A225158 -id:A225158 - cp's OEIS frontend

A225165 Denominators of the sequence s(n) of the sum resp. product of fractions f(n) defined recursively by f(1) = 6/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.

1, 5, 155, 176855, 265770796655, 679134511201261085170655, 4943777738415359153962876938905400001585992709055

Offset: 1

Martin Renner, Apr 30 2013

nonn

Numerators of the sequence s(n) of the sum resp. product of fractions f(n) is A165424(n+2), hence sum(A165424(i+1)/A225158(i),i=1..n) = product(A165424(i+1)/A225158(i),i=1..n) = A165424(n+2)/a(n) = A173501(n+2)/a(n).

			f(n) = 6, 6/5, 36/31, 1296/1141, ...
6 + 6/5 = 6 * 6/5 = 36/5; 6 + 6/5 + 36/31 = 6 * 6/5 * 36/31 = 1296/155; ...
s(n) = 1/b(n) = 6, 36/5, 1296/155, ...

Cf. A076628, A165424, A173501, A225158.

b:=proc(n) option remember; b(n-1)-b(n-1)^2; end:
b(1):=1/6;
a:=n->6^(2^(n-1))*b(n);
seq(a(i),i=1..8);

a(n) = 6^(2^(n-1))*b(n) where b(n)=b(n-1)-b(n-1)^2 with b(1)=1/6.

A225200 Triangle (read by rows) of coefficients of the polynomials (in ascending order) of the denominators of the generalized sequence of fractions f(n) defined recursively by f(1) = m/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.

Original entry on oeis.org

1, -1, 1, 1, -1, 1, 1, -2, 2, -1, 1, 1, -4, 8, -10, 9, -6, 3, -1, 1, 1, -8, 32, -84, 162, -244, 298, -302, 258, -188, 118, -64, 30, -12, 4, -1, 1, 1, -16, 128, -680, 2692, -8456, 21924, -48204, 91656, -152952, 226580, -300664, 359992, -391232, 387820, -352074, 293685, -225696, 160120, -105024, 63750, -35832, 18654, -8994, 4014, -1656, 630, -220, 70, -20, 5, -1, 1

Offset: 1

Martin Renner, May 01 2013

The degree of the polynomial in row n > 1 is 2^(n-2), hence the number of coefficients in row n > 1 is given by 2^(n-2) + 1 = A094373(n-1).
For n > 2 a new row always begins and ends with 1.
The sum and product of the generalized sequence of fractions given by m^(2^(n-2)) divided by the polynomial p(n) are equal, i.e.,
m + m/(m-1) = m * m/(m-1) = m^2/(m-1);
m + m/(m-1) + m^2/(m^2-m+1) = m * m/(m-1) * m^2/(m^2-m+1) = m^4/(m^3-2*m^2+2*m-1).

			The triangle T(n,k), k = 0..2^(n-1), begins
   1;
  -1,  1;
   1, -1, 1;
   1, -2, 2,  -1, 1;
   1, -4, 8, -10, 9, -6, 3, -1, 1;

Cf. A100441, A225156, A225157, A225158, A225159, A225160, A225161, A225162, A225201.

b:=n->m^(2^(n-2)); # n > 1
b(1):=m;
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(op(PolynomialTools[CoefficientList](a(i),m,termorder=forward)),i=1..7);

cp's OEIS Frontend

A225165 Denominators of the sequence s(n) of the sum resp. product of fractions f(n) defined recursively by f(1) = 6/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.

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