A372496 Integers of the form k^2 + 1, where k >= 1, that are the product of two other integers of the form k^2 + 1, where k >= 1.
10, 50, 170, 290, 325, 442, 962, 1850, 2210, 3250, 5330, 8282, 9802, 12322, 15130, 17425, 17690, 24650, 33490, 44522, 58082, 58565, 64010, 65026, 74530, 94250, 103685, 117650, 145162, 177242, 191845, 214370, 237170, 257050, 305810, 332930, 361202
Offset: 1
Keywords
Examples
50 is a term since 50 = 7^2 + 1 = 10 * 5 = (3^2 + 1)*(2^2 + 1).
Links
- Ely Golden, Table of n, a(n) for n = 1..10667 (terms <= 10^16)
Programs
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Mathematica
formQ[k_] := k >= 1 && IntegerQ@Sqrt[k - 1]; prodQ[k_] := AnyTrue[Divisors[k][[2 ;; -2]], formQ[#] && formQ[k/#]&]; okQ[k_] := formQ[k] && prodQ[k]; Select[Range[2, 10^6], okQ] (* Jean-François Alcover, May 10 2024 *)
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PARI
isok1(k) = issquare(k-1) && (k>1); isok2(k) = fordiv(k, d, if (isok1(d) && isok1(k/d), return(1))); isok(k) = isok1(k) && isok2(k); \\ Michel Marcus, May 04 2024
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Python
from math import isqrt def is_perfect_square(n): return isqrt(abs(n))**2 == n limit = 10**16 sequence_entries = set() for a in range(1, isqrt(isqrt(limit))+1): u = a**2 + 1 for b in range(a+1, isqrt(limit//u)+1): v = b**2 + 1 if(is_perfect_square(u*v - 1)): sequence_entries.add(u*v) sequence_entries = sorted(sequence_entries) for i, j in enumerate(sequence_entries, 1): print(i, j)
Comments