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.

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A226357 Number of ordered triples (i,j,k) with |i|,|j|,|k|,|i*j*k| <= n and gcd(i,j,k) <= 1.

Original entry on oeis.org

1, 27, 75, 147, 243, 363, 483, 651, 819, 1011, 1179, 1443, 1683, 1995, 2211, 2475, 2763, 3171, 3459, 3915, 4251, 4611, 4923, 5475, 5883, 6411, 6771, 7275, 7707, 8403, 8811, 9555, 10059, 10611, 11067, 11715, 12291, 13179, 13683, 14331, 14931, 15915, 16419
Offset: 0

Views

Author

Robert Price, Jun 04 2013

Keywords

Comments

Note that gcd(0,m) = m for any m.

Crossrefs

|i| + |j| + |k| <= n instead of |i*j*k| <= n: A100450.
This sequence (A226357) without the GCD qualifier: A226359.
Distinct sums i+j+k with the GCD qualifier: A222947.
Distinct sums i+j+k without the GCD qualifier: A222945.
Distinct products i*j*k with or without the GCD qualifier is 2n+1: A005408.
With the further restriction i,j,k >= 0 ...
Distinct sums i+j+k <= n with the GCD qualifier: A223133.
Distinct sums i+j+k <= n without the GCD qualifier: A223134.
Distinct products i*j*k with or without the GCD qualifier is n+1: A000217(n+1).
Distinct sums i+j+k with i*j*k = n with the GCD qualifier: A223135.
Distinct sums i+j+k with i*j*k = n without the GCD qualifier: A226378.
Distinct products i*j*k with i*j*k = n with or without the GCD qualifier is trivial and always 1: A000012.
Ordered triples with the product <= n with the GCD qualifier: A226001.
Ordered triples with the product <= n without the GCD qualifier: A226600.
Ordered triples with the product = n with the GCD qualifier: A226602.
Ordered triples with the product = n without the GCD qualifier: A007425.

Programs

  • Mathematica
    f[n_] := Length[Complement[Union[Flatten[Table[If[Abs[i*j*k] <=  n && GCD[i, j, k] <= 1, {i, j, k}], {i, -n, n}, {j, -n, n}, {k, -n, n}], 2]], {Null}]]; Table[f[n], {n, 0, 100}]

A280076 Numbers n such that Sum_{d|n} tau(d) = Product_{d|n} tau(d).

Original entry on oeis.org

1, 4, 9, 25, 49, 121, 169, 289, 361, 529, 841, 961, 1369, 1681, 1849, 2209, 2809, 3481, 3721, 4489, 5041, 5329, 6241, 6889, 7921, 9409, 10201, 10609, 11449, 11881, 12769, 16129, 17161, 18769, 19321, 22201, 22801, 24649, 26569, 27889, 29929, 32041, 32761, 36481
Offset: 1

Views

Author

Jaroslav Krizek, Dec 25 2016

Keywords

Comments

Union of 1 and A001248 (squares of primes).
Numbers n such that A007425(n) = A211776(n).
Numbers n such that A007425(n) = Sum_{d|n} tau(d) = A211776(n) = Product_{d|n} tau(d) = 6.
Also squares of noncomposite numbers (A008578).
Subsequence of A350343. - Lorenzo Sauras Altuzarra, Sep 18 2022

Examples

			9 is a term because Sum_{d|9} tau(d) = 1+2+3 = Product_{d|9} tau(d) = 1*2*3 = 6.
		

Crossrefs

Programs

  • Magma
    [n: n in [1..1000000] | &*[NumberOfDivisors(d): d in Divisors(n)]  eq &+[NumberOfDivisors(d): d in Divisors(n)]]
    
  • Mathematica
    Select[Range@ 37500, Total@ # == Times @@ # &@ Map[DivisorSigma[0, #] &, Divisors@ #] &] (* Michael De Vlieger, Dec 25 2016 *)
  • PARI
    isok(n) = my(d = divisors(n), nd = apply(numdiv, d)); vecsum(nd) == prod(k=1, #nd, nd[k]); \\ Michel Marcus, Jun 26 2017

Formula

A007425(a(n)) = A211776(a(n)) = 6.
Apparently, a(n) = A331294(n + 3) if n > 5. - Lorenzo Sauras Altuzarra, Sep 18 2022

A280473 G.f.: Product_{i>=1, j>=1, k>=1} (1 + x^(i*j*k)).

Original entry on oeis.org

1, 1, 3, 6, 12, 21, 43, 70, 127, 215, 364, 591, 989, 1562, 2515, 3954, 6194, 9538, 14754, 22349, 33926, 50910, 76102, 112721, 166747, 244205, 356984, 518344, 749924, 1078711, 1547668, 2207418, 3140135, 4446572, 6276657, 8823776, 12371487, 17275879, 24061878
Offset: 0

Views

Author

Vaclav Kotesovec, Jan 04 2017

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 50; CoefficientList[Series[Product[(1+x^(i*j*k)), {i, 1, nmax}, {j, 1, nmax/i}, {k, 1, nmax/i/j}], {x, 0, nmax}], x]
    nmax = 50; A007425 = Table[Sum[DivisorSigma[0, d], {d, Divisors[n]}], {n, 1, nmax}]; s = 1 + x; Do[s *= Sum[Binomial[A007425[[k]], j]*x^(j*k), {j, 0, nmax/k}]; s = Expand[s]; s = Take[s, Min[nmax + 1, Exponent[s, x] + 1, Length[s]]];, {k, 2, nmax}]; Take[CoefficientList[s, x], nmax + 1] (* Vaclav Kotesovec, Aug 30 2018 *)

Formula

G.f.: Product_{k>=1} (1 + x^k)^tau_3(k), where tau_3() = A007425. - Ilya Gutkovskiy, May 22 2018

A061202 (tau<=)_4(n).

Original entry on oeis.org

1, 5, 9, 19, 23, 39, 43, 63, 73, 89, 93, 133, 137, 153, 169, 204, 208, 248, 252, 292, 308, 324, 328, 408, 418, 434, 454, 494, 498, 562, 566, 622, 638, 654, 670, 770, 774, 790, 806, 886, 890, 954, 958, 998, 1038, 1054, 1058, 1198, 1208, 1248, 1264, 1304, 1308
Offset: 1

Views

Author

Vladeta Jovovic, Apr 21 2001

Keywords

Comments

(tau<=)_k(n) = |{(x_1,x_2,...,x_k): x_1*x_2*...*x_k <= n}|, i.e., (tau<=)_k(n) is number of solutions to x_1*x_2*...*x_k <= n, x_i > 0.
Partial sums of A007426.
Equals row sums of triangle A140703. - Gary W. Adamson, May 24 2008

Crossrefs

Cf. tau_2(n): A000005, tau_3(n): A007425, tau_4(n): A007426, tau_5(n): A061200, tau_6(n): A034695, (unordered) 2-factorizations of n: A038548, (unordered) 3-factorizations of n: A034836, A001055, (tau<=)_2(n): A006218, (tau<=)_3(n): A061201, (tau<=)_5(n): A061203, (tau<=)_6(n): A061204.
Equals left column of triangle A140705.
Cf. A140703.

Programs

  • Mathematica
    (* Asymptotics: *) n * (Log[n]^3/6 + (2*EulerGamma - 1/2)*Log[n]^2 + (6*EulerGamma^2 - 4*EulerGamma - 4*StieltjesGamma[1] + 1)*Log[n] + 4*EulerGamma^3 - 6*EulerGamma^2 + 4*EulerGamma + 4*StieltjesGamma[1]*(1 - 3*EulerGamma) + 2*StieltjesGamma[2] - 1) (* Vaclav Kotesovec, Sep 09 2018 *)

Formula

(tau<=)k(n) = Sum{i=1..n} tau_k(i).
a(n) = Sum_{k = 1..n} tau_{3}(k)*floor (n/k), where tau_{3} is A007425. - Enrique Pérez Herrero, Jan 23 2013
a(n) ~ n * (log(n)^3/6 + (2*g - 1/2)*log(n)^2 + (6*g^2 - 4*g - 4*g1 + 1)*log(n) + 4*g^3 - 6*g^2 + 4*g + 4*g1*(1 - 3*g) + 2*g2 - 1), where g is the Euler-Mascheroni constant A001620, g1 and g2 are the Stieltjes constants, see A082633 and A086279. - Vaclav Kotesovec, Sep 09 2018
a(n) = Sum_{i=1..n} tau(i)*A006218(floor(n/i)). - Ridouane Oudra, Sep 17 2021
a(n) = Sum_{i=1..n} Sum_{j=1..n} Sum_{k=1..n} floor(n/(i*j*k)). - Ridouane Oudra, Oct 31 2022

A140773 Consider the products of all pairs of (not necessarily distinct) positive divisors of n. a(n) is the number of these products that divide n. a(n) also is the number of the products that are divisible by n.

Original entry on oeis.org

1, 2, 2, 4, 2, 5, 2, 6, 4, 5, 2, 10, 2, 5, 5, 9, 2, 10, 2, 10, 5, 5, 2, 16, 4, 5, 6, 10, 2, 14, 2, 12, 5, 5, 5, 20, 2, 5, 5, 16, 2, 14, 2, 10, 10, 5, 2, 24, 4, 10, 5, 10, 2, 16, 5, 16, 5, 5, 2, 28, 2, 5, 10, 16, 5, 14, 2, 10, 5, 14, 2, 32, 2, 5, 10, 10, 5, 14, 2, 24, 9, 5, 2, 28, 5, 5, 5, 16, 2, 28, 5
Offset: 1

Views

Author

Leroy Quet, May 29 2008

Keywords

Comments

Number of 3D grids of n congruent boxes with two different edge lengths, in a box, modulo rotation (cf. A034836 for cubes instead of boxes and A007425 for boxes with three different edge lengths; cf. A000005 for the 2D case). - Manfred Boergens, Feb 25 2021
Number of distinct faces obtainable by arranging n unit cubes into a cuboid. - Chris W. Milson, Mar 14 2021

Examples

			The divisors of 20 are 1,2,4,5,10,20. There are 10 pairs of divisors whose product divides 20: 1*1=1, 1*2=2, 1*4=4, 1*5=5, 1*10=10, 1*20=20, 2*2=4, 2*5=10, 2*10=20, 4*5 = 20. Likewise, there are 10 products that are divisible by 20: 4*5=20, 2*10=20, 4*10=40, 10*10=100, 1*20=20, 2*20=40, 4*20=80, 5*20=100, 10*20=200, 20*20=400. So a(20) = 10.
		

Crossrefs

Cf. A140774.
Cf. A369255 (parity), A369256 (positions of odd terms), A378213 (Dirichlet inverse).

Programs

  • Mathematica
    (* first do *) Needs["Combinatorica`"] (* then *) f[n_] := Count[ n/Times @@@ Union[Sort /@ Tuples[Divisors@ n, 2]], Integer]; Array[f, 91] (* _Robert G. Wilson v, May 31 2008 *)
    d=Divisors[n]; r=Length[d]; Sum[Ceiling[Length[Divisors[d[[j]]]]/2],{j,r}] (* Manfred Boergens, Feb 25 2021 *)
  • PARI
    \\ Two implementations, after the two different interpretations given by the author of the sequence:
    A140773v1(n) = { my(ds = divisors(n),s=0); for(i=1,#ds,for(j=i,#ds,if(!(n%(ds[i]*ds[j])),s=s+1))); s; }
    A140773v2(n) = { my(ds = divisors(n),s=0); for(i=1,#ds,for(j=i,#ds,if(!((ds[i]*ds[j])%n),s=s+1))); s; }
    \\ Antti Karttunen, May 19 2017
    
  • Python
    # See C. W. Milson link.

Formula

a(n) = Sum_{m|n} A038548(m) = Sum_{m|n} ceiling(d(m)/2), where d(m) = number of divisors of m (A000005). - Manfred Boergens, Feb 25 2021
a(n) = Sum_{d|n} A135539(d,n/d). - Ridouane Oudra, Jul 10 2021
a(n) = (A007425(n) + A046951(n))/2. - Ridouane Oudra, Apr 10 2024
G.f.: Sum_{k>=1} Sum_{d|k} x^(k^2/d)/(1 - x^k). - Miles Wilson, Jun 12 2025

Extensions

Corrected and extended by Robert G. Wilson v, May 31 2008

A145511 Dirichlet g.f.: (1-2/2^s+7/4^s)*zeta(s)^3.

Original entry on oeis.org

1, 1, 3, 7, 3, 3, 3, 19, 6, 3, 3, 21, 3, 3, 9, 37, 3, 6, 3, 21, 9, 3, 3, 57, 6, 3, 10, 21, 3, 9, 3, 61, 9, 3, 9, 42, 3, 3, 9, 57, 3, 9, 3, 21, 18, 3, 3, 111, 6, 6, 9, 21, 3, 10, 9, 57, 9, 3, 3, 63, 3, 3, 18, 91, 9, 9, 3, 21, 9, 9, 3, 114, 3, 3, 18, 21, 9, 9, 3, 111, 15, 3, 3, 63, 9, 3, 9, 57, 3, 18, 9
Offset: 1

Views

Author

N. J. A. Sloane, Mar 14 2009

Keywords

Comments

Dirichlet convolution of [1,-2,0,7,0,0,0,0,...] and A007425. - R. J. Mathar, Feb 07 2011

Crossrefs

Programs

  • Maple
    read("transforms") :
    nmax := 100 :
    L := [1,-2,0,7,seq(0,i=1..nmax)] :
    MOBIUSi(%) :
    MOBIUSi(%) :
    MOBIUSi(%) ; # R. J. Mathar, Sep 25 2017
  • Mathematica
    f[p_, e_] := (e + 1)*(e + 2)/2; f[2, e_] := 3*(e - 1)*e + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 25 2022 *)
  • PARI
    up_to = 65537;
    t1 = direuler(p=2, up_to, 1/(1-X)^3);
    t3 = direuler(p=2, 2, 1-2*X^1+7*X^2, up_to);
    v145511 = dirmul(t1, t3);
    A145511(n) = v145511[n]; \\ Antti Karttunen, Sep 27 2018, after R. J. Mathar's PARI-code for A158327
    
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,1] == 2, 3*(f[i,2]-1)*f[i,2]+1, (f[i,2]+1)*(f[i,2]+2)/2)); } \\ Amiram Eldar, Oct 25 2022

Formula

From Amiram Eldar, Oct 25 2022: (Start)
Multiplicative with a(2^e) = 3*(e-1)*e+1 and a(p^e) = (e+1)*(e+2)/2 if p > 2.
Sum_{k=1..n} a(k) ~ (7/8)*n*log(n)^2 + c_1*n*log(n) + c_2*n, where c_1 = 21*gamma/4 - 5*log(2)/2 - 7/4 and c_2 = 7/4 + 21*gamma*(gamma-1)/4 - 15*gamma*log(2)/2 - 21*gamma_1/4 + 5*log(2)/2 + 3*log(2)^2, where gamma is Euler's constant (A001620) and gamma_1 is the 1st Stieltjes constant (A082633). (End)

A295931 Number of ways to write n in the form n = (x^y)^z where x, y, and z are positive integers.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 1, 3, 3, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Gus Wiseman, Nov 29 2017

Keywords

Comments

By convention a(1) = 1.
Values can be 1, 3, 6, 9, 10, 15, 18, 21, 27, 28, 30, 36, 45, 54, 60, 63, 84, 90, etc. - Robert G. Wilson v, Dec 10 2017

Examples

			The a(256) = 10 ways are:
(2^1)^8    (2^2)^4   (2^4)^2  (2^8)^1
(4^1)^4    (4^2)^2   (4^4)^1
(16^1)^2   (16^2)^1
(256^1)^1
		

Crossrefs

Programs

  • Maple
    f:= proc(n) local m,d,t;
      m:= igcd(seq(t[2],t=ifactors(n)[2]));
      add(numtheory:-tau(d),d=numtheory:-divisors(m))
    end proc:
    f(1):= 1:
    map(f, [$1..100]); # Robert Israel, Dec 19 2017
  • Mathematica
    Table[Sum[DivisorSigma[0,d],{d,Divisors[GCD@@FactorInteger[n][[All,2]]]}],{n,100}]

Formula

a(A175082(k)) = 1, a(A093771(k)) = 3.
a(n) = Sum_{d|A052409(n)} A000005(d).

A061391 a(n) = t(n,3) = Sum_{d|n} tau(d^3), where tau(n) = number of divisors of n, cf. A000005.

Original entry on oeis.org

1, 5, 5, 12, 5, 25, 5, 22, 12, 25, 5, 60, 5, 25, 25, 35, 5, 60, 5, 60, 25, 25, 5, 110, 12, 25, 22, 60, 5, 125, 5, 51, 25, 25, 25, 144, 5, 25, 25, 110, 5, 125, 5, 60, 60, 25, 5, 175, 12, 60, 25, 60, 5, 110, 25, 110, 25, 25, 5, 300, 5, 25, 60, 70, 25, 125, 5, 60, 25, 125, 5, 264
Offset: 1

Views

Author

Vladeta Jovovic, Apr 29 2001

Keywords

Comments

Inverse Mobius transform of A048785. - R. J. Mathar, Feb 09 2011

Crossrefs

Cf. t(n, 0) = A000005(n), t(n, 1) = A007425(n), t(n, 2) = A035116(n).
Cf. A048691.

Programs

  • Mathematica
    f[p_, e_] := (3*e^2 + 5*e + 2)/2; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 16 2020 *)
  • PARI
    A061391 = n -> sumdiv(n, d, numdiv(d^3));
    for(n=1, 10000, write("b061391.txt", n, " ", A061391(n)));
    \\ Antti Karttunen, Jan 17 2017
    
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, (1 + 2*X)/(1 - X)^3)[n], ", ")) \\ Vaclav Kotesovec, May 15 2021
    
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, (1 - 3*X^2 + 2*X^3)/(1 - X)^5)[n], ", ")) \\ Vaclav Kotesovec, Aug 20 2021

Formula

t(n, k) = Sum_{d|n} tau(d^k) is multiplicative: if the canonical factorization of n = Product p^e(p) over primes then t(n, k) = Product t(p^e(p), k), t(p^e(p), k) = (1/2) *(k*e(p)+2)*(e(p)+1).
For k=2 we get an interesting identity: Sum_{d|n} tau(d^2)=(tau(n))^2, cf. A048691, A035116.
a(n) = Sum_{d|n} tau(n*d). - Benoit Cloitre, Nov 30 2002
G.f.: Sum_{n>=1} tau(n^3)*x^n/(1-x^n). - Joerg Arndt, Jan 01 2011
Dirichlet g.f.: zeta(s)^3 * Product_{primes p} (1 + 2/p^s). - Vaclav Kotesovec, May 15 2021
Dirichlet g.f.: zeta(s)^5 * Product_{primes p} (1 - 3/p^(2*s) + 2/p^(3*s)). - Vaclav Kotesovec, Aug 20 2021

A127170 Triangle read by rows: T(n,k) is the number of divisors of n that are divisible by k, with 1 <= k <= n.

Original entry on oeis.org

1, 2, 1, 2, 0, 1, 3, 2, 0, 1, 2, 0, 0, 0, 1, 4, 2, 2, 0, 0, 1, 2, 0, 0, 0, 0, 0, 1, 4, 3, 0, 2, 0, 0, 0, 1, 3, 0, 2, 0, 0, 0, 0, 0, 1, 4, 2, 0, 0, 2, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 6, 4, 3, 2, 0, 2, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 4, 2, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1
Offset: 1

Views

Author

Gary W. Adamson, Jan 06 2007

Keywords

Comments

Column k lists the terms of A000005 interleaved with k - 1 zeros.
Eigensequence of the triangle = A007557; i.e., sequence A007557 shifts to the left upon multiplication by A127170. - Gary W. Adamson, Apr 27 2009

Examples

			First 10 rows of the triangle:
  1;
  2, 1;
  2, 0, 1;
  3, 2, 0, 1;
  2, 0, 0, 0, 1;
  4, 2, 2, 0, 0, 1;
  2, 0, 0, 0, 0, 0, 1;
  4, 3, 0, 2, 0, 0, 0, 1;
  3, 0, 2, 0, 0, 0, 0, 0, 1;
  4, 2, 0, 0, 2, 0, 0, 0, 0, 1;
  ...
		

Crossrefs

Programs

Formula

A007429(n) = Sum_{i=1..n} i*a(i).
T(n,k) = A000005(n/k), if k divides n, otherwise 0, with n >= 1 and 1 <= k <= n. - Omar E. Pol, Apr 01 2015

Extensions

8 terms taken from Example section and then corrected in Data section by Omar E. Pol, Mar 30 2015
Extended beyond a(21) by Omar E. Pol, Apr 01 2015
New name (which was a comment dated Mar 30 2015) from Omar E. Pol, Feb 16 2022

A145444 Dirichlet g.f.: (1+3/4^s+2/8^s)*zeta(s)^3.

Original entry on oeis.org

1, 3, 3, 9, 3, 9, 3, 21, 6, 9, 3, 27, 3, 9, 9, 39, 3, 18, 3, 27, 9, 9, 3, 63, 6, 9, 10, 27, 3, 27, 3, 63, 9, 9, 9, 54, 3, 9, 9, 63, 3, 27, 3, 27, 18, 9, 3, 117, 6, 18, 9, 27, 3, 30, 9, 63, 9, 9, 3, 81, 3, 9, 18, 93, 9, 27, 3, 27, 9, 27, 3, 126, 3, 9, 18, 27, 9, 27, 3, 117, 15, 9, 3, 81, 9, 9, 9, 63
Offset: 1

Views

Author

N. J. A. Sloane, Mar 14 2009

Keywords

Comments

Dirichlet convolution of [1,0,0,3,0,0,0,2,0,0,...] with A007425. - R. J. Mathar, Sep 25 2017

Crossrefs

Programs

  • Maple
    nmax := 10000 :
    L := [1,0,0,3,0,0,0,2,seq(0,i=1..nmax)] :
    MOBIUSi(%) :
    MOBIUSi(%) :
    MOBIUSi(%) ; # R. J. Mathar, Sep 25 2017
  • Mathematica
    f[p_, e_] := (e + 1)*(e + 2)/2; f[2, e_] := 3*(e - 1)*e + 3; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 25 2022 *)
  • PARI
    t1=direuler(p=2,200,1/(1-X)^3)
    t2=direuler(p=2,2,1+3*X^2+2*X^3,200)
    t3=dirmul(t1,t2)
    
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,1] == 2, 3*(f[i,2]-1)*f[i,2]+3, (f[i,2]+1)*(f[i,2]+2)/2)); } \\ Amiram Eldar, Oct 25 2022

Formula

From Amiram Eldar, Oct 25 2022: (Start):
Multiplicative with a(2^e) = 3*(e-1)*e+3 for e > 0, and a(p^e) = (e+1)*(e+2)/2 if p > 2.
Sum_{k=1..n} a(k) ~ n*log(n)^2 + c_1*n*log(n) + c_2*n, where c_1 = 6*gamma - 9*log(2)/4 - 2 and c_2 = 2 + 6*gamma*(gamma-1) - 27*gamma*log(2)/4 - 6*gamma_1 + 9*log(2)/4 + 21*log(2)^2/8, where gamma is Euler's constant (A001620) and gamma_1 is the 1st Stieltjes constant (A082633). (End)
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