cp's OEIS Frontend

a(n)+2 is prime for n=1,2. No others are prime for n <= 200. Compare A002110 and A078586. - T. D. Noe, Dec 01 2002

Also, a(n) is least hypotenuse of exactly A003462(n) Pythagorean triangles of which 2^(n-1) are primitive. - Lekraj Beedassy, Dec 06 2003

Also, a(n) are the record setting values of m, for the number of solutions to "m*k-1 is a square", for some k, 1 <= k < m. There is one solution for m=2, and for a given m = a(n) there are 2^n solutions. For a given m there also 2^(n-1) solutions for primitively representing m as x^2 + y^2. See A008782. Also compare with A102476, which applies to "m*k+1 is a square". - Richard R. Forberg, Mar 18 2016

a(n) is the smallest m such that A000089(m) = 2^n. Also, numbers k for which A000089(k) sets a new record. - Jianing Song, Apr 27 2019

Or, first occurrences of n in A193330.

Is the sequence finite?

a(n) exists for arbitrarily large n, and in particular a(n+k) < A185952(n) by the Chinese Remainder Theorem and the fact that -1 is a square mod the primes in A002313, for some k >= 0. Probably a(n) exists for each n. - Charles R Greathouse IV, Apr 21 2015

From Jon E. Schoenfield, Jun 14-15 2015: (Start)

Numbers of the form m^2+1 cannot be divisible by 3; they may be divisible by 2 (but not 2^2), and the only other numbers they can have as prime factors are 5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, ..., i.e., the terms of A002144. This explains why 5 tends to appear with such high multiplicity in the factorizations of a(n)^2+1, as numbers with a higher multiplicity of the prime factor 5 are more likely to be small enough to be the smallest integer with n prime factors than numbers whose n prime factors (counted with multiplicity) are mostly larger than 5. Having 2 as one of the n prime factors is also an advantage, which accounts for the predominance of odd terms, yielding even values of m^2+1.

For k>1, if m^2+1 is divisible by 5^k, then there are only two possible residues of m mod 5^k; e.g., if m^2+1 is divisible by 5^4, then m mod 5^4 must be either 182 or 443. Thus it is not coincidental that the last few digits of some terms also appear as the last few digits of other terms, e.g., terms ending in 443 or 443+500 = 943, or in 182+125 = 307 or 182+625 = 807. (End)

Least value of A225102 that is the product of n distinct primes.

From Jon E. Schoenfield, Mar 18 2017: (Start)

For each n, we can write a(n) = p^j + q^k where p and q are prime and 2 <= j <= k; since a(n) is squarefree, p and q are distinct.

Suppose j and k are both even. Then a(n) cannot have any prime factor f such that f == 3 (mod 4) (see A002145). Thus, a(n) is the product of n distinct terms of {2, 5, 13, 17, 29, 37, 41, ...} = A002313, so a(n) >= Product_{i=1..n} A002313(i) = A185952(n).

In fact, however, a(n) < A185952(n) for n = 4..15, and it seems nearly certain that this holds for all n > 3. In any case, if we search for a(n) by generating products of n distinct primes and, for each such product P, testing whether there exists a solution for P = p^j + q^k, then we need not consider solutions in which both j and k are even unless P >= A185952(n).

Additionally, since the sum of any two cubes that is divisible by 3 is also divisible by 9 (hence nonsquarefree), any P that is divisible by 3 cannot be the sum of two cubes, so the exponents j and k cannot both be divisible by 3. (Every P < 2*5*7*11*...*prime(n+1) = A002110(n+1)/3 is divisible by 3.) Thus, for every P that is divisible by 3 and < A185292(n), we can rule out every ordered pair (j,k) except (2,3) and (3,4) (which could be tested together by computing t = P - r^3 for each prime r < P^(1/3) and, if t is square, checking whether sqrt(t) is a prime or the square of a prime) and those with k >= 5 (which could be tested by checking whether t = P - q^k is a prime power for each prime power q^k that is less than P and has k >= 5). (End)

a(17) <= 63985284333636413237490 = 2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 * 23 * 29 * 37 * 41 * 43 * 59 * 61 * 103 * 409 = 10461281^3 + 250679912393^2. - Jon E. Schoenfield, Mar 31 2017