A360387 a(1) = 1, and for n > 1, a(n) is the number of ways that a(1..n-1) can be divided into contiguous subsequences of equal sum.

1, 1, 2, 2, 2, 3, 1, 3, 1, 3, 1, 2, 2, 4, 2, 2, 4, 3, 1, 4, 3, 1, 6, 1, 1, 3, 1, 4, 4, 1, 1, 1, 1, 5, 1, 2, 5, 1, 1, 1, 3, 1, 1, 1, 2, 6, 1, 1, 2, 1, 1, 3, 1, 4, 1, 2, 1, 5, 1, 1, 1, 5, 1, 1, 1, 3, 1, 2, 2, 7, 1, 2, 2, 3, 1, 6, 1, 1, 4, 2, 2, 4, 3, 1, 3, 1, 2

Offset

1,3

Comments

No divisions counts as 1 way of dividing the sequence.

Is the sequence unbounded?

Links

Samuel Harkness, Table of n, a(n) for n = 1..10000

Example

Initial terms with corresponding divisions (starting with single subsequence, then more):
 n  a(n)  Ways of dividing (subsequences separated by "|")
 -  ----  ------------------------------------------------
 1     1
 2     1  [1]
 3     2  [1,1]; [1|1]
 4     2  [1,1,2]; [1,1|2]
 5     2  [1,1,2,2]; [1,1|2|2]
 6     3  [1,1,2,2,2]; [1,1,2|2,2]; [1,1|2|2|2]
 7     1  [1,1,2,2,2,3]
 8     3  [1,1,2,2,2,3,1]; [1,1,2,2|2,3,1]; [1,1,2|2,2|3,1]
 9     1  [1,1,2,2,2,3,1,3]
10     3  [1,1,2,2,2,3,1,3,1]; [1,1,2,2,2|3,1,3,1]; [1,1,2|2,2|3,1|3,1]

Prog

(PARI) { m = s = 0; v = 1; for (n=1, 87, print1 (v", "); m += 2^s+=v; v = 0; fordiv (s, d, t = sum(i=1, d, 2^(i*s/d)); if (bitand(m, t)==t, v++))) } \\ Rémy Sigrist, Feb 09 2023

Crossrefs

Cf. A308746.

Sequence in context: A227738 A103960 A242626 * A306399 A240689 A233567

Adjacent sequences: A360384 A360385 A360386 * A360388 A360389 A360390

Keyword

nonn

Author

Neal Gersh Tolunsky, Feb 05 2023

Status

approved