A055573 -id:A055573 - cp's OEIS frontend

A070266 Maximum element in the simple continued fraction expansion of H(n) = 1+1/2+1/3+...+1/n, the n-th harmonic number.

1, 2, 5, 12, 8, 4, 5, 7, 7, 13, 50, 15, 6, 39, 43, 9, 14, 61, 25, 17, 36, 13, 5, 17, 18, 20, 68, 45, 25, 198, 88, 1090, 120, 244, 29, 29, 24, 111, 567, 22, 20, 37, 1812, 59, 41, 336, 121, 32, 37, 314, 104, 162, 289, 146, 1557, 42, 78, 639, 52, 812, 116, 32, 131, 236, 59

Offset: 1

Benoit Cloitre, May 09 2002

			The simple continued fraction expansion of H(10) is [2, 1, 13, 12, 1, 3, 1, 2], hence a(10) = 13.

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

Cf. A001008, A002805, A055573.

Table[ Max[ ContinuedFraction[ HarmonicNumber[n]]], {n, 1, 65}]

A091532 Where the number of terms in simple continued fraction for H(j) exceeds all H(i), j>i and H(k) is the k-th harmonic number.

Original entry on oeis.org

1, 2, 3, 5, 7, 8, 9, 13, 16, 17, 19, 23, 25, 26, 28, 29, 35, 36, 43, 45, 48, 49, 54, 57, 62, 72, 73, 79, 88, 90, 91, 99, 103, 108, 110, 113, 115, 116, 118, 125, 128, 148, 149, 157, 163, 168, 171, 172, 184, 193, 199, 205, 209, 234, 240, 243, 259, 265, 269, 270, 281, 283

Offset: 1

Robert G. Wilson v, Jan 19 2004

nonn

Where A055573 increases.

Amiram Eldar, Table of n, a(n) for n = 1..4000

Cf. A001008, A002805, A055573.

t = Table[ Length[ ContinuedFraction[ HarmonicNumber[n]]], {n, 1, 299}]; a = {1}; Do[ If[ t[[n]] > t[[a[[ -1]]]], AppendTo[a, n]], {n, 1, 299}]; a

A070985 Number of terms in the simple continued fraction for Sum_{k=1..n} 1/k^2.

Original entry on oeis.org

1, 2, 5, 7, 9, 7, 10, 20, 18, 14, 22, 19, 18, 24, 26, 24, 30, 30, 28, 37, 25, 30, 35, 35, 34, 38, 47, 52, 49, 54, 40, 49, 49, 69, 57, 67, 78, 67, 67, 68, 67, 64, 65, 86, 76, 81, 92, 79, 83, 83, 95, 82, 85, 80, 84, 95, 92, 91, 121, 105, 100, 108, 111, 109, 118, 105, 110, 88

Offset: 1

Benoit Cloitre, May 18 2002

Sum_{k>=1} 1/k^2 = zeta(2) = Pi^2/6.

			The simple continued fraction for Sum_{k=1..10} 1/k^2 is [1, 1, 1, 4, 1, 1, 10, 4, 1, 2, 5, 2, 1, 24] which contains 14 terms, hence a(10) = 14.

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

Cf. A013679, A055573, A007406, A007407.

lcf[f_] := Length[ContinuedFraction[f]]; lcf /@ Accumulate[Table[1/k^2, {k, 1, 100}]] (* Amiram Eldar, Apr 30 2022 *)

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

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

A091656 Least number k such that the continued fraction expansion of H(k) contains the numbers 1, 2, ..., n, where H(k) is the k-th Harmonic number.

Original entry on oeis.org

1, 2, 5, 9, 9, 13, 26, 63, 68, 68, 68, 87, 121, 121, 165, 207, 207, 221, 221, 287, 289, 325, 428, 440, 483, 544, 544, 544, 544, 544, 558, 558, 558, 966, 1035, 1035, 1146, 1146, 1332, 1332, 1332, 1665, 1665, 1665, 1665, 1665, 1727, 1727, 2052, 2157, 2331, 2331

Offset: 1

Robert G. Wilson v, Jan 26 2004

nonn

			a(6) = 13 because CF( H(13)) = 3 + [5, 1, 1, 4, 2, 1, 3, 2, 1, 3, 1, 4, 1, 6], the first six integers are present.

Charles R Greathouse IV, Table of n, a(n) for n = 1..250

Cf. A055573, A091655.

f[n_] := Block[{k = 1}, While[ StringPosition[ ToString[ Union[ ContinuedFraction[ Sum[1/i, {i, 1, k}]]]], StringDrop[ ToString[ Table[i, {i, n}]], -1]] == {}, k++ ]; k]; Table[ f[n], {n, 1, 52}]

list(lim)=my(v=vector(lim\1),n,t,H,i=1);while(1,H+=1/n++;t=vecsort(contfrac(H),,8);if(#t>=i&&t[i]==i,v[i]=n;print1(n":"i", ");if(i++>#v,return(v));H-=1/n;n--)) \\ Charles R Greathouse IV, Jan 25 2012

A353299 a(n) is the length of the continued fraction for the sum of the reciprocals of the first n primes.

Original entry on oeis.org

2, 3, 2, 5, 9, 10, 11, 16, 13, 20, 27, 27, 31, 43, 37, 41, 43, 47, 50, 58, 53, 57, 65, 83, 69, 62, 80, 84, 88, 93, 88, 110, 119, 117, 104, 111, 116, 126, 114, 140, 130, 164, 166, 132, 158, 154, 166, 168, 178, 178, 146, 176, 192, 188, 190, 203, 213, 191, 224, 236, 234, 238, 236, 236, 251

Offset: 1

Ilya Gutkovskiy, Apr 09 2022

nonn

			Sum_{k=1..2} 1/prime(k) = 1/2 + 1/3 = 5/6 = 0 + 1/(1 + 1/5), so a(2) = 3.
Sum_{k=1..4} 1/prime(k) = 1/2 + 1/3 + 1/5 + 1/7 = 247/210 = 1 + 1/(5 + 1/(1 + 1/(2 + 1/12))), so a(4) = 5.

Ilya Gutkovskiy, Scatterplot of a(n)/(n*log(n)) up to n=10000
Eric Weisstein's World of Mathematics, Continued Fraction
Eric Weisstein's World of Mathematics, Harmonic Series of Primes

Row lengths of A260615.

Cf. A002110, A024451, A055573.

Table[Length[ContinuedFraction[Sum[1/Prime[k], {k, 1, n}]]], {n, 1, 65}]

a(n) = #contfrac(sum(k=1, n, 1/prime(k))); \\ Michel Marcus, Apr 10 2022

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

A070986 Number of terms in the simple continued fraction for Sum_{k=1..n} 1/k^3.

Original entry on oeis.org

1, 2, 5, 9, 14, 10, 15, 16, 14, 20, 22, 19, 33, 25, 27, 31, 41, 33, 52, 47, 36, 54, 38, 56, 50, 56, 65, 68, 81, 76, 77, 84, 75, 80, 88, 76, 90, 89, 98, 89, 113, 100, 110, 101, 104, 103, 121, 122, 131, 121, 131, 127, 124, 134, 127, 132, 143, 131, 143, 144, 161, 141

Offset: 1

Benoit Cloitre, May 18 2002

Sum_{k>=1} 1/k^3 = zeta(3) = 1.2020... (A002117).

			The simple continued fraction for Sum_{k=1..10} 1/k^3 is [1, 5, 16, 135, 1, 5, 2, 3, 2, 1, 1, 1, 3, 3, 16, 1, 2, 1, 1, 2] which contains 20 terms, hence a(10)=20.

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

Cf. A002117, A055573.

lcf[f_] := Length[ContinuedFraction[f]]; lcf /@ Accumulate[Table[1/k^3, {k, 1, 100}]] (* Amiram Eldar, Apr 29 2022 *)

for(n=1,100,print1(length(contfrac(sum(i=1,n,1/i^3))),","))

Limit_{n ->infinity} a(n)/n = C = 2.5....

A070987 Number of terms in simple continued fraction for sum(k=1,n,1/k^4).

Original entry on oeis.org

1, 2, 8, 8, 9, 12, 22, 23, 27, 29, 33, 33, 49, 39, 48, 52, 58, 62, 65, 68, 73, 67, 75, 72, 80, 83, 87, 89, 100, 91, 93, 109, 113, 112, 101, 105, 107, 118, 123, 131, 118, 120, 123, 141, 151, 148, 157, 165, 157, 170, 180, 158, 187, 181, 181, 195, 187, 181, 194, 188

Offset: 1

Benoit Cloitre, May 18 2002

sum(k>=1,1/k^4)=zeta(4)=Pi^4/90

			The simple continued fraction for sum(k=1,10,1/k^4) is [1, 12, 5, 3, 1, 2, 10, 12, 1, 2, 4, 2, 2, 2, 1, 7, 11, 1, 1, 2, 5, 2, 2, 4, 3, 1, 1, 1, 2] which contains 29 terms, hence a(10)=29.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A055573.

Length[ContinuedFraction[#]]&/@Accumulate[1/Range[60]^4] (* Harvey P. Dale, Dec 20 2012 *)

for(n=1,100,print1(length(contfrac(sum(i=1,n,1/i^4))),","))

lim n ->infinity a(n)/n=C=3, 3....

A091655 Least number k such that the continued fraction expansion of H(k) contains n, where H(k) is the k-th Harmonic number.

Original entry on oeis.org

2, 5, 6, 3, 11, 8, 5, 12, 30, 17, 4, 10, 17, 12, 31, 15, 25, 46, 63, 66, 51, 35, 30, 48, 46, 44, 31, 39, 21, 42, 53, 14, 44, 45, 53, 15, 70, 28, 78, 87, 131, 74, 11, 108, 52, 75, 71, 50, 80, 78, 83, 44, 73, 18, 52, 73, 58, 142, 86, 77, 27, 138, 148, 84, 144, 81, 114, 73, 139, 132

Offset: 1

Robert G. Wilson v, Jan 26 2004

nonn

			a(15) = 12 because CF( H(12)) = 3 + [9, 1, 2, 4, 1, 1, 1, 15, 4], so 15 is present.

Cf. A055573, A091656.

f[n_] := Block[{k = 1}, While[ Position[ ContinuedFraction[ HarmonicNumber[k]], n] == {}, k++ ]; k]; Table[ f[n], {n, 1, 77}]

A336088 k such that L(H(k,2)) = 2*L(H(k,1)) where L(x) is the number of terms in the continued fraction of x and H(k,r) = Sum_{u=1..k} 1/u^r.

Original entry on oeis.org

28, 61, 90, 105, 121, 321, 339, 382, 408, 466, 602, 1079, 1121, 1596, 1782, 2067, 2104, 2170, 2220, 2250, 2435, 2456, 2884, 3141, 3242, 3321, 3328, 3435, 4195, 4323, 4348, 4497, 4766, 4914, 5241, 5526, 6290, 6581, 6597, 9306, 9734

Offset: 1

Benoit Cloitre, Oct 04 2020

nonn

Conjecture: this sequence is infinite. More generally for any fixed integers a,b,c,d >= 1, there are infinitely many k's such that c*d*L(H(k,a)^b) = a*b*L(H(k,c)^d) where L(x) is the number of terms in the continued fraction of x and H(k,r) = Sum_{u=1..k} 1/u^r. Here, (a,b,c,d) = (2,1,1,1).

Cf. A055573, A070985.

c[n_, r_] := Length @ ContinuedFraction @ HarmonicNumber[n, r]; Select[Range[10^4], c[#, 2] == 2 * c[#, 1] &] (* Amiram Eldar, Oct 04 2020 *)

H1=H2=1;for(n=2,10000,H1=H1+1/n;H2=H2+1/n^2;if(length(contfrac(H2))==2*length(contfrac(H1)),print1(n,",")))

cp's OEIS Frontend

A070266 Maximum element in the simple continued fraction expansion of H(n) = 1+1/2+1/3+...+1/n, the n-th harmonic number.

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A091532 Where the number of terms in simple continued fraction for H(j) exceeds all H(i), j>i and H(k) is the k-th harmonic number.

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A070985 Number of terms in the simple continued fraction for Sum_{k=1..n} 1/k^2.

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A091656 Least number k such that the continued fraction expansion of H(k) contains the numbers 1, 2, ..., n, where H(k) is the k-th Harmonic number.

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A353299 a(n) is the length of the continued fraction for the sum of the reciprocals of the first n primes.

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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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A070986 Number of terms in the simple continued fraction for Sum_{k=1..n} 1/k^3.

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A070987 Number of terms in simple continued fraction for sum(k=1,n,1/k^4).

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