Damon Lay - cp's OEIS frontend

User: Damon Lay

Damon Lay has authored 2 sequences.

A363072 Add primes until a perfect power appears. When a perfect power appears, that term is the smallest root of the perfect power. Then return to adding primes, beginning with the next prime.

2, 5, 10, 17, 28, 41, 58, 77, 10, 39, 70, 107, 148, 191, 238, 291, 350, 411, 478, 549, 622, 701, 28, 117, 214, 315, 418, 525, 634, 747, 874, 1005, 1142, 1281, 1430, 1581, 1738, 1901, 2068, 2241, 2420, 51, 242, 435, 632, 831, 1042, 1265, 1492, 1721, 1954, 2193

Offset: 1

Damon Lay, May 16 2023

			The first term is the first prime, p(1) = 2
a(1) = p(1) = 2
a(2) = a(1) + p(2) = 2 + 3 = 5
a(3) = a(2) + p(3) = 5 + 5 = 10
etc.
a(8) = 58 + 19 = 77
a(9) is determined:
a(8) + p(9) = 77 + 23 = 100, a perfect power.  10 is the smallest root of 100, therefore a(9) = 10
a(10) = 10 + p(10) = 10 + 29 = 39
and so on.

Cf. A001597.

root[n_] := Surd[n, GCD @@ FactorInteger[n][[;; , 2]]]; a[1] = 2; a[n_] := a[n] = root[a[n - 1] + Prime[n]]; Array[a, 100] (* Amiram Eldar, May 21 2023 *)

A362372 Inventory of powers. Initialize the sequence with '1'. Then record the number of powers of 1 thus far, then do the same for powers of 2 (2, 4, 8, ...), powers of 3, etc. When the count is zero, do not record a zero; rather start the inventory again with the powers of 1.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 1, 5, 1, 1, 7, 1, 1, 9, 1, 2, 10, 2, 2, 10, 4, 2, 1, 1, 12, 6, 2, 1, 1, 1, 1, 16, 8, 2, 2, 1, 1, 1, 1, 1, 2, 21, 12, 2, 2, 1, 1, 1, 1, 1, 2, 26, 15, 2, 2, 1, 1, 1, 1, 1, 2, 31, 18, 2, 2, 1, 1, 1, 1, 1, 2, 36, 21, 2, 2, 1, 2, 1, 1, 1

Offset: 0

Damon Lay, Apr 17 2023

A variant of the inventory sequence, A342585.
The graph exhibits sharp jumps followed by a rapid decline forming a periodic hockey stick pattern. Larger-scale, near-linear structures also appear.
Periodic patterns in the relative frequency of any given number also are present. For example, perform a rolling count of the number of times 2 appears in the previous 40 entries.
Open question: will all positive integers appear in the sequence?

			As an irregular triangle, the table begins:
   1;
   1;
   2, 1;
   3, 1, 1;
   5, 1, 1;
   7, 1, 1;
   9, 1, 2;
  10, 2, 2;
  10, 4, 2, 1, 1;
  12, 6, 2, 1, 1, 1, 1;
  16, 8, 2, 2, 1, 1, 1, 1, 1, 2;
  ...
Initialize the sequence with '1'.
Powers of 1 are counted in the first column, powers of 2 in the second, powers of 3 in the third, etc.

Michael S. Branicky, Table of n, a(n) for n = 0..10000 (first 4330 terms from Damon Lay)

Cf. A342585 and similar variants thereof: A345730, A347791, A348218, A352799, A353092.

from collections import Counter
from sympy import divisors, perfect_power
def powers_in(n):
    t = perfect_power(n) # False for n == 1
    return [n] if not t else [t[0]**d for d in divisors(t[1])]
def aupton(nn):
    num, alst, inventory = 1, [1], Counter([1])
    while len(alst) <= nn:
        c = inventory[num]
        if c == 0: num = 1
        else: num += 1; alst.append(c); inventory.update(powers_in(c))
    return alst
print(aupton(100)) # Michael S. Branicky, May 05 2023

cp's OEIS Frontend

User: Damon Lay

A363072 Add primes until a perfect power appears. When a perfect power appears, that term is the smallest root of the perfect power. Then return to adding primes, beginning with the next prime.

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica