A056240 Smallest number whose prime divisors (taken with multiplicity) add to n.

2, 3, 4, 5, 8, 7, 15, 14, 21, 11, 35, 13, 33, 26, 39, 17, 65, 19, 51, 38, 57, 23, 95, 46, 69, 92, 115, 29, 161, 31, 87, 62, 93, 124, 155, 37, 217, 74, 111, 41, 185, 43, 123, 86, 129, 47, 215, 94, 141, 188, 235, 53, 329, 106, 159, 212, 265, 59, 371, 61, 177, 122

Offset

2,1

Comments

a(n) is the index of first occurrence of n in A001414.

From David James Sycamore and Michel Marcus, Jun 16 2017, Jun 28 2017: (Start)

Recursive calculation of a(n):

For prime p, a(p) = p.

For even composite n, let P_n denote the largest prime < n-1 such that n-P_n is prime (except if n = 6).

For odd composite n, let P_n denote the largest prime < n-1 such that n-3-P_n is prime.

Conjecture: a(n) = min { q*a(n-q); q prime, P_n <= q < n-1 }.

Examples:

For n = 9998, P_9998 = 9967 and a(9998) = min { 9973*a(25), 9967*a(31) } = 9967*31 = 308977.

For n = 875, P_875 = 859 and a(875) = min { 863*a(12), 859*a(16) } = 863*35 = 30205.

Note: A000040 and A288313 are both subsequences of this sequence. (End)

Links

Reinhard Zumkeller, Table of n, a(n) for n = 2..10000

Hans Havermann, Tables of sum-of-prime-factors sequences (overview with links to the first 50000 sums).

Formula

Trivial but essential: a(n) >= n. - David A. Corneth, Mar 23 2018

a(n) >= n with equality iff n = 4 or n is prime. - M. F. Hasler, Jan 19 2019

Example

a(8) = 15 = 3*5 because 15 is the smallest number whose prime divisors sum to 8.
a(10000) = 586519: Let pp(n) be the largest prime < n and the candidate being the current value that might be a(10000). Then we see that pp(10000 - 1) = 9973, giving a candidate 9973 * a(10000 - 9973) = 9973 * 92. pp(9973) = 9967, giving a candidate 9967 * a(10000 - 9967) = 9967 * 62. pp(9967) = 9949, giving the candidate 9949 * a(10000 - 9949) = 9962 * 188. This is larger than our candidate so we keep 9967 * 62 as our candidate. pp(9949) = 9941, giving a candidate 9941 * pp(10000 - 9941) = 9941 * 59. We see that (n - p) * a(p) >= (n - p) * p > candidate = 9941 * 59 for p > 59 so we stop iterating to conclude a(10000) = 9941 * 59 = 586519. - David A. Corneth, Mar 23 2018, edited by M. F. Hasler, Jan 19 2019

Maple

A056240 := proc(n)
    local k ;
    for k from 1 do
        if A001414(k) = n then
            return k ;
        end if;
    end do:
end proc:
seq(A056240(n), n=2..80) ; # R. J. Mathar, Apr 15 2024

Mathematica

a = Table[0, {75}]; Do[b = Plus @@ Flatten[ Table[ #1, {#2}] & @@@ FactorInteger[n]]; If[b < 76 && a[[b]] == 0, a[[b]] = n], {n, 2, 1000}]; a (* Robert G. Wilson v, May 04 2002 *)
b[n_] := b[n] = Total[Times @@@ FactorInteger[n]];
a[n_] := For[k = 2, True, k++, If[b[k] == n, Return[k]]];
Table[a[n], {n, 2, 63}] (* Jean-François Alcover, Jul 03 2017 *)

Prog

(Haskell)
a056240 = (+ 1) . fromJust . (`elemIndex` a001414_list)
-- Reinhard Zumkeller, Jun 14 2012
(PARI) isok(k, n) = my(f=factor(k)); sum(j=1, #f~, f[j, 1]*f[j, 2]) == n;
a(n) = my(k=2); while(!isok(k, n), k++); k; \\ Michel Marcus, Jun 21 2017
(PARI) a(n) = {if(n < 7, return(n + 2*(n==6))); my(p = precprime(n), res); if(p == n, return(p), p = precprime(n - 2); res = p * a(n - p); while(res > (n - p) * p && p > 2, p = precprime(p - 1); res = min(res, a(n - p) * p)); res)} \\ David A. Corneth, Mar 23 2018
(PARI) A056240(n, p=n-1, m=oo)=if(n<6 || isprime(n), n, n==6, 8, until(p<3 || (n-p=precprime(p-1))*p >= m=min(m, A056240(n-p)*p), ); m) \\ M. F. Hasler, Jan 19 2019

Crossrefs

Cf. A000040, A001414, A064502, A288313.

First column of array A064364, n>=2.

See A000792 for the maximal numbers whose prime factors sums up to n.

Sequence in context: A074756 A240221 A075162 * A069968 A298882 A086931

Adjacent sequences: A056237 A056238 A056239 * A056241 A056242 A056243

Keyword

nonn,easy

Author

Adam Kertesz, Aug 19 2000

Extensions

More terms from James A. Sellers, Aug 25 2000

Status

approved