A069880 -id:A069880 - cp's OEIS frontend

A058299 Sum of terms in simple continued fraction for Sum_{k=0..n} 1/k!.

1, 2, 4, 5, 10, 15, 20, 16, 38, 30, 67, 49, 63, 80, 92, 139, 173, 99, 127, 159, 190, 198, 423, 198, 259, 221, 326, 631, 394, 273, 280, 341, 359, 397, 539, 418, 518, 533, 662, 3502, 735, 815, 701, 706, 611, 839, 793, 768, 781, 983, 858, 1035, 883, 3476, 1154

Offset: 0

Leroy Quet, Dec 07 2000

			a(3) = 2 + 1 + 2 = 5 because 1/0! + 1/1! + 1/2! + 1/3! = 8/3 = 2 + 1/(1 + 1/2).

Robert Israel, Table of n, a(n) for n = 0..3300

Cf. A069880 (number of summands).

seq(convert(numtheory:-cfrac(add(1/k!,k=0..n),quotients),`+`), n=0..100); # Robert Israel, Aug 29 2018

a(n) = vecsum(contfrac(sum(k=0, n, 1/k!))); \\ Michel Marcus, Aug 29 2018

A070267 Maximum element in the simple continued fraction expansion of e(n) = 1+1/2!+1/3!+...+1/n!.

Original entry on oeis.org

1, 2, 2, 3, 8, 5, 4, 14, 6, 29, 10, 16, 20, 18, 42, 59, 13, 14, 59, 35, 31, 184, 24, 65, 42, 64, 401, 71, 26, 24, 36, 31, 52, 187, 28, 41, 128, 177, 3041, 249, 315, 162, 118, 36, 101, 135, 86, 70, 194, 104, 274, 62, 2515, 305, 68, 59, 49, 88, 359, 280, 100, 702, 52

Offset: 1

Benoit Cloitre, May 09 2002

			The simple continued fraction expansion of e(10) is [1, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 11, 1, 1, 29, 1, 1, 2], hence a(10) = 29.

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

Cf. A061354, A061355, A069880, A070266.

Table[ Max[ ContinuedFraction[ Sum[1/i!, {i, 1, n}]]], {n, 1, 65}]

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

A071528 Number of 1's among the elements of the simple continued fraction for e(n)=sum(k=1,n,1/k!).

Original entry on oeis.org

1, 1, 2, 2, 4, 5, 6, 8, 7, 12, 11, 11, 15, 13, 13, 16, 19, 17, 18, 19, 23, 25, 25, 27, 29, 32, 32, 27, 40, 40, 46, 35, 44, 38, 41, 43, 40, 46, 45, 55, 54, 57, 62, 53, 57, 52, 59, 67, 61, 67, 66, 69, 74, 80, 79, 85, 77, 78, 76, 83, 85, 88, 96, 78, 101, 93, 89, 101, 88, 106, 95

Offset: 1

Benoit Cloitre, Jun 02 2002

It seems that lim n ->infinity a(n)/A069880(n) = C = 0.5... which is different from (log(4)-log(3))/log(2)=0.415... the expected density of 1's (cf. measure theory of continued fractions).

			e(10) has for continued fraction [1, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 11, 1, 1, 29, 1, 1, 2] which contains 12 "1's" hence a(10)=12.

for(n=1,150,if(prod(i=1,length(contfrac((1+1/n)^n)),n-component(contfrac((1+1/n)^n),i)) == 0,print1(n,",")))

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A058299 Sum of terms in simple continued fraction for Sum_{k=0..n} 1/k!.

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A070267 Maximum element in the simple continued fraction expansion of e(n) = 1+1/2!+1/3!+...+1/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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A071528 Number of 1's among the elements of the simple continued fraction for e(n)=sum(k=1,n,1/k!).

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