A309981 -id:A309981 - cp's OEIS frontend

A327265 a(n) is the smallest k such that A309981(k) = n.

1, 2, 5, 11, 19, 31, 51, 89, 123, 151, 179, 181, 180, 365, 634, 657, 656, 655

Offset: 0

Jon E. Schoenfield, Sep 15 2019

Consider the problem of deducing the identity of an unknown positive integer k given only its number of divisors tau(k) and those of its first m successors, i.e., tau(k+1), tau(k+2), ... tau(k+m). A309981(k) is the minimum m that allows unique identification of k, and a(n) is the smallest k for which that minimum m is n.

			n = 0: Using 0 successors, the fact that tau(k) = 1 is sufficient to deduce that k = 1; there is no other k whose identity can be deduced given only tau(k), so a(0) = 1.
n = 1: tau(k) = 2 for all primes k, but given tau(k) = 2 and tau(k+1) = 2, the only solution is k = 2 (since k = 2 and k + 1 = 3 are the only two consecutive integers that are both prime). Other than k = 1, whose identity can be deduced given only tau(k), k = 2 is the smallest k whose identity can be deduced given only tau(k) and tau(k+1), so a(1) = 2.
n = 2: There are many integers k such that tau(k) = 2 and tau(k+1) = 4, but the only k such that tau(k), tau(k+1), and tau(k+2) are 2, 4, and 2, respectively, is k = 5. Thus A309981(5) = 2. There is no number k < 5 for which A309981(k) = 2, so a(2) = 5.
There are numbers m != 89 such that tau(m+j) = tau(89+j) for all j in 0..6 (the first such number is 242510633), but there is no number m such that tau(m+j) = tau(89+j) for all j in 0..7, and 89 is the smallest k such that A309981(k)=7, so a(7)=89.
a(8)=123; the smallest m such that tau(m+j) = tau(123+j) for all j in 0..7 is apparently 476129486151666513937.
a(9)=151; the smallest m such that tau(m+j) = tau(151+j) for all j in 0..8 is 3579145012951.

Cf. A309981.

a(6) corrected by Jon E. Schoenfield, Dec 15 2019

A327909 a(n) is the smallest start of a run of n or more integers having a prime factor greater than n.

Original entry on oeis.org

2, 5, 13, 19, 55, 65, 113, 151, 151, 226, 364, 406, 736, 736, 1057, 1057, 1409, 1409, 2059, 2059, 2313, 2313, 2313, 2313, 2313, 2313, 2313, 6007, 6961, 6961, 10305, 12013, 12013, 12013, 12013, 12013, 12026, 12026, 17501, 17501, 17501, 17501, 20833, 20833

Offset: 1

Jon E. Schoenfield, Oct 06 2019

nonn

Is a(n) an upper bound on A327265(n)? A327265(n) = a(n) at n = 1, 2, 4, and 9.

			      |     prime     |
   k  | factorization | gpf(k) | tau(k)
  ----+---------------+--------+-------
  151 |      151      |   151  |   2
  152 |   2^3 * 19    |    19  |   8
  153 |   3^2 * 17    |    17  |   6
  154 |  2 * 7 * 11   |    11  |   8
  155 |    5 * 31     |    31  |   4
  156 | 2^2 * 3 * 13  |    13  |  12
  157 |      157      |   157  |   2
  158 |    2 * 79     |    79  |   4
  159 |    3 * 53     |    53  |   4

Thomas Garrison, Table of n, a(n) for n = 1..369

Cf. A006530 (greatest prime factor of n).

Cf. A309981, A327265.

A:= Vector(100): A[1]:= 2: count:= 1:
B:= Vector(100):
for i from 2 while count < 100 do
  p:= max(numtheory:-factorset(i));
  for j from 1 to min(p-1,100) do
    if B[j] = 0 then B[j]:= i fi
  od;
  for j from p to 100 do
    if B[j] > 0 and B[j] <= i-j and A[j] = 0 then A[j]:= B[j]; count:= count+1; fi
  od;
  if p <= 99 then B[p..100]:= 0 fi;
od:
convert(A,list); # Robert Israel, Jan 23 2023

a(n) = {my(k=1); x=0; while(xn, x++, x=0)); k-n+1;} \\ Jinyuan Wang, Oct 26 2019

A361088 Irregular table, read by rows, where row n holds the tau signature of n, i.e., the shortest sequence (tau(n+k), 0 <= k <= m) that uniquely identifies n; tau = A000005.

Original entry on oeis.org

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

Offset: 1

M. F. Hasler, Apr 07 2023

Row lengths are given by A309981(n) + 1; see there (and the OEIS wiki page) for examples.

			The first 20 rows read as follows:
   n | row n: tau-signature of n
  ---+--------------------------
   1 | [1]
   2 | [2, 2]
   3 | [2, 3]
   4 | [3, 2]
   5 | [2, 4, 2]
   6 | [4, 2, 4]
   7 | [2, 4, 3]
   8 | [4, 3]
   9 | [3, 4, 2]
  10 | [4, 2, 6]
  11 | [2, 6, 2, 4]
  12 | [6, 2, 4]
  13 | [2, 4, 4, 5]
  14 | [4, 4, 5]
  15 | [4, 5]
  16 | [5, 2]
  17 | [2, 6, 2, 6]
  18 | [6, 2, 6]
  19 | [2, 6, 4, 4, 2]
  20 | [6, 4, 4, 2, 8]
See the wiki page for proofs.

M. F. Hasler et al, tau signature, OEIS Wiki, April 2023.

Cf. A309981, A327265, A161460, A000005 (tau = numdiv).

signatures=Map(); LIMIT=10^5 /* This search limit should (possibly dynamically, or by hand) be increased as n grows beyond 100. As of today, the value for n=49 is not yet proven. */
A361088_row(n,s=0)={if(!s, s=iferr(mapget(signatures,n),E,[]); #s|| for(L=1,oo, s=concat(s,numdiv(n+L-1)); A361088_row(n,s)|| [mapput(signatures,n,[s,LIMIT]); return(s)]); s[2]>=LIMIT&& return(s[1]); s=s[1]; while(A361088_row(n,s), s=concat(s,numdiv(n+#r))); mapput(signatures,n,[s,LIMIT]); return(s)); my(r=iferr(mapget(signatures,s), E,[])); if(!r, r=[n,n], r[2]2, return(r[#r-1]), r[#r]>=LIMIT, return); for(j=max(r[2],n)+1,LIMIT, for(k=1,#s, numdiv(j+k-1)!=s[k]&& next(2)); mapput(signatures,s,[n,j,j]); return(j)); mapput(signatures,s,[n,LIMIT])}

cp's OEIS Frontend

A327265 a(n) is the smallest k such that A309981(k) = n.

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A327909 a(n) is the smallest start of a run of n or more integers having a prime factor greater than n.

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Maple

PARI

A361088 Irregular table, read by rows, where row n holds the tau signature of n, i.e., the shortest sequence (tau(n+k), 0 <= k <= m) that uniquely identifies n; tau = A000005.

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Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI