A103245 - OEIS

%I #23 Jan 05 2025 19:51:38

%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://web.archive.org/web/2024*/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