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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A103977 Zumkeller deficiency of n: Let d_1 ... d_k be the divisors of n. Then a(n) = min_{ e_1 = +-1, ... e_k = +-1 } | Sum_i e_i d_i |.

Original entry on oeis.org

1, 1, 2, 1, 4, 0, 6, 1, 5, 2, 10, 0, 12, 4, 6, 1, 16, 1, 18, 0, 10, 8, 22, 0, 19, 10, 14, 0, 28, 0, 30, 1, 18, 14, 22, 1, 36, 16, 22, 0, 40, 0, 42, 4, 12, 20, 46, 0, 41, 7, 30, 6, 52, 0, 38, 0, 34, 26, 58, 0, 60, 28, 22, 1, 46, 0, 66, 10, 42, 0, 70, 1, 72, 34, 26, 12, 58, 0, 78, 0
Offset: 1

Views

Author

Yasutoshi Kohmoto, Jan 01 2007

Keywords

Comments

Like the ordinary deficiency (A033879) obtains 0's only at perfect numbers (A000396), the Zumkeller deficiency obtains 0's only at integer-perfect numbers, A083207. See the formula section. Unlike the ordinary deficiency, this obtains only nonnegative values. See A378600 for another version. - Antti Karttunen, Dec 03 2024

Examples

			a(6) = 1 + 2 + 3 - 6 = 0.
		

Crossrefs

Cf. A125732, A125733, A005835, A023196, A033879, A083206, A083207 (positions of 0's), A263837, A378643 (Dirichlet inverse), A378644 (Möbius transform), A378645, A378646, A378647 (an analog of A000027), A378648 (an analog of sigma), A378649 (an analog of Euler phi), A379503 (positions of 1's), A379504, A379505.
Cf. A378600 (signed variant).
Cf. also A058377, A119347.

Programs

  • Maple
    A103977 := proc(n) local divs,a,acandid,filt,i,p,sigs ; divs := convert(numtheory[divisors](n),list) ; a := add(i,i=divs) ; for sigs from 0 to 2^nops(divs)-1 do filt := convert(sigs,base,2) ; while nops(filt) < nops(divs) do filt := [op(filt), 0] ; od ; acandid := 0 ; for p from 0 to nops(divs)-1 do if op(p+1,filt) = 0 then acandid := acandid-op(p+1,divs) ; else acandid := acandid+op(p+1,divs) ; fi ; od: acandid := abs(acandid) ; if acandid < a then a := acandid ; fi ; od: RETURN(a) ; end: seq(A103977(n),n=1..80) ; # R. J. Mathar, Nov 27 2007
    # second Maple program:
    a:= proc(n) option remember; local l, b; l, b:= [numtheory[divisors](n)[]],
          proc(s, i) option remember; `if`(i<1, s,
            min(b(s+l[i], i-1), b(abs(s-l[i]), i-1)))
          end: b(0, nops(l))
        end:
    seq(a(n), n=1..80);  # Alois P. Heinz, Dec 05 2024
  • Mathematica
    a[n_] := Module[{d = Divisors[n], c, p, m}, c = CoefficientList[Product[1 + x^i, {i, d}], x]; p = -1 + Position[c, ?(# > 0 &)] // Flatten; m = Length[p]; If[OddQ[m], If[(d = p[[(m + 1)/2]] - p[[(m - 1)/2]]) == 1, 0, d], p[[m/2 + 1]] - p[[m/2]]]]; Array[a, 100] (* _Amiram Eldar, Dec 11 2019 *)
  • PARI
    nonzerocoefpositions(p) = { my(v=Vec(p), lista=List([])); for(i=1,#v,if(v[i], listput(lista,i))); Vec(lista); }; \\ Doesn't need to be 0-based, as we use their differences only.
    A103977(n) = { my(p=1); fordiv(n, d, p *= (1 + 'x^d)); my(plist=nonzerocoefpositions(p), m = #plist, d); if(!(m%2), plist[1+(m/2)]-plist[m/2], d = plist[(m+1)/2]-plist[(m-1)/2]; if(1==d,0,d)); }; \\ Antti Karttunen, Dec 03 2024, after Mathematica-program by Amiram Eldar

Formula

If n=p (prime), then a(n)=p-1. If n=2^m, then a(n)=1. [Corrected by R. J. Mathar, Nov 27 2007]
a(n) = 0 iff n is a Zumkeller number (A083207). - Amiram Eldar, Jan 05 2020
From Antti Karttunen, Dec 03 2024: (Start)
a(n) = A033879(n) iff n is a non-abundant number (A263837).
a(n) = abs(A378600(n)).
a(n) = 2*A378647(n) - A378648(n). [Analogously to A033879(n) = 2*n - sigma(n)]
a(n) = 0 <=> A083206(n) > 0.
(End)
a(p^e) = p^e - (1+p+...+p^(e-1)) = (p^e*(p-2) + 1)/(p-1) for prime p. - Jianing Song, Dec 05 2024
a(n) = 1 <=> A379504(n) > 0. - Antti Karttunen, Jan 07 2025

Extensions

More terms from R. J. Mathar, Nov 27 2007
Name "Zumkeller deficiency" coined by Antti Karttunen, Dec 03 2024

A335431 Numbers of the form q*(2^k), where q is one of the Mersenne primes (A000668) and k >= 0.

Original entry on oeis.org

3, 6, 7, 12, 14, 24, 28, 31, 48, 56, 62, 96, 112, 124, 127, 192, 224, 248, 254, 384, 448, 496, 508, 768, 896, 992, 1016, 1536, 1792, 1984, 2032, 3072, 3584, 3968, 4064, 6144, 7168, 7936, 8128, 8191, 12288, 14336, 15872, 16256, 16382, 24576, 28672, 31744, 32512, 32764, 49152, 57344, 63488, 65024, 65528, 98304, 114688, 126976, 130048, 131056, 131071
Offset: 1

Views

Author

Antti Karttunen, Jun 28 2020

Keywords

Comments

Numbers of the form 2^k * ((2^p)-1), where p is one of the primes in A000043, and k >= 0.
Numbers k such that A000265(k) is in A000668.
Numbers k for which A331410(k) = 1.
Numbers k that themselves are not powers of two, but for which A335876(k) = k+A052126(k) is [a power of 2].
Conjecture: This sequence gives all fixed points of map n -> A332214(n) and its inverse n -> A332215(n). See also notes in A029747 and in A163511.

Crossrefs

Cf. A000043, A000396 (even terms form a subsequence), A000668 (primes present), A335882, A341622.
Row 1 of A335430.
Positions of 1's in A331410, in A364260, and in A364251 (characteristic function).
Subsequence of A054784.

Programs

  • Mathematica
    qs = 2^MersennePrimeExponent[Range[6]] - 1; max = qs[[-1]]; Reap[Do[n = 2^k*q; If[n <= max, Sow[n]], {k, 0, Log2[max]}, {q, qs}]][[2, 1]] // Union (* Amiram Eldar, Feb 18 2021 *)
  • PARI
    A000265(n) = (n>>valuation(n,2));
    isA000668(n) = (isprime(n)&&!bitand(n,1+n));
    isA335431(n) = isA000668(A000265(n));

Formula

A332214(a(n)) = A332215(a(n)) = a(n) for all n.
Sum_{n>=1} 1/a(n) = 2 * A173898 = 1.0329083578... - Amiram Eldar, Feb 18 2021

A133028 Even perfect numbers divided by 2.

Original entry on oeis.org

3, 14, 248, 4064, 16775168, 4294934528, 68719345664, 1152921504069976064, 1329227995784915872327346307976921088, 95780971304118053647396689042151819065498660774084608, 6582018229284824168619876730229361455111736159193471558891864064, 7237005577332262213973186563042994240786838745737417944533177174565599576064
Offset: 1

Views

Author

Omar E. Pol, Oct 20 2007, Apr 23 2008, Apr 28 2009

Keywords

Comments

a(13) has 314 digits and is too large to include. - R. J. Mathar, Oct 23 2007
Largest proper divisor of n-th even perfect number.
Also numbers k such that A000203(k) is divisible 24. - Ctibor O. Zizka, Jun 29 2009

Crossrefs

Programs

  • Maple
    a:=proc(n) if isprime(2^n-1)=true then 2^(n-2)*(2^n-1) else end if end proc: seq(a(n),n=1..120); # Emeric Deutsch, Oct 24 2007
  • Mathematica
    p = Select[2^Range[400] - 1, PrimeQ]; p*(p+1)/4 (* Vladimir Joseph Stephan Orlovsky, Feb 02 2012 *)
    Map[2^(#-2) * (2^# - 1) &, MersennePrimeExponent[Range[12]]] (* Amiram Eldar, Oct 21 2024 *)

Formula

a(n) = A000396(n)/2. - R. J. Mathar, Oct 23 2007 [Assuming there are no odd perfect numbers. - Jianing Song, Sep 17 2022]
a(n) = 2^(A000043(n) - 2) * A000668(n). - Omar E. Pol, Mar 01 2008
a(n) = A032742(A000396(n)), assuming there are no odd perfect numbers.

Extensions

More terms from R. J. Mathar and Emeric Deutsch, Oct 23 2007

A323653 Multiperfect numbers m such that sigma(m) is also multiperfect.

Original entry on oeis.org

1, 459818240, 51001180160, 13188979363639752997731839211623940096, 5157152737616023231698245840143799191339008, 54530444405217553992377326508106948362108928, 133821156044600922812153118065015159487725568, 4989680372093758991515359988337845750507257510078971904
Offset: 1

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Author

Jaroslav Krizek, Jan 21 2019

Keywords

Comments

Multiperfect numbers m such that sigma(m) divides sigma(sigma(m)).
Also k-multiperfect numbers m such that k*m is also multiperfect.
Corresponding values of numbers k(n) = sigma(a(n)) / a(n): 1, 3, 3, 5, 5, 5, 5, 5, ...
Corresponding values of numbers h(n) = sigma(k(n) * a(n)) / (k(n) * a(n)): 1, 4, 4, 6, 6, 6, 6, 6, ...
Number of k-multiperfect numbers m such that sigma(n) is also multiperfect for k = 3..6: 2, 0, 20, 0.
From Antti Karttunen, Mar 20 2021, Feb 18 2022: (Start)
Conjecture 1 (a): This sequence consists of those m for which sigma(m)/m is an integer (thus a term of A007691), and coprime with m. Or expressed in a slightly weaker form (b): {1} followed by those m for which sigma(m)/m is an integer, but not a divisor of m. In a slightly stronger form (c): For m > 1, sigma(m)/m is always the least prime not dividing m. This would imply both (a) and (b) forms.
Conjecture 2: This sequence is finite.
Conjecture 3: This sequence is the intersection of A007691 and A351458.
Conjecture 4: This is a subsequence of A349745, thus also of A351551 and of A351554.
Note that if there existed an odd perfect number x that were not a multiple of 3, then both x and 2*x would be terms in this sequence, as then we would have: sigma(x)/x = 2, sigma(2*x)/(2*x) = 3, sigma(6*x)/(6*x) = 4. See also the diagram in A347392 and A353365.
(End)
From Antti Karttunen, May 16 2022: (Start)
Apparently for all n > 1, A336546(a(n)) = 0. [At least for n=2..23], while A353633(a(n)) = 1, for n=1..23.
The terms a(1) .. a(23) are only cases present among the 5721 known and claimed multiperfect numbers with abundancy <> 2, as published 03 January 2022 under Flammenkamp's site, that satisfy the condition for inclusion in this sequence.
They are also the only 23 cases among that data such that gcd(n, sigma(n)/n) = 1, or in other words, for which the n and its abundancy are relatively prime, with abundancy in all cases being the least prime that does not divide n, A053669(n), which is a sufficient condition for inclusion in A351458.
(End)

Examples

			3-multiperfect number 459818240 is a term because number 3*459818240 = 1379454720 is a 4-multiperfect number.
		

Crossrefs

Programs

  • Magma
    [n: n in [1..10^6] | SumOfDivisors(n) mod n eq 0 and SumOfDivisors(SumOfDivisors(n)) mod SumOfDivisors(n) eq 0];
    
  • PARI
    ismulti(n) = (sigma(n) % n) == 0;
    isok(n) = ismulti(n) && ismulti(sigma(n)); \\ Michel Marcus, Jan 26 2019

A083209 Numbers whose divisors can be partitioned in exactly one way into two disjoint sets with the same sum.

Original entry on oeis.org

6, 12, 20, 28, 56, 70, 88, 104, 176, 208, 272, 304, 368, 464, 496, 550, 650, 736, 836, 928, 992, 1184, 1312, 1376, 1504, 1696, 1888, 1952, 2752, 3008, 3230, 3392, 3770, 3776, 3904, 4030, 4288, 4510, 4544, 4672, 5056, 5170, 5312, 5696, 5830, 6208, 6464
Offset: 1

Views

Author

Reinhard Zumkeller, Apr 22 2003

Keywords

Comments

A083206(a(n))=1; perfect numbers (A000396) are a subset; problem: are weird numbers (A006037) a subset?
The weird numbers A006037 are not a subset of this sequence. The first missing weird number is A006037(8) = 10430. - Alois P. Heinz, Oct 29 2009
All numbers of the form p*2^k are in this sequence for k>0 and odd primes p between 2^(k+1)/3 and 2^(k+1). - T. D. Noe, Jul 08 2010
"Numbers with exactly one subset of their sets of divisors such that the complement has the same sum." - This was the original name of the sequence, but strictly taken is incorrect, because there are always two subsets that satisfy this condition: the subset and its complement. - Antti Karttunen, Dec 02 2024

Examples

			n=20: 2+4+5+10 = 1+20, 20 is a term (A083206(20)=1).
		

Crossrefs

Subsequence of A083207, Zumkeller numbers.
Positions of 1's in A083206.
Cf. A005101, A005835, A064771, A337739 (terms with record number of divisors), A378449 (characteristic function), A378530 (subsequence).
Cf. also A378652, and A335143, A335199, A335202, A335219, A335217, A339980 for variants.

Programs

  • Maple
    with(numtheory): b:= proc(n,l) option remember; local m, ll, i; m:= nops(l); if n<0 then 0 elif n=0 then 1 elif m=0 or add(i, i=l)Alois P. Heinz, Oct 29 2009
  • Mathematica
    b[n_, l_] := b[n, l] = Module[{m, ll, i}, m = Length[l]; Which[n<0, 0, n == 0, 1, m == 0 || Total[l] Nothing]; b[n, ll] + b[n - l[[m]], ll]]]; a[n_] := a[n] = Module[{i, k, l, m, r}, For[k = If[n == 1, 1, a[n-1]+1], True, k++, l = Divisors[k]; {m, r} = QuotientRemainder[Total[l], 2]; If[r==0 && b[m, l]==2, Break[]]]; k]; Table[Print["a(", n, ") = ", a[n]]; a[n], {n, 1, 50}] (* Jean-François Alcover, Jan 31 2017, after Alois P. Heinz *)
  • PARI
    isA083209 = A378449; \\ Antti Karttunen, Nov 28 2024

Extensions

More terms from Alois P. Heinz, Oct 29 2009
Improved the definition, old name moved to the comments - Antti Karttunen, Dec 02 2024

A088012 Odd solutions to abs(sigma(k) - 2k) <= log(k). Numbers k whose abundance-radius does not exceed log(k).

Original entry on oeis.org

1155, 8925, 32445, 442365, 159030135, 815634435, 2586415095, 1956860570050575, 221753180448460815, 747406020889133775
Offset: 1

Views

Author

Keywords

Comments

This sequence should include odd perfect numbers too, if they exist.
From Walter Nissen, Dec 15 2005: (Start)
abundancy(k) k 2k sigma(k) abundance
1.99480519480519 1155 2310 2304 -6
2.00067226890756 8925 17850 17856 6
2.00018492834027 32445 64890 64896 6
2.00001356346004 442365 884730 884736 6
2.00000011318610 159030135 318060270 318060288 18
1.99999999264376 815634435 1631268870 1631268864 -6
2.00000000695943 2586415095 5172830190 5172830208 18
As it happens, abundance of these is -6, 6 or 18. This is not necessarily true for larger terms. (End)
See also A171929 and A188597 and A188263 for sequences of numbers (any / deficient / abundant) whose relative abundancy tends to 2. - M. F. Hasler, Feb 19 2017
3278298202600507814120339275775985 is also a term with abundance 30. In fact, it and 815634435 are the only odd terms known where abs(sigma(k)-2k) <= log_10(k). - Alexander Violette, Nov 05 2020; updated by Max Alekseyev, Jul 27 2025
Also includes 827880257692739174385 and 255286886041240176056063754225. - Max Alekseyev, Jul 27 2025

Examples

			1155 is in the sequence because sigma(1155) = 2304, giving 2*1155 - 2304 = 6, while natural log of 1155 is about 7.05.
From _M. F. Hasler_, Jul 18 2016: (Start)
We have the following factorizations:
1155 = 3 * 5 * 7 * 11,
8925 = 3 * 5^2 * 7 * 17,
32445 = 3^2 * 5 * 7 * 103,
442365 = 3 * 5 * 7 * 11 * 383,
159030135 = 3^5 * 5 * 11 * 73 * 163,
815634435 = 3 * 5 * 7 * 11 * 547 * 1291,
2586415095 = 3^2 * 5 * 11 * 31 * 41 * 4111.
The sequence appears to be a subsequence of A171929. (End)
		

Crossrefs

Programs

  • Mathematica
    abu[x_] := Abs[DivisorSigma[1, x]-2*x] Do[If[ !Greater[abu[n], Log[n]//N]&&OddQ[n], Print[n]], {n, 1, 100000}]
  • PARI
    is(n)=n%2 && abs(sigma(n)-2*n)<=log(n) \\ Charles R Greathouse IV, Feb 21 2017

Extensions

a(7) from Donovan Johnson, Dec 21 2008
a(9) from Alexander Violette confirmed and a(8), a(10) added by Max Alekseyev, Jul 27 2025

A133049 Squares of Mersenne primes A000668(n).

Original entry on oeis.org

9, 49, 961, 16129, 67092481, 17179607041, 274876858369, 4611686014132420609, 5316911983139663487003542222693990401, 383123885216472214589586755549637256619304505646776321
Offset: 1

Views

Author

Omar E. Pol, Oct 30 2007, Apr 23 2008

Keywords

Comments

Sum of last A000043(n) divisors of the n-th even perfect number. In other words; sum of divisors that are not powers of 2 of the n-th even perfect number, or sum of divisors that are multiples of the n-th Mersenne prime A000668(n) of the n-th even perfect number. See A139247 for more information.
See the structure of the divisors of perfect numbers in A135652, A135653, A135654 and A135655.

Examples

			a(3)=961 because the 3rd Mersenne prime is 31 and 31^2=961.
		

Crossrefs

Programs

  • Mathematica
    Select[2^Range[1000] - 1, PrimeQ]^2 (* G. C. Greubel, Oct 03 2017 *)
  • PARI
    forprime(p=2, 1000, if(ispseudoprime(2^p-1), print1((2^p-1)^2", "))) \\ G. C. Greubel, Oct 03 2017

Formula

a(n) = A000668(n)^2

Extensions

More terms from Olaf Voß, Feb 13 2008

A265640 Prime factorization palindromes (see comments for definition).

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 11, 12, 13, 16, 17, 18, 19, 20, 23, 25, 27, 28, 29, 31, 32, 36, 37, 41, 43, 44, 45, 47, 48, 49, 50, 52, 53, 59, 61, 63, 64, 67, 68, 71, 72, 73, 75, 76, 79, 80, 81, 83, 89, 92, 97, 98, 99, 100, 101, 103, 107, 108, 109, 112, 113, 116, 117, 121, 124, 125, 127, 128, 131, 137, 139, 144
Offset: 1

Views

Author

Vladimir Shevelev, Dec 11 2015

Keywords

Comments

a(66) is the first term at which this sequence differs from A119848.
A number N is called a prime factorization palindrome (PFP) if all its prime factors, taking into account their multiplicities, can be arranged in a row with central symmetry (see example). It is easy to see that every PFP-number is either a square or a product of a square and a prime. In particular, the sequence contains all primes.
Numbers which are both palindromes (A002113) and PFP are 1,2,3,4,5,7,9,11,44,99,101,... (see A265641).
If n is in the sequence, so is n^k for all k >= 0. - Altug Alkan, Dec 11 2015
The sequence contains all perfect numbers except 6 (cf. A000396). - Don Reble, Dec 12 2015
Equivalently, numbers that have at most one prime factor with odd multiplicity. - Robert Israel, Feb 03 2016
Numbers whose squarefree part is noncomposite. - Peter Munn, Jul 01 2020

Examples

			44 is a member, since 44=2*11*2.
52 is a member, since 52=2*13*2. [This illustrates the fact that the digits don't need to form a palindrome. This is not a base-dependent sequence. - _N. J. A. Sloane_, Oct 05 2024]
180 is a member, since 180=2*3*5*3*2.
		

Crossrefs

Cf. A000396, A000720, A002113, A265641, complement of A229153.
Disjoint union of A229125 and (A000290\{0}).
Cf. A013661 (zeta(2)).

Programs

  • Maple
    N:= 1000: # to get all terms <= N
    P:= [1,op(select(isprime, [2,seq(i,i=3..N,2)]))]:
    sort([seq(seq(p*x^2,x=1..floor(sqrt(N/p))),p=P)]); # Robert Israel, Feb 03 2016
  • Mathematica
    M = 200; P = Join[{1}, Select[Join[{2}, Range[3, M, 2]], PrimeQ]]; Sort[ Flatten[Table[Table[p x^2, {x, 1, Floor[Sqrt[M/p]]}], {p, P}]]] (* Jean-François Alcover, Apr 09 2019, after Robert Israel *)
  • PARI
    for(n=1, 200, if( ispseudoprime(core(n)) || issquare(n), print1(n, ", "))) \\ Altug Alkan, Dec 11 2015
    
  • Python
    from math import isqrt
    from sympy.ntheory.factor_ import core, isprime
    def ok(n): return isqrt(n)**2 == n or isprime(core(n))
    print([k for k in range(1, 145) if ok(k)]) # Michael S. Branicky, Oct 03 2024
    
  • Python
    from math import isqrt
    from sympy import primepi, mobius
    def A265640(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x):
            c = n-(a:=isqrt(x))
            for y in range(1,a+1):
                m = x//y**2
                c -= primepi(m)-sum(mobius(k)*(m//k**2) for k in range(1, isqrt(m)+1))
            return c
        return bisection(f,n,n) # Chai Wah Wu, Jan 30 2025

Formula

lim A(x)/pi(x) = zeta(2) where A(x) is the number of a(n) <= x and pi is A000720.

A324213 Number of k with 0 <= k <= sigma(n) such that n-k and 2n-sigma(n) are relatively prime.

Original entry on oeis.org

2, 4, 3, 8, 4, 2, 4, 16, 12, 9, 6, 14, 6, 12, 8, 32, 10, 26, 8, 21, 14, 18, 12, 20, 30, 16, 18, 2, 14, 24, 10, 64, 16, 24, 22, 88, 14, 30, 26, 36, 18, 32, 14, 42, 26, 28, 24, 54, 56, 80, 20, 32, 26, 40, 36, 60, 38, 42, 30, 56, 18, 42, 48, 128, 42, 48, 22, 50, 28, 72, 26, 122, 26, 54, 58, 46, 48, 56, 26, 86, 120, 60, 42, 96, 54
Offset: 1

Views

Author

Antti Karttunen and David A. Corneth, May 26 2019, with better name from Charlie Neder, Jun 02 2019

Keywords

Comments

Number of ways to form the sum sigma(n) = x+y so that n-x and n-y are coprime, with x and y in range 0..sigma(n).
From Antti Karttunen, May 28 - Jun 08 2019: (Start)
Empirically, it seems that a(n) >= A034444(n) and also that a(n) >= A034444(A000203(n)) unless n is in A000396.
Specifically, if it could be proved that a(n) >= A034444(n)/2 for n >= 2, which in turn would imply that a(n) >= A001221(n) for all n, then we would know that no odd perfect numbers could exist. Note that a(n) must be 2 on all perfect numbers, whether even or odd. See also A325819.
(End)

Examples

			For n=1, sigma(1) = 1, both gcd(1-0, 1-(1-0)) = gcd(1,0) = 1 and gcd(1-1, 1-(1-1)) = gcd(0,1) = 1, thus a(1) = 2.
--
For n=3, sigma(3) = 4, we have 5 cases to consider:
  gcd(3-0, 3-(4-0)) = 1 = gcd(3-4, 3-(4-4)),
  gcd(3-1, 3-(4-1)) = 2 = gcd(3-3, 3-(4-3)),
  gcd(3-2, 3-(4-2)) = 1,
of which three cases give 1 as a result, thus a(3) = 3.
--
For n=6, sigma(6) = 12, we have 13 cases to consider:
  gcd(6-0, 6-(12-0)) = 6 = gcd(6-12, 6-(12-12)),
  gcd(6-1, 6-(12-1)) = 5 = gcd(6-11, 6-(12-11)),
  gcd(6-2, 6-(12-2)) = 4 = gcd(6-10, 6-(12-10)),
  gcd(6-3, 6-(12-3)) = 3 = gcd(6-9, 6-(12-9)),
  gcd(6-4, 6-(12-4)) = 2 = gcd(6-8, 6-(12-8))
  gcd(6-5, 6-(12-5)) = 1 = gcd(6-7, 6-(12-7)),
  gcd(6-6, 6-(12-6)) = 0,
of which only two give 1 as a result, thus a(6) = 2.
--
For n=10, sigma(10) = 18, we have 19 cases to consider:
  gcd(10-0, 10-(18-0)) = 2 = gcd(10-18, 10-(18-18)),
  gcd(10-1, 10-(18-1)) = 1 = gcd(10-17, 10-(18-17)),
  gcd(10-2, 10-(18-2)) = 2 = gcd(10-16, 10-(18-16)),
  gcd(10-3, 10-(18-3)) = 1 = gcd(10-15, 10-(18-15)),
  gcd(10-4, 10-(18-4)) = 2 = gcd(10-14, 10-(18-14)),
  gcd(10-5, 10-(18-5)) = 1 = gcd(10-13, 10-(18-13)),
  gcd(10-6, 10-(18-6)) = 2 = gcd(10-12, 10-(18-12)),
  gcd(10-7, 10-(18-7)) = 1 = gcd(10-11, 10-(18-11)),
  gcd(10-8, 10-(18-8)) = 2 = gcd(10-10, 10-(18-10)),
  gcd(10-9, 10-(18-9)) = 1,
of which 9 cases give 1 as a result, thus a(10) = 9.
--
For n=15, sigma(15) = 24, we have 25 cases to consider:
  gcd(15-0, 15-(24-0)) = 3 = gcd(15-24, 15-(24-24)),
  gcd(15-1, 15-(24-1)) = 2 = gcd(15-23, 15-(24-23)),
  gcd(15-2, 15-(24-2)) = 1 = gcd(15-22, 15-(24-22)),
  gcd(15-3, 15-(24-3)) = 6 = gcd(15-21, 15-(24-21)),
  gcd(15-4, 15-(24-4)) = 1 = gcd(15-20, 15-(24-20)),
  gcd(15-5, 15-(24-5)) = 2 = gcd(15-19, 15-(24-19)),
  gcd(15-6, 15-(24-6)) = 3 = gcd(15-18, 15-(24-18)),
  gcd(15-7, 15-(24-7)) = 2 = gcd(15-17, 15-(24-17)),
  gcd(15-8, 15-(24-8)) = 1 = gcd(15-16, 15-(24-16)),
  gcd(15-9, 15-(24-9)) = 6 = gcd(15-15, 15-(24-15)),
  gcd(15-10, 15-(24-10)) = 1 = gcd(15-14, 15-(24-14)),
  gcd(15-11, 15-(24-11)) = 2 = gcd(15-13, 15-(24-13)),
  gcd(15-12, 15-(24-12)) = 3,
of which 2*4 = 8 cases give 1 as a result, thus a(15) = 8.
		

Crossrefs

Programs

  • Mathematica
    Array[Sum[Boole[1 == GCD[#1 - i, #1 - (#2 - i)]], {i, 0, #2}] & @@ {#, DivisorSigma[1, #]} &, 85] (* Michael De Vlieger, Jun 09 2019 *)
  • PARI
    A324213(n) = { my(s=sigma(n)); sum(i=0,s,(1==gcd(n-i,n-(s-i)))); };

Formula

a(n) = Sum_{i=0..sigma(n)} [1 == gcd(n-i,n-(sigma(n)-i))], where [ ] is the Iverson bracket and sigma(n) is A000203(n).
a(A000396(n)) = 2.
a(n) = A325815(n) + A034444(n).
a(n) = 1+A000203(n) - A325816(n).
a(A228058(n)) = A325819(n).

A332214 Mersenne-prime fixing variant of permutation A163511: a(n) = A332212(A163511(n)).

Original entry on oeis.org

1, 2, 4, 3, 8, 9, 6, 7, 16, 27, 18, 49, 12, 21, 14, 5, 32, 81, 54, 343, 36, 147, 98, 25, 24, 63, 42, 35, 28, 15, 10, 31, 64, 243, 162, 2401, 108, 1029, 686, 125, 72, 441, 294, 175, 196, 75, 50, 961, 48, 189, 126, 245, 84, 105, 70, 155, 56, 45, 30, 217, 20, 93, 62, 11, 128, 729, 486, 16807, 324, 7203, 4802, 625, 216, 3087, 2058, 875
Offset: 0

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Author

Antti Karttunen, Feb 09 2020

Keywords

Comments

Any Mersenne prime (A000668) times any power of 2, i.e., sequence A335431, is fixed by this map (note the indexing), including also all even perfect numbers. It is not currently known whether there are any additional fixed points.
Because a(n) has the same prime signature as A163511(n), it implies that applying A046523 and A052409 to this sequence gives the same results as with A163511, namely, sequences A278531 and A365805. - Antti Karttunen, Oct 09 2023

Crossrefs

Cf. A163511, A332211, A332212, A332215 (inverse permutation).
Cf. A278531 [= A046523(a(n))], A290251 [= A001222(a(n))], A365805 [= A052409(a(n))], A366372 [= a(n)-n], A366373 [= gcd(n,a(n))], A366374 (numerator of n/a(n)), A366375 (denominator of n/a(n)), A366376.
Cf. A000043, A000668, A000396, A324200, A335431 (conjectured to give all the fixed points).

Programs

  • PARI
    A332214(n) = A332212(A163511(n));
    
  • PARI
    \\ Needs precomputed data for A332211:
    v332211 = readvec("b332211_to.txt"); \\ Prepared with gawk ' { print $2 } ' < b332211.txt > b332211_to.txt
    A332211(n) = v332211[n];
    A332214(n) = if(!n, 1, my(i=1, p=A332211(i), t=1); while(n>1, if(!(n%2), (t*=p), i++; p=A332211(i)); n >>= 1); (t*p)); \\ Antti Karttunen, Oct 09 2023

Formula

a(n) = A332212(A163511(n)).
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