cp's OEIS Frontend

A159951 Fibonacci integral quotients associated with the dividends in A159950 and the divisors in A003481.

%I A159951 #16 Jun 02 2025 01:39:09
%S A159951 12,856800,139890541190400,50664770469826998541056000,
%T A159951 40527253814267058837705250384270510080000,
%U A159951 71554565901386985191123530075861409411081105273676595200000
%N A159951 Fibonacci integral quotients associated with the dividends in A159950 and the divisors in A003481.
%C A159951 The first example of an integral quotient in the Fibonacci sequence is 12 because 240/20=12. 240 is the product of terms through 8, and 20 the sum. Thereafter, with every other additional pair of terms in the Fibonacci sequence, another integral quotient occurs.
%C A159951 Let m be an even positive integer. Then the sequence defined by b_m(n) = Product_{k = 1..2*n+1} F(m*k) / Sum_{k = 1..2*n+1} F(m*k) appears to be integral. - _Peter Bala_, Nov 12 2021
%F A159951 a(n) = (Product_{k = 1..4*n+2} Fibonacci(k))/(Sum_{k = 1..4*n+2} Fibonacci(k)) = (Product_{k = 1..4*n+2} Fibonacci(k))/(Fibonacci(4*n+4) - 1) =  Fibonacci(2*n+1)/Fibonacci(2*n+3) * Product_{k = 1..4*n+1} Fibonacci(k), which shows a(n) is integral. Cf. A175553. - _Peter Bala_, Nov 11 2021
%e A159951 The first two integral quotients occur in the Fibonacci sequence as illustrated by the following: (1*1*2*3*5*8)/(1+1+2+3+5+8) = 240/20 = 12, integral; (1*1*2*3*5*8*13*21*34*55)/(1+1+2+3+5+8+13+21+34+55) = 122522400/143 = 856800, integral.
%p A159951 with(combinat):
%p A159951 seq(mul(fibonacci(k), k = 1..4*n+2)/(fibonacci(4*n+4) - 1), n = 1..10); # _Peter Bala_, Nov 04 2021
%o A159951 (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 A159951 Cf. A000045, A001519, A001906, A003481, A033890, A159950, A175553.
%K A159951 nonn,easy
%O A159951 1,1
%A A159951 _Enoch Haga_, Apr 27 2009