A061228 -id:A061228 - cp's OEIS frontend

A299143 a(n) is the least k > n such that gcd(k,n) > 1 and gcd(k+1,n+1) > 1.

8, 9, 14, 15, 20, 21, 14, 15, 32, 33, 38, 39, 20, 21, 50, 51, 56, 57, 26, 27, 68, 69, 34, 35, 32, 33, 86, 87, 92, 93, 38, 39, 44, 45, 110, 111, 44, 45, 122, 123, 128, 129, 50, 51, 140, 141, 62, 63, 56, 57, 158, 159, 64, 65, 62, 63, 176, 177, 182, 183, 68, 69

Offset: 2

Alex Ratushnyak, Feb 03 2018

nonn

			8 is the least k>2 such that gcd(8,2)>1 and gcd(9,3)>1. So a(2)=8.
15 is the least k>9 such that gcd(15,9)>1 and gcd(16,10)>1. Therefore a(9)=15.

Robert Israel, Table of n, a(n) for n = 2..10000

Cf. A172170.

Cf. A061228 or A159475 (when simply gcd(k,n) > 1).

f:= proc(n) local k;
      for k from n+1 do if igcd(k,n)>1 and igcd(k+1,n+1)>1 then return k fi od
end proc:
map(f, [$2..100]); # Robert Israel, Mar 08 2018

Array[Block[{k = # + 1}, While[Or[CoprimeQ[#, k], CoprimeQ[# + 1, k + 1]], k++]; k] &, 62, 2] (* Michael De Vlieger, Feb 03 2018 *)

a(n) = for (k=n+1, oo, if (gcd(n,k)>1 && gcd(n+1, k+1)>1, return (k))) \\ Rémy Sigrist, Feb 04 2018

From Rémy Sigrist, Feb 04 2018: (Start)
a(p) = 3 * p for any odd prime p.
a(2*k + 1) = a(2*k) + 1 for any k > 0.
a(n) = n + 2*A172170(n + 1) for any n > 1.
(End)

A322292 a(n) = Max_{c composite, c < n} (c + least prime factor of c).

Original entry on oeis.org

6, 6, 8, 8, 10, 12, 12, 12, 14, 14, 16, 18, 18, 18, 20, 20, 22, 24, 24, 24, 26, 30, 30, 30, 30, 30, 32, 32, 34, 36, 36, 40, 40, 40, 40, 42, 42, 42, 44, 44, 46, 48, 48, 48, 50, 56, 56, 56, 56, 56, 56, 60, 60, 60, 60, 60, 62, 62, 64, 66, 66, 70, 70, 70, 70, 72, 72, 72

Offset: 5

Michel Marcus, Dec 02 2018

nonn

a(n) is only defined for n >= 5, because for n < 5, the condition {c composite, c < n} results in the empty set.

			a(5) = 6 because the largest composite c < n = 5 is 4, which has the largest prime factor 2. Hence a(5) = 4 + 2 = 6. - _David A. Corneth_, Dec 03 2018

Robert Israel, Table of n, a(n) for n = 5..10000
Paul Erdos, Some unconventional problems in number theory, Acta Mathematica Hungarica, 33(1):71-80, 1979. See p. 73.

Cf. A061228, A159475, A322293.

N:= 100: # to get a(5)..a(N)
V:= Vector(N):
V[5]:= 6;
for n from 6 to N do
  if isprime(n-1) then V[n]:= V[n-1]
  else V[n]:= max(V[n-1],n-1+min(numtheory:-factorset(n-1)))
  fi
od:
convert(V[5..N],list); # Robert Israel, Dec 03 2018

a[n_] := Module[{smax = 0}, Do[If[CompositeQ[m],  smax = Max[smax, m + FactorInteger[m][[1, 1]]]], {m, 2, n-1}]; smax]; Array[a, 100, 5] (* Amiram Eldar, Dec 02 2018 *)

a(n) = {my(smax = 0); for(m=2, n-1, if (!isprime(m), smax = max(smax, m + factor(m)[1, 1]); )); smax; }

A322293 Integers k such that A322292(k) <= k.

Original entry on oeis.org

6, 8, 12, 14, 18, 20, 24, 30, 32, 42, 44, 48, 60, 62, 72, 74, 84, 90, 102, 104, 108, 110, 114, 132, 140, 168, 182, 198, 200, 234, 240, 242, 270, 272, 282, 284, 312, 314, 318, 354, 360, 390, 420, 422, 434, 462, 464, 468, 510, 572, 648, 660, 662, 762, 840, 884, 888, 942, 1064

Offset: 1

Michel Marcus, Dec 02 2018

nonn

Erdos conjectures that this sequence is finite.

Robert Israel, Table of n, a(n) for n = 1..100 (terms 1..98 from Michel Marcus)(all terms <= 10^8).
Paul Erdos, Some unconventional problems in number theory, Acta Mathematica Hungarica, 33(1):71-80, 1979. See p. 73.

Cf. A061228, A159475, A322292.

N:= 10^6: # to get all terms <= N
Res:= 6: v:= 6:
for n from 7 to N do
  if not isprime(n-1) then v:= max(v, n-1 + min(numtheory:-factorset(n-1))) fi;
if v <= n then Res:= Res, n fi;
od:
Res; # Robert Israel, Dec 03 2018

f[n_] := Module[{smax = 0}, Do[If[CompositeQ[m],  smax = Max[smax, m + FactorInteger[m][[1, 1]]]], {m, 2, n-1}]; smax];aQ[n_] := f[n]<=n; Select[Range[6, 1000], aQ] (* Amiram Eldar, Dec 02 2018 *)

f(n) = {my(smax = 0); forcomposite(m=1, n-1, smax = max(smax, m + factor(m)[1,1]);); smax;} \\ A322292
isok(n) = f(n) <= n;

A130904 Numbers n such that the trajectory of the map n -> (n + lpf(n)) / 2 reaches 3, where lpf(n) is the least prime factor of n (A020639).

Original entry on oeis.org

3, 4, 6, 9, 10, 15, 16, 18, 25, 27, 28, 30, 33, 34, 48, 49, 51, 52, 54, 55, 57, 58, 63, 64, 66, 91, 93, 94, 96, 99, 100, 102, 105, 106, 108, 111, 112, 114, 119, 121, 123, 124, 126, 129, 130, 169, 180, 183, 184, 186, 187, 189, 190, 195, 196, 198

Offset: 1

Grant Garcia, Jan 06 2011

nonn

The only prime term is 3 because all primes map to themselves.

			(15 + 3) / 2 = 9, (9 + 3) / 2 = 6, (6 + 2) / 2 = 4, and (4 + 2) / 2 = 3, thus 15 is in the sequence.
(21 + 3) / 2 = 12, (12 + 2) / 2 = 7, and (7 + 7) / 2 = 7, so 21 never reaches 3 and therefore is not in the sequence.

Cf. A177980, A020639, A061228.

A179329 Number of iterations of (n + lpf(n)) / 2 required to reach a prime, where lpf equals the least prime factor.

Original entry on oeis.org

0, 0, 1, 0, 2, 0, 1, 3, 3, 0, 1, 0, 2, 4, 4, 0, 4, 0, 1, 2, 2, 0, 1, 5, 3, 5, 5, 0, 5, 0, 1, 5, 5, 2, 1, 0, 2, 3, 3, 0, 3, 0, 1, 2, 2, 0, 6, 6, 4, 6, 6, 0, 6, 6, 1, 6, 6, 0, 1, 0, 2, 6, 6, 3, 6, 0, 3, 2, 2, 0, 1, 0, 3, 4, 4, 4, 4, 0, 1, 4, 4, 0, 1, 3, 2, 3, 3, 0, 3, 7, 1, 7, 7, 5, 7, 0, 5, 7, 7, 0, 7, 0, 1, 7, 7

Offset: 2

Grant Garcia, Jan 12 2011

nonn

Zeros indicate prime indices.

			a(15) gives (15 + 3) / 2 = 9, (9 + 3) / 2 = 6, (6 + 2) / 2 = 4, (4 + 2) / 2 = 3. Four iterations were required to reach a prime, so a(15) = 4.

Cf. A177980, A020639, A061228.

f[n_] := Length@ NestWhileList[(# + FactorInteger[#][[1, 1]])/2 &, n, ! PrimeQ@ # &] - 1; Array[f, 105, 2]

A263569 Number of distinct prime divisors p of 2n such that lpf(2n - p) = p, where lpf = least prime factor (A020639).

Original entry on oeis.org

0, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 2, 1, 2, 2, 1, 1, 2, 2, 2, 3, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 3, 2, 1, 3, 1, 2, 2, 2, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 2, 2, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 2, 1, 2, 3, 1, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 4

Offset: 1

Gionata Neri, Oct 21 2015

nonn

a(n) gives the number of times 2*n occurs in A061228.

			a(10) = 1 since the distinct prime divisors of 2*10 = 20 are 2 and 5, A020639(20 - 2) = 2 and A020639(20 - 5) = 3, so only prime 2 is to be considered.
a(15) = 3 since the distinct prime divisors of 2*15 = 30 are 2, 3 and 5, A020639(30 - 2) = 2 and A020639(30 - 3) = 3 and A020639(30 - 5) = 5, so all three prime 2, 3 and 5 are to be considered.

Antti Karttunen, Table of n, a(n) for n = 1..20000

Cf. A020639, A061228.

f[n_] := Select[First /@ FactorInteger[2 n], FactorInteger[2 n - #][[1, 1]] == # &]; Length /@ Table[f@ n, {n, 2, 105}] (* Michael De Vlieger, Oct 22 2015 *)

a(n) = {my(f=factor(2*n)); sum(k=1, #f~, p=f[k,1]; p == factor(2*n-p)[1,1]);} \\ Michel Marcus, Oct 31 2015

A179997 Where records occur in A179329.

Original entry on oeis.org

2, 4, 6, 9, 15, 25, 48, 91, 169, 336, 669, 1335, 2665, 5317, 10573, 21037, 42072, 84141, 168279, 336553, 673099, 1345639, 2691276, 5382549, 10765095, 21528959, 43057916, 86115821, 172231631, 344463251, 688926491, 1377847349, 2755694696, 5511375473, 11022750944

Offset: 1

Grant Garcia, Jan 13 2011

nonn

a(n + 1) / a(n) converges to 2.

Cf. A179329, A177980, A130904, A020639, A061228.

Corrected and extended by Grant Garcia, Jan 19 2011

a(30)-a(35) from Amiram Eldar, Oct 25 2024

cp's OEIS Frontend

A299143 a(n) is the least k > n such that gcd(k,n) > 1 and gcd(k+1,n+1) > 1.

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A322292 a(n) = Max_{c composite, c < n} (c + least prime factor of c).

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A322293 Integers k such that A322292(k) <= k.

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A130904 Numbers n such that the trajectory of the map n -> (n + lpf(n)) / 2 reaches 3, where lpf(n) is the least prime factor of n (A020639).

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A179329 Number of iterations of (n + lpf(n)) / 2 required to reach a prime, where lpf equals the least prime factor.

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Mathematica

A263569 Number of distinct prime divisors p of 2*n such that lpf(2*n - p) = p, where lpf = least prime factor (A020639).

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A179997 Where records occur in A179329.

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Extensions

A263569 Number of distinct prime divisors p of 2n such that lpf(2n - p) = p, where lpf = least prime factor (A020639).