cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

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.

Original entry on oeis.org

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

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Author

Jon E. Schoenfield, Aug 25 2019

Keywords

Comments

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)

Examples

			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).
		

Crossrefs

Formula

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

A118612 Denominator if the numerator and denominator of the continued fraction rational approximation of sqrt(2) are both prime.

Original entry on oeis.org

2, 5, 29, 44560482149, 13558774610046711780701
Offset: 1

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Author

Joshua Zucker, May 08 2006

Keywords

Comments

Next term, if it exists, is bigger than 489 digits (the 1279th convergent to sqrt(2)).

Crossrefs

A086397 has the numerators. This sequence is a subsequence of A000129, A086383 and A101411.

Programs

  • Mathematica
    For[n = 2, n < 1500, n++, a := Join[{1}, Table[2, {i, 2, n}]]; If[PrimeQ[Denominator[FromContinuedFraction[a]]], If[PrimeQ[Numerator[FromContinuedFraction[a]]], Print[Denominator[FromContinuedFraction[a]]]]]] (* Stefan Steinerberger, May 09 2006 *)

A163742 Prime pairs (p,q) of the form p=A002315(k), q=A001653(k) for some k.

Original entry on oeis.org

7, 5, 41, 29, 63018038201, 44560482149, 19175002942688032928599, 13558774610046711780701
Offset: 1

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Author

Vincenzo Librandi, Aug 03 2009

Keywords

Comments

By construction, all these pairs satisfy p^2 -2*q^2 = -1.
One can generate the combined A002315 and A001653 by a mixed recursion 3p+4q -> p; 2p+3q -> q, and then test after each step both p and q for primality.

Examples

			(A002315(1),A001653(1)) = (7,5) are both prime and define the first pair in the sequence.
(A002315(2),A001653(2)) = (41,29) are both prime and define the second pair in the sequence.
(A002315(3),A001653(3)) = (239,169=13^2) contain the composite 169 and do not contribute to the sequence.
		

Crossrefs

Bisections are (essentially): A086397, A118612.

Extensions

Definition clarified by R. J. Mathar, Aug 12 2009

A164986 Numbers of the form 2p^2 = q^2 + 1, where p and q are primes.

Original entry on oeis.org

50, 1682, 3971273138702695316402, 367680737852094722224630791187352516632102802
Offset: 1

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Author

Rick L. Shepherd, Sep 03 2009

Keywords

Comments

A079704 INTERSECT A002522. Subsequence of A088920 (Solutions k to the Diophantine equation k = 2n^2 = m^2+1): those terms for which associated m in A002315 and n in A001653 are both prime.
Corresponding p are prime Pell numbers (prime denominators of continued fraction convergents to sqrt(2)).
Corresponding q are prime numerators of the continued fraction convergents to sqrt(2).
Corresponding p, q, p^2, q^2, (p,q), (q,p), etc., form subsequences of many other OEIS sequences; see cross-references.
Any further terms are too large to include here.

Examples

			a(1) = 50 as 50 = 2*5^2 = 7^2 + 1, where 5 and 7 are prime.
		

Crossrefs

Cf. A088920, A118612, A086397, A086395, A002315 (NSW numbers), A088165 (prime NSW numbers = prime RMS numbers (A140480)), A001653, A000129 (Pell numbers), A086383, A101411, A079704, A002522, A008843, A104683, A163742, etc.

Formula

a(n) = 2*(A118612(n+1))^2 = (A086397(n+1))^2 + 1.
Showing 1-4 of 4 results.