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 A247221 Numbers n such that 2*n^2 + 1 divides 2^n + 1. 2
 0, 1, 67653, 2124804 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS Numbers n such that (2^n + 1)/(2*n^2 + 1) is an integer. a(5) > 2*10^10. - Chai Wah Wu, Dec 07 2014 LINKS Table of n, a(n) for n=1..4. EXAMPLE 0 is in this sequence because 2^0 + 1 = 2 divides 2*0^2 + 1 = 1, 1 is in this sequence because 2^1 + 1 = 3 divides 2*1^2 + 1 = 3. MATHEMATICA a247221[n_Integer] := Select[Range[n], Divisible[2^# + 1, 2*#^2 + 1] &]; a247221[2500000] (* Michael De Vlieger, Nov 30 2014 *) PROG (Magma) [n: n in [1..300000] | Denominator((2^n+1)/(2*n^2+1)) eq 1]; (PARI) for(n=0, 10^9, if(Mod(2, 2*n^2+1)^n==-1, print1(n, ", "))); \\ Joerg Arndt, Nov 30 2014 (Python) A247221_list = [n for n in range(10**6) if pow(2, n, 2*n*n+1) == 2*n*n] # Chai Wah Wu, Dec 07 2014 CROSSREFS Cf. A247205, A247220. Sequence in context: A249208 A234120 A101697 * A224655 A234655 A104948 Adjacent sequences: A247218 A247219 A247220 * A247222 A247223 A247224 KEYWORD nonn,more AUTHOR Juri-Stepan Gerasimov, Nov 30 2014 STATUS approved

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Last modified June 10 03:49 EDT 2023. Contains 363187 sequences. (Running on oeis4.)