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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A329610 a(n) = A329638(A324201(n)).

Original entry on oeis.org

1, 5, 269, 6181, 42467605, 6698339765, 137657144245, 2482519341068581109, 4650094754920132605500516092412538533, 172211278811469777927840379819793091108558586157350885, 6181597550035473098411130103418930148893343460909707255358224869, 15429082227467244202811178054212231454083499413948013030392138990797549495221
Offset: 1

Views

Author

Antti Karttunen, Nov 22 2019

Keywords

Comments

The first 12 terms are all squarefree (in A005117). Does this hold in general?

Crossrefs

Formula

a(n) = A329638(A324201(n)) = A329639(A324201(n)) = A329641(A324201(n)).

A000396 Perfect numbers k: k is equal to the sum of the proper divisors of k.

Original entry on oeis.org

6, 28, 496, 8128, 33550336, 8589869056, 137438691328, 2305843008139952128, 2658455991569831744654692615953842176, 191561942608236107294793378084303638130997321548169216
Offset: 1

Views

Author

Keywords

Comments

A number k is abundant if sigma(k) > 2k (cf. A005101), perfect if sigma(k) = 2k (this sequence), or deficient if sigma(k) < 2k (cf. A005100), where sigma(k) is the sum of the divisors of k (A000203).
The numbers 2^(p-1)*(2^p - 1) are perfect, where p is a prime such that 2^p - 1 is also prime (for the list of p's see A000043). There are no other even perfect numbers and it is believed that there are no odd perfect numbers.
Numbers k such that Sum_{d|k} 1/d = 2. - Benoit Cloitre, Apr 07 2002
For number of divisors of a(n) see A061645(n). Number of digits in a(n) is A061193(n). - Lekraj Beedassy, Jun 04 2004
All terms other than the first have digital root 1 (since 4^2 == 4 (mod 6), we have, by induction, 4^k == 4 (mod 6), or 2*2^(2*k) = 8 == 2 (mod 6), implying that Mersenne primes M = 2^p - 1, for odd p, are of the form 6*t+1). Thus perfect numbers N, being M-th triangular, have the form (6*t+1)*(3*t+1), whence the property N mod 9 = 1 for all N after the first. - Lekraj Beedassy, Aug 21 2004
The earliest recorded mention of this sequence is in Euclid's Elements, IX 36, about 300 BC. - Artur Jasinski, Jan 25 2006
Theorem (Euclid, Euler). An even number m is a perfect number if and only if m = 2^(k-1)*(2^k-1), where 2^k-1 is prime. Euler's idea came from Euclid's Proposition 36 of Book IX (see Weil). It follows that every even perfect number is also a triangular number. - Mohammad K. Azarian, Apr 16 2008
Triangular numbers (also generalized hexagonal numbers) A000217 whose indices are Mersenne primes A000668, assuming there are no odd perfect numbers. - Omar E. Pol, May 09 2008, Sep 15 2013
If a(n) is even, then 2*a(n) is in A181595. - Vladimir Shevelev, Nov 07 2010
Except for a(1) = 6, all even terms are of the form 30*k - 2 or 45*k + 1. - Arkadiusz Wesolowski, Mar 11 2012
a(4) = A229381(1) = 8128 is the "Simpsons's perfect number". - Jonathan Sondow, Jan 02 2015
Theorem (Farideh Firoozbakht): If m is an integer and both p and p^k-m-1 are prime numbers then x = p^(k-1)*(p^k-m-1) is a solution to the equation sigma(x) = (p*x+m)/(p-1). For example, if we take m=0 and p=2 we get Euclid's result about perfect numbers. - Farideh Firoozbakht, Mar 01 2015
The cototient of the even perfect numbers is a square; in particular, if 2^p - 1 is a Mersenne prime, cototient(2^(p-1) * (2^p - 1)) = (2^(p-1))^2 (see A152921). So, this sequence is a subsequence of A063752. - Bernard Schott, Jan 11 2019
Euler's (1747) proof that all the even perfect number are of the form 2^(p-1)*(2^p-1) implies that their asymptotic density is 0. Kanold (1954) proved that the asymptotic density of odd perfect numbers is 0. - Amiram Eldar, Feb 13 2021
If k is perfect and semiprime, then k = 6. - Alexandra Hercilia Pereira Silva, Aug 30 2021
This sequence lists the fixed points of A001065. - Alois P. Heinz, Mar 10 2024

Examples

			6 is perfect because 6 = 1+2+3, the sum of all divisors of 6 less than 6; 28 is perfect because 28 = 1+2+4+7+14.
		

References

  • Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 4.
  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 2d ed. 1966, pp. 11-23.
  • Stanley J. Bezuszka, Perfect Numbers (Booklet 3, Motivated Math. Project Activities), Boston College Press, Chestnut Hill MA, 1980.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 136-137.
  • Euclid, Elements, Book IX, Section 36, about 300 BC.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.3 Perfect and Amicable Numbers, pp. 82-83.
  • R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section B1.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 239.
  • T. Koshy, "The Ends Of A Mersenne Prime And An Even Perfect Number", Journal of Recreational Mathematics, Baywood, NY, 1998, pp. 196-202.
  • Joseph S. Madachy, Madachy's Mathematical Recreations, New York: Dover Publications, Inc., 1979, p. 149 (First publ. by Charles Scribner's Sons, New York, 1966, under the title: Mathematics on Vacation).
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 46-48, 244-245.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 83-87.
  • József Sándor and Borislav Crstici, Handbook of Number Theory, II, Springer Verlag, 2004.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Ian Stewart, L'univers des nombres, "Diviser Pour Régner", Chapter 14, pp. 74-81, Belin-Pour La Science, Paris, 2000.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, chapter 4, pages 127-149.
  • Horace S. Uhler, On the 16th and 17th perfect numbers, Scripta Math., Vol. 19 (1953), pp. 128-131.
  • André Weil, Number Theory, An approach through history, From Hammurapi to Legendre, Birkhäuser, 1984, p. 6.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, pp. 107-110, Penguin Books, 1987.

Crossrefs

See A000043 for the current state of knowledge about Mersenne primes.
Cf. A228058 for Euler's criterion for odd terms.
Positions of 0's in A033879 and in A033880.
Cf. A001065.

Programs

  • Haskell
    a000396 n = a000396_list !! (n-1)
    a000396_list = [x | x <- [1..], a000203 x == 2 * x]
    -- Reinhard Zumkeller, Jan 20 2012
    
  • Mathematica
    Select[Range[9000], DivisorSigma[1,#]== 2*# &] (* G. C. Greubel, Oct 03 2017 *)
    PerfectNumber[Range[15]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Dec 10 2018 *)
  • PARI
    isA000396(n) = (sigma(n) == 2*n);
    
  • Python
    from sympy import divisor_sigma
    def ok(n): return n > 0 and divisor_sigma(n) == 2*n
    print([k for k in range(9999) if ok(k)]) # Michael S. Branicky, Mar 12 2022

Formula

The perfect number N = 2^(p-1)*(2^p - 1) is also multiplicatively p-perfect (i.e., A007955(N) = N^p), since tau(N) = 2*p. - Lekraj Beedassy, Sep 21 2004
a(n) = 2^A133033(n) - 2^A090748(n), assuming there are no odd perfect numbers. - Omar E. Pol, Feb 28 2008
a(n) = A000668(n)*(A000668(n)+1)/2, assuming there are no odd perfect numbers. - Omar E. Pol, Apr 23 2008
a(n) = A000217(A000668(n)), assuming there are no odd perfect numbers. - Omar E. Pol, May 09 2008
a(n) = Sum of the first A000668(n) positive integers, assuming there are no odd perfect numbers. - Omar E. Pol, May 09 2008
a(n) = A000384(A019279(n)), assuming there are no odd perfect numbers and no odd superperfect numbers. a(n) = A000384(A061652(n)), assuming there are no odd perfect numbers. - Omar E. Pol, Aug 17 2008
a(n) = A006516(A000043(n)), assuming there are no odd perfect numbers. - Omar E. Pol, Aug 30 2008
From Reikku Kulon, Oct 14 2008: (Start)
A144912(2, a(n)) = 1;
A144912(4, a(n)) = -1 for n > 1;
A144912(8, a(n)) = 5 or -5 for all n except 2;
A144912(16, a(n)) = -4 or -13 for n > 1. (End)
a(n) = A019279(n)*A000668(n), assuming there are no odd perfect numbers and odd superperfect numbers. a(n) = A061652(n)*A000668(n), assuming there are no odd perfect numbers. - Omar E. Pol, Jan 09 2009
a(n) = A007691(A153800(n)), assuming there are no odd perfect numbers. - Omar E. Pol, Jan 14 2009
Even perfect numbers N = K*A000203(K), where K = A019279(n) = 2^(p-1), A000203(A019279(n)) = A000668(n) = 2^p - 1 = M(p), p = A000043(n). - Lekraj Beedassy, May 02 2009
a(n) = A060286(A016027(n)), assuming there are no odd perfect numbers. - Omar E. Pol, Dec 13 2012
For n >= 2, a(n) = Sum_{k=1..A065549(n)} (2*k-1)^3, assuming there are no odd perfect numbers. - Derek Orr, Sep 28 2013
a(n) = A275496(2^((A000043(n) - 1)/2)) - 2^A000043(n), assuming there are no odd perfect numbers. - Daniel Poveda Parrilla, Aug 16 2016
a(n) = A156552(A324201(n)), assuming there are no odd perfect numbers. - Antti Karttunen, Mar 28 2019
a(n) = ((2^(A000043(n)))^3 - (2^(A000043(n)) - 1)^3 - 1)/6, assuming there are no odd perfect numbers. - Jules Beauchamp, Jun 06 2025

Extensions

I removed a large number of comments that assumed there are no odd perfect numbers. There were so many it was getting hard to tell which comments were true and which were conjectures. - N. J. A. Sloane, Apr 16 2023
Reference to Albert H. Beiler's book updated by Harvey P. Dale, Jan 13 2025

A156552 Unary-encoded compressed factorization of natural numbers.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 8, 7, 6, 9, 16, 11, 32, 17, 10, 15, 64, 13, 128, 19, 18, 33, 256, 23, 12, 65, 14, 35, 512, 21, 1024, 31, 34, 129, 20, 27, 2048, 257, 66, 39, 4096, 37, 8192, 67, 22, 513, 16384, 47, 24, 25, 130, 131, 32768, 29, 36, 71, 258, 1025, 65536, 43, 131072, 2049, 38, 63, 68, 69, 262144
Offset: 1

Views

Author

Leonid Broukhis, Feb 09 2009

Keywords

Comments

The primes become the powers of 2 (2 -> 1, 3 -> 2, 5 -> 4, 7 -> 8); the composite numbers are formed by taking the values for the factors in the increasing order, multiplying them by the consecutive powers of 2, and summing. See the Example section.
From Antti Karttunen, Jun 27 2014: (Start)
The odd bisection (containing even terms) halved gives A244153.
The even bisection (containing odd terms), when one is subtracted from each and halved, gives this sequence back.
(End)
Question: Are there any other solutions that would satisfy the recurrence r(1) = 0; and for n > 1, r(n) = Sum_{d|n, d>1} 2^A033265(r(d)), apart from simple variants 2^k * A156552(n)? See also A297112, A297113. - Antti Karttunen, Dec 30 2017

Examples

			For 84 = 2*2*3*7 -> 1*1 + 1*2 + 2*4 + 8*8 =  75.
For 105 = 3*5*7 -> 2*1 + 4*2 + 8*4 = 42.
For 137 = p_33 -> 2^32 = 4294967296.
For 420 = 2*2*3*5*7 -> 1*1 + 1*2 + 2*4 + 4*8 + 8*16 = 171.
For 147 = 3*7*7 = p_2 * p_4 * p_4 -> 2*1 + 8*2 + 8*4 = 50.
		

Crossrefs

One less than A005941.
Inverse permutation: A005940 with starting offset 0 instead of 1.
Cf. also A297106, A297112 (Möbius transform), A297113, A153013, A290308, A300827, A323243, A323244, A323247, A324201, A324812 (n for which a(n) is a square), A324813, A324822, A324823, A324398, A324713, A324815, A324819, A324865, A324866, A324867.

Programs

  • Mathematica
    Table[Floor@ Total@ Flatten@ MapIndexed[#1 2^(#2 - 1) &, Flatten[ Table[2^(PrimePi@ #1 - 1), {#2}] & @@@ FactorInteger@ n]], {n, 67}] (* Michael De Vlieger, Sep 08 2016 *)
  • PARI
    a(n) = {my(f = factor(n), p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res}; \\ David A. Corneth, Mar 08 2019
    
  • PARI
    A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
    A156552(n) = if(1==n, 0, if(!(n%2), 1+(2*A156552(n/2)), 2*A156552(A064989(n)))); \\ (based on the given recurrence) - Antti Karttunen, Mar 08 2019
    
  • Perl
    # Program corrected per instructions from Leonid Broukhis. - Antti Karttunen, Jun 26 2014
    # However, it gives correct answers only up to n=136, before corruption by a wrap-around effect.
    # Note that the correct answer for n=137 is A156552(137) = 4294967296.
    $max = $ARGV[0];
    $pow = 0;
    foreach $i (2..$max) {
    @a = split(/ /, `factor $i`);
    shift @a;
    $shift = 0;
    $cur = 0;
    while ($n = int shift @a) {
    $prime{$n} = 1 << $pow++ if !defined($prime{$n});
    $cur |= $prime{$n} << $shift++;
    }
    print "$cur, ";
    }
    print "\n";
    (Scheme, with memoization-macro definec from Antti Karttunen's IntSeq-library, two different implementations)
    (definec (A156552 n) (cond ((= n 1) 0) (else (+ (A000079 (+ -2 (A001222 n) (A061395 n))) (A156552 (A052126 n))))))
    (definec (A156552 n) (cond ((= 1 n) (- n 1)) ((even? n) (+ 1 (* 2 (A156552 (/ n 2))))) (else (* 2 (A156552 (A064989 n))))))
    ;; Antti Karttunen, Jun 26 2014
    
  • Python
    from sympy import primepi, factorint
    def A156552(n): return sum((1<Chai Wah Wu, Mar 10 2023

Formula

From Antti Karttunen, Jun 26 2014: (Start)
a(1) = 0, a(n) = A000079(A001222(n)+A061395(n)-2) + a(A052126(n)).
a(1) = 0, a(2n) = 1+2*a(n), a(2n+1) = 2*a(A064989(2n+1)). [Compare to the entanglement recurrence A243071].
For n >= 0, a(2n+1) = 2*A244153(n+1). [Follows from the latter clause of the above formula.]
a(n) = A005941(n) - 1.
As a composition of related permutations:
a(n) = A003188(A243354(n)).
a(n) = A054429(A243071(n)).
For all n >= 1, A005940(1+a(n)) = n and for all n >= 0, a(A005940(n+1)) = n. [The offset-0 version of A005940 works as an inverse for this permutation.]
This permutations also maps between the partition-lists A112798 and A125106:
A056239(n) = A161511(a(n)). [The sums of parts of each partition (the total sizes).]
A003963(n) = A243499(a(n)). [And also the products of those parts.]
(End)
From Antti Karttunen, Oct 09 2016: (Start)
A161511(a(n)) = A056239(n).
A029837(1+a(n)) = A252464(n). [Binary width of terms.]
A080791(a(n)) = A252735(n). [Number of nonleading 0-bits.]
A000120(a(n)) = A001222(n). [Binary weight.]
For all n >= 2, A001511(a(n)) = A055396(n).
For all n >= 2, A000120(a(n))-1 = A252736(n). [Binary weight minus one.]
A252750(a(n)) = A252748(n).
a(A250246(n)) = A252754(n).
a(A005117(n)) = A277010(n). [Maps squarefree numbers to a permutation of A003714, fibbinary numbers.]
A085357(a(n)) = A008966(n). [Ditto for their characteristic functions.]
For all n >= 0:
a(A276076(n)) = A277012(n).
a(A276086(n)) = A277022(n).
a(A260443(n)) = A277020(n).
(End)
From Antti Karttunen, Dec 30 2017: (Start)
For n > 1, a(n) = Sum_{d|n, d>1} 2^A033265(a(d)). [See comments.]
More linking formulas:
A106737(a(n)) = A000005(n).
A290077(a(n)) = A000010(n).
A069010(a(n)) = A001221(n).
A136277(a(n)) = A181591(n).
A132971(a(n)) = A008683(n).
A106400(a(n)) = A008836(n).
A268411(a(n)) = A092248(n).
A037011(a(n)) = A010052(n) [conjectured, depends on the exact definition of A037011].
A278161(a(n)) = A046951(n).
A001316(a(n)) = A061142(n).
A277561(a(n)) = A034444(n).
A286575(a(n)) = A037445(n).
A246029(a(n)) = A181819(n).
A278159(a(n)) = A124859(n).
A246660(a(n)) = A112624(n).
A246596(a(n)) = A069739(n).
A295896(a(n)) = A053866(n).
A295875(a(n)) = A295297(n).
A284569(a(n)) = A072411(n).
A286574(a(n)) = A064547(n).
A048735(a(n)) = A292380(n).
A292272(a(n)) = A292382(n).
A244154(a(n)) = A048673(n), a(A064216(n)) = A244153(n).
A279344(a(n)) = A279339(n), a(A279338(n)) = A279343(n).
a(A277324(n)) = A277189(n).
A037800(a(n)) = A297155(n).
For n > 1, A033265(a(n)) = 1+A297113(n).
(End)
From Antti Karttunen, Mar 08 2019: (Start)
a(n) = A048675(n) + A323905(n).
a(A324201(n)) = A000396(n), provided there are no odd perfect numbers.
The following sequences are derived from or related to the base-2 expansion of a(n):
A000265(a(n)) = A322993(n).
A002487(a(n)) = A323902(n).
A005187(a(n)) = A323247(n).
A324288(a(n)) = A324116(n).
A323505(a(n)) = A323508(n).
A079559(a(n)) = A323512(n).
A085405(a(n)) = A323239(n).
The following sequences are obtained by applying to a(n) a function that depends on the prime factorization of its argument, which goes "against the grain" because a(n) is the binary code of the factorization of n, which in these cases is then factored again:
A000203(a(n)) = A323243(n).
A033879(a(n)) = A323244(n) = 2*a(n) - A323243(n),
A294898(a(n)) = A323248(n).
A000005(a(n)) = A324105(n).
A000010(a(n)) = A324104(n).
A083254(a(n)) = A324103(n).
A001227(a(n)) = A324117(n).
A000593(a(n)) = A324118(n).
A001221(a(n)) = A324119(n).
A009194(a(n)) = A324396(n).
A318458(a(n)) = A324398(n).
A192895(a(n)) = A324100(n).
A106315(a(n)) = A324051(n).
A010052(a(n)) = A324822(n).
A053866(a(n)) = A324823(n).
A001065(a(n)) = A324865(n) = A323243(n) - a(n),
A318456(a(n)) = A324866(n) = A324865(n) OR a(n),
A318457(a(n)) = A324867(n) = A324865(n) XOR a(n),
A318458(a(n)) = A324398(n) = A324865(n) AND a(n),
A318466(a(n)) = A324819(n) = A323243(n) OR 2*a(n),
A318467(a(n)) = A324713(n) = A323243(n) XOR 2*a(n),
A318468(a(n)) = A324815(n) = A323243(n) AND 2*a(n).
(End)

Extensions

More terms from Antti Karttunen, Jun 28 2014

A323244 a(1) = 0; and for n > 1, a(n) = A033879(A156552(n)).

Original entry on oeis.org

0, 1, 1, 2, 1, 4, 1, 6, 0, 5, 1, 10, 1, 16, 2, 6, 1, 12, 1, 18, -3, 18, 1, 22, -4, 46, 4, 22, 1, 10, 1, 30, 14, 82, -2, 14, 1, 256, -12, 22, 1, 36, 1, 66, 8, 226, 1, 46, -12, 19, 8, 130, 1, 28, -19, 70, -12, 748, 1, 42, 1, 1362, 16, 22, 10, 42, 1, 214, 254, 40, 1, 38, 1, 3838, 10, 406, -10, 106, 1, 78, -12, 5458, 1, 26, -72, 12250, -348, 30, 1, 12
Offset: 1

Views

Author

Antti Karttunen, Jan 10 2019

Keywords

Comments

After a(1) = 0, the other zeros occur for k >= 1, at A005940(1+A000396(k)), which, provided no odd perfect numbers exist, is equal to A324201(k) = A062457(A000043(k)): 9, 125, 161051, 410338673, ..., etc.
There are 2321 negative terms among the first 10000 terms.

Crossrefs

Cf. A324201 (positions of zeros, conjectured), A324551 (of negative terms), A324720 (of nonnegative terms), A324721 (of positive terms), A324731, A324732.
Cf. A329644 (Möbius transform).
Cf. A323174, A324055, A324185, A324546 for other permutations of deficiency, and also A324574, A324575, A324654.

Programs

  • Mathematica
    Array[2 # - If[# == 0, 0, DivisorSigma[1, #]] &@ Floor@ Total@ Flatten@ MapIndexed[#1 2^(#2 - 1) &, Flatten[Table[2^(PrimePi@ #1 - 1), {#2}] & @@@ FactorInteger@ #]] &, 90] (* Michael De Vlieger, Apr 21 2019 *)
  • PARI
    A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
    A156552(n) = if(1==n, 0, if(!(n%2), 1+(2*A156552(n/2)), 2*A156552(A064989(n))));
    A323244(n) = if(1==n, 0, my(k=A156552(n)); (2*k)-sigma(k));
    
  • Python
    from sympy import divisor_sigma, primepi, factorint
    def A323244(n): return (lambda n: (n<<1)-divisor_sigma(n))(sum((1< 1 else 0 # Chai Wah Wu, Mar 10 2023

Formula

a(n) = 2*A156552(n) - A323243(n).
a(1) = 0; and for n > 1, a(n) = A033879(A156552(n)).
a(n) = A323248(n) + A001222(n) = (A323247(n) - A323243(n)) + A001222(n).
From Antti Karttunen, Mar 12 2019 & Nov 23 2019: (Start)
a(n) = Sum_{d|n} (2*A297112(d) - A324543(d)) = Sum_{d|n} A329644(d).
A002487(a(n)) = A324115(n).
a(n) = A329638(n) - A329639(n).
a(n) = A329645(n) - A329646(n).
(End)

A324543 Möbius transform of A323243, where A323243(n) = sigma(A156552(n)).

Original entry on oeis.org

0, 1, 3, 3, 7, 2, 15, 4, 9, 5, 31, 3, 63, 2, 8, 16, 127, -1, 255, 4, 21, 16, 511, 8, 21, 20, 12, 27, 1023, 6, 2047, 8, 20, 48, 20, 20, 4095, 2, 78, 32, 8191, -6, 16383, 17, 9, 288, 32767, 8, 45, -3, 122, 45, 65535, 4, 53, 20, 270, 278, 131071, 2, 262143, 688, 12, 72, 56, 23, 524287, 125, 260, -8, 1048575, 20, 2097151, 260, 3, 363, 44, -7, 4194303
Offset: 1

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Author

Antti Karttunen, Mar 07 2019

Keywords

Comments

The first four zeros after a(1) occur at n = 192, 288, 3645, 6075.
There are 1562 negative terms among the first 10000 terms.
Applying this function to the divisors of the first four terms of A324201 reveals the following pattern:
----------------------------------------------------------------------------------
A324201 divisors a(n) applied to each: Sum
9: [1, 3, 9] -> [0, 3, 9] 12 = 2*6
125: [1, 5, 25, 125] -> [0, 7, 21, 28] 56 = 2*28
161051: [1, 11, 121, 1331, 14641, 161051] -> [0, 31, 93, 124, 496, 248] 992 = 2*496
410338673: [1, 17, 289, 4913, 83521, 1419857, 24137569, 410338673]
-> [0, 127, 381, 508, 2032, 1016, 9144, 3048] 16256 = 2*8128
The second term (the first nonzero) of the latter list = A000668(n), and the sum is always twice the corresponding perfect number, which forces either it or at least many of its divisors to be present. For example, in the fourth case, although 8128 = A000396(4) itself is not present, we still have 127, 508, 1016 and 2032 in the list. See also A329644.

Crossrefs

Programs

  • Mathematica
    Table[DivisorSum[n, MoebiusMu[n/#] If[# == 1, 0, DivisorSigma[1, Floor@ Total@ Flatten@ MapIndexed[#1 2^(#2 - 1) &, Flatten[Table[2^(PrimePi@ #1 - 1), {#2}] & @@@ FactorInteger@ #]]]] &], {n, 79}] (* Michael De Vlieger, Mar 11 2019 *)
  • PARI
    A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
    A156552(n) = if(1==n, 0, if(!(n%2), 1+(2*A156552(n/2)), 2*A156552(A064989(n))));
    memoA323243 = Map();
    A323243(n) = if(1==n, 0, my(v); if(mapisdefined(memoA323243,n,&v),v, v=sigma(A156552(n)); mapput(memoA323243,n,v); (v)));
    A324543(n) = sumdiv(n,d,moebius(n/d)*A323243(d));

Formula

a(n) = Sum_{d|n} A008683(n/d) * A323243(d).
a(A000040(n)) = A000225(n).
a(A001248(n)) = A173033(n) - A000225(n) = A068156(n) = 3*(2^n - 1).
a(2*A000040(n)) = A324549(n).
a(A002110(n)) = A324547(n).
a(n) = 2*A297112(n) - A329644(n), and for n > 1, a(n) = 2^A297113(n) - A329644(n). - Antti Karttunen, Dec 08 2019

A329644 Möbius transform of A323244, the deficiency of A156552(n).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 4, -1, 3, 1, 5, 1, 14, 0, 0, 1, 9, 1, 12, -5, 16, 1, 8, -5, 44, 4, 5, 1, 2, 1, 24, 12, 80, -4, -4, 1, 254, -14, 0, 1, 22, 1, 47, 7, 224, 1, 24, -13, 19, 6, 83, 1, 12, -21, 44, -14, 746, 1, 14, 1, 1360, 20, -8, 8, 9, 1, 131, 252, 24, 1, 12, 1, 3836, 13, 149, -12, 71, 1, 56, -16, 5456, 1, -21, -74, 12248, -350, -40, 1
Offset: 1

Views

Author

Antti Karttunen, Nov 21 2019

Keywords

Comments

The first eleven zeros occur at n = 1, 15, 16, 40, 96, 119, 120, 160, 893, 2464, 6731. There are 3091 negative terms among the first 10000 terms.
Applying this function to the divisors of the first four terms of A324201 reveals the following pattern:
------------------------------------------------------------------------------------
A324201(n) divisors a(n) applied Sum of positive
to each: terms, A329610
9: [1, 3, 9] -> [0, 1, -1] 1
125: [1, 5, 25, 125] -> [0, 1, -5, 4] 5
161051: [1, 11, 121, 1331, 14641, 161051] -> [0, 1, -29, 4, -240, 264] 269
410338673: [1, 17, 289, 4913, 83521, 1419857, 24137569, 410338673]
-> [0, 1, -125, 4, -1008, 1032, -5048, 5144] 6181
The positive and negative terms seem to alternate, and the fourth term (from case n=125 onward) is always 4. See also array A329637.

Crossrefs

Programs

  • PARI
    A156552(n) = {my(f = factor(n), p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res}; \\ From A156552
    A323244(n) = if(1==n, 0, my(k=A156552(n)); (2*k)-sigma(k));
    A329644(n) = sumdiv(n,d,moebius(n/d)*A323244(d));

Formula

a(n) = Sum_{d|n} A008683(n/d) * A323244(d).
a(n) = Sum_{d|n} A008683(n/d) * (2*A156552(d) - A323243(d)).
a(1) = 0; for n > 1, a(n) = 2*A297112(n) - A324543(n) = 2^A297113(n) - A324543(n).
a(n) = A329642(n) - A329643(n).
For all n >= 1, a(A000040(n)^2) = A323244(A000040(n)^2)-1 = -A036563(n).
For all primes p, a(p^3) = A323244(p^3) - A323244(p^2) = 4.

A300827 Lexicographically earliest sequence such that a(i) = a(j) => A324193(i) = A324193(j) for all i, j >= 1.

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 4, 5, 6, 2, 7, 2, 8, 9, 10, 2, 11, 2, 12, 13, 14, 2, 15, 16, 17, 9, 18, 2, 19, 2, 20, 21, 22, 23, 24, 2, 25, 26, 27, 2, 28, 2, 29, 30, 31, 2, 32, 33, 34, 35, 36, 2, 37, 38, 39, 40, 41, 2, 42, 2, 43, 44, 45, 46, 47, 2, 48, 49, 50, 2, 51, 2, 52, 53, 54, 55, 56, 2, 57, 58, 59, 2, 60, 61, 62, 63, 64, 2, 65, 66, 67
Offset: 1

Views

Author

Antti Karttunen, Mar 13 2018

Keywords

Comments

Apart from primes, the sequence contains duplicate values at points p*q and p^3, where p*q are the product of two successive primes, with p < q (sequences A006094, A030078). Question: are there any other cases where a(x) = a(y), with x < y ?
The reason why this is not equal to A297169: Even though A297112 contains only powers of two after the initial zero, as A297112(n) = 2^A033265(A156552(d)) for n > 1, and A297168(n) is computed as Sum_{d|n, dA297112(d), still a single 1-bit in binary expansion of A297168(n) might be formed as a sum of several terms of A297112(d), i.e., could be born of carries.
From Antti Karttunen, Feb 28 2019: (Start)
A297168(n) = Sum_{d|n, dA297112(d) will not produce any carries (in base-2) if and only if n is a power of prime. Only in that case the number of summands (A000005(n)-1) is equal to the number of prime factors counted with multiplicity, A001222(n) = A000120(A156552(n)). (A notable subset of such numbers is A324201, numbers that are mapped to even perfect numbers by A156552). Precisely because there are so few points with duplicate values (apart from primes), this sequence is not particularly good for filtering other sequences, because the number of false positives is high. Any of the related sequences like A324203, A324196, A324197 or A324181 might work better in that respect. In any case, the following implications hold (see formula section of A324193 for the latter): (End)
For all i, j:
a(i) = a(j) => A297168(i) = A297168(j). (The same holds for A297169).
a(i) = a(j) => A324181(i) = A324181(j) => A324120(i) = A324120(j).

Examples

			For n = 15, with proper divisors 3 and 5, we have f(n) = prime(1+A297167(3)) * prime(1+A297167(5)) = prime(2)*prime(3) = 3*5 = 15.
For n = 27, with proper divisors 3 and 9, we have f(n) = prime(1+A297167(3)) * prime(1+A297167(9)) = prime(2)*prime(3) = 3*5 = 15.
Because f(15) = f(27), the restricted growth sequence transform allots the same number (in this case 9) for both, so a(15) = a(27) = 9.
		

Crossrefs

Programs

  • PARI
    up_to = 65537;
    rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; };
    write_to_bfile(start_offset,vec,bfilename) = { for(n=1, length(vec), write(bfilename, (n+start_offset)-1, " ", vec[n])); }
    A061395(n) = if(1==n, 0, primepi(vecmax(factor(n)[, 1]))); \\ After M. F. Hasler's code for A006530.
    A297167(n) = if(1==n, 0, (A061395(n) + (bigomega(n)-omega(n)) -1));
    Aux300827(n) = { my(m=1); if(n<=2, n-1, fordiv(n,d,if((d>1)&(dA297167(d)))); (m)); };
    write_to_bfile(1,rgs_transform(vector(up_to,n,Aux300827(n))),"b300827.txt");

Formula

Restricted growth sequence transform of sequence f, defined as f(1) = 0, f(2) = 1, and for n > 2, f(n) = Product_{d|n, 1A297167(d)).
a(p) = 2 for all primes p.
a(A006094(n)) = a(A030078(n)), for all n >= 1.

Extensions

Name changed by Antti Karttunen, Feb 21 2019

A332223 a(1) = 1, and for n > 1, a(n) = A005940(1+sigma(A156552(n))).

Original entry on oeis.org

1, 2, 4, 5, 8, 9, 16, 7, 25, 18, 32, 25, 64, 21, 21, 49, 128, 27, 256, 35, 40, 121, 512, 49, 125, 385, 49, 121, 1024, 13, 2048, 13, 225, 1573, 105, 77, 4096, 57, 187, 343, 8192, 63, 16384, 65, 55, 4693, 32768, 121, 625, 32, 15625, 85, 65536, 81, 180, 91, 253, 9945, 131072, 175, 262144, 508079, 625, 847, 729, 169, 524288, 2057, 2601, 105, 1048576
Offset: 1

Views

Author

Antti Karttunen, Feb 12 2020

Keywords

Comments

From Antti Karttunen, Jul 31 - Aug 06 2020: (Start)
As a curiosity, like with sigma, also here a(14) = a(15). [Cf. also A003973 and A341512]
Question: is it possible that a(k) = 2*k for any k? If not, then the deficiency (A033879) cannot be -1, and there are no quasiperfect numbers. If there were such cases, then A156552(k) = q would be an instance of quasiperfect number, which should also be an odd square, thus k would need to be of the form 4u+2.
In range n <= 10000, a(n) is a nontrivial multiple of n only at n = [25, 35, 343, 539, 847, 3315] with a(n) = [125, 105, 2401, 2695, 2541, 9945]. The quotients are thus also odd: 5, 3, 7, 5, 3, 3.
This rather meager empirical evidence motivates a conjecture that no quotient a(n)/n may be an even integer, and particularly, never a power of 2 larger than one, which (when translated back to the ordinary, unconjugated sigma) claims that it is not possible that sigma(n) = 2^k * n + 2^k - 1, for any n > 1, k > 0. See also A336700 and A336701, where this leads to a rather surprising empirical observation.
(End)

Crossrefs

Cf. A003961, A332449, A332450, A332451, A332460 (for other functions similarly conjugated).

Programs

  • PARI
    A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); t }; \\ From A005940
    A156552(n) = {my(f = factor(n), p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res}; \\ From A156552
    A332223(n) = if(1==n,n,A005940(1+sigma(A156552(n))));
    
  • PARI
    A332223(n) = if(1==n,n,A005940(1+sumdiv(A156552(n),d,d))); \\ Antti Karttunen, Aug 04 2020

Formula

For n > 1, a(n) = A005940(1+A000203(A156552(n))) = A005940(1+A323243(n)).
a(A324201(n)) = A003961(A324201(n)). [It's an open problem whether A324201 gives all such solutions]
For n > 1, a(n) = A005940(1 + (Sum_{d|A156552(n)} d)). - Antti Karttunen, Aug 04 2020

A324120 Binary weight of SumXOR variant of A297168: a(n) = A000120(A324180(n)).

Original entry on oeis.org

0, 0, 0, 1, 0, 2, 0, 2, 1, 2, 0, 2, 0, 2, 2, 3, 0, 2, 0, 2, 2, 2, 0, 2, 1, 2, 2, 2, 0, 2, 0, 4, 2, 2, 2, 3, 0, 2, 2, 4, 0, 2, 0, 2, 2, 2, 0, 2, 1, 2, 2, 2, 0, 2, 2, 4, 2, 2, 0, 4, 0, 2, 2, 5, 2, 2, 0, 2, 2, 2, 0, 2, 0, 2, 2, 2, 2, 2, 0, 4, 3, 2, 0, 4, 2, 2, 2, 4, 0, 2, 2, 2, 2, 2, 2, 2, 0, 2, 2, 3, 0, 2, 0, 4, 2
Offset: 1

Views

Author

Antti Karttunen, Feb 19 2019

Keywords

Crossrefs

Programs

Formula

a(n) = A000120(A324180(n)).
a(n) <= A324190(n).
a(p^k) = k-1 for all primes p and exponents k >= 1.

A324200 a(n) = 2^(A000043(n)-1) * ((2^A059305(n)) - 1), where A059305 gives the prime index of the n-th Mersenne prime, while A000043 gives its exponent.

Original entry on oeis.org

6, 60, 32752, 137438953408
Offset: 1

Views

Author

Antti Karttunen, Feb 18 2019

Keywords

Comments

If there are no odd perfect numbers then these are the positions of zeros in A324185.
The next term has 314 digits:
11781361728633673532894774498354952494238773929196300355071513798753168641589311119865182769801300280680127783231251635087526446289021607771691249214388576215221396663491984443067742263787264024212477244347842938066577043117995647400274369612403653814737339068225047641453182709824206687753689912418253153056583680.

Crossrefs

Programs

Formula

a(n) = ((2^A000720(A000668(n)))-1) * 2^(A000043(n)-1) = ((2^A059305(n)) - 1) * 2^(A000043(n)-1).
If no odd perfect numbers exist, then a(n) = A243071(A000396(n)), and thus A007814(a(n)) = A007814(A000396(n)).
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