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A276175
a(n) = (a(n-1)+1)*(a(n-2)+1)*(a(n-3)+1)/a(n-4) with a(0) = a(1) = a(2) = a(3) = 1.
5
1, 1, 1, 1, 8, 36, 666, 222111, 685187756, 2819713283228248, 644335913093223286486628176, 5604757351123068775966272886689217889936356651, 14861563788248216173988661093334637018340529129342104300621091389266132702213641
OFFSET
0,5
COMMENTS
Conjecture: a(n) is integer for all n >= 0. It has been checked by computer for n <= 40. A proof was proposed by 'mercio' as an answer to the MSE question, which however lacks details and heavily relies on computation. [Updated by Max Alekseyev, May 07 2023]
LINKS
Math.StackExchange.com, Is A276175 integer-only? Aug 27 2016.
FORMULA
Let b(n) = (b(n-1)*b(n-2)*b(n-3)+1)/b(n-4) with b(0) = 1/2, b(1) = 4, b(2) = b(3) = 1/2, then a(n) = b(n)*b(n+1)*b(n+2). - Seiichi Manyama, Sep 03 2016
The sequence 4*b(n) is given by A362884. Correspondingly, a(n) = A362884(n) * A362884(n+1) * A362884(n+2) / 64. - Max Alekseyev, May 07 2023
a(n) = a(3-n), 0 = a(n)*a(n+4)*(a(n+4)+1) - a(n+5)*a(n+1)*(a(n+1)+1) for all n in Z. - Michael Somos, Feb 23 2019
log(a(n)) ~ c * A289917^n, where c = 0.26774381278698... - Vaclav Kotesovec, Aug 27 2021
MATHEMATICA
RecurrenceTable[{a[n] == (a[n - 1] + 1) (a[n - 2] + 1) (a[n - 3] + 1)/a[n - 4],
a[0] == a[1] == a[2] == a[3] == 1}, a, {n, 0, 12}] (* Michael De Vlieger, Aug 25 2016 *)
a[ n_] := With[{m = Max[3 - n, n]}, If[ m < 4, 1, (a[m - 1] + 1) (a[m - 2] + 1) (a[m - 3] + 1)/a[m - 4]]]; (* Michael Somos, Jun 02 2019 *)
PROG
(PARI) a(n) = if (n <=3, 1, (a(n-1)+1)*(a(n-2)+1)*(a(n-3)+1)/a(n-4)); \\ Michel Marcus, Aug 23 2016
(Ruby)
def A(m, n)
a = Array.new(m, 1)
ary = [1]
while ary.size < n + 1
i = a[1..-1].inject(1){|s, i| s * (i + 1)}
break if i % a[0] > 0
a = *a[1..-1], i / a[0]
ary << a[0]
end
ary
end
def A276175(n)
A(4, n)
end # Seiichi Manyama, Aug 23 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
Bruno Langlois, Aug 23 2016
STATUS
approved