A327265 -id:A327265 - cp's OEIS frontend

A309981 a(n) is the smallest number k such that the value of n can be deduced given only the values tau(n), tau(n+1), ..., tau(n+k), where tau is the number of divisors function.

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

Offset: 1

Jon E. Schoenfield, Aug 25 2019

nonn

For every number n in 2..87, there exists at least one other number m in 2..72970801 such that tau(m+j) = tau(n+j) for each nonnegative j less than the listed term, which establishes a lower bound for the term. (For the smallest such m corresponding to each term in the Data, see the Links.) That each value listed in the Data is the exact value was proved using a variety of methods such as those in the Example section and some of the methods in the R. J. Mathar link at A161460.
From Jon E. Schoenfield, Sep 23 2019: (Start)
It follows from the definition of this sequence that a(n) >= a(n-1) - 1, which provides an easy lower bound on a(n) (given a(n-1)). An easy (but weak) upper bound is provided by a(n) <= (ceiling(sqrt(n))+1)^2 - n, since any counterexample would mean that there exists a number n for which a list of the tau values of all the consecutive integers from n through (ceiling(sqrt(n))+1)^2 would not be sufficient to allow the deduction of the identity of n, even though those consecutive integers would include at least two consecutive squares (immediately identifiable as squares by their odd tau values), and the only pair of consecutive squares with difference d is (floor(d/2)^2, ceiling(d/2)^2).
For the indices at which the values 0..17 first appear in this sequence, see A327265.
Do all nonnegative integers eventually appear? (End)

			a(1) = 0 is the only occurrence of 0 since n = 1 is the only number that has only 1 divisor and therefore tau(n) = 1 uniquely identifies n = 1. For any other number we need more tau values (tau(n+j), 0 <= j <= k) in order to identify n through these k+1 values.
a(2) = 1: n = 2 is the only prime such that n+1 is also prime, therefore tau(n) = 2 and tau(n+1) = 2 implies n = 2 (and only tau(n) = 2 is not enough).
a(3) = 1: n = 3 is the only prime such that n+1 is the square of a prime, i.e., n+1 has 3 divisors, so tau(n) = 2 and tau(n+1) = 3 uniquely identify n = 3.
a(4) = 1: n = 4 is the only square of a prime such that n+1 is a prime. (For all other n = p^2 with p prime, n+1 is an even composite.)
a(5) = 2: the numbers tau(n) and tau(n+1) are 2 and 4, respectively, for n = 5, 7, 13, 37, 61, ..., but 5 is the only number n such that (tau(n), tau(n+1), tau(n+2)) = (2, 4, 2). (Indeed this means that n, n+2 are twin primes, separated by a semiprime n+1. But all larger twin primes are of the form 6m+-1, so the intermediate number is a larger multiple of 6 with more than 4 divisors.)
From _M. F. Hasler_, Mar 31 2023: (Start)
a(6) = 2: Here tau(n) = 4, tau(n+1) = 2, tau(n+2) = 4 mean that n is a semiprime or p^3, followed by a prime and then another semiprime or q^3. For larger n, prime powers are excluded through parity ((n, n+1) = (p^3, p') and/or (n+1, n+2) = (p', q^3) are all odd), and one can't have two even semiprimes separated by 2, n = 2p, n+2 = 2q. So n = 6 is the only such number. (End)
a(49) = 2: (tau(n), tau(n+1)) = (3,6) for n = 49, 1681, and only two other known values (i.e., for squares of terms > 3 in A086397), but (tau(n), tau(n+1), tau(n+2)) = (3, 6, 4) only for n = 49 (which is the only square of a prime p such that both sqrt((p^2 + 1)/2) and (p^2 + 2)/3 are also prime).

Jon E. Schoenfield, Table of n, m(n) for n = 1..87, where m(n) is the smallest nonnegative integer (other than n) such that tau(m+j) = tau(n+j) for all j in 0..a(n)-1 (corrected Apr 20 2023, with thanks to M. F. Hasler for pointing out that there were errors in the original file).

Cf. A000005, A086397, A161460, A327265.

a(1)=0; for n > 1, a(n) = 1 iff n is a term of A161460.

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

A309981 a(n) is the smallest number k such that the value of n can be deduced given only the values tau(n), tau(n+1), ..., tau(n+k), where tau is the number of divisors function.

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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

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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

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Examples

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Crossrefs

Programs

PARI