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.
Original entry on oeis.org
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
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.
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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)
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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 *)
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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
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
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)
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
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)
A338759
a(n+1) is the maximum number of groups which can be built from the terms in this sequence so far and using each term only once which result in a(n) as their product with a(1) = 1.
Original entry on oeis.org
1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 2, 2, 3, 2, 4, 4, 5, 2, 5, 3, 3, 4, 6, 6, 7, 1, 7, 2, 6, 8, 5, 4, 8, 7, 3, 5, 5, 6, 10, 7, 4, 9, 3, 6, 12, 13, 1, 8, 9, 5, 7, 5, 8, 10, 8, 11, 1, 9, 6, 13, 2, 7, 6, 14, 7, 7, 8, 12, 15, 7, 9, 7, 10, 10, 11, 2, 8, 13, 3, 7, 11, 3, 8, 14
Offset: 1
To get a(n+1), count how many times a(n) appears in the sequence.
For 1 and primes, this is already a(n+1).
For prime squares, also count how many times the prime factor appears in the sequence, divide it by 2 and round it down.
For example, the next term after a(43) = 9 is 3, because 9 appeared 1 time (at a(43) itself) and 3 appeared 5 times, which can arranged in 2 groups of 3 X 3.
For semiprimes, count how many times the semiprime itself appears in the sequence. Then count how many times the 2 factors appear and add the smallest number.
For example, the next term after a(30) = 6 is 8, because 6 appeared 4 times and the factors 2 and 3 appeared 6 and 4 times. We can build 4 groups of 2 X 3 of them.
A338775
a(n+1) is the number of times a(n) is the least common multiple of one or more consecutive terms in this sequence so far with a(1) = 1.
Original entry on oeis.org
1, 1, 3, 3, 7, 1, 4, 2, 1, 5, 2, 3, 8, 1, 6, 3, 9, 2, 4, 8, 5, 3, 10, 5, 4, 9, 3, 11, 1, 7, 4, 10, 7, 5, 5, 7, 6, 6, 8, 6, 9, 5, 8, 7, 7, 9, 6, 10, 8, 8, 10, 9, 7, 10, 10, 12, 1, 8, 12, 3, 12, 7, 11, 3, 13, 1, 9, 9, 12, 8, 13, 3, 14, 1, 10, 14, 3, 15, 3, 16, 1
Offset: 1
For n = 2:
- a(1) = lcm(a(1)),
- so a(2) = 1.
For n = 3:
- a(2) = lcm(a(1)) = lcm(a(2)) = lcm(a(1), a(2)),
- so a(3) = 3.
For n = 4:
- a(3) = lcm(a(3)) = lcm(a(2), a(3)) = lcm(a(1), a(2), a(3)),
- so a(4) = 3.
Showing 1-5 of 5 results.
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