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
Keywords
Examples
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]
Links
- Samuel Harkness, Table of n, a(n) for n = 1..10000
Crossrefs
Cf. A308746.
Programs
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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
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