A378505 a(0) = 0, a(n) = 0 where a(n-1) is nonzero and divisible by n. Otherwise a(n) = n + a(n-1).

0, 1, 3, 0, 4, 9, 15, 22, 30, 39, 49, 60, 0, 13, 27, 42, 58, 75, 93, 112, 132, 153, 175, 198, 222, 247, 273, 300, 328, 357, 387, 418, 450, 483, 517, 552, 588, 625, 663, 0, 40, 81, 123, 166, 210, 255, 301, 348, 396, 445, 495, 546, 598, 651, 705, 760, 816

Offset: 0

Author

Stuart Coe, Nov 29 2024

Keywords

• nonn
• easy
• look

Comments

From Robert Israel, Nov 29 2024: (Start)

There are an infinite number of zeros in the sequence. If not, suppose a(m) was the last one. Then for j > m, we would have a(j-1) = (m+1) + (m+2) + ... + (j-1) = (j+m)*(j-m-1)/2. In particular, a(m*(m+1)-1) = (m-1)*m*(m+1)*(m+2)/2, which is divisible by m*(m+1) since m-1 or m+2 is even, and that would mean a(m*(m+1)) = 0, contradiction. (End)

From David A. Corneth, Nov 30 2024: (Start)

For some z_0 > 0 such that a(z_0) = 0 we can find the next z_1 > z_0 such that a(z_1) = 0 but for any z_0 < t < z_1 we have a(t) > 0. Then a(t) = binomial(t+1, 2) - binomial(z0+1, 2). Since a(z_1) = 0 we have z_1 | (binomial(z_1 -1 +1, 2) - binomial(z0+1, 2)) = z_1 * (z_1 - 1) / 2 - z_0 * (z_0 + 1)/2.

This means z_1 | 2*(z_1 * (z_1 - 1) / 2 - z_0 * (z_0 + 1)/2) = (z_1 * (z_1 - 1) - z_0 * (z_0 + 1)) where (z_1 * (z_1 - 1) - z_0 * (z_0 + 1)) > 0. As z_1 | z_1 * (z_1 - 1) we also have z_1 | (z_0 * (z_0 + 1)). By the proof of Robert Israel above such z_1 must exist. (End)

Examples

			a(11) = 49 + 11 = 60, since a(10) = 49, which does not exactly divide by 11.
a(12) = 0, since a(11) = 60, a nonzero number that exactly divides by 12.
a(13) = 0 + 13 = 13, since a(12) = 0 (so is not nonzero).
From _David A. Corneth_, Nov 30 2024: (Start)
The next 0 after n = 12 is a divisor of 12 * (12 + 1) = 156. The divisors of 156 that are larger than 12 + 1 = 13 are 26, 39, 52, 78, 156. Of these, 39 is the smallest z_1 such that z_1 | z_1 * (z_1 - 1) / 2 - z_0 * (z_0 + 1)/2. Therefore the next 0 after n = 12 is at n = 39.
The first few zeros are at 0, 3, 12, 39, 65, 78, 158, 237, ... (= A379013). As 100 is between the consecutive zeros 78 and 158 there is no zero strictly between 78 and 158 and so a(100) = 100*(100+1) / 2 - 78*(78+1)/2 = 1969. (End)
		

Links

• Paolo Xausa, Table of n, a(n) for n = 0..10000
• David A. Corneth, PARI programs
• Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^20, showing a(n) = 0 instead as 1/2 for visibility.

Crossrefs

Cf. A068627, A379013 (positions of zeros).

Programs

Maple

a:= proc(n) option remember; `if`(n=0, 0, (t->
     `if`(t>0 and irem(t,n)=0, 0, t+n))(a(n-1)))
    end:
seq(a(n), n=0..56);  # Alois P. Heinz, Nov 29 2024

Mathematica

Module[{n = 0}, NestList[If[Divisible[#, ++n] && # > 0, 0, # + n] &, 0, 100]] (* Paolo Xausa, Dec 09 2024 *)

PARI

\\ See Corneth link

Python

from itertools import count, islice
def agen(): # generator of terms
    an = 0
    for n in count(1):
        yield an
        an = 0 if an > 0 and an%n == 0 else an + n
print(list(islice(agen(), 70))) # Michael S. Branicky, Nov 29 2024