A355636 a(1) = a(2) = 1; for n > 2, a(n) is the smallest positive number that has not yet appeared that has the same number of divisors as the sum a(n-2) + a(n-1) but does not equal the sum.
1, 1, 3, 9, 18, 6, 30, 100, 24, 12, 196, 48, 20, 28, 80, 60, 72, 84, 90, 40, 42, 8, 32, 54, 10, 729, 2, 14, 81, 15, 108, 21, 22, 5, 26, 7, 27, 33, 96, 34, 56, 126, 66, 320, 35, 38, 11, 4, 39, 13, 44, 46, 132, 51, 55, 57, 162, 58, 140, 150, 70, 156, 62, 65, 17, 69, 74, 77, 19, 160, 23, 82, 78
Offset: 1
Keywords
Examples
a(6) = 6 as a(4) + a(5) = 9 + 18 = 27 which has four divisors, and 6 is the smallest unused number that does not equal 27 and has four divisors.
Links
- Scott R. Shannon, Image of the first 250000 terms. The green line is y = n.
Programs
-
PARI
listm(nn) = my(va = vector(nn)); va[1] = 1; va[2] = 1; my(m = Map()); mapput(m, 1, 1); for (n=3, nn, my(s=va[n-2]+va[n-1], d=numdiv(s), k=1, vs=Vec(va, n-1)); while (mapisdefined(m, k) || (k==s) || (numdiv(k)!=d), k++); va[n] = k; mapput(m, k, n);); va; \\ Michel Marcus, Jul 11 2022
-
Python
from sympy import divisor_count from itertools import count, islice def agen(): anm1, an, mink, seen = 1, 1, 2, {1} yield 1 for n in count(2): yield an k, target, tsum = mink, divisor_count(anm1+an), anm1+an while k in seen or k == tsum or divisor_count(k) != target: k += 1 while mink in seen: mink += 1 anm1, an = an, k seen.add(an) print(list(islice(agen(), 73))) # Michael S. Branicky, Jul 26 2022
Comments