cp's OEIS Frontend

6 is the only term that is in A104210, the absolute values of all other terms residing in its complement, A319630, thus 3 occurs only once in A379231. Proof: If k is not a multiple of 3 and k is in A104210, then there are primes p > 3 and q = nextprime(p) that both divide k and q also divides A003961(k). However, q does not divide 2k +- 3, therefore the equation 2k +- 3 = A003961(k) is unsolvable in these cases. So let's assume that k is a multiple of 3, which immediately entails that k must be also even, for A003961(k) to be a multiple of 3. Let x = k/6; then the equation can be rewritten as 2*6*x +- 3 = A003961(6)*A003961(x) <=> 12x +- 3 = 15*A003961(x) <=> 3*(4x +- 1) = 3*5*A003961(x). The only value of x that satisfies the equation is x=1 (as A003961(n)>n for all n>1), hence k=6.

If it exists, abs(a(15)) > 2^32.

15 is the only term that is in A104210, the absolute values of all other terms residing in its complement, A319630, thus 5 occurs only once in A379231. Proof: If k is not a multiple of 5 and k is in A104210, then there are primes p (either p=2 or p > 5 and q = nextprime(p) that both divide k and q also divides A003961(k). However, q does not divide 2k +- 5, therefore the equation 2k +- 5 = A003961(k) is unsolvable in these cases. So let's assume that k is a multiple of 5, which immediately entails that k must be also a multiple of 3, for A003961(k) to be a multiple of 5. Let x = k/15; then the equation can be rewritten as 2*15*x +- 5 = A003961(15)*A003961(x) <=> 30x +- 5 = 35*A003961(x) <=> 5*(6x +- 1) = 5*7*A003961(x). The only value of x that satisfies the equation is x=1 (as A003961(n)>n for all n>1), hence k=15.

If it exists, abs(a(14)) > 2^32.

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.

If it exists, abs(a(21)) > 2^32.

Numbers k for which A246277(2k-1) = A246277(k). This in turn implies a looser condition A046523(2k-1) = A046523(k).

Nonsquarefree terms are rare: 25, 841 (= 29^2), 970225 ( = 5^2 * 197^2), ..., also 414690595, which is not a square. Some of these are also terms of A048674. Compare to A348511.