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A103245
Triangle read by rows: T(n,k) = binomial(2n+1, n-k)*Fibonacci(2k+1), 0 <= k <= n.
2
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
KEYWORD
nonn,tabl
AUTHOR
Emeric Deutsch, Mar 19 2005
STATUS
approved