A331614 a(n+1) is the number of times a(n) is the sum of one or more consecutive terms in this sequence so far with a(1) = 1.

1, 1, 2, 2, 3, 2, 4, 3, 3, 4, 4, 5, 4, 6, 4, 7, 5, 5, 6, 5, 7, 6, 6, 7, 7, 8, 3, 5, 8, 5, 9, 7, 9, 8, 6, 8, 7, 10, 7, 11, 8, 8, 9, 9, 10, 8, 10, 9, 11, 9, 12, 6, 9, 13, 8, 11, 10, 10, 11, 11, 12, 7, 12, 8, 12, 9, 14, 10, 12, 10, 13, 9, 15, 7, 13, 10, 14, 11, 13, 11, 14, 12, 11, 15

Offset: 1

Author

S. Brunner, Jan 22 2020

Keywords

• nonn

Examples

			For example we look for the next 2 terms after a(7) = 4:
The sequence so far: 1, 1, 2, 2, 3, 2, 4.
We count how many times we can sum up consecutive terms to get 4 as result (and include all 4's already in the sequence).
There are 3 ways to get a sum of 4: 1 + 1 + 2, 2 + 2 and 4. This gives us a(8) = 3.
For the next term we count all sums of 3 we can get: 1 + 2, 3, 3. This means there are 3 ways and a(9) = 3.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10000
• Samuel B. Reid, Density plot of one billion terms. This plot is normalized by column.

Crossrefs

Cf. A332518.

Programs

Excel

Cell A1: 1
Cell A2: =countif(A$1:AZ1;A1)
Cell B2: =if(A1="";"";A1+$A2)
Copy B2 and paste into area B2:AZ2
Copy row 2 and paste down (5000 lines worked, more could be slow)

Mathematica

a[1] = 1; a[n_] := a[n] = Block[{c = 0, j, s}, Do[j = i; s = 0; While[j < n && s < a[n - 1], s += a[j]; j++]; If[s == a[n - 1], c++], {i, n - 1}]; c]; Array[a, 84] (* Giovanni Resta, Jan 23 2020 *)
(* Second program needing version >= 10.1 *)
a[n_] := a[n] = If[n == 1, 1, SequenceCount[Array[a, n-1], s_ /; Total[s] == a[n-1], Overlaps -> True]];
Array[a, 100] (* Jean-François Alcover, Feb 15 2020 *)

PARI

for (n=1, #a=vector(#t=vector(84)), print1 (a[n]=if(n==1, 1, t[a[n-1]])", "); s=0; forstep (k=n, 1, -1, if (#tRémy Sigrist, Feb 14 2020