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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A159950 Dividends where Fibonacci products/sums yield integral quotients.

Original entry on oeis.org

240, 122522400, 137932073613734400, 342696507457909818131702784000, 1879127177606120717127879344567470740879360000, 22740756589119797763590969093409514524935686067027158720512000000
Offset: 1

Views

Author

Enoch Haga, Apr 27 2009

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

Cf. A159951, A001519, A001906, A003481, A033890, A003266, A052449.

Programs

  • Maple
    seq(mul(fibonacci(k), k = 1..4*n+2), n = 1..10); # Peter Bala, Nov 04 2021
  • 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

Formula

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.
a(n) = Product_{k = 1..4*n+2} Fibonacci(k) = A003266(4*n+2) = A052449(4*n+2) - 1. - Peter Bala, Nov 04 2021

A349272 a(n) = Product_{k = 1..2*n+1} Fibonacci(2*k) / Sum_{k = 1..2*n+1} Fibonacci(2*k).

Original entry on oeis.org

1, 2, 315, 2471040, 918185538816, 16047302734562299200, 13178031727820369629763174400, 508406658175888466343652105865846784000, 921456090985190879093613420564815806955580862464000, 78458394721620642094151397745899367347021362840662985785265356800
Offset: 0

Views

Author

Peter Bala, Nov 12 2021

Keywords

Comments

Let m be an even positive integer. We conjecture that the sequence defined by b_m(n) = Product_{k = 1..2*n+1} Fibonacci(m*k) / Sum_{k = 1..2*n+1} Fibonacci(m*k) is integral. The formula given below proves the conjecture in the present case m = 2. The cases m = 4 and m = 6 of the conjecture can be proved in a similar manner.
More generally, if F(n,x) denotes the n-th Fibonacci polynomial we conjecture that, for each n, the rational function Product_{k = 1..2*n+1} F(m*k,x) / Sum_{k = 1..2*n+1} F(m*k,x) is an integral polynomial.

Links

Crossrefs

Cf. A000045, A159951, A175553.
Cf. A003266, A194157.

Programs

  • Maple
    with(combinat):
seq(mul(fibonacci(2*k), k = 1..2*n+1)/add(fibonacci(2*k), k = 1..2*n+1), n = 0..10);
  • Mathematica
    Table[Product[ Fibonacci[2k],{k,2n+1}]/Sum[Fibonacci[2k],{k,2n+1}],{n,0,9}] (* Stefano Spezia, Nov 13 2021 *)
  • PARI
    a(n) = prod(k = 1, 2*n+1, fibonacci(2*k)) / sum(k = 1, 2*n+1, fibonacci(2*k)); \\ Michel Marcus, Nov 12 2021

Formula

a(n) = F(2*n+1)/F(2*n+2) * Product_{k = 1..2*n} Fibonacci(2*k), shows a(n) to be integral. Cf. A159951.
a(n) ~ A194159 * phi^(4*n^2 + 2*n - 1) / 5^n, where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Aug 31 2023
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