A369351 The smallest number such that n or more numbers k exist such that a(n) - k = sopfr(a(n)) + sopfr(k), where sopfr(m) is the sum of the primes dividing m, with repetition.

6, 35, 77, 169, 287, 1147, 1517, 1517, 4352, 4352, 4352, 14647, 55488, 55488, 114091, 121673, 167137, 206837, 277928, 277928, 277928, 277928, 277928, 722473, 2165407, 2498227, 2498227, 2498227, 5271391, 5770603, 8321771, 8321771, 9983680, 9983680, 9983680, 9983680, 28277536, 28277536, 28277536, 28277536, 28277536

Offset

1,1

Comments

From David A. Corneth, Feb 11 2024:

To find terms we can do the following:

a(n) - k = sopfr(a(n)) + sopfr(k) can be rewritten as

a(n) - sopfr(a(n)) = k + sopfr(k)

So one side depends on a(n) and the other side depends on k.

Now to find terms <= u, make a list V of values m + sopfr(m) where m <= 2*precprime(u). So like for 59 we get those values are 38, 44, 48 from V.

We look up 59 in a table W of values m - sopfr(m) and find that 77 is the smallest such value. Hence a(3) <= 77. By looking through all m <= 77 in V and then referring to W we find a(3) = 77.

Links

Michael S. Branicky, Table of n, a(n) for n = 1..60

David A. Corneth, PARI program

Example

a(1) = 6 as 6 is the smallest number to have one number (k = 1) such that 6 - 1 = 5 = sopfr(6) + sopfr(1) = 5 + 0 = 5.
a(2) = 35 as 35 is the smallest number to have two numbers (k = 14, 15) such that 35 - 14 = 21 = sopfr(35) + sopfr(14) = 12 + 9 = 21, and 35 - 15 = 20 = sopfr(35) + sopfr(15) = 12 + 8 = 20.
a(3) = 77 as 77 is the smallest number to have three numbers (k = 38, 44, 48) such that 77 - 38 = 39 = sopfr(77) + sopfr(38) = 18 + 21 = 39, 77 - 44 = 33 = sopfr(77) + sopfr(44) = 18 + 15 = 33, and  77 - 48 = 29 = sopfr(77) + sopfr(48) = 18 + 11 = 29.

Prog

(Python)
from sympy import factorint
from functools import cache
from itertools import count, islice
from collections import Counter
kcount, kmax = Counter(), 0
@cache
def sopfr(n): return sum(p*e for p, e in factorint(n).items())
def f(n):   # revised based on comment by David A. Corneth, Feb 11 2024
    global kcount, kmax
    target = n - sopfr(n)
    for k in range(kmax+1, target+1):
        kcount[k+sopfr(k)] += 1
        kmax += 1
    return kcount[target]
def agen(): # generator of terms
    adict, n = dict(), 1
    for m in count(2):
        v = f(m)
        if v not in adict: adict[v] = m
        for i in range(n, v+1): yield m; n += 1
print(list(islice(agen(), 12))) # Michael S. Branicky, Feb 09 2024

Crossrefs

Cf. A001414, A369348, A369349, A369350, A369352, A369353, A370091.

Sequence in context: A046025 A198327 A009583 * A370091 A033578 A101077

Adjacent sequences: A369348 A369349 A369350 * A369352 A369353 A369354

Keyword

nonn

Author

Scott R. Shannon, Jan 21 2024

Extensions

a(24) from Michael S. Branicky, Feb 08 2024

a(25) from Michael S. Branicky, Feb 11 2024

More terms from David A. Corneth, Feb 11 2024

Status

approved