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%I #13 Nov 14 2021 06:26:27
%S 240,122522400,137932073613734400,342696507457909818131702784000,
%T 1879127177606120717127879344567470740879360000,
%U 22740756589119797763590969093409514524935686067027158720512000000
%N Dividends where Fibonacci products/sums yield integral quotients
%C In looking at the Fibonacci sequence I happened to notice that after each pair of terms >1 the product of terms divided by the sum of terms produced an integral quotient every other time. Example 240/20=12, integral.
%H Carlos Rivera, <a href="http://www.primepuzzles.net/problems/prob_055.htm">Problem 55. Fibonacci dividing terms</a>, The Prime Puzzles & Problems Connection.
%F a(1)=240 because in the Fibonacci sequence up to 8 : 1 1 2 3 5 8, the product is 240 1*1*2*3*5*8. The sum is 1+1+2+3+5+8=20 (see A003481). The integral quotient is 12. From then on, every other pair produces an integral quotient.
%F a(n) = Product_{k = 1..4*n+2} Fibonacci(k) = A003266(4*n+2) = A052449(4*n+2) - 1. - _Peter Bala_, Nov 04 2021
%e This table illustrates the alternating nature of the first three integral quotients: 1 1 2 3 -- 6/7=.85+ 5 8 -- 240/20=12 Integral 13 21 -- 65520/54=1213.33+ 34 55 -- 122522400/143=856800 Integral 89 144 -- 1570247078400/376=4176189038.29+ 233 377 -- 137932073613734400/986=139890541190400 Integral etc.
%p seq(mul(fibonacci(k), k = 1..4*n+2), n = 1..10); # _Peter Bala_, Nov 04 2021
%o (UBASIC) 10 'Fibo 20 'R=SUM:S=PRODUCT 30 'T integral every other pair 40 A=1:S=1:print A;:S=S*1 50 B=1:print B;:S=S*B 60 C=A+B:print C;:R=R+C:S=S*C 70 D=B+C:print D;:R=R+D:R=R+2:print R:S=S*D:print S 80 T=S/R:if T=int(S/R) then print T:stop 90 A=C:B=D:R=R-2:goto 60
%Y Cf. A159951, A001519, A001906, A003481, A033890, A003266, A052449.
%K nonn,easy
%O 1,1
%A _Enoch Haga_, Apr 27 2009
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