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A246362 Numbers n such that if 2n-1 = Product_{k >= 1} (p_k)^(c_k), then n < Product_{k >= 1} (p_{k-1})^(c_k), where p_k indicates the k-th prime, A000040(k).

%I A246362 #21 May 20 2025 12:43:40
%S A246362 4,6,7,9,10,12,15,16,19,20,21,22,24,27,29,30,31,34,35,36,37,40,42,44,
%T A246362 45,46,47,48,49,51,52,54,55,56,57,60,62,64,65,66,67,69,70,71,72,75,76,
%U A246362 78,79,80,81,82,84,85,87,89,90,91,92,96,97,99,100,101,102,103,105,106,107,108,109,110,111,112,114,115
%N A246362 Numbers n such that if 2n-1 = Product_{k >= 1} (p_k)^(c_k), then n < Product_{k >= 1} (p_{k-1})^(c_k), where p_k indicates the k-th prime, A000040(k).
%C A246362 Numbers n such that A064216(n) > n.
%C A246362 Numbers n such that A064989(2n-1) > n.
%C A246362 The sequence grows as:
%C A246362       a(100) = 148
%C A246362      a(1000) = 1449
%C A246362     a(10000) = 14264
%C A246362    a(100000) = 141259
%C A246362   a(1000000) = 1418197
%C A246362 and the powers of 10 occur at:
%C A246362         a(5) = 10
%C A246362        a(63) = 100
%C A246362       a(701) = 1000
%C A246362      a(6973) = 10000
%C A246362     a(70845) = 100000
%C A246362    a(705313) = 1000000
%C A246362 suggesting that the ratio a(n)/n is converging to a constant and an arbitrary natural number is more than twice as likely to be here than in the complement A246361. Compare this to the ratio present in the "inverse" case A246282.
%H A246362 Antti Karttunen, <a href="/A246362/b246362.txt">Table of n, a(n) for n = 1..10000</a>
%e A246362 4 is present, as 2*4 - 1 = 7 = p_4, and p_{4-1} = p_3 = 5 > 4.
%e A246362 5 is not present, as 2*5 - 1 = 9 = p_2 * p_2, and p_1 * p_1 = 4, with 4 < 5.
%e A246362 6 is present, as 2*6 - 1 = 11 = p_5, and p_{5-1} = p_4 = 7 > 6.
%e A246362 35 is present, as 2*35 - 1 = 69 = 3*23 = p_2 * p_9, and p_1 * p_8 = 2*19 = 38 > 35.
%o A246362 (PARI)
%o A246362 default(primelimit, 2^30);
%o A246362 A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
%o A246362 A064216(n) = A064989((2*n)-1);
%o A246362 isA246362(n) = (A064216(n) > n);
%o A246362 n = 0; i = 0; while(i < 10000, n++; if(isA246362(n), i++; write("b246362.txt", i, " ", n)));
%o A246362 (Scheme)
%o A246362 ;; With _Antti Karttunen_'s IntSeq-library.
%o A246362 (define A246362 (MATCHING-POS 1 1 (lambda (n) (> (A064216 n) n))))
%Y A246362 Complement: A246361.
%Y A246362 Setwise difference of A246372 and A048674.
%Y A246362 Cf. A000040, A064216, A064989, A246282.
%K A246362 nonn
%O A246362 1,1
%A A246362 _Antti Karttunen_, Aug 24 2014