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A177378 a(n) is the smallest prime p>2 such that there are 2*n or 2*n+1 positive integers m for which the exponents of 2 and p in the prime power factorization of m! are both powers of 2.

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%I A177378 #10 Mar 24 2017 00:33:14
%S A177378 11,13,3,29,31,251,127,509,1021,4091,4093,65519,8191,131063,262133,
%T A177378 262139,131071,1048571,524287,8388593,4194301,67108837,16777213,
%U A177378 67108861,1073741789,2147483587,2147483629,536870909
%N A177378 a(n) is the smallest prime p>2 such that there are 2*n or 2*n+1 positive integers m for which the exponents of 2 and p in the prime power factorization of m! are both powers of 2.
%H A177378 V. Shevelev, <a href="http://journals.impan.gov.pl/aa/Inf/126-3-1.html">Compact integers and factorials</a>, Acta Arithmetica 126 (2007), no. 3, 195-236.
%F A177378 For sufficiently large n, 2^n - 1 <= a(n) <= 2^ceiling(40*n/19). Let k >= n. Put g = g(n,k) = min{odd j >= 2^(k-n): 2^k - j is prime} and h(n) = min{k: k - n = floor(log_2(g))}. Then a(n) = 2^h(n) - g(n,h(n)).
%e A177378 By the formula, for n=6, consider k >= 6. If k=6, then g(6,6) = 3, but 6 does not equal to 6 - floor(log_2(3)); if k=7, then g=15, but 6 does not equal to 7 - floor(log_2(15)); if k=8, then g=5 and we see that 6 = 8 - floor(log_2(5)). Therefore a(6) = 2^8 - 5 = 251.
%Y A177378 Cf. A000142, A050376, A169655, A169661, A177349, A177355, A177436, A177458, A177459, A177498.
%K A177378 nonn
%O A177378 1,1
%A A177378 _Vladimir Shevelev_, May 07 2010