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A358033 a(1) = 2; a(n) - a(n-1) = A093803(a(n-1)), the largest odd proper divisor of a(n-1).

%I A358033 #35 Oct 27 2022 12:49:23
%S A358033 2,3,4,5,6,9,12,15,20,25,30,45,60,75,100,125,150,225,300,375,500,625,
%T A358033 750,1125,1500,1875,2500,3125,3750,5625,7500,9375,12500,15625,18750,
%U A358033 28125,37500,46875,62500,78125,93750,140625,187500,234375,312500,390625,468750
%N A358033 a(1) = 2; a(n) - a(n-1) = A093803(a(n-1)), the largest odd proper divisor of a(n-1).
%F A358033 a(n+1) - a(n) = A056487(floor((n-2)/3)), for n > 2. This works because A056487(n+3) = A056487(n+2)*A056487(n+1)/A056487(n). - _Thomas Scheuerle_, Oct 26 2022
%e A358033 a(1) = 2.
%e A358033 a(2) = 3. The only proper divisor of 2 is 1; 2 + 1 = 3.
%e A358033 a(3) = 4. The only proper divisor of 3 is 1; 3 + 1 = 4.
%e A358033 ...
%e A358033 a(8) = 15.
%e A358033 a(9) = 20. Proper divisors of 15 are 1, 3, 5; largest odd proper divisor = 5; 15 + 5 = 20.
%o A358033 (Python)
%o A358033 a_n = 2
%o A358033 result = [2]
%o A358033 for n in range(30):
%o A358033     temp = []
%o A358033     for i in range(1, a_n):
%o A358033         if a_n % i == 0:
%o A358033             if (i % 2 != 0) and (i != a_n):
%o A358033                 temp.append(i)
%o A358033     result.append(a_n + max(temp))
%o A358033     a_n = a_n + max(temp)
%o A358033 print(result)
%o A358033 (PARI) f(n) = my(x=if(n==1, 1, n/factor(n)[1, 1])); x >> valuation(x, 2); \\ _Michel Marcus_, Oct 26 2022
%o A358033 lista(nn) = my(va = vector(nn)); va[1] = 2; for (n=2, nn, va[n] = va[n-1] + f(va[n-1]);); va; \\ _Michel Marcus_, Oct 26 2022
%Y A358033 Cf. A093803, A000792 (with largest proper divisor instead).
%Y A358033 Cf. A027750, A038754, A056487, A356639.
%K A358033 nonn,easy
%O A358033 1,1
%A A358033 _Eric Angelini_ and _Gavin Lupo_, Oct 25 2022