A225159 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 7/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.

1, 6, 43, 2143, 5211907, 30351298460743, 1016966398053911225889737707, 1130815308619683511655208290917557601522304473342184143

Offset: 1

Author

Martin Renner, Apr 30 2013

Keywords

• nonn

Comments

Numerators of the sequence of fractions f(n) is A165425(n+1), hence sum(A165425(i+1)/a(i),i=1..n) = product(A165425(i+1)/a(i),i=1..n) = A165425(n+2)/A225166(n).

Examples

			f(n) = 7, 7/6, 49/43, 2401/2143, ...
7 + 7/6 = 7 * 7/6 = 49/6; 7 + 7/6 + 49/43 = 7 * 7/6 * 49/43 = 2401/258; ...
		

Crossrefs

Cf. A100441, A165425, A225166.

Programs

Maple

b:=n->7^(2^(n-2)); # n > 1
b(1):=7;
p:=proc(n) option remember; p(n-1)*a(n-1); end;
p(1):=1;
a:=proc(n) option remember; b(n)-p(n); end;
a(1):=1;
seq(a(i),i=1..9);

Formula

a(n) = 7^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.

a(n) = 7^(2^(n-2)) - p(n) with a(1) = 1 and p(n) = p(n-1)*a(n-1) with p(1) = 1.