A308190 -id:A308190 - cp's OEIS frontend

A308191 a(n) = smallest m such that A308190(m) = n, or -1 if no such m exists.

5, 6, 8, 7, 10, 16, 17, 30, 29, 54, 53, 102, 101, 198, 197, 390, 389, 774, 773, 1542, 3080, 3079, 6154, 12304, 24604, 36901, 73798, 147592, 295180, 295517, 591030, 1182056, 1574849, 3149694, 4728211, 6299383, 12598762, 25197520, 25197533, 50395062, 100790120, 100790119, 201580234, 403160464, 806320924, 1232145821, 2464291638

Offset: 0

N. J. A. Sloane, Jun 14 2019

nonn

It seems plausible that m exists for all n >= 0.
From Chai Wah Wu, Jun 14 2019: (Start)
All terms are even or prime. If a(n+1) is even, then 2*a(n)-a(n+1) = 4. a(n+1) <= 2*(a(n)-2) and thus m exists for all n >= 0. The proof in the comments of A308193 is applicable for this sequence as well.
If a(n) is prime, then a(n-1) <= a(n) + 1. For the prime terms 7, 17, 29, 53, 101, 197, 389, 773, 3079, 100790119, a(n-1) = a(n) + 1.
(End)

Chai Wah Wu, Table of n, a(n) for n = 0..54

Cf. A111234, A308190.

a(24)-a(41) from Chai Wah Wu, Jun 14 2019
a(42)-a(44) from Chai Wah Wu, Jun 15 2019
a(45)-a(46) from Chai Wah Wu, Jun 16 2019

A308193 Indices of records in A308190.

Original entry on oeis.org

5, 6, 7, 10, 16, 17, 29, 53, 101, 197, 389, 773, 1542, 3079, 6154, 12304, 24604, 36901, 73798, 147592, 295180, 295517, 591030, 1182056, 1574849, 3149694, 4728211, 6299383, 12598762, 25197520, 25197533, 50395062, 100790119, 201580234, 403160464, 806320924, 1232145821, 2464291638

Offset: 1

N. J. A. Sloane, Jun 14 2019

nonn

For terms a(1) through a(16), with one exception, 2*a(n) - a(n+1) is either 4 or 5. Does this pattern continue, and if so, why?
From Chai Wah Wu, Jun 14 2019: (Start)
The pattern does not continue. a(17) = 24604, a(18) = 36901.
Theorem:
1. All terms are even or prime.
2. If a(n+1) is even, then 2*a(n)-a(n+1) = 4.
3. a(n+1) <= 2*(a(n)-2).
Proof: If a(n+1) = x is even, then A111234(x) = 2+x/2 = y. If we assume that x >= 6, then y < x. Thus A308190(x) = A308190(y)+1, i.e., a(n) <= y. If a(n) < y, then A308190(2*(a(n)-2)) = A308190(a(n)) + 1.
Since a(n) is a record value, this means that the next record value is at most at 2*(a(n)-2), i.e., 2*(a(n)-2) < x = a(n+1), a contradiction.
Thus we have shown that if a(n+1) is even, then 2*a(n) = a(n+1)+4.
If a(n+1) = x is an odd composite with smallest prime factor p > 2, then A308190(x) = A308190(y)+1 where y = p+x/p. On the other hand, A308190(2*(y-2)) = A308190(y)+1. Since 2*(y-2) < x, this contradicts the fact that a(n+1) = x is a record value.
(End)

Chai Wah Wu, Table of n, a(n) for n = 1..45

Cf. A111234, A308190, A308191, A308192.

a(17)-a(36) from Chai Wah Wu, Jun 14 2019

a(37)-a(38) from Chai Wah Wu, Jun 16 2019

A308192 Record values in A308190.

Original entry on oeis.org

0, 1, 3, 4, 5, 6, 8, 10, 12, 14, 16, 18, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 54

Offset: 1

N. J. A. Sloane, Jun 14 2019

Cf. A111234, A308190, A308191, A308193.

a(17)-a(36) from Chai Wah Wu, Jun 14 2019
a(37)-a(38) from Chai Wah Wu, Jun 16 2019
a(39)-a(41) from Chai Wah Wu, Jun 17 2019
a(42)-a(45) from Chai Wah Wu, Jun 24 2019

A308194 Number of steps to reach 5 when iterating x -> A063655(x) starting at x=n.

Original entry on oeis.org

0, 1, 3, 2, 2, 4, 5, 4, 4, 3, 3, 3, 4, 3, 4, 3, 5, 5, 6, 5, 5, 4, 5, 6, 7, 6, 6, 5, 4, 5, 5, 5, 7, 6, 4, 5, 6, 5, 5, 4, 4, 6, 5, 4, 4, 4, 4, 5, 5, 4, 4, 4, 6, 7, 5, 4, 6, 5, 4, 4, 4, 5, 7, 6, 5, 5, 6, 5, 6, 5, 4, 7, 4, 5, 5, 4, 4, 6, 6, 5, 6, 5, 6, 5, 6, 5, 4, 6, 6, 5, 6, 4, 7, 6, 4, 4, 8, 7, 7, 6, 6, 5

Offset: 5

N. J. A. Sloane, Jun 14 2019

nonn

It is easy to show that every number n >= 5 eventually reaches 5. This was conjectured by Ali Sada. For A111234 sends a composite n > 5 to a smaller number, and sends a prime > 5 to a smaller number in two steps. Furthermore no number >= 5 can reach a number less than 5. So all numbers >= 5 eventually reach 5.

Ali Sada, Email to N. J. A. Sloane, Jun 13 2019.

Rémy Sigrist, Table of n, a(n) for n = 5..10000

Cf. A063655, A308190, A308195.

b(n) = { my(c=1); fordiv(n, d, if((d*d)>=n, if((d*d)==n, return(2*d), return(c+d))); c=d); (0); } \\ after A063655
a(n) = for (k=0, oo, if (n==5, return (k), n=b(n))) \\ Rémy Sigrist, Jun 14 2019

from sympy import divisors
def A308194(n):
    c, x = 0, n
    while x != 5:
        d = divisors(x)
        l = len(d)
        x = d[(l-1)//2] + d[l//2]
        c += 1
    return c # Chai Wah Wu, Jun 14 2019

A308725 Number of steps to reach 6 or 7 when iterating x -> A227215(x) starting at x=n, where A227215(n) gives the smallest such sum a+b+c of three positive integers for which abc = n.

Original entry on oeis.org

3, 3, 2, 2, 1, 0, 0, 1, 1, 2, 4, 1, 3, 3, 2, 2, 7, 2, 6, 2, 5, 4, 6, 2, 5, 3, 2, 5, 5, 3, 4, 3, 3, 3, 4, 3, 9, 5, 8, 5, 7, 2, 6, 3, 5, 4, 4, 5, 3, 2, 6, 8, 9, 2, 8, 4, 7, 4, 6, 2, 5, 4, 4, 2, 7, 3, 4, 6, 3, 4, 6, 4, 5, 6, 4, 7, 7, 3, 4, 4, 3, 4, 8, 4, 7, 5, 4, 8, 7, 4, 6, 3, 5, 3, 6, 4, 9, 3, 8, 4

Offset: 1

Ali Sada, Jun 20 2019

nonn

Starting from n, choose factorization n = m1*m2*m3 so that the sum x = m1+m2+m3 is minimal, then set n = x and repeat. a(n) gives the number of steps needed to reach either 6 or 7. The process is guaranteed to reach either term, because we only use factorization n = n*1*1 when n is either 1 or a prime number, that are the only cases (apart from A227215(4)=5) for which A227215(n) > n as then A227215(n) = n+2. Moreover, for n > 3, at least one of n, n+2, n+4 is composite, leading to a further significant drop in the trajectory after at most two consecutive +2 steps. - Comment clarified by Antti Karttunen, Jul 12 2019

Records: 3, 4, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, ..., occur at: n = 1, 11, 17, 37, 107, 233, 307, 1289, 3986, 6637, 14347, 69029, .... - Antti Karttunen, Jul 12 2019

			    1 = 1*1*1  --> 1 + 1 + 1  = 3
    3 = 1*1*3  --> 1 + 1 + 3  = 5
    5 = 1*1*5  --> 1 + 1 + 5  = 7, thus a(1) = 3.
.
    4 = 1*2*2  --> 1 + 2 + 2  = 5,
    5 = 1*1*5  --> 1 + 1 + 5  = 7, thus a(4) = 2.
.
  560 = 7*8*10 --> 7 + 8 + 10 = 25
   25 = 1*5*5  --> 1 + 5 +  5 = 11
   11 = 1*1*11 --> 1 + 1 + 11 = 13
   13 = 1*1*13 --> 1 + 1 + 13 = 15
   15 = 1*3*5  --> 1 + 3 +  5 =  9
    9 = 1*3*3  --> 3 + 3 +  1 =  7, thus a(560) = 6.
.
   84 = 3*4*7  --> 3 + 4 + 7 = 14
   14 = 1*2*7  --> 1 + 2 + 7 = 10
   10 = 1*2*5  --> 1 + 2 + 5 =  8
    8 = 2*2*2  --> 2 + 2 + 2 =  6, thus a(84) = 4.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A227215, A308190.

maxTerm = 99 (* Should be increased if output -1 appears. *);
f[m_] := Module[{m1, m2, m3, factors}, factors = {m1, m2, m3} /. {ToRules[ Reduce[1 <= m1 <= m2 <= m3 && m == m1 m2 m3, {m1, m2, m3}, Integers]]}; SortBy[factors, Total] // First];
a[n_] := Module[{cnt = 0, m = n, fm, step}, While[!(m == 6 || m == 7), step = {fm = f[m], m = Total[fm]}; (* Print[n," ",step]; *) cnt++; If[cnt > maxTerm, Return[-1]]]; cnt];
Array[a, 100] (* Jean-François Alcover, Jul 03 2019 *)

A227215(n) = { my(ms=3*n); fordiv(n, i, for(j=i, (n/i), if(!(n%j),for(k=j, n/(i*j), if(i*j*k==n, ms = min(ms,(i+j+k))))))); (ms); }; \\ Like code in A227215.
A308725(n) = if((6==n)||(7==n),0,1+A308725(A227215(n)));
\\ Memoized implementation:
memoA308725 = Map();
A308725(n) = if((6==n)||(7==n), 0, my(v); if(mapisdefined(memoA308725,n,&v), v, v = 1+A308725(A227215(n)); mapput(memoA308725,n,v); (v))); \\ Antti Karttunen, Jul 12 2019

If n is 6 or 7, a(n) = 0, otherwise a(n) = 1 + a(A227215(n)). - Antti Karttunen, Jul 11 2019

cp's OEIS Frontend

A308191 a(n) = smallest m such that A308190(m) = n, or -1 if no such m exists.

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A308193 Indices of records in A308190.

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A308192 Record values in A308190.

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A308194 Number of steps to reach 5 when iterating x -> A063655(x) starting at x=n.

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A308725 Number of steps to reach 6 or 7 when iterating x -> A227215(x) starting at x=n, where A227215(n) gives the smallest such sum a+b+c of three positive integers for which a*b*c = n.

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A308725 Number of steps to reach 6 or 7 when iterating x -> A227215(x) starting at x=n, where A227215(n) gives the smallest such sum a+b+c of three positive integers for which abc = n.