2019
MATHEMATICA
Table[Fibonacci[k]*Binomial[n, k], {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Oct 30 2019 *)
PROG
(PARI) T(n, k) = binomial(n, k)*fibonacci(k);
for(n=1, 12, for(k=1, n, print1(T(n, k), ", "))) \\ G. C. Greubel, Oct 30 2019
(Magma) [Binomial(n, k)*Fibonacci(k): k in [1..n], n in [1..12]]; // G. C. Greubel, Oct 30 2019
(Sage) [[binomial(n, k)*fibonacci(k) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Oct 30 2019
(GAP) Flat(List([1..12], n-> List([1..n], k-> Binomial(n, k)*Fibonacci(k) ))); # G. C. Greubel, Oct 30 2019
KEYWORD
nonn,tabl
AUTHOR
Clark Kimberling, May 03 2004
STATUS
approved