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 A350228 Multiplicative Van Eck sequence: for n >= 2, if there exists an m < n such that a(m) = a(n), take the largest such m and set a(n+1) = (n-m)*a(n); otherwise a(n+1) = 1. Start with a(1)=1 and a(2)=0. 8
 1, 0, 1, 2, 1, 2, 4, 1, 3, 1, 2, 10, 1, 3, 15, 1, 3, 9, 1, 3, 9, 27, 1, 4, 68, 1, 3, 21, 1, 3, 9, 90, 1, 4, 40, 1, 3, 21, 210, 1, 4, 28, 1, 3, 21, 147, 1, 4, 28, 196, 1, 4, 16, 1, 3, 33, 1, 3, 9, 252, 1, 4, 40, 1120, 1, 4, 16, 224, 1, 4, 16, 64, 1, 4, 16, 64, 256, 1, 5, 1, 2, 140 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS Theorem 1: There are infinitely many 1's. Proof: Suppose not. At each successive term, the sequence either grows or returns a 1. If no 1's occur after a certain term called x, then the sequence is monotonically increasing after x. It can be shown that the series grows at least as fast as the factorial between ones. Then for some sufficiently large n > x, a(n) must be greater than any term to appear in the sequence thus far. And a 1 would be returned as the next term. Theorem 2: If a(n)=k, and k != 1, then an upper bound for the number of terms from the previous 1 to the n-th term is the length of the longest factoring of k such that the m-th factor is at least equal to m. Call this f(n). If a(n)=1, then the last 1 is at most f(a(n-1))+1 terms away. Proof: This is because the term a(n) can be represented as x(1)*x(2)*...*x(m), with x(i) being the last time a(n-m+i-1) was seen in the sequence. Because you must cross at least the last 1 to get to the previous a(n), x(i) >= i. So the longest string of x(i)'s that can exist is one where the i-th factor is greater than or equal to i. The factors are not necessarily prime. The "longest factoring" (f(n)) refers to the longest string of numbers (x(1)*x(2)*...*x(m)) that can be multiplied to arrive at n. Theorem 3: If a(n) is prime then a(n-2) is appearing in the sequence for the first time. Proof: Suppose a(n) is prime. Then a(n-1) must be 1 or it must be a nontrivial divisor of a(n), unless a(n) = 1 (or the trivial case that a(n) = 0). There are no nontrivial divisors of prime numbers so a(n-1) must equal 1. It follows then that a(n-2) is appearing for the first time in the sequence because a(n-1) = 1. LINKS Michael De Vlieger, Table of n, a(n) for n = 1..10000 Michael De Vlieger, Log-log scatterplot of a(n) for n = 1..2^16, labeling and showing records in red, and labeling and showing first instances of differences of positions of 1's in blue. MATHEMATICA f[1]=1; f[n_]:=0; f2[n_]:=0; a:=(Block[{q=f2[x]}, If[q!=0, s[#]=(#-1-q)*(x), s[#]=1]])&; s[1]=1; s[2]=0; x=0; data=Reap[Sow[1]; Sow[0]; Do[Sow[x=a[n]]; f2[x]=f[x]; f[x]=n, {n, 3, 1000000}]][[2, 1]] PROG (Python) from itertools import islice, count def A350228gen(): yield from (1, 0) b, bdict = 0, {1:(1, ), 0:(2, )} for n in count(3): if len(l := bdict[b]) > 1: m = (n-1-l[-2])*b if m in bdict: bdict[m] = (bdict[m][-1], n) else: bdict[m] = (n, ) b = m else: bdict[1] = (bdict[1][-1], n) b = 1 yield b A350228_list = list(islice(A350228gen(), 20)) # Chai Wah Wu, Dec 21 2021 (PARI) findm(list, n) = {forstep (m=n-1, 1, -1, if (list[m] == list[n], return(m))); return(0); } lista(nn) = {my(list = List([1, 0])); for (n=3, nn, my(m = findm(list, n-1)); if (m, listput(list, (n-1-m)*list[n-1]), listput(list, 1); ); ); Vec(list); } \\ Michel Marcus, Jan 16 2022 CROSSREFS Cf. A181391, A350231, A350234, A350244. Sequence in context: A125694 A136678 A110162 * A199087 A306913 A087704 Adjacent sequences: A350225 A350226 A350227 * A350229 A350230 A350231 KEYWORD nonn AUTHOR Jasmine Miller, Dec 20 2021 STATUS approved

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Last modified February 28 21:38 EST 2024. Contains 370400 sequences. (Running on oeis4.)