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

2, 3, 4, 5, 6, 9, 12, 15, 20, 25, 30, 45, 60, 75, 100, 125, 150, 225, 300, 375, 500, 625, 750, 1125, 1500, 1875, 2500, 3125, 3750, 5625, 7500, 9375, 12500, 15625, 18750, 28125, 37500, 46875, 62500, 78125, 93750, 140625, 187500, 234375, 312500, 390625, 468750

Offset

1,1

Links

Andrew Howroyd, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,5).

Formula

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

G.f.: x*(2 + 3*x + 4*x^2 + 5*x^3 + 6*x^4 + 9*x^5 + 2*x^6)/(1 - 5*x^6). - Andrew Howroyd, Nov 23 2025

Example

a(1) = 2.
a(2) = 3. The only proper divisor of 2 is 1; 2 + 1 = 3.
a(3) = 4. The only proper divisor of 3 is 1; 3 + 1 = 4.
...
a(8) = 15.
a(9) = 20. Proper divisors of 15 are 1, 3, 5; largest odd proper divisor = 5; 15 + 5 = 20.

Prog

(Python)
a_n = 2
result = [2]
for n in range(30):
    temp = []
    for i in range(1, a_n):
        if a_n % i == 0:
            if (i % 2 != 0) and (i != a_n):
                temp.append(i)
    result.append(a_n + max(temp))
    a_n = a_n + max(temp)
print(result)
(PARI) f(n) = my(x=if(n==1, 1, n/factor(n)[1, 1])); x >> valuation(x, 2); \\ Michel Marcus, Oct 26 2022
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

Crossrefs

Cf. A093803, A000792 (with largest proper divisor instead).

Cf. A027750, A038754, A056487, A356639.

Sequence in context: A133463 A187550 A307818 * A057492 A178715 A018123

Adjacent sequences: A358030 A358031 A358032 * A358034 A358035 A358036

Keyword

nonn,easy

Author

Eric Angelini and Gavin Lupo, Oct 25 2022

Status

approved