cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Miko Labalan

Miko Labalan's wiki page.

Miko Labalan has authored 5 sequences.

A379561 a(n) = A003418(n+1)*H(n), where H(n) = 1 + 1/2 + ... + 1/n is the n-th harmonic number.

Original entry on oeis.org

2, 9, 22, 125, 137, 1029, 2178, 6849, 7129, 81191, 83711, 1118273, 1145993, 1171733, 2391514, 41421503, 42142223, 813635157, 825887397, 837527025, 848612385, 19761458895, 19994251455, 101086721625, 102157567401, 309561680403, 312536252003, 9146733078187
Offset: 1

Views

Author

Miko Labalan, Dec 26 2024

Keywords

Comments

abs(log(a(n)) - n - log(log(n))) < c*sqrt(n)*log(n)^(-1/2), where constant c = (2+A206431)*Pi/4. This also gives the upper bound of the squared error, (log(a(n)) - n - log(log(n)))^2 < (c^2)*n*log(n)^(-1).
A slightly better absolute error bound could be achieved by using the imaginary part of the nontrivial zeros of the Riemann zeta function, (zetazero(n)-1/2)/sqrt(-1) ~ (2*Pi)*n*LambertW(n/exp(1))^(-1). That bound would be, abs(log(a(n)) - n - log(log(n))) < sqrt(k)*sqrt(n)*LambertW(n/exp(1))^(-1/2), where constant k = 4*Pi/(1+2*A206431). This also gives the upper bound of the squared error, (log(a(n)) - n - log(log(n)))^2 < k*n*LambertW(n/exp(1))^(-1). The midline of the squared error would run along (4/(4+A206431))*n*LambertW(n/exp(1))^(-1).
Another slightly better absolute error bound but without relying on the properties of the zeta zeros would be, abs(log(a(n)) - n - log(log(n))) < n^(3/(9-10*A077761)).
log(a(n))-c*sqrt(n)*log(n)^(-1/2) is a lower bound of sigma_1(n) = A000203(n). Such that, n+log(log(n))-c*sqrt(n)*log(n)^(-1/2) < sigma_1(n) < H(n)+exp(H(n))*log(H(n)).
a(n) gives the total number of ordered pairs (k,m) where k in set {1,2,...,n}, m in set {1,2,...,A003418(n+1)}, and k divides m. Example: For n = 3, there are 22 ordered pairs (k,m) where k is {1,2,3} and m is a multiple of k up to 12. For k = 1, every m is a multiple of 1, m is {1,2,3,4,5,6,7,8,9,10,11,12} so there are 12 pairs. For k = 2, every m is a multiple of 2, m is {2,4,6,8,10,12} so there are 6 pairs. For k = 3, every m is a multiple of 3, m is {3,6,9,12} so there are 4 pairs. So the total ordered pairs is 12 + 6 + 4 = 22 = a(3). Each ordered pair (k,m) also represents an edge in a bipartite graph. Counting all such pairs gives the total number of edges in a graph.

Examples

			a(n)/A025558(n) = [ 2/1, 9/4, 22/9, 125/48, 137/50, 1029/360, 2178/735, ... ]
To evaluate the integral:
For n = 1: Integral_{x=0..1} Li_1(x^(1/2))/x^(1/2) dx = Integral_{x=0..1} -log(1-x^(1/2))/x^(1/2) dx = -2 * -(Sum_{x=1..oo} 1/(x*(x+1))) = -2 * -1 = 2.
For n = 2: Integral_{x=0..1} Li_1(x^(1/3))/x^(1/3) dx = Integral_{x=0..1} -log(1-x^(1/3))/x^(1/3) dx = -3 * -(Sum_{x=1..oo} 1/(x*(x+2))) = -3 * -((1/2)*(1+1/2)) = -3 * -3/4 = 9/4.
For n = 3: Integral_{x=0..1} Li_1(x^(1/4))/x^(1/4) dx = Integral_{x=0..1} -log(1-x^(1/4))/x^(1/4) dx = -4 * -(Sum_{x=1..oo} 1/(x*(x+3))) = -4 * -((1/3)*(1+1/2+1/3)) = -4 * -11/18 = 22/9.

Links

Crossrefs

Cf. A001008/A002805 (harmonic numbers).
Cf. A003418 (lcm).
Cf. A025558 (denominator).
Cf. A193758 (very similar sequence).
Cf. A000203, A025529, A027457, A077761, A152648, A206431.

Programs

  • PARI
    a(n) = lcm(vector(n+1, i, i))*sum(i=1, n, 1/i); \\ Michel Marcus, Dec 28 2024

Formula

a(n) = A025558(n)*(Integral_{x=0..1} Li_1(x^(1/(n+1)))/x^(1/(n+1)) dx).
a(n) = A025558(n) + A027457(n+1).
Integral_{x=0..1} Li_1(x^(1/(n+1)))/x^(1/(n+1)) dx = ((n+1)/n)*H(n) = a(n)/A025558(n).
((n+1)/n)*H(n) ~ log(n) + gamma + (log(n)+gamma+1/2)/n + O(1/n^2).
log(a(n)) ~ n + log(log(n)) + O(c*sqrt(n)*log(n)^(-1/2)), (See comments for constant c).
G.f. for ((n+1)/n)*H(n): G(x) = Li_2(x)+(1/2)*log(1-x)^2-log(1-x)/(1-x), the lim_{x->oo} G(x) = -zeta(2).
Hyperbolic l.g.f. for ((n+1)/n)*H(n): LH(x) = Li_2(x)+(1/2)*log(1-x)^2+Li_3(x)-Li_3(1-x)+Li_2(1-x)*log(1-x)+(1/2)*log(x)*log(1-x)^2+zeta(3), the Integral_{x=0..1} LH(x) dx = 2*zeta(3) = A152648.
Dirichlet g.f. for ((n+1)/n)*H(n): zeta(s+1)*(zeta(s)+zeta(s+2)).

A367481 Primitive practical numbers of the form 2 * 3^i * prime(k).

Original entry on oeis.org

30, 42, 66, 78, 306, 342, 414, 522, 558, 666, 2214, 2322, 2538, 2862, 3186, 3294, 3618, 3834, 3942, 4266, 4482, 4806, 5238, 5454, 5562, 5778, 5886, 6102, 20574, 21222, 22194, 22518, 24138, 24462, 25434, 26406, 27054, 28026, 28998, 29322, 30942, 31266, 31914
Offset: 1

Views

Author

Miko Labalan, Nov 19 2023

Keywords

Comments

This sequence and A308710 are both non-overlapping subsets of A267124.
a(n) is a number of the form 2 * 3^i * prime(k) for i > 0 and b(i) < k <= b(i+1) where b(n) = Sum_{m=2..n+1} A233919(m).
Terms are pseudoperfect numbers, A005835, but are not primitive pseudoperfect numbers, A006036.
Since no term is a square or twice a square, there are no terms k such that sigma(k) is odd. Therefore, according to Proposition 10 by Rao/Peng (see their JNT paper at A083207) all terms are Zumkeller numbers. - Ivan N. Ianakiev, Nov 28 2023

Crossrefs

Cf. A000040, A000244, A005153, A005835, A006036, A233919, A267124, A308710.

Programs

  • Mathematica
    a[n_]:=2*3^(Floor[Log[2*Prime[n+2]]/Log[3]]-1)*Prime[n+2]; Array[a,43] (* Stefano Spezia, Nov 19 2023 *)

Formula

a(n) = 2 * 3^(floor(log_3(2*prime(n+2)))-1) * prime(n+2).

A367520 a(n) = binomial(A007310(n+1),n)/A007310(n+1).

Original entry on oeis.org

1, 1, 3, 15, 55, 364, 1428, 10659, 43263, 345345, 1430715, 11920740, 50067108, 429757960, 1822766520, 15991836267, 68328754959, 609599676595, 2619631042665, 23682737320935, 102240109897695, 934272656316720, 4048514844039120, 37325172122483700
Offset: 0

Views

Author

Miko Labalan, Nov 21 2023

Keywords

Comments

This sequence is related to sequence A001764.

Crossrefs

Cf. A001764, A007310, A235534, A367521.

Programs

  • PARI
    f(n) = n\2*6-(-1)^n; \\ A007310
a(n) = my(x=f(n+1)); binomial(x,n)/x; \\ Michel Marcus, Nov 22 2023

Formula

Limit_{n->oo, n odd} A001764(n)/a(n) = 2/3.
a(n) = A001764(n) iff n is 1 or even.
G.f.: ((1 - 4F3(-1/6,1/6,1/3,2/3; 1/4,1/2,3/4; (729*x^2)/16)) / 3*x) + 4F3(1/6,1/3,2/3,5/6; 1/2,3/4,5/4; (729*x^2)/16).
E.g.f.: 4F5(1/6,1/3,2/3,5/6; 1/2,1/2,3/4,1,5/4; (729*x^2)/64) + x * 4F5(5/6,7/6,4/3,5/3; 5/4,3/2,3/2,7/4,2; (729*x^2)/64).

A367521 a(n) = binomial(A007310(n+1),n+1)/A007310(n+1).

Original entry on oeis.org

1, 2, 5, 30, 99, 728, 2652, 21318, 81719, 690690, 2731365, 23841480, 96282900, 859515920, 3524015272, 31983672534, 132638171391, 1219199353190, 5101386767295, 47365474641870, 199611643133595, 1868545312633440, 7921007303554800, 74650344244967400
Offset: 0

Views

Author

Miko Labalan, Nov 21 2023

Keywords

Comments

This sequence is related to sequence A006013.

Crossrefs

Cf. A006013, A007310, A367520.

Programs

  • PARI
    f(n) = n\2*6-(-1)^n; \\ A007310
a(n) = my(x=f(n+1)); binomial(x,n+1)/x; \\ Michel Marcus, Nov 22 2023

Formula

Limit_{n->oo, n even} a(n)/A006013(n) = 2/3.
a(n) = A006013(n) iff n is 0 or odd.
G.f.: 4F3(1/6,1/3,2/3,5/6; 1/4,3/4,3/2; (729*x^2)/16) - (2 * (4F3(-1/6,1/6,1/3,2/3; 1/4,1/2,3/4; (729*x^2)/16) - 1) / 3*x).
E.g.f.: 4F5(1/6,1/3,2/3,5/6; 1/4,1/2,3/4,1,3/2; (729*x^2)/64) + 2 * x * 4F5(5/6,7/6,4/3,5/3; 5/4,3/2,3/2,7/4,2; (729*x^2)/64).

A308710 Primitive practical numbers of the form 2^i * prime(k).

Original entry on oeis.org

6, 20, 28, 88, 104, 272, 304, 368, 464, 496, 1184, 1312, 1376, 1504, 1696, 1888, 1952, 4288, 4544, 4672, 5056, 5312, 5696, 6208, 6464, 6592, 6848, 6976, 7232, 8128, 16768, 17536, 17792, 19072, 19328, 20096, 20864, 21376, 22144, 22912, 23168, 24448, 24704, 25216
Offset: 1

Views

Author

Miko Labalan, Jun 19 2019

Keywords

Comments

Intersection of A267124 and A100368.
a(n) is a number of the form 2^i * prime(k) for i > 0 and A007053(i) < k <= A007053(i+1).
Terms are pseudoperfect numbers, A005835 and are also primitive pseudoperfect numbers, A006036.

Links

Crossrefs

Cf. A000040, A000079, A005153, A005835, A006036, A007053, A100368, A267124.

Programs

Formula

a(n) = 2^floor(log_2(prime(n+1))) * prime(n+1).

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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