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

%I

%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 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

%Y Column 0 is A001700. Column 1 is A024483. T(n, n)=A001519(n) (the odd-subscripted Fibonacci numbers). Row sums are the powers of 5 (A000351). Alternating row sums yield A054108.

%K nonn,tabl

%O 0,2

%A _Emeric Deutsch_, Mar 19 2005

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Last modified May 21 11:54 EDT 2019. Contains 323443 sequences. (Running on oeis4.)