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.

Showing 1-3 of 3 results.

A050999 Sum of squares of odd divisors of n.

Original entry on oeis.org

1, 1, 10, 1, 26, 10, 50, 1, 91, 26, 122, 10, 170, 50, 260, 1, 290, 91, 362, 26, 500, 122, 530, 10, 651, 170, 820, 50, 842, 260, 962, 1, 1220, 290, 1300, 91, 1370, 362, 1700, 26, 1682, 500, 1850, 122, 2366, 530, 2210, 10, 2451, 651, 2900, 170, 2810, 820, 3172, 50, 3620, 842, 3482
Offset: 1

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Author

Keywords

Comments

Denoted by Delta_2(n) in Glaisher 1907. - Michael Somos, May 17 2013
The sum of squares of even divisors of 2*k = 4*A001157(k), and the sum of squares of even divisors of 2*k-1 vanishes, for k >= 1. - Wolfdieter Lang, Jan 07 2017

Examples

			x + x^2 + 10*x^3 + x^4 + 26*x^5 + 10*x^6 + 50*x^7 + x^8 + 91*x^9 + 26*x^10 + ...
		

References

  • J. W. L. Glaisher, On the representations of a number as the sum of two, four, six, eight, ten, and twelve squares, Quart. J. Math. 38 (1907), 1-62 (see p. 4).

Crossrefs

Programs

  • Haskell
    a050999 = sum . map (^ 2) . a182469_row
    -- Reinhard Zumkeller, May 01 2012
    
  • Mathematica
    a[n_] := 1/2*Sum[(1 - (-1)^d)*d^2, {d, Divisors[n]}]; Table[a[n], {n, 1, 59}] (* Jean-François Alcover, Oct 23 2012, from 2nd formula *)
    a[ n_] := If[ n < 1, 0, Sum[ Mod[ d, 2] d^2, {d, Divisors@n}]] (* Michael Somos, May 17 2013 *)
    f[p_, e_] := If[p == 2, 1, (p^(2*e + 2) - 1)/(p^2 - 1)]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Nov 22 2020 *)
    Table[Total[Select[Divisors[n],OddQ]^2],{n,80}] (* Harvey P. Dale, Jul 19 2024 *)
  • PARI
    a(n)=sumdiv(n,d, if(d%2==1, d^2, 0 ) );  /* Joerg Arndt, Oct 07 2012 */
    
  • Python
    from sympy import divisor_sigma
    def A050999(n): return int(divisor_sigma(n>>(~n&n-1).bit_length(),2)) # Chai Wah Wu, Jul 16 2022

Formula

From Vladeta Jovovic, Sep 10 2001: (Start)
Multiplicative with a(p^e) = 1 if p = 2, (p^(2e+2)-1)/(p^2-1) if p > 2.
a(n) = (1/2)*Sum_{d|n} (1-(-1)^d)*d^2.
a(2n) = sigma_2(2n) - 4*sigma_2(n), a(2n+1) = sigma_2(2n+1), where sigma_2(n) is sum of squares of divisors of n (A001157).
More generally, if b(n, k) is the sum of k-th powers of odd divisors of n then b(2n, k) = sigma_k(2n)-2^k*sigma_k(n), b(2n+1, k) = sigma_k(2n+1). b(n, k) is multiplicative with a(p^e) = 1 if p = 2, (p^(k*e+k)-1)/(p^k-1) if p > 2. (End)
G.f. for b(n, k): Sum_{m>0} m^k*x^m*(1-(2^k-1)*x^m)/(1-x^(2*m)). - Vladeta Jovovic, Oct 19 2002
Dirichlet g.f. (1-2^(2-s))*zeta(s)*zeta(s-2). - R. J. Mathar, Apr 06 2011
Dirichlet convolution of A001157 with [1,-4,0,0,0,0...]. Dirichlet convolution of [1,-3,1,-3,1,-3,..] with A000290. Dirichlet convolution of [1,0,9,0,25,0,49,0,81,...] with A000012 (or A057427). - R. J. Mathar, Jun 28 2011
a(n) = sum(A182469(n,k)^2: k=1..A001227(n)). [Reinhard Zumkeller, May 01 2012]
Sum_{k=1..n} a(k) ~ zeta(3) * n^3 / 6. - Vaclav Kotesovec, Nov 09 2018
G.f.: Sum_{n >= 1} x^n*(1 + 6*x^(2*n) + x^(4*n))/(1 - x^(2*n))^3. - Peter Bala, Dec 19 2021
Sum_{k=1..n} (-1)^(k+1) * a(k) ~ zeta(3) * n^3 / 8. - Vaclav Kotesovec, Aug 07 2022

A352038 Sum of the 10th powers of the odd proper divisors of n.

Original entry on oeis.org

0, 1, 1, 1, 1, 59050, 1, 1, 59050, 9765626, 1, 59050, 1, 282475250, 9824675, 1, 1, 3486843451, 1, 9765626, 282534299, 25937424602, 1, 59050, 9765626, 137858491850, 3486843451, 282475250, 1, 576660215300, 1, 1, 25937483651, 2015993900450, 292240875, 3486843451
Offset: 1

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Author

Wesley Ivan Hurt, Mar 01 2022

Keywords

Examples

			a(10) = 9765626; a(10) = Sum_{d|10, d<10, d odd} d^10 = 1^10 + 5^10 = 9765626.
		

Crossrefs

Sum of the k-th powers of the odd proper divisors of n for k=0..10: A091954 (k=0), A091570 (k=1), A351647 (k=2), A352031 (k=3), A352032 (k=4), A352033 (k=5), A352034 (k=6), A352035 (k=7), A352036 (k=8), A352037 (k=9), this sequence (k=10).

Programs

  • Mathematica
    f[2, e_] := 1; f[p_, e_] := (p^(10*e+10) - 1)/(p^10 - 1); a[1] = 0; a[n_] := Times @@ f @@@ FactorInteger[n] - If[OddQ[n], n^10, 0]; Array[a, 60] (* Amiram Eldar, Oct 11 2023 *)
  • Python
    from math import prod
    from sympy import factorint
    def A352038(n): return prod((p**(10*(e+1))-1)//(p**10-1) for p, e in factorint(n).items() if p > 2) - (n**10 if n % 2 else 0) # Chai Wah Wu, Mar 01 2022

Formula

a(n) = Sum_{d|n, d
G.f.: Sum_{k>=1} (2*k-1)^10 * x^(4*k-2) / (1 - x^(2*k-1)). - Ilya Gutkovskiy, Mar 02 2022
From Amiram Eldar, Oct 11 2023: (Start)
a(n) = A321814(n) - n^10*A000035(n).
Sum_{k=1..n} a(k) ~ c * n^11, where c = (zeta(11)-1)/22 = 0.0000224631... . (End)

A352056 Sum of the 10th powers of the divisor complements of the odd proper divisors of n.

Original entry on oeis.org

0, 1024, 59049, 1048576, 9765625, 60467200, 282475249, 1073741824, 3486843450, 10000001024, 25937424601, 61918412800, 137858491849, 289254656000, 576660215299, 1099511627776, 2015993900449, 3570527693824, 6131066257801, 10240001048576, 16680163512499
Offset: 1

Author

Wesley Ivan Hurt, Mar 01 2022

Keywords

Examples

			a(10) = 10^10 * Sum_{d|10, d<10, d odd} 1 / d^10 = 10^10 * (1/1^10 + 1/5^10) = 10000001024.
		

Crossrefs

Sum of the k-th powers of the divisor complements of the odd proper divisors of n for k=0..10: A091954 (k=0), A352047 (k=1), A352048 (k=2), A352049 (k=3), A352050 (k=4), A352051 (k=5), A352052 (k=6), A352053 (k=7), A352054 (k=8), A352055 (k=9), this sequence (k=10).

Programs

  • Mathematica
    A352056[n_]:=DivisorSum[n,1/#^10&,#A352056,50] (* Paolo Xausa, Aug 10 2023 *)
    a[n_] := DivisorSigma[-10, n/2^IntegerExponent[n, 2]] * n^10 - Mod[n, 2]; Array[a, 100] (* Amiram Eldar, Oct 13 2023 *)
  • PARI
    a(n) = n^10 * sigma(n >> valuation(n, 2), -10) - n % 2; \\ Amiram Eldar, Oct 13 2023

Formula

a(n) = n^10 * Sum_{d|n, d
G.f.: Sum_{k>=2} k^10 * x^k / (1 - x^(2*k)). - Ilya Gutkovskiy, May 19 2023
From Amiram Eldar, Oct 13 2023: (Start)
a(n) = A321814(n) * A006519(n)^10 - A000035(n).
Sum_{k=1..n} a(k) = c * n^11 / 11, where c = 2047*zeta(11)/2048 = 1.00000566605... . (End)
Showing 1-3 of 3 results.