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 1-10 of 15 results. Next

A276634 Sum of cubes of proper divisors of n.

Original entry on oeis.org

0, 1, 1, 9, 1, 36, 1, 73, 28, 134, 1, 316, 1, 352, 153, 585, 1, 981, 1, 1198, 371, 1340, 1, 2556, 126, 2206, 757, 3160, 1, 4752, 1, 4681, 1359, 4922, 469, 8605, 1, 6868, 2225, 9710, 1, 12600, 1, 12052, 4257, 12176, 1, 20476, 344, 16759, 4941, 19846, 1, 26496, 1457, 25624, 6887, 24398, 1
Offset: 1

Views

Author

Ilya Gutkovskiy, Sep 08 2016

Keywords

Comments

More generally, the Dirichlet generating function for the sum of k-th powers of proper divisors of n is zeta(s-k)*(zeta(s) - 1).

Examples

			a(10) = 1^3 + 2^3 + 5^3 = 134, because 10 has 3 proper divisors {1,2,5}.
a(11) = 1^3 = 1, because 11 has 1 proper divisor {1}.
		

Crossrefs

Programs

  • Magma
    [DivisorSigma(3, n) - n^3: n in [1..70]]; // Vincenzo Librandi, Sep 09 2016
  • Mathematica
    Table[DivisorSigma[3, n] - n^3, {n, 70}]
  • PARI
    a(n) = sigma(n, 3) - n^3; \\ Michel Marcus, Sep 08 2016
    

Formula

a(n) = 1 if n is prime.
a(p^k) = (p^(3*k) - 1)/(p^3 - 1) for p prime.
Dirichlet g.f.: zeta(s-3)*(zeta(s) - 1).
a(n) = A001158(n) - A000578(n).
A000035(a(n)) = A053867(n).
Sum_{n=1..k} a(n) ~ k^2*(Pi^4*k^2/90 - (k + 1)^2)/4.
G.f.: -x*(1 + 4*x + x^2)/(1 - x)^4 + Sum_{k>=1} k^3*x^k/(1 - x^k). - Ilya Gutkovskiy, Mar 17 2017

A279363 Sum of 4th powers of proper divisors of n.

Original entry on oeis.org

0, 1, 1, 17, 1, 98, 1, 273, 82, 642, 1, 1650, 1, 2418, 707, 4369, 1, 7955, 1, 10898, 2483, 14658, 1, 26482, 626, 28578, 6643, 41090, 1, 62644, 1, 69905, 14723, 83538, 3027, 133923, 1, 130338, 28643, 174994, 1, 236692, 1, 249170, 57893, 279858, 1, 423794, 2402, 401267, 83603, 485810, 1, 644372, 15267, 659842, 130403, 707298, 1, 1053636
Offset: 1

Views

Author

Ilya Gutkovskiy, Dec 10 2016

Keywords

Examples

			a(10) = 1^4 + 2^4 + 5^4 = 642, because 10 has 3 proper divisors {1,2,5}.
a(11) = 1^4 = 1, because 11 has 1 proper divisor {1}.
		

Crossrefs

Programs

  • Mathematica
    Table[DivisorSigma[4, n] - n^4, {n, 60}]
  • PARI
    for(n=1, 60, print1(sigma(n, 4) - n^4,", ")) \\ Indranil Ghosh, Mar 18 2017
    
  • Python
    from sympy.ntheory import divisor_sigma
    print([divisor_sigma(n,4) - n**4 for n in range(1,61)]) # Indranil Ghosh, Mar 18 2017

Formula

a(n) = 1 if n is prime.
a(p^k) = (p^(4*k) - 1)/(p^4 - 1) when p is prime.
Dirichlet g.f.: zeta(s-4)*(zeta(s) - 1).
a(n) = A001159(n) - A000583(n).
G.f.: -x*(1 + 11*x + 11*x^2 + x^3)/(1 - x)^5 + Sum_{k>=1} k^4 x^k/(1 - x^k). - Ilya Gutkovskiy, Mar 18 2017
Sum_{k=1..n} a(k) ~ (Zeta(5) - 1)*n^5 / 5. - Vaclav Kotesovec, Feb 02 2019

A279847 a(n) = Sum_{k=1..n} k^2*(floor(n/k) - 1).

Original entry on oeis.org

0, 1, 2, 7, 8, 22, 23, 44, 54, 84, 85, 151, 152, 206, 241, 326, 327, 458, 459, 605, 664, 790, 791, 1065, 1091, 1265, 1356, 1622, 1623, 2023, 2024, 2365, 2496, 2790, 2865, 3480, 3481, 3847, 4026, 4636, 4637, 5373, 5374, 6000, 6341, 6875, 6876, 7982, 8032, 8787, 9086, 9952, 9953, 11137, 11284
Offset: 1

Views

Author

Ilya Gutkovskiy, Dec 20 2016

Keywords

Comments

Sum of all squares of proper divisors of all positive integers <= n.
Total volume of all rectangular prisms with dimensions (x, x, z) and integers x and y, such that x + y = n, 0 < x <= y, and z = floor(y/x). - Wesley Ivan Hurt, Dec 21 2020

Examples

			For n = 7 the proper divisors of the first seven positive integers are {0}, {1}, {1}, {1, 2}, {1}, {1, 2, 3}, {1} so a(7) = 0^2 + 1^2 + 1^2 + 1^2 + 2^2 + 1 ^2 + 1^2 + 2^2 + 3^2 + 1^2 = 23.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[k^2 (Floor[n/k] - 1), {k, 1, n}], {n, 55}]
    Table[Sum[DivisorSigma[2, k] - k^2, {k, 1, n}], {n, 55}]
  • PARI
    a(n) = sum(k=1, n, k^2*(floor(n/k)-1)) \\ Felix Fröhlich, Dec 20 2016
    
  • Python
    from math import isqrt
    def A279847(n): return (-n*(n+1)*(2*n+1)-(s:=isqrt(n))**2*(s+1)*(2*s+1) + sum((q:=n//k)*(6*k**2+q*(2*q+3)+1) for k in range(1,s+1)))//6 # Chai Wah Wu, Oct 21 2023

Formula

G.f.: -x*(1 + x)/(1 - x)^4 + (1/(1 - x))*Sum_{k>=1} k^2*x^k/(1 - x^k).
a(n) = A064602(n) - A000330(n).
a(n) = Sum_{k=1..n} A067558(k).
a(n) = Sum_{k=1..n} (A001157(k) - A000290(k)).
a(p^k) = a(p^k-1) + (p^(2*k) - 1)/(p^2 - 1), when p is prime.
a(n) ~ ((zeta(3) - 1)/3)*n^3.
a(n) = Sum_{k=1..floor(n/2)} k^2 * floor((n-k)/k). - Wesley Ivan Hurt, Dec 21 2020

A321258 Square array A(n,k), n >= 1, k >= 0, read by antidiagonals: A(n,k) = sigma_k(n) - n^k.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 1, 2, 0, 1, 1, 3, 1, 0, 1, 1, 5, 1, 3, 0, 1, 1, 9, 1, 6, 1, 0, 1, 1, 17, 1, 14, 1, 3, 0, 1, 1, 33, 1, 36, 1, 7, 2, 0, 1, 1, 65, 1, 98, 1, 21, 4, 3, 0, 1, 1, 129, 1, 276, 1, 73, 10, 8, 1, 0, 1, 1, 257, 1, 794, 1, 273, 28, 30, 1, 5
Offset: 1

Views

Author

Ilya Gutkovskiy, Nov 01 2018

Keywords

Comments

A(n,k) is the sum of k-th powers of proper divisors of n.

Examples

			Square array begins:
  0,  0,   0,   0,   0,    0,  ...
  1,  1,   1,   1,   1,    1,  ...
  1,  1,   1,   1,   1,    1,  ...
  2,  3,   5,   9,  17,   33,  ...
  1,  1,   1,   1,   1,    1,  ...
  3,  6,  14,  36,  98,  276,  ...
		

Crossrefs

Columns k=0..5 give A032741, A001065, A067558, A276634, A279363, A279364.
Cf. A109974, A285425, A286880, A321259 (diagonal).

Programs

  • Mathematica
    Table[Function[k, DivisorSigma[k, n] - n^k][i - n], {i, 0, 12}, {n, 1, i}] // Flatten
    Table[Function[k, SeriesCoefficient[Sum[j^k x^(2 j)/(1 - x^j), {j, 1, n}], {x, 0, n}]][i - n], {i, 0, 12}, {n, 1, i}] // Flatten

Formula

G.f. of column k: Sum_{j>=1} j^k*x^(2*j)/(1 - x^j).
Dirichlet g.f. of column k: zeta(s-k)*(zeta(s) - 1).
A(n,k) = 1 if n is prime.

A339353 G.f.: Sum_{k>=1} k^2 * x^(k*(k + 1)) / (1 - x^k).

Original entry on oeis.org

0, 1, 1, 1, 1, 5, 1, 5, 1, 5, 1, 14, 1, 5, 10, 5, 1, 14, 1, 21, 10, 5, 1, 30, 1, 5, 10, 21, 1, 39, 1, 21, 10, 5, 26, 30, 1, 5, 10, 46, 1, 50, 1, 21, 35, 5, 1, 66, 1, 30, 10, 21, 1, 50, 26, 70, 10, 5, 1, 91, 1, 5, 59, 21, 26, 50, 1, 21, 10, 79, 1, 130, 1, 5, 35, 21, 50, 50, 1, 110
Offset: 1

Views

Author

Ilya Gutkovskiy, Dec 01 2020

Keywords

Comments

Sum of squares of divisors of n that are smaller than sqrt(n).

Crossrefs

Programs

  • Mathematica
    nmax = 80; CoefficientList[Series[Sum[k^2 x^(k (k + 1))/(1 - x^k), {k, 1, nmax}], {x, 0, nmax}], x] // Rest
    Table[DivisorSum[n, #^2 &, # < Sqrt[n] &], {n, 80}]
  • PARI
    a(n) = sumdiv(n, d, if (d^2 < n, d^2)); \\ Michel Marcus, Dec 02 2020

A140362 Semiprimes pq that divide the sum of the squares of their divisors, 1+p^2+q^2+(pq)^2.

Original entry on oeis.org

10, 65, 20737
Offset: 1

Views

Author

Mohamed Bouhamida, Jul 22 2008, Jul 27 2008

Keywords

Comments

6 is the smallest integer n which is the product of two distinct primes and which divides the sum of the cubes of the divisors of n. Are there other numbers with this property?
Using Pell equations and a Fibonacci identity, Max Alekseyev and I have shown that all terms are the product of prime Fibonacci numbers whose indices are twin primes. The first three terms are Fib(3)*Fib(5), Fib(5)*Fib(7) and Fib(11)*Fib(13). The other two known terms are Fib(431)*Fib(433) and Fib(569)*Fib(571), huge numbers that are in the b-file. The sequence probably has no additional terms. - T. D. Noe, Jul 27 2008
Let a, b, c and d be consecutive odd-indexed Fibonacci numbers. Then it can be proved that 1 + b^2 + c^2 + (bc)^2 = abcd, which shows that bc divides 1 + b^2 + c^2 + (bc)^2. Hence if b and c are prime, then bc is in this sequence. - T. D. Noe, Jul 27 2008
Empirical search suggests that A067558(a(n))/A032741(a(n)) = a(n). A032741(a(n)) = 3 for all n by definition of semiprime. A067558(a(n)) must also then be divisible by 3. a(n) can be called the n-th "perfect mean square aliquot number". - William Krier, Dec 16 2024

Examples

			10 divides (1^2 + 2^2 + 5^2) giving 3 - the number of proper divisors of semiprime 10.
65 divides (1^2 + 5^2 + 13^2) giving 3 - the number of proper divisors of semiprime 65.
20737 divides (1^2 + 89^2 + 233^2) giving 3 - the number of proper divisors of semiprime 20737.
		

Crossrefs

Programs

  • PARI
    isok(n) = sigma(n, 2) - n^2 == 3*n; \\ Michel Marcus, Jun 24 2014

A279364 Sum of 5th powers of proper divisors of n.

Original entry on oeis.org

0, 1, 1, 33, 1, 276, 1, 1057, 244, 3158, 1, 9076, 1, 16840, 3369, 33825, 1, 67101, 1, 104182, 17051, 161084, 1, 290676, 3126, 371326, 59293, 555688, 1, 870552, 1, 1082401, 161295, 1419890, 19933, 2206525, 1, 2476132, 371537, 3336950, 1, 4646784, 1, 5315740, 821793, 6436376, 1, 9301876, 16808, 9868783
Offset: 1

Views

Author

Ilya Gutkovskiy, Dec 10 2016

Keywords

Examples

			a(10) = 1^5 + 2^5 + 5^5 = 3158, because 10 has 3 proper divisors {1,2,5}.
a(11) = 1^5 = 1, because 11 has 1 proper divisor {1}.
		

Crossrefs

Programs

  • Mathematica
    Table[DivisorSigma[5, n] - n^5, {n, 50}]
  • PARI
    for(n=1, 50, print1(sigma(n, 5) - n^5,", ")) \\ Indranil Ghosh, Mar 18 2017
    
  • Python
    from sympy.ntheory import divisor_sigma
    print([divisor_sigma(n,5) - n**5 for n in range(1,51)]) # Indranil Ghosh, Mar 18 2017

Formula

a(n) = 1 if n is prime.
a(p^k) = (p^(5*k) - 1)/(p^5 - 1) when p is prime.
Dirichlet g.f.: zeta(s-5)*(zeta(s) - 1).
a(n) = A001160(n) - A000584(n).
G.f.: -x*(1 + 26*x + 66*x^2 + 26*x^3 + x^4)/(1 - x)^6 + Sum_{k>=1} k^5 x^k/(1 - x^k). - Ilya Gutkovskiy, Mar 18 2017
Sum_{k=1..n} a(k) ~ (zeta(6) - 1) * n^6 / 6. - Amiram Eldar, Jan 11 2025

A321259 a(n) = sigma_n(n) - n^n.

Original entry on oeis.org

0, 1, 1, 17, 1, 794, 1, 65793, 19684, 9766650, 1, 2194095090, 1, 678223089234, 30531927033, 281479271743489, 1, 150196195641350171, 1, 100000096466944316978, 558545874543637211, 81402749386839765307626, 1, 79501574308536809523296482, 298023223876953126
Offset: 1

Views

Author

Ilya Gutkovskiy, Nov 01 2018

Keywords

Comments

a(n) is the sum of n-th powers of proper divisors of n.

Crossrefs

Programs

  • Magma
    [DivisorSigma(n, n) - n^n: n in [1..30]]; // Vincenzo Librandi, Nov 02 2018
    
  • Mathematica
    Table[DivisorSigma[n, n] - n^n, {n, 25}]
    nmax = 25; Rest[CoefficientList[Series[Sum[(k x)^(2 k)/(1 - (k x)^k), {k, 1, nmax}], {x, 0, nmax}], x]]
  • PARI
    a(n) = sigma(n, n) - n^n; \\ Michel Marcus, Nov 02 2018

Formula

G.f.: Sum_{k>=1} (k*x)^(2*k)/(1 - (k*x)^k).
a(n) = A023887(n) - A000312(n).
a(n) = A321258(n,n).
a(n) = 1 if n is prime.

A177404 Numbers n such that n^2 minus (sum of the squares of the proper divisors of n) is a prime number.

Original entry on oeis.org

2, 4, 8, 9, 18, 25, 32, 49, 64, 72, 81, 100, 121, 162, 242, 256, 289, 361, 392, 400, 484, 512, 529, 576, 648, 729, 784, 1058, 1296, 1352, 1444, 1681, 1849, 1922, 2048, 2116, 2500, 3136, 3872, 4418, 4608, 4624, 5184, 5776, 5832, 6889, 7225, 7396, 7744
Offset: 1

Views

Author

Claudio Meller, Dec 10 2010

Keywords

Comments

Sum of squares of proper divisors of n is given by A067558.

Examples

			9 has proper divisors 3 and 1; it is in the list because 9^2-(3^2+1^2) = 71 is a prime number.
		

Crossrefs

Cf. A067558.

Programs

  • Magma
    a067558:=func< n | n le 1 select 0 else &+[ D[k]^2: k in [1..#D-1] ] where D is Divisors(n) >; [ n: n in [1..10000] | IsPrime(n^2-a067558(n)) ];
  • Mathematica
    Select[Range[8000],PrimeQ[#^2-Total[Most[Divisors[#]]^2]]&] (* Harvey P. Dale, Dec 14 2011 *)

A201879 Numbers n such that sigma_2(n) - n^2 is a square.

Original entry on oeis.org

1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 30, 31, 37, 41, 43, 47, 53, 59, 61, 67, 70, 71, 73, 79, 83, 89, 97, 101, 102, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257
Offset: 1

Views

Author

Michel Lagneau, Dec 06 2011

Keywords

Comments

Numbers n such that sum of the square of proper (or aliquot) divisors of n is a square.
All primes are in this sequence. Nonprimes in the sequence are 1, 30, 70, 102, 282, 286, 646, 730, 920, 1242, ... - Charles R Greathouse IV, Dec 06 2011

Examples

			a(12)=30 because the aliquot divisors of 30 are  1, 2, 3, 5, 6, 10, 15, the sum of whose squares is 1^2 + 2^2 + 3^2 + 5^2 + 6^2 + 10^2 + 15^2 = 400 = 20^2.
		

Crossrefs

Programs

  • Maple
    A067558 := proc(n)
        numtheory[sigma][2](n)-n^2 ;
    end proc:
    isA201879 := proc(n)
        issqr(A067558(n)) ;
    end proc:
    for n from 1 to 300 do
        if isA201879(n) then
            printf("%d,",n);
        end if;
    end do: # R. J. Mathar, Dec 07 2011
  • Mathematica
    Select[Range[400], IntegerQ[Sqrt[DivisorSigma[2, #]-#^2]]&]
  • PARI
    is(n)=issquare(sigma(n,2)-n^2) \\ Charles R Greathouse IV, Dec 06 2011

Formula

{n: A067558(n) in A000290}. - R. J. Mathar, Dec 07 2011
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