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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A092509 Möbius transform of sequence A008475.

Original entry on oeis.org

0, 2, 3, 2, 5, 0, 7, 4, 6, 0, 11, 0, 13, 0, 0, 8, 17, 0, 19, 0, 0, 0, 23, 0, 20, 0, 18, 0, 29, 0, 31, 16, 0, 0, 0, 0, 37, 0, 0, 0, 41, 0, 43, 0, 0, 0, 47, 0, 42, 0, 0, 0, 53, 0, 0, 0, 0, 0, 59, 0, 61, 0, 0, 32, 0, 0, 67, 0, 0, 0, 71, 0, 73, 0, 0, 0, 0, 0, 79, 0, 54, 0, 83, 0, 0, 0, 0, 0, 89, 0, 0, 0, 0
Offset: 1

Views

Author

Leroy Quet, Dec 31 2004

Keywords

Crossrefs

Programs

  • Mathematica
    f[n_] := Plus @@ Map[Power @@ #1 &, FactorInteger[n]]; mt[n_] := Block[{d = Divisors[n]}, Plus @@ (MoebiusMu /@ (n/d)*f /@ d)]; Table[ mt[n], {n, 93}] (* Robert G. Wilson v, Jan 12 2005 *)
  • PARI
    A008475(n) = { my(f=factor(n)); vecsum(vector(#f~,i,f[i,1]^f[i,2])); };
    A092509(n) = sumdiv(n,d,moebius(d)*A008475(n/d)); \\ Antti Karttunen, Nov 17 2017

Formula

a(n) = 0 if n is 1 or not a power of a prime;
a(n) = n if n is a prime;
a(n) = n*(1 -1/p) if n = p^k, k>= 2, p=prime.
a(n) = Sum_{d|n} A008683(d)*A008475(n/d). - Antti Karttunen, Nov 17 2017

Extensions

More terms from Robert G. Wilson v, Jan 13 2005

A167515 The sum over the divisors of n, except the maximum-prime-power divisors collected in A008475.

Original entry on oeis.org

1, 1, 1, 3, 1, 7, 1, 7, 4, 11, 1, 21, 1, 15, 16, 15, 1, 28, 1, 33, 22, 23, 1, 49, 6, 27, 13, 45, 1, 62, 1, 31, 34, 35, 36, 78, 1, 39, 40, 77, 1, 84, 1, 69, 64, 47, 1, 105, 8, 66, 52, 81, 1, 91, 56, 105, 58, 59, 1, 156, 1, 63, 88, 63, 66, 128, 1, 105, 70
Offset: 1

Views

Author

Jaroslav Krizek, Dec 15 2010

Keywords

Comments

If n = Product (p_j^k_j) is the standard prime power decomposition of n, there is a set of size A001221(n) which contains the divisors which are largest powers of primes, {p_1^k_1, p_2^k_2, ..., p_j^k_j}. a(n) sums all the divisors not in this set. If p, q, ..., z are distinct primes, k are natural numbers (A000027), p^k prime powers (A000961), the following formulas hold: a(p) = 1. a(pq) = pq+1. a(pq...z) = (p+1)* (q+1)* ... *(z+1) - (p+q+ ...+z). a(p^k) = (p^k-1)/(p-1).

Examples

			For n = 12, set of prime-power-factor divisors of 12: {3, 4}, set of non-(prime-power-factor) divisors on 12: {1, 2, 6, 12}. a(12) = 1+2+6+12=21.
		

Crossrefs

Programs

Formula

a(n) = A000203(n) - A008475(n).
a(n) = A178636(n) + 1.

A333801 Numbers k such that A008475(k)+1 = A008475(k+1).

Original entry on oeis.org

2, 3, 4, 7, 8, 16, 20, 31, 35, 127, 143, 208, 256, 650, 1479, 2464, 2623, 4233, 4345, 5183, 8099, 8191, 9424, 11024, 11919, 12099, 14905, 16159, 20220, 20800, 21716, 22194, 24335, 26123, 27335, 27390, 30457, 34945, 38180, 40425, 52206, 56563, 65536, 67123, 68264
Offset: 1

Views

Author

Amiram Eldar, Apr 05 2020

Keywords

Comments

A variation of A064111 and A228126 with unitary prime-power divisors instead of prime divisors.

Examples

			4 is a term since A008475(4) + 1 = 4 + 1 = 5 = A008475(5).
		

Crossrefs

Programs

  • Mathematica
    s[1] = 0; s[n_] := Plus @@ (Power @@@ FactorInteger[n]); seq = {}; s1 = 0; Do[s2 = s[n]; If[s1 + 1 == s2, AppendTo[seq, n - 1]]; s1 = s2, {n, 2, 10^5}]; seq

A002497 Numbers N in A002809 such that there is rho > 0 such that for all A > 0, A008475(A)-A008475(N) >= rho*log(A/N).

Original entry on oeis.org

3, 12, 60, 420, 4620, 60060, 180180, 360360, 6126120, 116396280, 2677114440, 77636318760, 2406725881560, 89048857617720, 3651003162326520, 156993135980040360, 313986271960080720, 14757354782123793840, 14757354782123793840, 782139803452561073520, 46146248403701103337680
Offset: 1

Views

Author

Keywords

Comments

The numbers contain the starred entries on pp. 187-190 of Nicolas. It is a subsequence of A002809 by selecting only elements of a set/property "G" (page 150). G contains all N such that a real, strictly positive rho exists such that for all strictly positive integers A we have l(A)-l(N) >= rho*log(A/N). The function l()=A008475() is defined on page 139. - R. J. Mathar, Mar 23 2012
These numbers were named superior l-composite numbers (nombres l-composes superieurs, the function l(n) is A002809) by Massias, in analogy to Ramanujan's superior highly composite numbers (A002201). Deléglise and Nicolas named these numbers l-superchampion numbers. They are used by Deléglise et al. in calculating values of Landau's function g(n) (A000793). - Amiram Eldar, Aug 23 2019

References

  • 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).

Crossrefs

Extensions

Edited by M. F. Hasler, Mar 29 2015
a(16)-a(21) from the paper by Massias added by Amiram Eldar, Aug 23 2019

A114521 a(n) = A008475(A114520(n)).

Original entry on oeis.org

5, 7, 7, 11, 13, 11, 11, 19, 13, 13, 19, 17, 29, 31, 17, 23, 43, 19, 29, 31, 23, 61, 73, 41, 37, 83, 19, 47, 31, 67, 53, 103, 29, 17, 109, 37, 127, 71, 23, 139, 41, 151, 83, 31, 43, 181, 193, 131, 23, 101, 23, 199, 29, 41, 107, 19, 61, 43, 37, 113, 71, 229, 23, 67, 241
Offset: 1

Views

Author

Leroy Quet, Dec 05 2005

Keywords

Examples

			24 (which is composite) is the sixth term of sequence A114520. 24 = 2^3 * 3 and 2^3 + 3 = 11 (which is prime). So a(6) = 11.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Plus @@ Power @@@ FactorInteger[n]; f /@ Select[Select[Range[500], PrimeQ[f[ # ]] &], ! PrimeQ[ # ] &] (* Ray Chandler, Dec 07 2005 *)
  • PARI
    A008475(n)=local(t);if(n<1,0,t=factor(n);sum(k=1,matsize(t)[1],t[k,1]^t[k,2])); for(i=1,900,if(!isprime(i)&&isprime(A008475(i)),print1(A008475(i),","))) (Herrgesell)

Extensions

Extended by Ray Chandler and Lambert Herrgesell (zero815(AT)googlemail.com), Dec 07 2005

A163408 Positive integers n such that A008475(n) is composite.

Original entry on oeis.org

4, 8, 9, 14, 15, 16, 20, 21, 25, 26, 27, 30, 32, 33, 35, 38, 39, 42, 44, 45, 46, 49, 50, 51, 55, 56, 57, 60, 62, 63, 64, 65, 66, 68, 69, 70, 74, 75, 77, 78, 80, 81, 84, 85, 86, 87, 90, 91, 92, 93, 94, 95, 96, 98, 99, 102, 104, 105, 106, 110, 111, 114, 115, 116
Offset: 1

Views

Author

Leroy Quet, Jul 27 2009

Keywords

Comments

Every integer >=2 is exclusively either in this sequence or in sequence A114518.

Crossrefs

Extensions

More terms from Franklin T. Adams-Watters, Aug 25 2011

A163409 Subsequence of composite terms of A008475.

Original entry on oeis.org

4, 8, 9, 9, 8, 16, 9, 10, 25, 15, 27, 10, 32, 14, 12, 21, 16, 12, 15, 14, 25, 49, 27, 20, 16, 15, 22, 12, 33, 16, 64, 18, 16, 21, 26, 14, 39, 28, 18, 18, 21, 81, 14, 22, 45, 32, 16, 20, 27, 34, 49, 24, 35, 51, 20, 22, 21, 15, 55, 18, 40, 24, 28, 33, 22, 24, 16, 121, 63, 44, 35, 125, 18, 128, 46, 20
Offset: 1

Views

Author

Leroy Quet, Jul 27 2009

Keywords

Crossrefs

Formula

a(n) = A008475(A163408(n)).

Extensions

More terms from Max Alekseyev, Oct 14 2012

A038701 Prime powers q for which f(g(m(q))) = m(q), where f = A051703, g = A008475 and m = A003418.

Original entry on oeis.org

2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17, 19, 23, 25, 29, 31, 32, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 103, 107, 109, 113
Offset: 1

Views

Author

Vladeta Jovovic, May 01 2000

Keywords

Comments

These functions are defined for all natural numbers > 1 by: g(x) = Sum (p_j^k_j) where x = Product (p_j^k_j) is prime factorization of x (A008475); f(n) = max{x:g(x)=n} (A051703); m(n) = lcm(1,2,3,...,n) (A003418).
There are no more prime powers in the list <= 199. Conjecture: The sequence is finite, i.e., f(g(m(q))) > m(q) for sufficiently great prime powers q.
No other terms below 409. - Max Alekseyev, Sep 05 2023

Examples

			27 is not in the list because m(27) = 2^4*3^3*5^2*7*11*13*17*19*23, g(m(27))=158, f(158) = 3*5*7*11*13*17*19*23*29*31 > m(27).
		

Crossrefs

Extensions

Offset changed to 1 by Jinyuan Wang, Mar 16 2020

A082085 Fixed points when A008475 is iterated started at factorials of prime numbers.

Original entry on oeis.org

2, 5, 16, 37, 11, 23, 31, 67, 25, 193, 7, 7, 19, 19, 5939, 27, 13, 11, 11, 503, 15889, 37, 11, 4651, 52960025378359863409578953, 8, 13, 11, 25, 79, 19, 25, 56707367, 7, 103, 23, 9, 61
Offset: 1

Views

Author

Labos Elemer, Apr 08 2003

Keywords

Examples

			n=25: p(25)=97, start with 97!, end at a large prime: 52960025378359863409578953=a(25); it seems difficult to predict magnitude of fixed point.
		

Crossrefs

Formula

a(n) = A082081(A000142(A000040(n))).

A354424 Numbers k for which the ratio A008475(k)/k reaches a record low.

Original entry on oeis.org

2, 6, 10, 12, 15, 20, 28, 30, 40, 42, 56, 60, 84, 105, 120, 140, 168, 180, 210, 252, 280, 315, 330, 360, 385, 390, 420, 616, 630, 660, 770, 780, 840, 924, 1092, 1155, 1260, 1540, 1820, 1848, 1980, 2184, 2310, 2520, 2730, 3080, 3465, 3640, 3960, 4095, 4290, 4620, 5460, 6552, 6930
Offset: 1

Views

Author

Chris Grossack, Jul 11 2022

Keywords

Comments

Sequence gives the numbers k for which m/k reaches a record low, where m is minimal so that the symmetric group S_m has an element of order k.

Examples

			First, an element of order 2 shows up in S_2, so the smallest ratio we've seen so far is 1. This is the smallest ratio we see until we reach 6, since there's an element of order 6 in S_5. Next is 10, since there's an element of order 10 in S_7, and 7/10 is the next ratio smaller than 5/6. Then comes 12, since S_7 also has an element of order 12, and 7/12 is the next ratio less than 7/10, etc.
		

Crossrefs

Cf. A008475.

Programs

  • Mathematica
    s = {}; fm = 2; Do[If[(f = Plus @@ Power @@@ FactorInteger[n]/n) < fm, fm = f; AppendTo[s, n]], {n, 2, 7000}]; s (* Amiram Eldar, Jul 12 2022 *)
  • PARI
    b(n) = my(f=factor(n)); vecsum(vector(#f~, i, f[i, 1]^f[i, 2])); \\ A008475
    lista(nn) = my(m=oo, list=List(), x); for (n=2, nn, if ((x=b(n)/n) < m, m = x; listput(list, n););); Vec(list); \\ Michel Marcus, Jul 12 2022
  • Sage
    memo = {1: (2,1)}
    def a(n):
        if n in memo.keys(): return memo[n]
        _ = a(n-1)
        prev, prevRatio = memo[n-1]
        ratio = 1
        N = prev
        while ratio >= prevRatio:
            N += 1
            # compute m so that S_m has an element of order N
            principalDivisors = list(factor(N))
            m = sum([a^b for (a,b) in principalDivisors])
            ratio = m/N
        memo[n] = (N, ratio)
        return N
    
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