A069887 -id:A069887 - cp's OEIS frontend

A069655 Maximum element in the simple continued fraction expansion for (1+1/n)^n.

2, 4, 3, 3, 20, 16, 11, 34, 18, 128, 56, 168, 39, 21, 162, 116, 37, 113, 72, 105, 73, 245, 244, 74, 159, 187, 253, 663, 101, 166, 34, 41, 87, 71, 46, 449, 181, 1874, 130, 215, 457, 317, 196, 256, 160, 107, 72, 147, 209, 114, 2632, 134, 252, 844, 1285, 341, 656

Offset: 1

Benoit Cloitre, May 09 2002

Limit_{n -> infinity} (1+1/n)^n = e.

			The simple continued fraction expansion of (1+1/10)^10 is [2, 1, 1, 2, 5, 1, 128, 1, 2, 12, 5, 3, 46, 1, 11, 7], hence a(10) = 128.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001113, A069887.

Table[ Max[ ContinuedFraction[ (1 + 1/n)^n]], {n, 1, 60}]

A071529 Number of 1's among the elements of the simple continued fraction for (1+1/n)^n.

Original entry on oeis.org

0, 0, 1, 1, 1, 4, 7, 7, 12, 5, 8, 10, 23, 18, 25, 14, 18, 17, 14, 24, 22, 22, 35, 15, 21, 30, 29, 33, 37, 30, 27, 47, 49, 44, 54, 55, 53, 51, 46, 46, 43, 60, 64, 65, 79, 64, 64, 67, 73, 66, 79, 68, 60, 76, 86, 85, 85, 83, 86, 74, 90, 84, 93, 106, 90, 85, 98, 107, 88, 104, 86

Offset: 1

Benoit Cloitre, Jun 02 2002

It seems that lim_{n->infinity} a(n)/A069887(n) = C = 0.41..., which is close to (log(4)-log(3))/log(2)=0.415..., the expected density of 1's (cf. measure theory of continued fractions).

			(1+1/14)^14 has for continued fraction [2, 1, 1, 1, 2, 6, 1, 7, 1, 6, 2, 1, 4, 21, 1, 1, 7, 1, 1, 1, 3, 2, 7, 2, 7, 1, 2, 4, 1, 3, 2, 1, 1, 1, 5, 1, 2, 5, 1, 2] which contains 18 "1's" hence a(14)=18.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Table[Count[ContinuedFraction[(1+1/n)^n],1],{n,80}] (* Harvey P. Dale, Mar 11 2013 *)

for(n=1,100,s=(1+1/n)^n; print1(sum(i=1,length(contfrac(s)),if(1-component(contfrac(s),i),0,1)),","))

A070154 Number of terms in the simple continued fraction expansion of Sum_{k=0..n}(-1)^k/(2k+1), the Leibniz-Gregory series for Pi/4.

Original entry on oeis.org

1, 3, 4, 9, 5, 9, 14, 10, 10, 19, 16, 21, 22, 22, 24, 20, 19, 24, 28, 28, 29, 30, 39, 31, 44, 40, 44, 33, 41, 47, 44, 48, 54, 48, 60, 49, 63, 51, 65, 72, 64, 70, 78, 64, 79, 77, 74, 87, 75, 86, 82, 94, 88, 106, 106, 94, 104, 108, 87, 107, 86, 106, 98, 110, 115, 110, 105, 115

Offset: 0

Benoit Cloitre, May 06 2002

Pi/4 = Sum_{k=>0} (-1)^k/(2k+1).

			The simple continued fraction for Sum(k=0,10,(-1)^k/(2k+1)) is [0, 1, 4, 4, 1, 3, 54, 1, 2, 1, 1, 4, 11, 1, 2, 2] which contains 16 elements, hence a(10)=16.

Amiram Eldar, Table of n, a(n) for n = 0..10000

Cf. A003881, A007509, A055573, A069880, A069887.

lcf[f_] := Length[ContinuedFraction[f]]; lcf /@ Accumulate[Table[(-1)^k/(2*k + 1), {k, 0, 100}]] (* Amiram Eldar, Apr 29 2022 *)

for(n=1,100,print1( length(contfrac(sum(i=0,n,(-1)^i/(2*i+1)))),","))

Limit_{n -> infinity} a(n)/n = C = 1.6...

Offset changed to 0 and a(0) inserted by Amiram Eldar, Apr 29 2022

A071599 Product of elements in the simple continued fraction for (1+1/n)^n.

Original entry on oeis.org

2, 8, 24, 216, 4320, 19008, 103488, 1292544, 1548288, 3264307200, 24710676480, 54623600640, 16562257920, 3345695539200, 8216950210560, 33673108783104000, 205682009702400000, 15655109317676236800, 12302792042521559040000

Offset: 1

Benoit Cloitre, Jun 01 2002

It appears that lim_{n->oo} a(n)^(1/A069887(n)) = A002210 (Khinchin constant = 2.68...). - Benoit Cloitre, Jan 29 2006

			The continued fraction for (1+1/5)^5 is [2, 2, 20, 1, 9, 2, 3] and 2*2*20*1*9*2*3=4320 hence a(5)=4320

Harvey P. Dale, Table of n, a(n) for n = 1..449

Table[Times@@ContinuedFraction[(1+1/n)^n],{n,20}] (* Harvey P. Dale, May 02 2019 *)

a(n) = prod(i=1, length(contfrac((1+1/n)^n)), component(contfrac((1+1/n)^n), i));

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A069655 Maximum element in the simple continued fraction expansion for (1+1/n)^n.

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A071529 Number of 1's among the elements of the simple continued fraction for (1+1/n)^n.

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A070154 Number of terms in the simple continued fraction expansion of Sum_{k=0..n}(-1)^k/(2k+1), the Leibniz-Gregory series for Pi/4.

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A071599 Product of elements in the simple continued fraction for (1+1/n)^n.

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