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 1-7 of 7 results.

A050449 a(n) = Sum_{d|n, d == 1 (mod 4)} d.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 1, 10, 6, 1, 1, 14, 1, 6, 1, 18, 10, 1, 6, 22, 1, 1, 1, 31, 14, 10, 1, 30, 6, 1, 1, 34, 18, 6, 10, 38, 1, 14, 6, 42, 22, 1, 1, 60, 1, 1, 1, 50, 31, 18, 14, 54, 10, 6, 1, 58, 30, 1, 6, 62, 1, 31, 1, 84, 34, 1, 18, 70, 6, 1, 10, 74, 38, 31, 1
Offset: 1

Views

Author

N. J. A. Sloane, Dec 23 1999

Keywords

Comments

Not multiplicative: a(3)*a(7) != a(21), for example. - R. J. Mathar, Dec 20 2011

Crossrefs

Cf. Sum_{d|n, d==1 (mod k)} d: A000593 (k=2), A078181 (k=3), this sequence (k=4), A284097 (k=5), A284098 (k=6), A284099 (k=7), A284100 (k=8).

Programs

  • Maple
    A050449 := proc(n)
            a := 0 ;
            for d in numtheory[divisors](n) do
                    if d mod 4 = 1 then
                            a := a+d ;
                    end if;
            end do:
            a;
    end proc:
    seq(A050449(n),n=1..40) ; # R. J. Mathar, Dec 20 2011
  • Mathematica
    a[n_] := DivisorSum[n, Boole[Mod[#, 4] == 1]*#&]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Jan 30 2018 *)
  • PARI
    a(n) = sumdiv(n, d, d*((d % 4) == 1)); \\ Michel Marcus, Jan 30 2018

Formula

G.f.: Sum_{n>=0} (4*n+1)*x^(4*n+1)/(1-x^(4*n+1)). - Vladeta Jovovic, Nov 14 2002
a(n) = A000593(n) - A050452(n). - Reinhard Zumkeller, Apr 18 2006
G.f.: Sum_{n >= 1} x^n*(1 + 3*x^(4*n))/(1 - x^(4*n))^2. - Peter Bala, Dec 19 2021
Sum_{k=1..n} a(k) = c * n^2 + O(n*log(n)), where c = Pi^2/48 = 0.205616... (A245058). - Amiram Eldar, Nov 26 2023

Extensions

More terms from Vladeta Jovovic, Nov 14 2002
More terms from Reinhard Zumkeller, Apr 18 2006

A078181 a(n) = Sum_{d|n, d == 1 (mod 3)} d.

Original entry on oeis.org

1, 1, 1, 5, 1, 1, 8, 5, 1, 11, 1, 5, 14, 8, 1, 21, 1, 1, 20, 15, 8, 23, 1, 5, 26, 14, 1, 40, 1, 11, 32, 21, 1, 35, 8, 5, 38, 20, 14, 55, 1, 8, 44, 27, 1, 47, 1, 21, 57, 36, 1, 70, 1, 1, 56, 40, 20, 59, 1, 15, 62, 32, 8, 85, 14, 23, 68, 39, 1, 88, 1, 5, 74, 38, 26, 100, 8, 14, 80, 71, 1
Offset: 1

Views

Author

Vladeta Jovovic, Nov 21 2002

Keywords

Crossrefs

Cf. Sum_{d|n, d==1 mod k} d: A000593 (k=2), this sequence (k=3), A050449 (k=4), A284097 (k=5), A284098 (k=6), A284099 (k=7), A284100 (k=8).

Programs

  • Maple
    A078181 := proc(n)
        a := 0 ;
        for d in numtheory[divisors](n) do
            if modp(d,3) =1 then
                a :=a+d ;
            end if;
        end do:
        a;
    end proc: # R. J. Mathar, May 11 2016
  • Mathematica
    a[n_] := Plus @@ Select[Divisors[n], Mod[#, 3] == 1 &]; Array[a, 100] (* Giovanni Resta, May 11 2016 *)

Formula

G.f.: Sum_{n>=0} (3*n+1)*x^(3*n+1)/(1-x^(3*n+1)).
G.f.: -q*P'/P where P = Product_{n>=0} (1 - q^(3*n+1)). - Joerg Arndt, Aug 03 2011
Conjecture. If a(n)=n+1 then n==1 (mod 3). (Is this easy to settle? It has been verified for n=1,2,3,...,2000.) - John W. Layman, Apr 03 2006
The conjecture is false. The first and only counterexample below 10^8 is a(6800) = 6801 and 6800 == 2 (mod 3). - Lambert Herrgesell (zero815(AT)googlemail.com), May 06 2008
Equals A051731 * [1, 0, 0, 4, 0, 0, 7, 0, 0, 10, ...]. - Gary W. Adamson, Nov 06 2007
A272027(n/3) + a(n) + A078182(n) = A000203(n). - R. J. Mathar, May 25 2020
G.f.: Sum_{n >= 1} x^n*(1 + 2*x^(3*n))/(1 - x^(3*n))^2. - Peter Bala, Dec 19 2021
Sum_{k=1..n} a(k) = c * n^2 + O(n*log(n)), where c = Pi^2/36 = 0.274155... (A353908). - Amiram Eldar, Nov 26 2023

A284097 a(n) = Sum_{d|n, d == 1 (mod 5)} d.

Original entry on oeis.org

1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 12, 7, 1, 1, 1, 17, 1, 7, 1, 1, 22, 12, 1, 7, 1, 27, 1, 1, 1, 7, 32, 17, 12, 1, 1, 43, 1, 1, 1, 1, 42, 28, 1, 12, 1, 47, 1, 23, 1, 1, 52, 27, 1, 7, 12, 57, 1, 1, 1, 7, 62, 32, 22, 17, 1, 84, 1, 1, 1, 1, 72, 43, 1, 1, 1, 77, 12, 33, 1
Offset: 1

Views

Author

Seiichi Manyama, Mar 20 2017

Keywords

Crossrefs

Cf. Sum_{d|n, d=1 mod k} d: A000593 (k=2), A078181 (k=3), A050449 (k=4), this sequence (k=5), A284098 (k=6), A284099 (k=7), A284100 (k=8).

Programs

  • Mathematica
    Table[Sum[If[Mod[d, 5] == 1, d, 0], {d, Divisors[n]}], {n, 80}] (* Indranil Ghosh, Mar 21 2017 *)
  • PARI
    for(n=1, 82, print1(sumdiv(n, d, if(Mod(d, 5)==1, d, 0)), ", ")) \\ Indranil Ghosh, Mar 21 2017
    
  • Python
    from sympy import divisors
    def a(n): return sum([d for d in divisors(n) if d%5==1]) # Indranil Ghosh, Mar 21 2017

Formula

G.f.: Sum_{k>=0} (5*k + 1)*x^(5*k+1)/(1 - x^(5*k+1)). - Ilya Gutkovskiy, Mar 21 2017
Sum_{k=1..n} a(k) = c * n^2 + O(n*log(n)), where c = Pi^2/60 = 0.164493... (A013661 / 10). - Amiram Eldar, Nov 26 2023

A284099 a(n) = Sum_{d|n, d == 1 (mod 7)} d.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 16, 9, 1, 1, 1, 1, 1, 23, 1, 9, 1, 1, 1, 1, 30, 16, 1, 9, 1, 1, 1, 37, 1, 1, 1, 9, 1, 1, 44, 23, 16, 1, 1, 9, 1, 51, 1, 1, 1, 1, 1, 9, 58, 30, 1, 16, 1, 1, 1, 73, 1, 23, 1, 1, 1, 1, 72, 45, 1, 1, 16, 1, 1, 79, 1, 9, 1, 1
Offset: 1

Views

Author

Seiichi Manyama, Mar 20 2017

Keywords

Crossrefs

Cf. A109703.
Cf. Sum_{d|n, d == 1 (mod k)} d: A000593 (k=2), A078181 (k=3), A050449 (k=4), A284097 (k=5), A284098 (k=6), this sequence (k=7), A284100 (k=8).
Cf. Sum_{d|n, d == k (mod 7)} d: this sequence (k=1), A284443 (k=2), A284444 (k=3), A284445 (k=4), A284446 (k=5), A284105 (k=6).

Programs

  • Mathematica
    Table[Sum[If[Mod[d, 7] == 1, d, 0], {d, Divisors[n]}], {n, 82}] (* Indranil Ghosh, Mar 21 2017 *)
    Table[DivisorSum[n,#&,Mod[#,7]==1&],{n,90}] (* Harvey P. Dale, Aug 08 2021 *)
  • PARI
    for(n=1, 82, print1(sumdiv(n, d, if(Mod(d, 7)==1, d, 0)), ", ")) \\ Indranil Ghosh, Mar 21 2017
    
  • Python
    from sympy import divisors
    def a(n): return sum([d for d in divisors(n) if d%7==1]) # Indranil Ghosh, Mar 21 2017

Formula

G.f.: Sum_{k>=0} (7*k + 1)*x^(7*k+1)/(1 - x^(7*k+1)). - Ilya Gutkovskiy, Mar 21 2017
G.f.: Sum_{n >= 1} x^n*(1 + 6*x^(7*n))/(1 - x^(7*n))^2. - Peter Bala, Dec 19 2021
Sum_{k=1..n} a(k) = c * n^2 + O(n*log(n)), where c = Pi^2/84 = 0.117495... . - Amiram Eldar, Nov 26 2023

A284098 a(n) = Sum_{d|n, d == 1 (mod 6)} d.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 14, 8, 1, 1, 1, 1, 20, 1, 8, 1, 1, 1, 26, 14, 1, 8, 1, 1, 32, 1, 1, 1, 8, 1, 38, 20, 14, 1, 1, 8, 44, 1, 1, 1, 1, 1, 57, 26, 1, 14, 1, 1, 56, 8, 20, 1, 1, 1, 62, 32, 8, 1, 14, 1, 68, 1, 1, 8, 1, 1, 74, 38, 26, 20, 8, 14, 80, 1, 1
Offset: 1

Views

Author

Seiichi Manyama, Mar 20 2017

Keywords

Crossrefs

Cf. Sum_{d|n, d==1 (mod k)} d: A000593 (k=2), A078181 (k=3), A050449 (k=4), A284097 (k=5), this sequence (k=6), A284099 (k=7), A284100 (k=8).

Programs

  • Mathematica
    Table[Sum[If[Mod[d, 6] == 1, d, 0], {d, Divisors[n]}], {n, 80}] (* Indranil Ghosh, Mar 21 2017 *)
  • PARI
    for(n=1, 82, print1(sumdiv(n, d, if(Mod(d, 6)==1, d, 0)), ", ")) \\ Indranil Ghosh, Mar 21 2017
    
  • Python
    from sympy import divisors
    def a(n): return sum([d for d in divisors(n) if d%6==1]) # Indranil Ghosh, Mar 21 2017

Formula

G.f.: Sum_{k>=0} (6*k + 1)*x^(6*k+1)/(1 - x^(6*k+1)). - Ilya Gutkovskiy, Mar 21 2017
G.f.: Sum_{n >= 1} x^n*(1 + 5*x^(6*n))/(1 - x^(6*n))^2. - Peter Bala, Dec 19 2021
Sum_{k=1..n} a(k) = c * n^2 + O(n*log(n)), where c = Pi^2/72 = 0.137077... (A086729). - Amiram Eldar, Nov 26 2023

A277090 Expansion of Product_{k>=0} 1/(1 - x^(8*k+1)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 4, 4, 4, 4, 4, 4, 4, 5, 6, 7, 7, 7, 7, 7, 7, 8, 10, 11, 12, 12, 12, 12, 12, 13, 15, 17, 18, 19, 19, 19, 19, 20, 23, 26, 28, 29, 30, 30, 30, 31, 34, 38, 41, 43, 44, 45, 45, 46, 50, 55, 60, 63, 65, 66, 67, 68, 72, 79, 85, 90, 93, 95, 96, 98, 103, 111, 120, 127, 132, 135, 137, 139, 145
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 29 2016

Keywords

Comments

Number of partitions of n into parts congruent to 1 mod 8.
More generally, the ordinary generating function for the number of partitions of n into parts congruent to 1 mod m (for m>0) is Product_{k>=0} 1/(1 - x^(m*k+1)).

Examples

			a(10) = 2, because we have [9, 1] and [1, 1, 1, 1, 1, 1, 1, 1, 1, 1].
		

Crossrefs

Cf. similar sequences of number of partitions of n into parts congruent to 1 mod m: A000009 (m=2), A035382 (m=3), A035451 (m=4), A109697 (m=5), A109701 (m=6), A109703 (m=7).

Programs

  • Mathematica
    CoefficientList[Series[QPochhammer[x, x^8]^(-1), {x, 0, 90}], x]

Formula

G.f.: Product_{k>=0} 1/(1 - x^(8*k+1)).
a(n) ~ exp((Pi*sqrt(n))/(2*sqrt(3)))*Gamma(1/8)/(4*3^(1/16)*(2*Pi)^(7/8)*n^(9/16)).
a(n) = (1/n)*Sum_{k=1..n} A284100(k)*a(n-k), a(0) = 1. - Seiichi Manyama, Mar 20 2017

A357912 a(n) = Sum_{d|n, d==1 (mod 11)} d.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 24, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 35, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 46, 24, 1, 13, 1, 1, 1, 1, 1, 1, 1, 57, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 68, 35, 24, 1, 1, 13, 1, 1, 1, 1, 1, 79, 1, 1, 1, 1, 1, 13, 1
Offset: 1

Views

Author

Seiichi Manyama, Jan 17 2023

Keywords

Crossrefs

Cf. Sum_{d|n, d==1 (mod k)} d: A000593 (k=2), A078181 (k=3), A050449 (k=4), A284097 (k=5), A284098 (k=6), A284099 (k=7), A284100 (k=8), this sequence (k=11).
Cf. A357911.

Programs

  • Mathematica
    a[n_] := DivisorSum[n, # &, Mod[#, 11] == 1 &]; Array[a, 100] (* Amiram Eldar, Aug 09 2023 *)
  • PARI
    a(n) = sumdiv(n, d, (Mod(d, 11)==1)*d);
    
  • PARI
    my(N=100, x='x+O('x^N)); Vec(sum(k=0, N, (11*k+1)*x^(11*k+1)/(1-x^(11*k+1))))

Formula

G.f.: Sum_{k>=0} (11*k+1) * x^(11*k+1)/(1 - x^(11*k+1)).
Showing 1-7 of 7 results.