A328096 -id:A328096 - cp's OEIS frontend

A328342 a(0) = 0; for n > 1, if a(n-1) has appeared three or more times then a(n) = total number of terms between the last and second-last appearance of a(n-1), minus the total number of terms between the second-last and third-last appearance of a(n-1). If a(n-1) has appeared two times, a(n) = total number of terms between the last and second-last appearance of a(n-1). If a(n-1) has appeared once, a(n) = 0.

0, 0, 1, 0, 1, 2, 0, 1, 1, -2, 0, 1, 2, 7, 0, 0, -3, 0, 1, 4, 0, 1, -4, 0, 0, -2, 16, 0, 2, 9, 0, 0, -2, -9, 0, 2, -9, 3, 0, 1, 15, 0, -1, 0, -1, 2, 3, 9, 18, 0, 4, 31, 0, -3, 37, 0, 0, -2, 18, 10, 0, 3, 6, 0, -1, 18, -3, -24, 0, 2, 14, 0, -2, -10, 0, 0, -2, -11, 0, 2, -14

Offset: 0

Scott R. Shannon, Oct 12 2019

sign

For the first 10 million terms the largest positive value is 9845628 while the largest negative value is -9748780. The largest positive term that has not appeared is 30832 and the largest negative term that has not appeared is -33426. It is likely that all integers eventually appear in the sequence, although this is unproven.

			a(1) = 0 as a(1-1) = a(0) = 0 has only appeared once.
a(2) = 1 as a(2-1) = a(1) = 0 has appeared twice, and the number of terms between the last and second-last appearance of 0, a(0) and a(1), is 1
a(4) = 1. a(4-1) = a(3) = 0, which has appeared three times. The number of terms between the last and second-last appearance of 0, a(3) and a(1), is 2. The number of terms between the second-last and third-last appearance of 0, a(1) and a(0), is 1. Thus a(4) = 2 - 1 = 1.
a(9) = -2. a(9-1) = a(8) = 1, which has appeared three times. The number of terms between the last and second-last appearance of 1, a(8) and a(7), is 1. The number of terms between the second-last and third-last appearance of 1, a(7) and a(4), is 3. Thus a(9) = 1 - 3 = -2.

Scott R. Shannon, Java code to produce the sequence.

Cf. A181391, A328096.

s = {0}; Do[d = Differences[Position[s, ?(# == s[[-1]] &)] // Flatten]; a = Switch[Length[d], 0, 0, 1, d[[1]], , d[[-1]] - d[[-2]]]; AppendTo[s, a], {80}]; s (* Amiram Eldar, Oct 13 2019 *)

A353778 a(1)=1. If a(n) is a novel term (seen for first time), a(n+1) = number of prior terms a(j) > a(n), 1 <= j <= n-1. If a(n) is a repeat of a term last seen at a(m), m < n, then a(n+1) = n-m-1.

Original entry on oeis.org

1, 0, 1, 1, 0, 2, 0, 1, 3, 0, 2, 4, 0, 2, 2, 0, 2, 1, 9, 0, 3, 11, 0, 2, 6, 2, 1, 8, 2, 2, 0, 7, 3, 11, 11, 0, 4, 24, 0, 2, 9, 21, 1, 15, 2, 4, 8, 18, 2, 3, 16, 3, 1, 9, 12, 5, 16, 5, 1, 5, 1, 1, 0, 23, 1, 2, 16, 9, 13, 8, 22, 2, 5, 12, 18, 26, 0, 13, 8, 8, 0

Offset: 1

David James Sycamore, May 07 2022

nonn

a(n) <= n for all n, with equality when n=1. Similar to A328096 initially, because novel terms are all records to begin with, but diverges when non-record novel terms appear, which produce nonzero following terms.

			a(1)=1 is a novel record term, hence a(2)=0. a(3)=1 because a(2)=0 is a novel (non-record) term and only one term, a(1), exceeds it.
a(25) is the first occurrence of 6, so a(26)=2 because two prior terms (9 and 11) are greater than 6.
a(37) is the second occurrence of 4, first seen at a(12), therefore a(38)=37-12-1=24.

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A328096.

Block[{a, c, k, m, s, nn}, nn = 81; c[] = -1; a[1] = c[1] = 1; s = {}; Do[m = LengthWhile[s, # > a[i - 1] &]; Set[k, If[c[#] == -1, m, i - c[#] - 1]] &@ a[i - 1]; Set[{a[i], c[a[i - 1]]}, {k, i}]; s = Insert[s, a[i - 1], 1 + m], {i, 2, nn}]; Array[a, nn] ] (* _Michael De Vlieger, May 08 2022 *)

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