A257366 Smallest integer m such that m^2 + 1 has exactly n prime factors, counted with multiplicity.

1, 3, 7, 43, 57, 307, 1068, 2943, 12943, 45807, 110443, 670807, 2733307, 25670807, 113561432, 123327057, 657922943, 17213170807, 7200891693, 148802454193, 1120482141693

Offset: 1

Author

Zak Seidov, Apr 21 2015

Keywords

• nonn
• more

Comments

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)

Examples

			a(1)=1 because 1^2+1=2(prime),
a(2)=3 because 3^2+1=10=2*5,
a(3)=7 because 7^2+1=50=2*2*5,
...............
a(14)=25670807 because 25670807^2+1=2*5^11*149*45289.
		

Crossrefs

Cf. A002144, A180278, A193330.

Programs

Mathematica

Table[m = 1; While[PrimeOmega[m^2 + 1] != n, m++]; m, {n, 12}] (* Michael De Vlieger, Apr 21 2015 *)

PARI

a(n)=my(m); while(bigomega(m++^2+1)!=n, ); m \\ Charles R Greathouse IV, Apr 21 2015

Extensions

a(15)-a(17) from Jon E. Schoenfield, Jun 14 2015

a(18)-a(19) from Jon E. Schoenfield, Jun 15 2015

a(20) from Jon E. Schoenfield, Jul 10 2015

a(21) from Max Alekseyev, Jan 08 2025