A046343 -id:A046343 - cp's OEIS frontend

A154367 Numbers k such that the sum of the prime factors of composite(k) (with multiplicity) is prime and lpf(composite(k)) + gpf(composite(k)) is composite.

18, 30, 36, 39, 44, 53, 54, 73, 76, 86, 112, 113, 116, 126, 132, 134, 141, 160, 163, 175, 191, 194, 197, 211, 214, 219, 231, 233, 250, 258, 265, 276, 279, 294, 295, 301, 308, 311, 312, 320, 325, 331, 333, 335, 338, 340, 341, 350, 351, 361, 376, 383, 385, 394

Offset: 1

Juri-Stepan Gerasimov, Jan 08 2009

nonn

			18 is a term because composite(18) = 28 = 2*2*7, 2 + 2 + 7 = 11 is prime, and 2 + 7 = 9 is composite.
30 is a term because composite(30) = 45 = 3*3*5, 3 + 3 + 5 = 11 is prime, and 3 + 5 = 8 is composite.

Cf. A000040 (primes), A002808 (composites).

isA002808 := proc(n) n >= 4 and not isprime(n) ; end proc:
A046343 := proc(n) pss(A002808(n)) ; end proc:
A020639 := proc(n) numtheory[factorset](n) ; min(op(%)) ; end proc:
A006530 := proc(n) numtheory[factorset](n) ; max(op(%)) ; end proc:
for n from 1 to 500 do if isprime(A046343(n)) and isA002808( A020639(A002808(n)) + A006530(A002808(n)) ) then printf("%d,",n) ; end if; end do: # R. J. Mathar, May 05 2010

Corrected (44 inserted, 120 removed, 146 removed) and extended by R. J. Mathar, May 05 2010

Name and Example section edited by Jon E. Schoenfield, Feb 11 2019

A273777 Consider all ways of writing the n-th composite number as the product of two divisors d1d2 = d3d4 = ... where each divisor is larger than 1; a(n) is the maximum of the sums {d1 + d2, d3 + d4, ...}.

Original entry on oeis.org

4, 5, 6, 6, 7, 8, 9, 8, 10, 11, 12, 10, 13, 14, 10, 15, 12, 16, 17, 18, 14, 19, 12, 20, 21, 16, 22, 23, 24, 18, 25, 26, 14, 27, 20, 28, 29, 16, 30, 22, 31, 32, 33, 24, 34, 18, 35, 36, 26, 37, 38, 39, 28, 40, 18, 41, 42, 30, 43, 44, 22, 45, 32, 46, 47, 20, 48

Offset: 1

Michel Lagneau, May 30 2016

nonn

The divisors must be > 1 and < n.

For the minimum sums see A273227.

			a(14) = 14 because A002808(14) = 24 = 2*12 = 3*8 = 4*6 and 2+12 = 14 is the maximum sum.

Cf. A002808, A020639, A046343, A063655, A273227.

with(numtheory):nn:=100:lst:={}:
for n from 1 to nn do:
it:=0:lst:={}:
d:=divisors(n):n0:=nops(d):
  if n0>2 then
  for i from 2 to n0-1 do:
   p:=d[i]:
    for j from i to n0-1 do:
      q:=d[j]:
       if p*q=n then
        lst:=lst union {p+q}:
        else
       fi:
     od:
    od:
    n0:=nops(lst):printf(`%d, `, lst[n0]):
   fi:
   od:

Function[n, Max@ Map[Plus[#, n/#] &, Rest@ Take[#, Ceiling[Length[#]/2]]] &@ Divisors@ n] /@ Select[Range@ 120, CompositeQ] (* Michael De Vlieger, May 30 2016 *)

lista(nn) = {forcomposite(n=2, nn, m = 0; fordiv(n, d, if ((d != 1) && (d != n), m = max(m, d+n/d));); print1(m, ", "););} \\ Michel Marcus, Sep 13 2017

Let m = A002808(n). Then a(n) = A020639(m) + m / A020639(m).

Name edited by Jon E. Schoenfield, Sep 12 2017

cp's OEIS Frontend

A154367 Numbers k such that the sum of the prime factors of composite(k) (with multiplicity) is prime and lpf(composite(k)) + gpf(composite(k)) is composite.

Views

Author

Keywords

Examples

Crossrefs

Programs

Maple

Extensions

A273777 Consider all ways of writing the n-th composite number as the product of two divisors d1d2 = d3d4 = ... where each divisor is larger than 1; a(n) is the maximum of the sums {d1 + d2, d3 + d4, ...}.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

cp's OEIS Frontend

A154367 Numbers k such that the sum of the prime factors of composite(k) (with multiplicity) is prime and lpf(composite(k)) + gpf(composite(k)) is composite.

Views

Author

Keywords

Examples

Crossrefs

Programs

Maple

Extensions

A273777 Consider all ways of writing the n-th composite number as the product of two divisors d1*d2 = d3*d4 = ... where each divisor is larger than 1; a(n) is the maximum of the sums {d1 + d2, d3 + d4, ...}.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A273777 Consider all ways of writing the n-th composite number as the product of two divisors d1d2 = d3d4 = ... where each divisor is larger than 1; a(n) is the maximum of the sums {d1 + d2, d3 + d4, ...}.