%I #20 Jan 13 2024 11:50:13
%S 1,0,1,2,2,2,2,3,2,1,0,1,0,0,2,2,1,3,2,4,5,2,4,9,5,2,7,8,6,6,4,7,5,3,
%T 8,5,3,2,4,8,2,0,4,4,7,0,0,4,3,4,2,1,2,3,2,1,3,1,1,5,2,2,6,4,3,5,4,5,
%U 7,2,3,10,5,4,10,7,5,7,7,10,9,2,5,17,9,5
%N a(n) is the number of ways n can be reached by the following method: we start with 1, then add or multiply alternately, and each operand must be 3 or 4 or 5.
%C We may start either with multiplication or summation. After summation the next step will be multiplication or vice versa.
%C 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.
%C 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.
%C 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)
%C The graphs represent a fluctuating upward trend. This is caused by an "outlier" at a(2) = 0, whose effect is amplified with recursion.
%H Yifan Xie, <a href="/A368876/a368876.png">Graph of the terms n = 1..10000</a>.
%e There are 3 ways reaching 8: 1*3+5=8, 1*4+4=8 and 1*5+3=8, so a(8)=3.
%p b:= proc(n, t) option remember; `if`(n=1, 1, add(`if`(t=1 and i<n,
%p b(n-i, 1-t), `if`(t=0 and irem(n, i)=0, b(n/i, 1-t), 0)), i=3..5))
%p end:
%p a:= n-> `if`(n=1, 1, add(b(n, i), i=0..1)):
%p seq(a(n), n=1..86); # _Alois P. Heinz_, Jan 12 2024
%o (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)", "); );
%Y Cf. A358094.
%K nonn,easy
%O 1,4
%A _Yifan Xie_, Jan 08 2024