OFFSET
1,4
COMMENTS
We may start either with multiplication or summation. After summation the next step will be multiplication or vice versa.
The only zeros in this sequence are at a(n) where n=2, 11, 13, 14, 42, 46, 47, 142, 146, 442; the only ones in this sequence are at a(n) where n=1, 3, 10, 12, 17, 52, 56, 58, 59, 182.
Proof: (Start) Let k be any number greater than 1334. If k == 0 (mod 3) subtract 3, if k == 1 subtract 4, if k == 2 subtract 5, then divide by 3. Repeat this process until k < 1335. Obviously we will get some number between 443 and 1334. By computation it is known that all these numbers can be reached, so all numbers > 442 can be reached, thus the zeros in this sequence can be verified.
Similarly, if k > 1334 can only be reached in one way, some number between 443 and 1334 can be reached in no more than one way, which is a contradiction, thus the ones in this sequence can also be verified. (End)
The graphs represent a fluctuating upward trend. This is caused by an "outlier" at a(2) = 0, whose effect is amplified with recursion.
LINKS
EXAMPLE
There are 3 ways reaching 8: 1*3+5=8, 1*4+4=8 and 1*5+3=8, so a(8)=3.
MAPLE
b:= proc(n, t) option remember; `if`(n=1, 1, add(`if`(t=1 and i<n,
b(n-i, 1-t), `if`(t=0 and irem(n, i)=0, b(n/i, 1-t), 0)), i=3..5))
end:
a:= n-> `if`(n=1, 1, add(b(n, i), i=0..1)):
seq(a(n), n=1..86); # Alois P. Heinz, Jan 12 2024
PROG
(PARI) for (n=1, #a=vector(#m=vector(86)), if (n==1, a[n] = m[n] = 1, if (n-3>0, a[n] += m[n-3]; ); if (n-4>0, a[n] += m[n-4]; ); if (n-5>0, a[n] += m[n-5]; ); if (n%3==0, m[n] += a[n/3]; ); if (n%4==0, m[n] += a[n/4]; ); if (n%5==0, m[n] += a[n/5]; ); ); print1(a[n]+m[n]-(n==1)", "); );
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Yifan Xie, Jan 08 2024
STATUS
approved