A052449 -id:A052449 - cp's OEIS frontend

A103815 a(n) = -1 + Product_{k=1..n} Fibonacci(k).

0, 0, 1, 5, 29, 239, 3119, 65519, 2227679, 122522399, 10904493599, 1570247078399, 365867569267199, 137932073613734399, 84138564904377983999, 83044763560621070207999, 132622487406311849122175999, 342696507457909818131702783999, 1432814097681520949608649339903999

Offset: 1

Jonathan Vos Post, Mar 29 2005

a(n) asymptotic to Phi^A000217(n). Prime for n = 4, 5, 6, 7, 8, 14, 15. Semiprime for n = 9, 10, 11, 20.

Thus, it is not until the 12th element in the sequence that we get number with more than 2 prime factors: 1570247078399 = 37 * 59 * 16349 * 43997. - Jonathan Vos Post, Dec 08 2012

			a(15) = 1 * 1 * 2 * 3 * 5 * 8 * 13 * 21 * 34 * 55 * 89 * 144 * 233 * 377 * 610 - 1 = 84138564904377983999 is prime.

Harvey P. Dale, Table of n, a(n) for n = 1..99

Cf. A000045, A000217, A003266, A052449, A053413, A053408.

F:= n-> (<<0|1>, <1|1>>^n)[1, 2]:
a:= n-> -1 + mul(F(i), i=1..n):
seq(a(n), n=1..20);  # Alois P. Heinz, Aug 09 2018

FoldList[Times,Fibonacci[Range[20]]]-1 (* Harvey P. Dale, Aug 29 2021 *)

a(n) = Product[Fibonacci[k], {k, 1, n}]-1 = Product[A000045[k], {k, 1, n}]-1.

a(n) = A003266(n) - 1. - Alois P. Heinz, Aug 09 2018

A159950 Dividends where Fibonacci products/sums yield integral quotients.

Original entry on oeis.org

240, 122522400, 137932073613734400, 342696507457909818131702784000, 1879127177606120717127879344567470740879360000, 22740756589119797763590969093409514524935686067027158720512000000

Offset: 1

Enoch Haga, Apr 27 2009

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.

			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.

Carlos Rivera, Problem 55. Fibonacci dividing terms, The Prime Puzzles & Problems Connection.

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

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

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

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

A103845 Product of first n Lucas numbers, plus one.

Original entry on oeis.org

1, 2, 4, 13, 85, 925, 16633, 482329, 22669417, 1722875617, 211913700769, 42170826452833, 13579006117811905, 7074662187380001985, 5963940223961341672513, 8134814465483270041306369

Offset: 0

Jonathan Vos Post, Mar 30 2005

Prime for n = 1, 3, 6; semiprime for n = 2, 4, 7, 8, 9, 10, 13, 14, 15, 23, 26. a(n) asymptotic to Phi^A000217(n).

			a(10) = 1 * 3 * 4 * 7 * 11 * 18 * 29 * 47 * 76 * 123 + 1 = 211913700769 = 349 * 607202581 which is semiprime.

Cf. A000045, A000204, A000217, A052449, A103815.

Join[{1},Rest[FoldList[Times,1,LucasL[Range[20]]]]+1] (* Harvey P. Dale, Oct 25 2011 *)

a(n) = Product[Lucasi[k], {k, 1, n}]+1. a(n) = Product[A000217(k), {k, 1, n}]+1.

a(n) = A070825(n)+1, n>0. - R. J. Mathar, Dec 22 2010

cp's OEIS Frontend

A103815 a(n) = -1 + Product_{k=1..n} Fibonacci(k).

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A159950 Dividends where Fibonacci products/sums yield integral quotients.

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A103845 Product of first n Lucas numbers, plus one.

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