|
%I #22 Dec 03 2019 04:03:23
%S 1,3,2,10,10,5,35,42,35,13,126,168,180,117,34,462,660,825,715,374,89,
%T 1716,2574,3575,3718,2652,1157,233,6435,10010,15015,17745,15470,9345,
%U 3495,610,24310,38896,61880,80444,80920,60520,31688,10370,1597,92378
%N Triangle read by rows: T(n,k) = binomial(2n+1, n-k)*Fibonacci(2k+1), 0 <= k <= n.
%D S. G. Guba, Problem No. 174, Issue No. 4, July-August 1965, p. 73 of Matematika v Skole.
%H Michael De Vlieger, <a href="/A103245/b103245.txt">Table of n, a(n) for n = 0..11475</a> (rows 0 <= n <= 150, flattened).
%H V. E. Hoggatt, Jr. and L. Carlitz, <a href="https://www.fq.math.ca/Scanned/5-3/advanced5-3.pdf">Problem H-77</a>, The Fibonacci Quarterly, 5, No. 3, 1967, 256-258.
%F T(n, k) = binomial(2n+1, n-k)*Fibonacci(2k+1), 0 <= k <= n.
%e Triangle begins:
%e 1;
%e 3, 2;
%e 10, 10, 5;
%e 35, 42, 35, 13;
%e 126, 168, 180, 117, 34;
%p 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
%t Table[Binomial[2 n + 1, n - k] Fibonacci[2 k + 1], {n, 0, 9}, {k, 0, n}] // Flatten (* _Michael De Vlieger_, Dec 01 2019 *)
%Y Column 0 is A001700.
%Y Column 1 is A024483.
%Y T(n, n) = A001519(n+1) (the odd-indexed Fibonacci numbers).
%Y Row sums are the powers of 5 (A000351).
%Y Alternating row sums yield A054108.
%K nonn,tabl
%O 0,2
%A _Emeric Deutsch_, Mar 19 2005
|
