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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A201880 Numbers n such that sigma_2(n) - n^2 is prime.

Original entry on oeis.org

4, 18, 21, 33, 39, 72, 93, 99, 100, 159, 171, 177, 189, 207, 213, 231, 245, 249, 261, 275, 291, 297, 303, 333, 338, 357, 369, 381, 399, 400, 453, 471, 475, 477, 484, 495, 537, 539, 543, 561, 609, 633, 648, 657, 669, 681, 711, 717, 783, 801, 833, 861, 909, 927
Offset: 1

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Author

Michel Lagneau, Dec 06 2011

Keywords

Comments

Numbers n such that sum of the squares of the proper (or aliquot) divisors of n is a prime number.

Examples

			a(3)=21 because the aliquot divisors of 21 are 1, 3, 7, the sum of whose squares is 1^2 + 3^2 + 7^2 = 59, prime.
		

Crossrefs

Programs

  • Maple
    A067558 := proc(n)
        numtheory[sigma][2](n)-n^2 ;
    end proc:
    isA201880 := proc(n)
        isprime(A067558(n)) ;
    end proc:
    for n from 1 to 1000 do
        if isA201880(n) then
            printf("%d,",n);
        end if;
    end do: # R. J. Mathar, Dec 07 2011
  • Mathematica
    Select[Range[400], PrimeQ[DivisorSigma[2, #]-#^2]&]
  • PARI
    is(n)=isprime(sigma(n,2)-n^2) \\ Charles R Greathouse IV, Dec 06 2011

Formula

{n: A067558(n) in A000040} - R. J. Mathar, Dec 07 2011

A279814 Numbers n such that the average of the squares of the proper divisors of n is an integer.

Original entry on oeis.org

2, 3, 5, 7, 8, 9, 10, 11, 13, 14, 17, 19, 22, 23, 25, 26, 29, 31, 34, 35, 37, 38, 41, 43, 46, 47, 49, 50, 53, 55, 58, 59, 61, 62, 65, 67, 71, 73, 74, 75, 77, 79, 80, 81, 82, 83, 85, 86, 89, 91, 94, 95, 97, 101, 103, 106, 107, 109, 113, 115, 118, 119, 121, 122, 125, 127, 131, 133, 134, 137, 139, 140, 142, 143, 145, 146, 149
Offset: 1

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Author

Ilya Gutkovskiy, Dec 19 2016

Keywords

Comments

Numbers n such that number of proper divisors of n (A032741) divides sum of squares of proper divisors of n (A067558).
All the prime numbers are in this sequence.

Examples

			8 is in the sequence because 8 has 3 proper divisors {1,2,4}, 1^2 + 2^2 + 4^2 = 21 and 3 divides 21.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[150], Mod[DivisorSigma[2, #1] - #1^2, DivisorSigma[0, #1] - 1] == 0 &]
    Select[Range[200],IntegerQ[Mean[Most[Divisors[#]]^2]]&] (* Harvey P. Dale, Jul 26 2019 *)
  • PARI
    is(n)=my(d=divisors(n)); d=apply(k->k^2, d[1..#d-1]); n>1 && sum(i=1,#d,d[i])%#d==0 \\ Charles R Greathouse IV, Dec 19 2016

A291108 Expansion of Sum_{k>=2} k^2*x^(2*k)/(1 - x^k).

Original entry on oeis.org

0, 0, 0, 4, 0, 13, 0, 20, 9, 29, 0, 65, 0, 53, 34, 84, 0, 130, 0, 145, 58, 125, 0, 273, 25, 173, 90, 265, 0, 399, 0, 340, 130, 293, 74, 614, 0, 365, 178, 609, 0, 735, 0, 625, 340, 533, 0, 1105, 49, 754, 298, 865, 0, 1183, 146, 1113, 370, 845, 0, 1859, 0, 965, 580, 1364, 194, 1743, 0, 1465, 538, 1599, 0, 2550, 0, 1373, 884
Offset: 1

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Author

Ilya Gutkovskiy, Aug 17 2017

Keywords

Comments

Sum of squares of divisors of n except 1 and n^2 (sum of squares of nontrivial divisors of n).

Examples

			a(6) = 13 because 6 has 4 divisors {1, 2, 3, 6} among which 2 are nontrivial {2, 3} and 2^2 + 3^2 = 13.
		

Crossrefs

Programs

  • Mathematica
    nmax = 75; Rest[CoefficientList[Series[Sum[k^2 x^(2 k)/(1 - x^k), {k, 2, nmax}], {x, 0, nmax}], x]]
    Join[{0}, Table[DivisorSigma[2, n] - n^2 - 1, {n, 2, 75}]]
  • PARI
    A291108(n) = sumdiv(n,d,if((1==d)||(n==d),0,d^2)); \\ Antti Karttunen, Jan 22 2025

Formula

G.f.: Sum_{k>=2} k^2*x^(2*k)/(1 - x^k).
a(n) = A001157(n) - A000290(n) - 1 for n > 1.
a(n) = A067558(n) - 1 for n > 1.
a(n) = A005063(n) if n is a semiprime (A001358).
a(n) = 0 if n is a prime or 1 (A008578).
a(n) = n if n is a square of prime (A001248).
a(p^k) = (p^(2*k) - p^2)/(p^2 - 1) for p is a prime and k > 0.

A301482 Composite numbers whose sum of aliquot parts divide the sum of the squares of their aliquot parts.

Original entry on oeis.org

8, 22, 27, 32, 77, 125, 128, 243, 343, 494, 512, 611, 660, 1073, 1281, 1331, 1425, 2033, 2048, 2187, 2197, 2332, 3125, 4172, 4565, 4913, 5293, 6031, 6859, 8192, 9983, 12167, 13969, 15818, 15947, 16807, 17485, 19683, 23489, 23840, 24389, 25241, 25389, 29791, 32768
Offset: 1

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Author

Paolo P. Lava, Mar 22 2018

Keywords

Comments

Semiprimes in the sequence: 22, 77, 611, 1073, 2033, 5293, 6031, 9983, 13969, 15947, 23489, 25241, 40301, 49901, 50249, 51101, 56759, 65017, 71677, 85079, 97217, 98099, 99101, .... - Robert Israel, Mar 29 2018
2^k is a term for all odd k > 1. - Michael S. Branicky, Aug 22 2021

Examples

			Aliquot parts of 77 are 1, 7, 11. Then (1^2 + 7^2 + 11^2)/(1 + 7 + 11) = 171/19 = 9.
		

Crossrefs

Programs

  • Maple
    with(numtheory): P:=proc(n)
    if not isprime(n) and frac((add(p^2,p=divisors(n))-n^2)/(sigma(n)-n))=0
    then n; fi; end: seq(P(i),i=2..35*10^3);
  • Mathematica
    aQ[n_] := CompositeQ[n] && Divisible[DivisorSigma[2, n] - n^2, DivisorSigma[1, n] - n]; Select[Range[33000], aQ] (* Amiram Eldar, Aug 17 2019 *)
  • PARI
    isok(n) = (n!=1) && !isprime(n) && (((sigma(n,2) - n^2) % (sigma(n) - n)) == 0); \\ Michel Marcus, Mar 23 2018
    
  • Python
    from sympy import divisors
    def ok(n):
        divs = divisors(n)[:-1]
        return len(divs) > 1 and sum(d**2 for d in divs)%sum(divs) == 0
    print(list(filter(ok, range(4, 32769)))) # Michael S. Branicky, Aug 22 2021

A374539 The sum of the squares of the infinitary divisors of n.

Original entry on oeis.org

1, 5, 10, 17, 26, 50, 50, 85, 82, 130, 122, 170, 170, 250, 260, 257, 290, 410, 362, 442, 500, 610, 530, 850, 626, 850, 820, 850, 842, 1300, 962, 1285, 1220, 1450, 1300, 1394, 1370, 1810, 1700, 2210, 1682, 2500, 1850, 2074, 2132, 2650, 2210, 2570, 2402, 3130, 2900
Offset: 1

Views

Author

Amiram Eldar, Jul 11 2024

Keywords

Comments

Also the sum of the infinitary divisors of n^2.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^(2^Position[Reverse@IntegerDigits[e, 2], ?(# == 1 &)]); a[1] = 1; a[n] := Times @@ (Flatten@(f @@@ FactorInteger[n]) + 1); Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n), b); prod(i = 1, #f~, b = binary(2 * f[i, 2]); prod(k=1, #b, if(b[k], 1+f[i, 1]^(2^(#b-k)), 1)));}
    
  • Python
    from math import prod
    from sympy import factorint
    def A374539(n): return prod(p**(1<Chai Wah Wu, Jul 11 2024

Formula

a(n) = A049417(n^2).
a(n) <= A001157(n), with equality if and only if n is in A036537.
Multiplicative with a(p^e) = Product{k>=1, e_k=1} (p^(2^(k+1)) + 1), where e = Sum_{k} e_k * 2^k is the binary representation of e, i.e., e_k is bit k of e.
Sum_{k=1..n} a(k) ~ c * n^3 / 3, where c = Product_{P} (1 + 1/(P^2*(P+1))) = 1.14142906130350119631..., and P are numbers of the form p^(2^k) where p is prime and k >= 0 (A050376).
Previous Showing 11-15 of 15 results.