A358094 -id:A358094 - cp's OEIS frontend

A358095 a(n) is the number of ways n can be reached in the algorithm explained in A358094 if the last operation is summation.

1, 0, 1, 2, 2, 1, 0, 1, 1, 2, 3, 3, 2, 3, 3, 0, 2, 3, 1, 2, 2, 2, 2, 3, 3, 4, 4, 2, 3, 4, 3, 5, 5, 0, 3, 5, 2, 6, 6, 1, 3, 4, 2, 5, 5, 2, 5, 5, 2, 3, 3, 3, 5, 6, 4, 7, 7, 2, 3, 4, 3, 6, 6, 3, 5, 7, 5, 7, 7, 0, 2, 5, 3, 8, 8, 2, 5, 9, 6, 10

Offset: 1

Yifan Xie, Nov 01 2022

From Yifan Xie, Jan 07 2025: (Start)
The following identities can be proved by induction (k, t are nonnegative integers):
a(n) = 0 iff n = 2 or n = 9*2^k - 2.
a(n) = 1 iff n = 1, 3, 6, 8, 9 or n = 21*2^k - 2.
If n > 71, a(n) = 2 iff n = m*3*2^k - 2, where m is in the set {26, 30, 32, 34, 40, 49}.
If n > 85, a(n) = 3 iff n = 108*2^k - 2, 108*2^k - 3, 108*2^k - 4, 108*2^k - 5, 123*2^k - 2, 126*2^k - 2, 132*2^k - 2, 150*2^k - 2, 174*2^k - 2, 210*2^k - 2, (108*2^t - 3)*2^k - 2.
If n > 60, a(n) = 4 iff n = 114*2^k - 2. (End)

			There are 3 ways to reach 11: (1*2+2)*2+3=11, (1+3)*2+3=11 and (1+2)*3+2=11.

Yifan Xie, Table of n, a(n) for n = 1..10000

Cf. A358094, A358096.

a(n) = A358096(n-2) + A358096(n-3) for n > 3.

a(6k) = a(2k-1) + a(3k-1); a(6k+1) = a(3k-1); a(6k+2) = a(6k+3) = a(2k) + a(3k); a(6k+4) = a(3k+1); a(6k+5) = a(2k+1) + a(3k+1). - Yifan Xie, Jan 07 2025

A358096 a(n) is the number of ways n can be reached in the algorithm explained in A358094 if the last operation is multiplication.

Original entry on oeis.org

1, 1, 1, 0, 0, 1, 0, 2, 1, 2, 0, 3, 0, 0, 2, 1, 0, 2, 0, 2, 0, 3, 0, 4, 0, 2, 1, 3, 0, 5, 0, 0, 3, 2, 0, 6, 0, 1, 2, 2, 0, 5, 0, 2, 3, 2, 0, 3, 0, 3, 2, 4, 0, 7, 0, 2, 1, 3, 0, 6, 0, 3, 2, 5, 0, 7, 0, 0, 2, 3, 0, 8, 0, 2, 3, 6, 0, 10, 0, 1

Offset: 1

Yifan Xie, Nov 01 2022

			There are 3 ways to reach 12: (1*3+3)*2=12, (1*2+2)*3=12 and (1+3)*3=12.

Yifan Xie, Table of n, a(n) for n = 1..10000

Cf. A358094, A358095.

a(n) = A358095(n/2) + A358095(n/3) if n == 0 (mod 6);
a(n) = A358095(n/2) if n == 2 or 4 (mod 6);
a(n) = A358095(n/3) if n == 3 (mod 6);
a(n) = 0 if n == 1 or 5 (mod 6).

A368876 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.

Original entry on oeis.org

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, 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, 7, 2, 3, 10, 5, 4, 10, 7, 5, 7, 7, 10, 9, 2, 5, 17, 9, 5

Offset: 1

Yifan Xie, Jan 08 2024

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.

			There are 3 ways reaching 8: 1*3+5=8, 1*4+4=8 and 1*5+3=8, so a(8)=3.

Yifan Xie, Graph of the terms n = 1..10000.

Cf. A358094.

b:= proc(n, t) option remember; `if`(n=1, 1, add(`if`(t=1 and i `if`(n=1, 1, add(b(n, i), i=0..1)):
seq(a(n), n=1..86);  # Alois P. Heinz, Jan 12 2024

b[n_, t_] := b[n, t] = If[n == 1, 1, Sum[If[t == 1 && i < n,
   b[n-i, 1-t], If[t == 0 && Mod[n, i] == 0, b[n/i, 1-t], 0]], {i, 3, 5}]];
a[n_] := If[n == 1, 1, Sum[b[n, i], {i, 0, 1}]];
Table[a[n], {n, 1, 86}] (* Jean-François Alcover, Feb 03 2025, after Alois P. Heinz *)

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)", "); );

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A358095 a(n) is the number of ways n can be reached in the algorithm explained in A358094 if the last operation is summation.

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A358096 a(n) is the number of ways n can be reached in the algorithm explained in A358094 if the last operation is multiplication.

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Formula

A368876 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.

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