A359634 -id:A359634 - cp's OEIS frontend

A358537 For n > 0, a(n) is the total number of terms in all contiguous subsequences of the terms up to a(n-1) that sum to n; a(0) = 1.

1, 1, 2, 2, 5, 4, 4, 2, 2, 5, 7, 8, 6, 11, 10, 16, 5, 22, 6, 19, 15, 22, 20, 9, 18, 5, 14, 16, 23, 9, 8, 11, 16, 12, 19, 21, 0, 21, 8, 20, 11, 17, 25, 28, 4, 18, 4, 30, 23, 40, 7, 20, 18, 18, 14, 9, 40, 9, 29, 32, 23, 6, 17, 23, 16, 8, 26, 32, 35, 27, 64, 10

Offset: 0

Neal Gersh Tolunsky, Dec 18 2022

			To find a(4), we look at the sequence so far (1, 1, 2, 2) to find contiguous subsequences that sum to 4: (1, 1, 2) and (2, 2). This is five terms in total, so a(4) = 5. Notice that the two subsequences overlap.
a(40) is 11 because the following contiguous subsequences sum to 40: (6, 19, 15); (23, 9, 8); (19, 21); (19, 21, 0). This is a total of 11 terms.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A359034, A331614, A359634.

N:= 100: V:= Array(0..N):
V[0]:= 1:
for n from 0 to N-1 do
  s:= 0;
  for j from n to 0 by -1 do
    s:= s + V[j];
    if s > N then break fi;
    if s > n then V[s]:= V[s] + n-j+1 fi;
  od;
od:
convert(V,list); # Robert Israel, Feb 16 2023

{ for (n=1, #a=m=vector(72), print1 (a[n] = if (n==1, 1, m[n-1])", "); s = w = 0; forstep (k=n, 1, -1, w++; if ((s += a[k]) > #m, break, s, m[s] += w))) } \\ Rémy Sigrist, Feb 09 2023

Data edited by Yifan Xie, Feb 08 2023

More terms from Rémy Sigrist, Feb 09 2023

A363279 a(0)=1; a(1)=2. For n>1, a(n) is the number of contiguous groups in the sequence thus far whose sum is n.

Original entry on oeis.org

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

Offset: 0

Neal Gersh Tolunsky, May 25 2023

nonn

			a(2)=1 because in the sequence thus far (1, 2), there is only one contiguous subsequence that sums to n=2: (2).
a(7)=3 because in the sequence thus far (1, 2, 1, 2, 1, 1, 2), there are three groups of consecutive terms that sum to n=7: (1, 2, 1, 2, 1); (2, 1, 2, 1, 1); (1, 2, 1, 1, 2).

Neal Gersh Tolunsky, Table of n, a(n) for n = 0..10000
Neal Gersh Tolunsky, Graph of first 10000 terms
Neal Gersh Tolunsky, Graph of first 100000 terms

Cf. A331614, A359634, A358537.

from collections import Counter
from itertools import count, islice
def agen(): # generator of terms
    yield from [1, 2]
    sumsn, c =  [2, 3], Counter([1, 2, 3])
    for n in count(2):
        an = c[n]
        yield an
        sumsn = [an] + [s + an for s in sumsn]
        c.update(sumsn)
print(list(islice(agen(), 86))) # Michael S. Branicky, May 25 2023

cp's OEIS Frontend

A358537 For n > 0, a(n) is the total number of terms in all contiguous subsequences of the terms up to a(n-1) that sum to n; a(0) = 1.

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