A058312 -id:A058312 - cp's OEIS frontend

A079542 Triangular array: T(n,1) = T(n,n) = n and T(n,k) = lcm(T(n-1,k-1), T(n-1,k)) for 1 < k < n.

1, 2, 2, 3, 2, 3, 4, 6, 6, 4, 5, 12, 6, 12, 5, 6, 60, 12, 12, 60, 6, 7, 60, 60, 12, 60, 60, 7, 8, 420, 60, 60, 60, 60, 420, 8, 9, 840, 420, 60, 60, 60, 420, 840, 9, 10, 2520, 840, 420, 60, 60, 420, 840, 2520, 10, 11, 2520, 2520, 840, 420, 60, 420, 840, 2520, 2520, 11, 12

Offset: 1

Reinhard Zumkeller, Jan 22 2003

T(2*n-1,n) = A058312(n-1) for n <= 13.

			Triangle begins:
  1
  2  2
  3  2  3
  4  6  6  4
  5 12  6 12  5
  6 60 12 12 60  6
  7 60 60 12 60 60  7

Cf. A080046.

T(n, 2) = A003418(n-1) for n > 2.

T(n,k) = if ((k==1) || (k==n), n, lcm(T(n-1,k-1), T(n-1,k)));
tabl(nn) = for (n=1, nn, for (k=1, n, print1(T(n,k), ", ")); print); \\ Michel Marcus, Apr 25 2018

A115388 Numerator of rational part of raw moment n of the line point picking problem.

Original entry on oeis.org

-1, 3, -4, 17, -41, 42, -289, 1171, -1739, 1753, -19157, 19262, -249251, 250241, -249383, 200107, -1696405, 1700409, -32239703, 161504821, -161227687, 161479627, -3708740681, 3713590526, -18545643343, 18566236531, -55641506293, 55694623643, -230529988171

Offset: 1

Eric W. Weisstein, Jan 21 2006

			-1 + 2*log(2), 3 - 4*log(2), -4 + 6*log(2), 17/3 - 8*log(2), -41/6 + 10*log(2), ...
The above sequence of numbers is given by 4*Integral_{x = 0..Pi/4} tan(x)^(2*n+1) * cos(x)^2 dx for n >= 1, or, equivalently, by Integral_{y = 0..1} Integral_{x = 0..1} 2*n*(x*y)^n/(x + y)^2 dx dy for n >= 1. - _Peter Bala_, Jan 04 2023

Eric Weisstein's World of Mathematics, Line Point Picking

Cf. A058312, A058313, A115389.

a := n -> numer(1 + 2*n*add((-1)^(n+k+1)/k, k = 1..n)):
seq(a(n), n = 1..28); # Peter Bala, Jan 05 2023
# Alternative:
a := n -> 2*n*((-1)^n*log(2) - LerchPhi(-1, 1, n + 1)) + 1:
seq(numer(simplify(a(n))), n = 1..29); # Peter Luschny, Jan 05 2023

From Pontus von Brömssen, Nov 03 2019: (Start)
For even n, a(n)/A115389(n) = 2*n*Sum_{k = n/2..n-1} 1/k - 1.
For odd n >= 3, a(n)/A115389(n) = -2*n*((Sum_{k = (n-1)/2..n-2} 1/k) - 1/(n-1)) - 1. (End)
a(n) = numerator of 1 + (2*n)*Sum_{k = 1..n} (-1)^(n+k+1)/k. - Peter Bala, Jan 05 2023
a(n) = numerator of 2*n*((-1)^n*log(2) - LerchPhi(-1, 1, n + 1)) + 1. - Peter Luschny, Jan 05 2023

A115389 Denominator of rational part of raw moment n of the line point picking problem.

Original entry on oeis.org

1, 1, 1, 3, 6, 5, 30, 105, 140, 126, 1260, 1155, 13860, 12870, 12012, 9009, 72072, 68068, 1225224, 5819814, 5542680, 5290740, 116396280, 111546435, 535422888, 514829700, 1487285800, 1434168450, 5736673800, 5545451340, 166363540200, 644658718275, 312561802800

Offset: 1

Eric W. Weisstein, Jan 21 2006

			-1 + 2*log(2), 3 - 4*log(2), -4 + 6*log(2), 17/3 - 8*log(2), -41/6 + 10*log(2), ...

Eric Weisstein's World of Mathematics, Line Point Picking

Cf. A058312, A058313, A115388.

a := n -> denom(1 + 2*n*add((-1)^(n+k+1)/k, k = 1..n)):
seq(a(n), n = 1..30); # Peter Bala, Jan 05 2023
# Alternative:
a := n -> 2*n*((-1)^n*log(2) - LerchPhi(-1, 1, n + 1)) + 1:
seq(denom(simplify(a(n))), n = 1..33); # Peter Luschny, Jan 05 2023

a(n) = denominator of 1 + (2*n)*Sum_{k = 1..n} (-1)^(n+k+1)/k. - Peter Bala, Jan 05 2023

a(n) = denominator of 2*n*((-1)^n*log(2) - LerchPhi(-1, 1, n + 1)) + 1. - Peter Luschny, Jan 05 2023

A119955 Numbers n such that denominator of n-th Harmonic Number equals denominator of n-th Alternative Harmonic Number.

Original entry on oeis.org

1, 2, 3, 4, 5, 9, 10, 11, 12, 13, 14, 27, 49, 50, 51, 52, 53, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 289, 290, 291, 292, 293, 841, 842, 843, 844

Offset: 1

Alexander Adamchuk, Aug 02 2006

nonn

Up to n=14 A002805[n] coincides with A058312[n]. a(n) up to a(12)=27 coincides with A096304[n].

			Denominators of Harmonic Number (H[n] = Sum[1/i, {i, n}]) are A002805[n] = {1,2,6,12,60,20,140,280,2520,2520,27720,27720,360360,360360,360360,...}.
Denominators of Alternative Harmonic Number (H'[n] = Sum[(-1)^(i+1)*1/i, {i, n}]) are A058312[n] = {1,2,6,12,60,60,420,840,2520,2520,27720,27720,360360,360360,72072,...}.
a(1) = 1 because A002805[1] = A058312[1].
15 is not in a(n) because A002805[15] = 360360 is not equal to A058312[15] = 72072.

Eric Weisstein's World of Mathematics, Harmonic Number.

Cf. A002805, A058312, A096304.

Do[s1=Denominator[Sum[(-1)^(i+1)*1/i, {i, n}]]; s2=Denominator[Sum[1/i, {i, n}]]; If[Equal[s2, s1], Print[n]], {n, 1, 1500}]

A262136 Number of distinct fractional parts of the numbers Sum_{i=j..k} (-1)^i/i with 1 <= j <= k <= n, where the fractional part of x is given by x - floor(x).

Original entry on oeis.org

1, 2, 4, 7, 11, 14, 20, 27, 35, 44, 54, 64, 76, 89, 103, 118, 134, 151, 169, 186, 206, 227, 249, 272, 296, 321, 347, 374, 402, 430, 460, 491, 523, 556, 590, 625, 661, 698, 736, 775, 815, 854, 896, 939, 983, 1028, 1074, 1121, 1169, 1218, 1268, 1319, 1371, 1424, 1478, 1532, 1588, 1645, 1703, 1762

Offset: 1

Zhi-Wei Sun, Sep 11 2015

nonn

Note that (-1)^n/n+(-1)^(n+1)/(n+1) = (-1)^n/(n*(n+1)) for any n > 0.
Conjecture: (i) Suppose that Sum_{i=j..k} (-1)^i/i and Sum_{r=s..t} (-1)^r/r with 0 < min{2,k} <= j <= k, 0 < min{2,t} <= s <= t and j <= s have the same fractional part, but the ordered pairs (j,k) and (s,t) are different. Then Sum_{i=j..k} (-1)^i/i = Sum_{r=s..t} (-1)^r/r. Moreover, if j is odd, then j > 1, k = j*(j+1) and (s,t) = (j+2,j*(j+1)-1); if j is even, then either (k = j+1 and s = t = j*(j+1)), or (k = j*(j+1)-1 and (s,t) = (j+2,j*(j+1))).
(ii) Let a > b >= 0 and m > 0 be integers with gcd(a,b) = 1 < max{a,m}. For each r = 0,1, the numbers Sum_{i=j..k} (-1)^(i-r*j)/(a*i-b)^m with 1 <= j <= k and (j > 1 if k > a-b = 1) have pairwise distinct fractional parts.
This is an analog of the conjecture in A261878. Part (i) of the conjecture implies that a(n) = n*(n-1)/2 + 2 - floor((sqrt(4n+1)-1)/2) - floor((sqrt(4n+1)-1)/4) for all n > 1.

			a(6) = 14 since the sums (-1)^j/j+...+(-1)^k/k with 0 < min{k,2} <= j <= k <= 6 and (j,k) different from (4,6) and (6,6) have pairwise distinct fractional parts, but (-1)^6/6 = (-1)^2/2+(-1)^3/3 and 1/4-1/5+1/6 = 1/2-1/3+1/4-1/5.

Zhi-Wei Sun, Table of n, a(n) for n = 1..1000

Cf. A058312, A058313, A261878.

frac[x_]:=x-Floor[x]
u[0]:=0
u[n_]:=u[n-1]+(-1)^n/n
S[n_]:=Table[frac[u[n]-u[m-1]],{m,Min[2,n],n}]
T[1]:=S[1]
T[n_]:=Union[T[n-1],S[n]]
Do[Print[n," ",Length[T[n]]],{n,1,60}]

A373893 a(n) is the length of the simple continued fraction for the n-th alternating harmonic number.

Original entry on oeis.org

0, 1, 2, 4, 7, 7, 5, 9, 8, 12, 9, 12, 11, 12, 13, 12, 18, 12, 17, 15, 15, 15, 19, 21, 18, 13, 21, 23, 25, 23, 26, 32, 28, 25, 24, 24, 31, 32, 33, 36, 41, 38, 38, 37, 44, 41, 37, 39, 47, 48, 42, 43, 43, 44, 46, 42, 44, 51, 45, 49, 52, 53, 62, 50, 57, 48, 55, 60, 52, 58, 70, 58, 60, 73, 67

Offset: 1

Ilya Gutkovskiy, Jun 21 2024

nonn

By "simple continued fraction" is meant a continued fraction whose terms are positive integers and the final term is >= 2.

			Sum_{k=1..7} (-1)^(k+1)/k = 319/420 = 1/(1 + 1/(3 + 1/(6 + 1/(3 + 1/5)))), so a(7) = 5.

Cf. A055573, A058312, A058313.

Table[Length[ContinuedFraction[Sum[(-1)^(k + 1)/k, {k, 1, n}]]] - 1, {n, 1, 75}]

from fractions import Fraction
from sympy.ntheory.continued_fraction import continued_fraction
def A373893(n): return len(continued_fraction(sum(Fraction(1 if k&1 else -1,k) for k in range(1,n+1))))-1 # Chai Wah Wu, Jun 27 2024

A173756 Partial sums of A058313.

Original entry on oeis.org

1, 2, 7, 14, 61, 98, 417, 950, 2829, 4456, 24873, 42980, 306091, 543462, 595741, 691290, 2459767, 4092108, 37557035, 193242042, 360012409, 516201296, 4341338257, 7943382348, 27024448579, 45075855410, 102204647503, 109956141102, 346222803073, 571398562364

Offset: 1

Jonathan Vos Post, Feb 23 2010

Partial sum of the numerator of the n-th alternating harmonic number.

Cf. A001008 (numerator of the n-th harmonic number), A025530, A058312 (denominators of the underlying sequence), A058313, A075830.

a := proc(n) local i, k:
add(numer(add((-1)^(k + 1)/k, k = 1 .. i)), i = 1 .. n): end proc:
seq(a(n), n = 1 .. 40); # Petros Hadjicostas, May 06 2020

a(n) = sum(i=1, n, numerator(sum(k=1, i, (-1)^(k+1)/k))); \\ Michel Marcus, May 07 2020

a(n) = Sum_{i=1..n} A058313(i) = Sum_{i=1..n} numerator(Sum_{k=1..i} (-1)^(k+1)/k). [Corrected by Petros Hadjicostas and Michel Marcus, May 06 2020]

Data corrected and extended by Petros Hadjicostas, May 06 2020

A305128 Expansion of e.g.f. Product_{k>=1} 1/(1 - x^k)^AH(k), where AH(k) is the k-th alternating harmonic number.

Original entry on oeis.org

1, 1, 3, 14, 79, 539, 4663, 42468, 457945, 5433281, 71036231, 994289658, 15544425103, 253283689619, 4489180389835, 84521336758904, 1687130833152561, 35365641206048129, 790065486354237643, 18340253632236738022, 449655289227002010351, 11492300073384698090795, 306803167368168113022271

Offset: 0

Ilya Gutkovskiy, May 26 2018

nonn

a(n)/n! is the Euler transform of [1, 1 - 1/2, 1 - 1/2 + 1/3, 1 - 1/2 + 1/3 - 1/4, 1 - 1/2 + 1/3 - 1/4 + 1/5, ...].

N. J. A. Sloane, Transforms

Cf. A028342, A058312, A058313, A168243, A303970.

nmax = 22; CoefficientList[Series[Product[1/(1 - x^k)^Sum[(-1)^(j + 1)/j, {j, 1, k}], {k, 1, nmax}], {x, 0, nmax}], x] Range[0, nmax]!
a[n_] := a[n] = If[n == 0, 1, Sum[Sum[d ((-1)^(d + 1) LerchPhi[-1, 1, d + 1] + Log[2]), {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[n! a[n], {n, 0, 22}]

E.g.f.: Product_{k>=1} 1/(1 - x^k)^(A058313(k)/A058312(k)).

A334750 Total number of swaps needed to sort all n! permutations of n elements by the optimal dual-pivot quicksort "Count".

Original entry on oeis.org

0, 0, 4, 16, 103, 711, 5526, 48066, 463248, 4908816, 56749536, 711299232, 9609618816, 139252708224, 2154724104960, 35464952597760, 618712803717120, 11405648080834560, 221541001069731840, 4522678391979417600, 96811891510299033600, 2168416142767145779200

Offset: 0

Petros Hadjicostas, May 09 2020

nonn

The dual-pivot quicksort "Count" differs slightly from the original version of the dual-pivot quicksort algorithm of Aumüller et al. (2016, 2019). See Algorithm 1 in Neininger and Straub (2017, 2018).
By the design of the algorithm, for n >= 2, Neininger and Straub always require at least two swaps "at the end in order to bring the [two] pivots to their final positions".
Here a(n) is n! times the average number of swaps of the dual-pivot quicksort "Count" when sorting a random permutation of length n.

M. Aumüller, M. Dietzfelbinger, C. Heuberger, D. Krenn, and H. Prodinger, Dual-Pivot Quicksort: Optimality, Analysis and Zeros of Associated Lattice Paths, arXiv:1611.00258 [math.CO], 2016.
M. Aumüller, M. Dietzfelbinger, C. Heuberger, D. Krenn, and H. Prodinger, Dual-Pivot Quicksort: Optimality, Analysis and Zeros of Associated Lattice Paths, Combin. Probab. Comput. 28(4) (2019), 485-518.
R. Neininger and J. Straub, Probabilistic analysis of the dual-pivot quicksort "count", arXiv:1710.07505 [cs.DS], 2017.
R. Neininger and J. Straub, Probabilistic analysis of the dual-pivot quicksort "count", 2018 Proceedings of the Meeting of Analytic Algorithmics and Combinatorics, New Orleans, Louisiana, USA, 7pp.

Cf. A058312, A058313, A288964, A288965, A288970, A288971.

lista(nn) = { nn = max(nn, 3); my(va = vector(nn)); va[1] = 0; va[2] = 4; for(n=3, nn, va[n] = n!*(6*sum(k=1, n-2, (n-k-1)*va[k]/k!)/(n*(n-1)) + 5*n/8 + 13/16 - 1/(16*(n - (1 + (-1)^n)/2)))); concat(0, va); };

H1(n) = sum(k=1, n, 1/k);
H2(n) = sum(k=1, n, (-1)^k/k);
H3(n) = if(0 == (n % 2), - (1/320)*(1/(n - 3) + 3/(n - 1)), (1/320)*(3/(n - 2) + 1/n));
lista(nn) = { nn = max(nn, 3); my(va = vector(nn)); va[1] = 0; va[2] = 4; va[3] = 16; for(n=4, nn, va[n] = n!*((3/4)*n*H1(n) + (1/20)*n*H2(n) - (4/5)*n + (3/4)*H1(n) + (1/20)*H2(n) - 23/160 - (-1)^n/40 + H3(n))); concat(0,va); };

Recurrence: a(n)/n! = (6/(n*(n - 1)))*(Sum_{k=0..n-2} (n - k - 1)*a(k)/k!) + 5*n/8 + 13/16 - 1/(16*(n - I[n is even])) for n >= 2 with a(0) = a(1) = 0, where I(condition) = 1 if the condition holds, and = 0 otherwise. [See p. 7 in Neininger and Straub (2017).]

a(n) = n!*((3/4)*n*Harmonic(n) + (1/20)*n*Harmonic_alt(n) - (4/5)*n + (3/4)*Harmonic(n) + (1/20)*Harmonic_alt(n) - 23/160 - (-1)^n/40 - ([n even]/320)*(1/(n - 3) + 3/(n - 1)) + ([n odd]/320)*(3/(n - 2) + 1/n)) for n >= 4, where Harmonic_alt(n) = Sum_{k=1..n} (-1)^n/n = -A058313(n)/A058312(n). [See Theorem 2.1 in Neininger and Straub (2017, 2018).]

A347978 E.g.f.: 1/(1 + x)^(1/(1 - x)).

Original entry on oeis.org

1, -1, 0, -3, 4, -30, 186, -630, 11600, -26712, 1005480, -2581920, 117196872, -485308824, 17734457664, -131070696120, 3387342915840, -43890398953920, 801577841697216, -17363169328243392, 233460174245351040, -7968629225100337920, 84363134551361043840

Offset: 0

Ilya Gutkovskiy, Sep 22 2021

sign

Cf. A005727, A007120, A008405, A024167, A058312, A058313, A073478, A087761.

nmax = 22; CoefficientList[Series[1/(1 + x)^(1/(1 - x)), {x, 0, nmax}], x] Range[0, nmax]!
A024167[n_] := n! Sum[(-1)^(k + 1)/k, {k, 1, n}]; a[0] = 1; a[n_] := a[n] = -Sum[Binomial[n - 1, k - 1] A024167[k] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 22}]

my(x='x+O('x^30)); Vec(serlaplace(1/(1+x)^(1/(1-x)))) \\ Michel Marcus, Sep 22 2021

E.g.f.: exp( Sum_{k>=1} x^k * Sum_{j=1..k} (-1)^j / j ).
a(0) = 1; a(n) = -Sum_{k=1..n} binomial(n-1,k-1) * A024167(k) * a(n-k).
a(0) = 1; a(n) = -Sum_{k=1..n} binomial(n,k) * A073478(k) * a(n-k).

cp's OEIS Frontend

A079542 Triangular array: T(n,1) = T(n,n) = n and T(n,k) = lcm(T(n-1,k-1), T(n-1,k)) for 1 < k < n.

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A115388 Numerator of rational part of raw moment n of the line point picking problem.

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A115389 Denominator of rational part of raw moment n of the line point picking problem.

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A119955 Numbers n such that denominator of n-th Harmonic Number equals denominator of n-th Alternative Harmonic Number.

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A262136 Number of distinct fractional parts of the numbers Sum_{i=j..k} (-1)^i/i with 1 <= j <= k <= n, where the fractional part of x is given by x - floor(x).

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A373893 a(n) is the length of the simple continued fraction for the n-th alternating harmonic number.

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A173756 Partial sums of A058313.

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A305128 Expansion of e.g.f. Product_{k>=1} 1/(1 - x^k)^AH(k), where AH(k) is the k-th alternating harmonic number.

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