cp's OEIS Frontend

A379237 Numbers k such that A003961(k) = 2k +- 7, multiplied by the sign of difference A003961(k)-2k, where A003961 is fully multiplicative with a(prime(i)) = prime(i+1).

%I A379237 #10 Dec 23 2024 09:42:31
%S A379237 9,35,-38,39,-51,69,-374,-4521,7869,10426,12639,-16094,-29354,102579,
%T A379237 -103881,1295206,-3298514,4267318,478642449,-2120241621
%N A379237 Numbers k such that A003961(k) = 2k +- 7, multiplied by the sign of difference A003961(k)-2k, where A003961 is fully multiplicative with a(prime(i)) = prime(i+1).
%C A379237 35 is the only term that is in A104210, the absolute values of all other terms residing in its complement, A319630, thus 7 occurs only once in A379231. Proof: If k is not a multiple of 7 and k is in A104210, then there are primes p (either p=2, p=3 or p > 7 and q = nextprime(p) that both divide k and q also divides A003961(k). However, q does not divide 2k +- 7, therefore the equation 2k +- 7  = A003961(k) is unsolvable in these cases. So let's assume that k is a multiple of 7, which immediately entails that k must be also a multiple of 5, for A003961(k) to be a multiple of 7. Let x = k/35; then the equation can be rewritten as 2*35*x +- 7 = A003961(35)*A003961(x) <=> 70x +- 7 = 77*A003961(x) <=> 7*(10x +- 1) = 7*11*A003961(x). The only value of x that satisfies the equation is x=1 (as A003961(n)>n for all n>1), hence k=35.
%C A379237 If it exists, abs(a(21)) > 2^32.
%H A379237 <a href="/index/Pri#prime_indices">Index entries for sequences related to prime indices in the factorization of n</a>.
%F A379237 {sign(A252748(k)) * k, for k such that abs(A252748(k)) = 7}.
%Y A379237 Cf. A003961, A104210, A252748, A319630.
%Y A379237 Cf. also A048674, A348514, A378980, A379231, A379233, A379235.
%K A379237 sign,hard,more
%O A379237 1,1
%A A379237 _Antti Karttunen_, Dec 23 2024