A057467 -id:A057467 - cp's OEIS frontend

A083550 Product of 2 consecutive prime differences of two successive terms of A001223.

2, 4, 8, 8, 8, 8, 8, 24, 12, 12, 24, 8, 8, 24, 36, 12, 12, 24, 8, 12, 24, 24, 48, 32, 8, 8, 8, 8, 56, 56, 24, 12, 20, 20, 12, 36, 24, 24, 36, 12, 20, 20, 8, 8, 24, 144, 48, 8, 8, 24, 12, 20, 60, 36, 36, 12, 12, 24, 8, 20, 140, 56, 8, 8, 56, 84, 60, 20, 8, 24, 48, 48, 36, 24, 24, 48

Offset: 1

Labos Elemer, May 22 2003

nonn

Hauke Löffler, Table of n, a(n) for n = 1..10000

Cf. A001223, A083538-A083555, A057467.

f[x_] := Prime[x+1]-Prime[x]
Table[f[w+1]*f[w], {w, 1, 128}]

a(n) = A001223(n)*A001223(n+1) = (prime(n+1)-prime(n))*(prime(n+2)-prime(n+1)).

A083551 Least common multiple of 2 consecutive prime differences, of two successive terms of A001223.

Original entry on oeis.org

2, 2, 4, 4, 4, 4, 4, 12, 6, 6, 12, 4, 4, 12, 6, 6, 6, 12, 4, 6, 12, 12, 24, 8, 4, 4, 4, 4, 28, 28, 12, 6, 10, 10, 6, 6, 12, 12, 6, 6, 10, 10, 4, 4, 12, 12, 12, 4, 4, 12, 6, 10, 30, 6, 6, 6, 6, 12, 4, 10, 70, 28, 4, 4, 28, 42, 30, 10, 4, 12, 24, 24, 6, 12, 12, 24, 8, 8, 40, 10, 10, 10, 6, 12

Offset: 1

Labos Elemer, May 22 2003

nonn

Vincenzo Librandi, Table of n, a(n) for n = 1..10000

Cf. A001223, A083538..A083555, A057467.

f[x_] := Prime[x+1]-Prime[x];  Table[LCM[f[w+1], f[w]], {w, 1, 128}]
Table[LCM[(Prime[n + 1] - Prime[n]), Prime[n + 2] - Prime[n + 1]], {n, 100}] (* Vincenzo Librandi, Mar 15 2018 *)
LCM@@#&/@Partition[Differences[Prime[Range[90]]],2,1] (* Harvey P. Dale, Oct 11 2020 *)

a(n) = lcm(A001223(n), A001223(n+1)).

A083552 Quotient when LCM of 2 consecutive prime differences is divided by GCD of the same two differences.

Original entry on oeis.org

2, 1, 2, 2, 2, 2, 2, 6, 3, 3, 6, 2, 2, 6, 1, 3, 3, 6, 2, 3, 6, 6, 12, 2, 2, 2, 2, 2, 14, 14, 6, 3, 5, 5, 3, 1, 6, 6, 1, 3, 5, 5, 2, 2, 6, 1, 3, 2, 2, 6, 3, 5, 15, 1, 1, 3, 3, 6, 2, 5, 35, 14, 2, 2, 14, 21, 15, 5, 2, 6, 12, 12, 1, 6, 6, 12, 2, 2, 20, 5, 5, 5, 3, 6, 6, 12, 2, 2, 2, 3, 6, 2, 2, 2, 6, 2, 6, 9

Offset: 1

Labos Elemer, May 22 2003

nonn

Conjecture: Every positive integer appears infinitely many times in this sequence. Example: a(834) = a(909) = ... = a(9901) = ... = 4. - Jerzy R Borysowicz, Dec 22 2018

All terms of this sequence are integers because gcd(r,s) divides lcm(r,s) for any r and s. - Jerzy R Borysowicz, Jan 05 2019

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001223, A083538-A083555, A057467.

f[x_] := Prime[x+1]-Prime[x]; Table[LCM[f[w+1], f[w]]/GCD[f[w+1], f[w]], {w, 1, 128}]

a(n) = my(da=prime(n+2)-prime(n+1), db=prime(n+1)-prime(n)); lcm(da, db)/gcd(da, db) \\ Felix Fröhlich, Jan 05 2019

a(n) = lcm(A001223(n), A001223(n+1))/gcd(A001223(n), A001223(n+1));

a(n) = A083551(n)/A057467(n).

A054682 a(n) = smallest prime p = prime(k) such that gcd( prime(k+1) - prime(k), prime(k+2) - prime(k+1) ) is a multiple of 2n.

Original entry on oeis.org

3, 89, 47, 1823, 1627, 199, 5939, 5591, 15823, 83117, 259033, 16763, 365851, 1074167, 69593, 1625027, 2541289, 255767, 11772613, 3312227, 247099, 23374859, 25767389, 3565931, 21369059, 15340943, 6314393, 59859131, 101996837, 4911251, 70136597, 166185431, 12012677, 198429983, 247837313, 23346737, 298626077, 1321272031, 43607351, 464208809

Offset: 1

Jeff Burch, Apr 18 2000

nonn

Different from A070018.

for(n=1,50,p=2: np=3: while((np-p)%(2*n)||(nextprime(np+2)-np)%(2*n),p=np: np=nextprime(np+2)): print1(p","))

a(n)=Min{x : A057467(x) is a multiple of 2n}

More terms from Larry Reeves (larryr(AT)acm.org), Nov 09 2000
Corrected and extended by Ralf Stephan, Feb 23 2004
More terms from Olaf Voß, Feb 17 2008

A070010 GCD of consecutive values of sum-of-proper divisors.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 11, 1, 1, 1, 6, 2, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 1, 1, 1, 23, 1, 1, 1, 1, 3, 1, 1, 5, 25, 1, 1, 1, 1, 3, 1, 1, 1, 1

Offset: 1

Labos Elemer, Apr 11 2002

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A000203, A001065; GCD of various consecutive function values: A048586, A057467, A058515, A060778, A060780, A069896.

GCD[DivisorSigma[1, n+1]-(n+1), DivisorSigma[1, n]-n]

A001065(n) = (sigma(n) - n);
A070010(n) = gcd(A001065(n),A001065(1+n)); \\ Antti Karttunen, Oct 05 2017

a(n) = gcd(A001065(n+1), A001065(n)).

A070018 a(n) = smallest prime p = prime(k) such that gcd( prime(k+1) - prime(k), prime(k+2) - prime(k+1) ) = 2n.

Original entry on oeis.org

3, 89, 47, 1823, 1627, 199, 5939, 5591, 15823, 83117, 259033, 16763, 365851, 1074167, 69593, 1625027, 2541289, 255767, 11772613, 3312227, 247099, 23374859, 25767389, 3565931, 21369059, 15340943, 6314393, 59859131, 101996837, 4911251, 70136597, 166185431, 12012677, 198429983, 247837313, 23346737, 298626077

Offset: 1

Labos Elemer and Benoit Cloitre, Apr 12 2002

nonn

			n=21: a(21)=247099, the consecutive prime triple {247099,247141,247183} determines {42,42} successive differences, the GCD of which is 2n=42.

Cf. A001223, A057467.

Different from A054682?

f[x_] := GCD[Prime[x+1]-Prime[x], Prime[x+2]-Prime[x+1]]; t = Table[0, {256} ]; Do[ c = f[n]; If[c <257 && t[[b]] == 0, t[[c]] = n], {n, 2, 1000000} ]; t Prime[t]

fp(n, vp) = {for (k=1, #vp-2, if (gcd(vp[k+1] - vp[k], vp[k+2] - vp[k+1]) == 2*n, return (vp[k])););}
lista(nn) = {my(vp = primes(10000)); for (n=1, nn, my(p = fp(n, vp)); if (p, print1(p, ", "), break););} \\ Michel Marcus, Aug 29 2019

a(n) = Min{x : A057467(x)=2n}.

Corrected and extended by Michel Marcus, Aug 29 2019

cp's OEIS Frontend

A083550 Product of 2 consecutive prime differences of two successive terms of A001223.

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A083551 Least common multiple of 2 consecutive prime differences, of two successive terms of A001223.

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A083552 Quotient when LCM of 2 consecutive prime differences is divided by GCD of the same two differences.

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A054682 a(n) = smallest prime p = prime(k) such that gcd( prime(k+1) - prime(k), prime(k+2) - prime(k+1) ) is a multiple of 2n.

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A070010 GCD of consecutive values of sum-of-proper divisors.

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A070018 a(n) = smallest prime p = prime(k) such that gcd( prime(k+1) - prime(k), prime(k+2) - prime(k+1) ) = 2n.

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