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 1-5 of 5 results.

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

Views

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)).

A104210 Positive integers divisible by at least 2 consecutive primes.

Original entry on oeis.org

6, 12, 15, 18, 24, 30, 35, 36, 42, 45, 48, 54, 60, 66, 70, 72, 75, 77, 78, 84, 90, 96, 102, 105, 108, 114, 120, 126, 132, 135, 138, 140, 143, 144, 150, 154, 156, 162, 165, 168, 174, 175, 180, 186, 192, 195, 198, 204, 210, 216, 221, 222, 225, 228, 231, 234, 240
Offset: 1

Views

Author

Leroy Quet, Mar 13 2005

Keywords

Comments

If a perfect square is in this sequence, then so is its square root (e.g., 144 and 12). - Alonso del Arte, May 07 2012
The numbers of terms not exceeding 10^k, for k=1,2,..., are 1, 22, 242, 2456, 24632, 246414, 2464272, 24643281, 246433426, ... Apparently, the asymptotic density of this sequence is 0.24643... - Amiram Eldar, Apr 10 2021

Examples

			35 is divisible by both 5 and 7, and 5 and 7 are consecutive primes.
77 is divisible by both 7 and 11, and 7 and 11 are consecutive primes.
110 is not in the sequence because, although it is divisible by 2, 5 and 11, it is not divisible by 3 or 7.
		

Crossrefs

Cf. A003961, A296210 (characteristic function), A319630 (complement), A379230 [= A252748(a(n))].
Positions of terms larger than 1 in A300820 and in A322361.
Subsequences: A006094, A349169 (conjectured, after its initial 1), A349176, A355527 (squarefree terms), A372566, A378884, A379232.

Programs

  • Maple
    N:= 1000: # for terms <= N
    R:= {}:
    p:= 2:
    do
      q:= p; p:= nextprime(p);
      if p*q > N then break fi;
      R:= R union {seq(i,i=p*q..N,p*q)}
    od:
    sort(convert(R,list)); # Robert Israel, Apr 13 2020
  • Mathematica
    fQ[n_] := Block[{lst = PrimePi /@ Flatten[ Table[ #[[1]], {1}] & /@ FactorInteger[n]]}, Count[ Drop[lst, 1] - Drop[lst, -1], 1] > 0]; Select[ Range[244], fQ[ # ] &] (* Robert G. Wilson v, Mar 16 2005 *)
  • PARI
    A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    is_A104210(n) = (gcd(n,A003961(n))>1); \\ Antti Karttunen, Dec 24 2024

Formula

{k such that gcd(k, A003961(k)) > 1}. - Antti Karttunen, Dec 24 2024

Extensions

More terms from Robert G. Wilson v, Mar 16 2005

A379233 Numbers k such that A003961(k) = 2k +- 3, 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

-5, 6, -7, -161, 1045, -2525, 2795, 4825, 9725, -159115, 307993, -359315, -18377525, 25484825
Offset: 1

Views

Author

Antti Karttunen, Dec 23 2024

Keywords

Comments

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.

Crossrefs

Formula

{sign(A252748(k)) * k, for k such that abs(A252748(k)) = 3}.

A379232 Terms k of A104210 for which abs(A003961(k)-(2*k)) is prime, where A003961 is fully multiplicative with a(prime(i)) = prime(i+1), and A104210 gives numbers m for which gcd(m,A003961(m)) > 1.

Original entry on oeis.org

6, 15, 35, 77, 646, 1334, 2294, 5681, 6293, 78793, 600677, 2255002, 2482402, 4564166, 8702275, 16798717, 17747807, 22197526, 236981285, 302764377, 1287983157, 2111098886, 2295003595
Offset: 1

Views

Author

Antti Karttunen, Dec 23 2024

Keywords

Comments

The definition entails that both abs(A379231(n)) and A151799(abs(A379231(n))) divide a(n).
If it exists, a(24) > 2^32.

Crossrefs

A379229 Difference A003961(k)-(2*k) computed for the natural numbers k for which k and A003961(k) are relatively prime, where A003961 is fully multiplicative with a(prime(i)) = prime(i+1).

Original entry on oeis.org

-1, -1, -1, 1, -3, -3, 11, 7, 1, -9, -9, 5, 49, -15, -15, 23, 13, -5, -17, -1, -1, 71, 43, -27, -25, 179, -1, -11, -33, -7, 7, 109, -39, -39, 29, -5, -41, 23, 47, -7, 49, -47, -19, 185, 1, -23, -57, -55, -13, 149, 601, -11, -63, 35, 7, -69, -67, -25, 55, -75, 407, 463, -35, -77, -37, -31, -19, 175, -81, 5, 77, -1
Offset: 1

Views

Author

Antti Karttunen, Dec 23 2024

Keywords

Crossrefs

Cf. also A379230.

Programs

  • PARI
    A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    A252748(n) = (A003961(n) - (2*n));
    is_A319630(n) = (1==gcd(n,A003961(n)));
    k=0; n=0; while(k<200, n++; if(is_A319630(n), print1(A252748(n),", ")));

Formula

a(n) = A252748(A319630(n)).
Showing 1-5 of 5 results.