A331614 -id:A331614 - cp's OEIS frontend

A332518 a(n+1) is the number of times a(n) is the product of one or more consecutive terms in this sequence so far with a(1) = 1.

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

Offset: 1

Rémy Sigrist, Feb 15 2020

nonn

This sequence is a multiplicative variant of A331614.

This sequence is unbounded.

			The first terms, alongside the corresponding products, are:
  n   a(n)  Products
  --  ----  --------
   1     1  a(1)
   2     1  a(1),           a(1)*a(2),  a(2)
   3     3  a(1)*a(2)*a(3), a(2)*a(3),  a(3)
   4     3  a(1)*a(2)*a(3), a(2)*a(3),  a(3),        a(4)
   5     4  a(5)
   6     1  a(1),           a(1)*a(2),  a(2),        a(6)
   7     4  a(5),           a(5)*a(6),  a(6)*a(7),   a(7)
   8     4  a(5),           a(5)*a(6),  a(6)*a(7),   a(7), a(8)
   9     5  a(9)
  10     1  a(1),           a(1)*a(2),  a(2),        a(6), a(10)
  11     5  a(9),           a(9)*a(10), a(10)*a(11), a(11)
  12     4  a(5),           a(5)*a(6),  a(6)*a(7),   a(7), a(8), a(12)

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, C program for A332518

Cf. A331614.

C
```
// See Links section.
```

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.

Original entry on oeis.org

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

A358799 a(0) = 0, and for any n >= 0, a(n+1) is the number of ways to write a(n) = a(i) XOR ... XOR a(j) with 0 <= i <= j <= n (where XOR denotes the bitwise XOR operator).

Original entry on oeis.org

0, 1, 2, 1, 3, 4, 2, 5, 4, 5, 6, 8, 2, 11, 2, 13, 6, 14, 10, 9, 9, 12, 14, 16, 2, 24, 6, 29, 5, 23, 3, 27, 12, 23, 9, 26, 17, 13, 26, 19, 15, 32, 4, 46, 2, 51, 1, 45, 6, 48, 6, 49, 7, 41, 9, 47, 10, 49, 17, 37, 21, 38, 23, 36, 24, 49, 30, 48, 24, 52, 22, 45

Offset: 0

Rémy Sigrist, Dec 06 2022

This sequence is a variant of A331614 and A332518; here we use binary XOR, there addition and multiplication, respectively.

This sequence is unbounded (if the sequence was bounded, with greatest value m, then, by the pigeonhole principle, some value, say v, would appear infinitely many times, and the next value after the (m+1)-th occurrence of v would be > m, a contradiction).

			The first terms, alongside the corresponding pairs (i,j)'s, are:
  n   a(n)  (i,j)'s
  --  ----  ---------------------------------------------------------
   0     0  N/A
   1     1  (0,0)
   2     2  (0,1), (1,1)
   3     1  (2,2)
   4     3  (0,1), (1,1), (3,3)
   5     4  (0,2), (1,2), (2,3), (4,4)
   6     2  (2,5), (5,5)
   7     5  (0,3), (1,3), (2,2), (3,4), (6,6)
   8     4  (0,5), (1,5), (4,6), (7,7)
   9     5  (2,5), (3,6), (4,8), (5,5), (8,8)
  10     6  (0,5), (1,5), (3,8), (4,6), (7,7), (9,9)
  11     8  (0,8), (1,8), (2,6), (3,5), (3,10), (5,6), (6,9), (10,10)
  12     2  (6,11), (11,11)

Rémy Sigrist, Table of n, a(n) for n = 0..10000
Rémy Sigrist, C program
Rémy Sigrist, Scatterplot of the first 250000 terms

Cf. A331614, A332518.

C
```
See Links section.
```

A359634 a(0)=1 and thereafter a(n) is the length of the longest contiguous group of terms in the sequence thus far that add up to n; if no such group exists, set a(n)=0.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 4, 3, 4, 5, 4, 5, 6, 4, 5, 6, 7, 6, 7, 8, 5, 7, 8, 9, 7, 6, 8, 9, 10, 6, 9, 10, 11, 9, 8, 10, 11, 12, 9, 10, 9, 11, 12, 13, 7, 12, 13, 14, 12, 11, 13, 14, 15, 11, 13, 11, 14, 15, 16, 13, 6, 14, 13, 15, 16, 17, 13, 15, 12, 16, 17, 18, 15, 8, 16, 14, 17, 18, 19, 15, 16, 12, 17, 14, 18, 19, 20

Offset: 0

Neal Gersh Tolunsky, Jan 08 2023

nonn

If a zero appears, it is not counted as a term in a contiguous grouping. For example, if (10, 30, 0, 60) is our longest group to sum to 100, this counts as 3 terms, not 4. However, in 50 million terms (computed by Kevin Ryde), a zero has not appeared. Why is this?

How does the lower envelope of this sequence behave?

			a(6) is 4 because in the sequence thus far (1,1,2,2,3,3), the longest run of consecutive terms that sums to 6 is (1,1,2,2), which is 4 terms.

Rémy Sigrist, Table of n, a(n) for n = 0..10000
Rémy Sigrist, C program

Cf. A331614, A358537. a(1-16) in A138099 are the same.

C
```
See Links section.
```

A359034 a(n+1) is the sum of the number of terms in all groups of contiguous terms that add up to a(n); a(1)=1.

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 4, 5, 3, 5, 4, 6, 6, 7, 7, 8, 10, 11, 4, 7, 9, 9, 10, 12, 13, 14, 13, 15, 8, 11, 7, 10, 13, 16, 19, 18, 18, 19, 19, 20, 7, 11, 8, 12, 14, 14, 15, 9, 11, 9, 12, 15, 10, 14, 16, 20, 14, 17, 17, 18, 22, 22, 23, 22, 24, 23, 23, 24, 24, 25, 28, 27, 22

Offset: 1

Neal Gersh Tolunsky, Dec 12 2022

nonn

If strongly smoothened, this sequence displays growth. This growth appears to be caused by the number of groups which is increasing by growing length of the sequence roughly proportional to n^(1/2). But the length of the groups appears to be nearly uninfluenced by this. - Thomas Scheuerle, Dec 14 2022

			a(17) is 10 because in the sequence so far (1, 1, 2, 3, 3, 4, 4, 5, 3, 5, 4, 6, 6, 7, 7, 8), these are the ways of adding contiguous terms to get a(16)=8: (2, 3, 3); (4, 4); (5, 3); (3, 5); (8). This is 10 terms in total, so a(17) is 10. Notice groups (5,3) and (3,5) overlap.

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
Thomas Scheuerle, Scatter plot of a(n) per number of groups found. For n = 1...20000. This is the mean number of contiguous terms for each n.
Thomas Scheuerle, Scatter plot, number of groups of contiguous terms for n = 1...20000

Cf. A331614, A358919. Begins the same as A124056 (until a(13)).

function a = A359034( max_n )
    a = [1 1];
    for n = 3:max_n
        s = 1; e = 1; sm = 1; c = 0;
        while e < n-1
            while sm < a(n - 1) && e < (n - 1)
                e = e + 1; sm = sm + a(e);
            end
            if sm == a(n - 1)
                c = c + (e - s) + 1;
            end
            s = s + 1;
            e = s; sm = a(s);
        end
        a(n) = c + 1;
    end
end % Thomas Scheuerle, Dec 14 2022

A336037 a(n+1) is the number of times the binary representation of a(n) is the concatenation of that of one or more consecutive terms in this sequence so far with a(1) = 1.

Original entry on oeis.org

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

Offset: 1

Rémy Sigrist, Jul 07 2020

This sequence has similarities with A331614 and A332518; here we concatenate consecutive terms, there we add and multiply them, respectively.

This sequence is unbounded.

			The first terms, alongside their binary representation and the corresponding concatenations, are:
  n   a(n)  bin(a(n))  Concatenations
  --  ----  ---------  --------------
   1     1          1  a(1)
   2     1          1  a(1), a(2)
   3     2         10  a(3)
   4     1          1  a(1), a(2), a(4)
   5     3         11  a(1)|a(2), a(5)
   6     2         10  a(3), a(6)
   7     2         10  a(3), a(6), a(7)
   8     3         11  a(1)|a(2), a(5), a(8)
   9     3         11  a(1)|a(2), a(5), a(8), a(9)
  10     4        100  a(10)

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, Density plot of the first 100000000 terms
Rémy Sigrist, PARI program for A336037

Cf. A331614, A332518.

PARI
```
See Links section.
```

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

A332518 a(n+1) is the number of times a(n) is the product of one or more consecutive terms in this sequence so far with a(1) = 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

C

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.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

PARI

Extensions

A358799 a(0) = 0, and for any n >= 0, a(n+1) is the number of ways to write a(n) = a(i) XOR ... XOR a(j) with 0 <= i <= j <= n (where XOR denotes the bitwise XOR operator).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

C

A359634 a(0)=1 and thereafter a(n) is the length of the longest contiguous group of terms in the sequence thus far that add up to n; if no such group exists, set a(n)=0.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

C

A359034 a(n+1) is the sum of the number of terms in all groups of contiguous terms that add up to a(n); a(1)=1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

MATLAB

A336037 a(n+1) is the number of times the binary representation of a(n) is the concatenation of that of one or more consecutive terms in this sequence so far with a(1) = 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

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.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Python