A103245 Triangle read by rows: T(n,k) = binomial(2n+1, n-k)*Fibonacci(2k+1), 0 <= k <= n.

1, 3, 2, 10, 10, 5, 35, 42, 35, 13, 126, 168, 180, 117, 34, 462, 660, 825, 715, 374, 89, 1716, 2574, 3575, 3718, 2652, 1157, 233, 6435, 10010, 15015, 17745, 15470, 9345, 3495, 610, 24310, 38896, 61880, 80444, 80920, 60520, 31688, 10370, 1597, 92378

Offset

0,2

References

S. G. Guba, Problem No. 174, Issue No. 4, July-August 1965, p. 73 of Matematika v Skole.

Links

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened).

V. E. Hoggatt, Jr. and L. Carlitz, Problem H-77, The Fibonacci Quarterly, 5, No. 3, 1967, 256-258.

Formula

T(n, k) = binomial(2n+1, n-k)*Fibonacci(2k+1), 0 <= k <= n.

Example

Triangle begins:
    1;
    3,   2;
   10,  10,   5;
   35,  42,  35,  13;
  126, 168, 180, 117,  34;

Maple

with(combinat): T:=(n, k)->binomial(2*n+1, n-k)*fibonacci(2*k+1): for n from 0 to 9 do seq(T(n, k), k=0..n) od; # yields sequence in triangular form

Mathematica

Table[Binomial[2 n + 1, n - k] Fibonacci[2 k + 1], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 01 2019 *)

Crossrefs

Column 0 is A001700.

Column 1 is A024483.

T(n, n) = A001519(n+1) (the odd-indexed Fibonacci numbers).

Row sums are the powers of 5 (A000351).

Alternating row sums yield A054108.

Sequence in context: A214844 A214966 A367300 * A019242 A064367 A113980

Adjacent sequences: A103242 A103243 A103244 * A103246 A103247 A103248

Keyword

nonn,tabl,changed

Author

Emeric Deutsch, Mar 19 2005

Status

approved