cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A379230 Difference A003961(k)-(2*k) computed for the natural numbers k for which gcd(k,A003961(k)) > 1, where A003961 is fully multiplicative with a(prime(i)) = prime(i+1).

Original entry on oeis.org

3, 21, 5, 39, 87, 45, 7, 153, 81, 85, 309, 267, 195, 63, 91, 531, 95, -11, 99, 327, 345, 1023, 81, 175, 909, 117, 705, 573, 321, 605, 159, 413, -65, 1737, 435, 121, 453, 1551, 125, 1149, 117, 189, 1215, 183, 3261, 205, 579, 447, 735, 2943, -119, 171, 775, 579, 253, 807, 2355, 357, 153, 1971, 155, 189, 1227, 2085
Offset: 1

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Author

Antti Karttunen, Dec 22 2024

Keywords

Comments

Each of the primes 3, 5, 7 occur here just once, while some other primes may occur multiple times, although very rare in general. See A379231 and A379232.

Crossrefs

Cf. A379231 (terms whose absolute value is a prime), A379232 (corresponding terms of A104210).
Cf. also A379233, A379235, A379237.

Formula

a(n) = A252748(A104210(n)).

A379231 Terms of A379230 whose absolute value is prime.

Original entry on oeis.org

3, 5, 7, -11, 19, 29, -37, -23, 31, -13, -11, 37, -311, 317, 593, 23, -11, -929, -53, 41, -59, -643, 29
Offset: 1

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Author

Antti Karttunen, Dec 23 2024

Keywords

Comments

Terms 3, 5, and 7 occur only once in this sequence, see A379233, A379235 and A379237 for the proofs.
Both abs(a(n)) and A151799(abs(a(n))) divide A379232(n).

Crossrefs

Subsequence of A379230.
Cf. A151799, A252748, A379232 (corresponding terms of A104210), A379233, A379235, A379237.

Formula

a(n) = A252748(A379232(n)).

A379235 Numbers k such that A003961(k) = 2k +- 5, multiplied by the sign of difference A003961(k)-2k, where A003961 is fully multiplicative with a(prime(i)) = prime(i+1).

Original entry on oeis.org

14, 15, -22, -46, 91, -2782, -269434, -1056574, 14129726, -25652506, 26594126, 34233062, 147087493
Offset: 1

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Author

Antti Karttunen, Dec 23 2024

Keywords

Comments

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.

Crossrefs

Formula

{sign(A252748(k)) * k, for k such that abs(A252748(k)) = 5}.

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).

Original entry on oeis.org

9, 35, -38, 39, -51, 69, -374, -4521, 7869, 10426, 12639, -16094, -29354, 102579, -103881, 1295206, -3298514, 4267318, 478642449, -2120241621
Offset: 1

Views

Author

Antti Karttunen, Dec 23 2024

Keywords

Comments

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.

Crossrefs

Formula

{sign(A252748(k)) * k, for k such that abs(A252748(k)) = 7}.
Showing 1-4 of 4 results.