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 A298669 Let b(n) = 2^n with n >= 2, and let c = k*b(n) + 1 for k >= 1; then a(n) is the smallest k such that c is prime and such that A007814(r(n)) = A007814(k) + n where r(n) is the remainder of 2^(b(n)/4) mod c, or 0 if no such k exists. 1
 0, 0, 1, 8, 1024, 5, 1071, 6443, 52743, 1184, 11131, 39, 7, 856079, 3363658, 9264, 3150, 1313151, 13, 33629, 555296667, 534689, 8388607, 5, 512212693, 193652, 286330, 282030, 7224372579, 1120049, 149041 (list; graph; refs; listen; history; text; internal format)
 OFFSET 2,4 COMMENTS a(n-2) <= A007117(n). a(33) <= 5463561471303. LINKS Wilfrid Keller, Fermat factoring status FORMULA For n >= 1, a(A204620(n)) = 3; a(A226366(n)) = 5; a(A280003(n)) = 7. PROG (PARI) print1(0, ", "0", "); for(n=4, 32, b=2^n; k=1; t=0; while(t<1, c=k*b+1; if(isprime(c), r=Mod(2, c)^(b/4); if(lift(r/b)<=k, if(valuation(lift(r), 2)==valuation(k, 2)+n, t=1; print1(k, ", ")))); k++)); CROSSREFS Cf. A007117, A007814, A298670. Sequence in context: A046242 A260028 A176367 * A007117 A291831 A085533 Adjacent sequences:  A298666 A298667 A298668 * A298670 A298671 A298672 KEYWORD nonn AUTHOR Arkadiusz Wesolowski, Jan 24 2018 STATUS approved

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Last modified June 21 19:59 EDT 2021. Contains 345365 sequences. (Running on oeis4.)