A100441 -id:A100441 - cp's OEIS frontend

A076628 a(n) = 2^(2^(n-1))*b(n) where b(1) = 1/2 and b(n+1) = b(n) - b(n)^2.

1, 1, 3, 39, 8463, 483008799, 1841209495473815103, 30574267942866411492610617989187955839, 9469098402231365955845075782127013884605836706970528486326625808482327697663

Offset: 1

Benoit Cloitre, Oct 22 2002

Michael Somos found that b(n) has some nice properties (see link titled "A Somos-Rusin recursion").
From Shai Covo (green355(AT)netvision.net.il), Mar 17 2010: (Start)
Define a sequence of fractions by b'(1) = 1/2 and b'(n+1) = b'(n) - b'(n)^2/2. Then, b'(n) = 2*b(n+1), n >= 1. The sequence b'(n) is the complement to 1 of the sequence f(n) defined in A167424 by f(1) = 1/2 and f(n+1) = (f(n)^2 + 1)/2, i.e., b'(n) + f(n) = 1. The numerator of b'(n) is given by a'(n) = 2^(2^n-1)*b'(n).
Thus a'(n) = a(n+1), n >= 1. The sequences a'(n) and A167424(n) are related by a'(n) + A167424(n) = 2^(2^n-1). (End)
From Shai Covo (green355(AT)netvision.net.il), Mar 25 2010: (Start)
The following is the counterpart of a comment in A167424. Suppose that U_1, U_2, ... is a sequence of independent uniform (0,1) random variables, and define random variables X_1, X_2, ... as follows: X_1 = U_1, and, for n >= 1, X_{n+1} = X_n or U_{n+1} according as U_{n+1} > E(X_n) or U_{n+1} < E(X_n), respectively, where E() denotes expectation. Then, the sequence E(X_n) is identical to the sequence b'(n) introduced in the comment dated Mar 17 2010. Sketch of proof. E(X_1) = 1/2; for n >= 1, by the law of total expectation, we have E(X_{n+1}) = (1 - E(X_n))*E(X_n) + E(X_n)*E(X_n)/2. Hence E(X_{n+1}) = E(X_n) - E(X_n)^2/2. (End)
b(n) = sqrt(p(n-1)/2) with p(n) as defined in A187131, so Sum_n b(n)^2 = 1/2 - Henry Bottomley, Mar 05 2011

Michael Somos, A Somos-Rusin recursion

Cf. A100441, A167424, A187131.

[1] cat [n eq 1 select 1 else Self(n-1)*2^(2^(n-1))-Self(n-1)^2: n in [1..8]]; // Vincenzo Librandi, Jun 17 2015

a[ n_] := If[ n < 2, Boole[n == 1], a[n - 1] (2^(2^(n - 2)) - a[n - 1])]; (* Michael Somos, Jul 24 2018 *)

b(n)=if(n<2,1/2,b(n-1)-b(n-1)^2); a(n)=numerator(b(n))

a(n) = 2^(2^(n-1))*[1/(n + log(n) + (1-e(oo)) + (log(n)/n)*(1+o(1)))], where 1 - e(oo) is a constant equal to about 1 - 0.232006 = 0.767994. The expression in the square brackets is Dave Rusin's "best estimate" for b(n), taken from the link "Michael Somos, A Somos-Rusin recursion" (where b(n) is denoted a(n)). - Shai Covo (green355(AT)netvision.net.il), Mar 15 2010
a(n+1) = a(n) * 2^(2^(n-1)) - a(n)^2 starting from a(1) = 1, also a(n) = sqrt(A187131(n-1)) - Henry Bottomley, Mar 05 2011
It appears that a(n) = Product_{k = 1..n} A100441(k). - Peter Bala, Feb 11 2015

Typo in definition corrected by Shai Covo (green355(AT)netvision.net.il), Mar 14 2010

A225156 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 3/1; 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, 2, 7, 67, 5623, 37772347, 1653794703916063, 3104205768420613437667191487267, 10767416908549848056705041797805600349527548164015760674541223

Offset: 1

Martin Renner, Apr 30 2013

Numerators of the sequence of fractions f(n) is A165421(n+1), hence sum(A165421(i+1)/a(i),i=1..n) = product(A165421(i+1)/a(i),i=1..n) = A165421(n+2)/A225163(n) = A011764(n-1)/A225163(n).

			f(n) = 3, 3/2, 9/7, 81/67, ...
3 + 3/2 = 3 * 3/2 = 9/2; 3 + 3/2 + 9/7 = 3 * 3/2 * 9/7 = 81/14; ...

Paul Yiu, Recreational Mathematics, Department of Mathematics, Florida Atlantic University, 2003, Chapter 5.4, p. 207 (Project).

Cf. A011764, A100441, A165421, A225163.

b:=n->3^(2^(n-2)); # n > 1
b(1):=3;
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);

a(n) = 3^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.

a(n) = 3^(2^(n-2)) - p(n) with a(1) = 1 and p(n) = p(n-1)*a(n-1) with p(1) = 1.

A225157 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 5/1; 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, 4, 21, 541, 345181, 136901485261, 21135572172649245550621, 496712610012943408146407697714437299262548141, 271328559212953102170688304392824035451911661168940831351173011072850527195615099225368381

Offset: 1

Martin Renner, Apr 30 2013

Numerators of the sequence of fractions f(n) is A165423(n+1), hence sum(A165423(i+1)/a(i),i=1..n) = product(A165423(i+1)/a(i),i=1..n) = A165423(n+2)/A225164(n) = A176594(n-1)/A225164(n).

			f(n) = 5, 5/4, 25/21, 625/541, ...
5 + 5/4 = 5 * 5/4 = 25/4; 5 + 5/4 + 25/21 = 5 * 5/4 * 25/21 = 625/84; ...

Cf. A100441, A165423, A176594, A225164.

b:=n->5^(2^(n-2)); # n > 1
b(1):=5;
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);

a(n) = 5^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.

a(n) = 5^(2^(n-2)) - p(n) with a(1) = 1 and p(n) = p(n-1)*a(n-1) with p(1) = 1.

A225158 Denominators of the sequence 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.

Original entry on oeis.org

1, 5, 31, 1141, 1502761, 2555339110801, 7279526598745139799221281, 58396508924557918552199410007906486608310469119041, 3723292553725227196293782783863296586090351965218332181732394788182320381276998127547535467381368961

Offset: 1

Martin Renner, Apr 30 2013

Numerators of the sequence of fractions f(n) is A165424(n+1), hence sum(A165424(i+1)/a(i),i=1..n) = product(A165424(i+1)/a(i),i=1..n) = A165424(n+2)/A225165(n) = A173501(n+2)/A225165(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; ...

Cf. A100441, A165424, A173501, A225165.

b:=n->6^(2^(n-2)); # n > 1
b(1):=6;
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);

a(n) = 6^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.

a(n) = 6^(2^(n-2)) - p(n) with a(1) = 1 and p(n) = p(n-1)*a(n-1) with p(1) = 1.

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.

Original entry on oeis.org

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

Offset: 1

Martin Renner, Apr 30 2013

nonn

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).

			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; ...

Cf. A100441, A165425, A225166.

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);

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.

A225160 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 8/1; 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, 7, 57, 3697, 15302113, 258902783918017, 73384158961115901868286873473, 5848244449673109813614947741525727934929692392922517757697

Offset: 1

Martin Renner, Apr 30 2013

nonn

Numerators of the sequence of fractions f(n) is A165426(n+1), hence sum(A165426(i+1)/a(i),i=1..n) = product(A165426(i+1)/a(i),i=1..n) = A165426(n+2)/A225167(n) = A167182(n+2)/A225167(n).

			f(n) = 8, 8/7, 64/57, 4096/3697, ...
8 + 8/7 = 8 * 8/7 = 64/7; 8 + 8/7 + 64/57 = 8 * 8/7 * 64/57 = 4096/399; ...

Cf. A100441, A165426, A167182, A225167.

b:=n->8^(2^(n-2)); # n > 1
b(1):=8;
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);

a(n) = 8^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.

a(n) = 8^(2^(n-2)) - p(n) with a(1) = 1 and p(n) = p(n-1)*a(n-1) with p(1) = 1.

A225161 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 9/1; 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, 8, 73, 5977, 39556153, 1714946746986937, 3196895220321005409761642330233, 11033196234263169646028268239301916905952651329069957632398777

Offset: 1

Martin Renner, Apr 30 2013

nonn

Numerators of the sequence of fractions f(n) is A165427(n+1), hence sum(A165427(i+1)/a(i),i=1..n) = product(A165427(i+1)/a(i),i=1..n) = A165427(n+2)/A225168(n) = A165421(n+3)/A225168(n) = A011764(n)/A225168(n).

			f(n) = 9, 9/8, 81/73, 6561/5977, ...
9 + 9/8 = 9 * 9/8 = 81/8; 9 + 9/8 + 81/73 = 9 * 9/8 * 81/73 = 6561/584; ...

Cf. A011764, A100441, A165421, A165427, A225168.

b:=n->9^(2^(n-2)); # n > 1
b(1):=9;
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..8);

a(n) = 9^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.

a(n) = 9^(2^(n-2)) - p(n) with a(1) = 1 and p(n) = p(n-1)*a(n-1) with p(1) = 1.

A225162 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 10/1; 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, 9, 91, 9181, 92480761, 9304615055139121, 93529710772930377727152664652641, 9394835719974970982728198049552322910011762062750179997188274881

Offset: 1

Martin Renner, Apr 30 2013

nonn

Numerators of the sequence of fractions f(n) is A165428(n+1), hence sum(A165428(i+1)/a(i),i=1..n) = product(A165428(i+1)/a(i),i=1..n) = A165428(n+2)/A225169(n) = A220812(n-1)/A225169(n).

			f(n) = 10, 10/9, 100/91, 10000/9181, ...
10 + 10/9 = 10 * 10/9 = 100/9; 10 + 10/9 + 100/91 = 10 * 10/9 * 100/91 = 10000/819; ...

Cf. A100441, A165428, A220812, A225169.

b:=n->10^(2^(n-2)); # n > 1
b(1):=10;
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..8);

a(n) = 10^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.

a(n) = 10^(2^(n-2)) - p(n) with a(1) = 1 and p(n) = p(n-1)*a(n-1) with p(1) = 1.

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

A076628 a(n) = 2^(2^(n-1))*b(n) where b(1) = 1/2 and b(n+1) = b(n) - b(n)^2.

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A225156 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 3/1; 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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A225157 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 5/1; 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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A225158 Denominators of the sequence 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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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.

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A225160 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 8/1; 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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A225161 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 9/1; 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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A225162 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 10/1; 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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