A076252 -id:A076252 - cp's OEIS frontend

A076251 Numbers m such that omega(m) = omega(m-1) + omega(m-2), where omega(m) is the number of distinct prime factors of m.

3, 6, 10, 18, 30, 33, 42, 60, 66, 84, 90, 102, 105, 110, 114, 126, 129, 130, 138, 150, 165, 168, 174, 180, 195, 198, 210, 228, 234, 252, 264, 270, 273, 285, 290, 294, 315, 318, 330, 345, 348, 354, 360, 385, 399, 402, 420, 434, 450, 462, 465, 468, 480, 504

Offset: 1

Joseph L. Pe, Nov 04 2002

nonn

			omega(18) = 2 = 1 + 1 = omega(17) + omega(16), so 18 belongs to the sequence.

G. C. Greubel, Table of n, a(n) for n = 1..10000

Cf. A001221, A076252, A076253.

omega[n_] := Length[FactorInteger[n]]; Select[Range[3, 10^3], omega[ # ] == omega[ # - 1] + omega[ # - 2] &]

lista(kmax) = {my(o1 = omega(1), o2 = omega(2), o3); for(k = 3, kmax, o3 = omega(k); if(o3 == o1 + o2, print1(k, ", ")); o1 = o2; o2 = o3);} \\ Amiram Eldar, Sep 18 2024

A076253 a(n) = the least positive integer solution of the "n-th omega recurrence" omega(k) = omega(k-1) + ... + omega(k-n), if such k exists; = 0 otherwise. (omega(n) denotes the number of distinct prime factors of n.)

Original entry on oeis.org

3, 3, 2310, 746130, 601380780, 89419589469210, 489423552293946270

Offset: 1

Joseph L. Pe, Nov 04 2002

Question: Is a(n) > 0 for all n, i.e. can the n-th omega recurrence be solved for all n?

Note that 601380780 is not squarefree. Using primorials, I easily found candidates up to a(8). - Lambert Klasen (lambert.klasen(AT)gmx.net), Nov 05 2005

			k=3 is the smallest solution of omega(k)=omega(k-1), so a(1)=3.
k=3 is the smallest solution of omega(k)=omega(k-1)+omega(k-2), so a(2)=3.
k=2310 is the smallest solution of omega(k)=omega(k-1)+omega(k-2)+omega(k-3), so a(3)=2310.

Cf. A001221, A006049, A076251, A076252.

(*code to find a(4)*) omega[n_] := Length[FactorInteger[n]]; ub = 2*10^6; For[i = 2, i <= ub, i++, a[i] = omega[i]]; start = 5; For[j = start, j <= ub, j++, If[a[j] == a[j - 1] + a[j - 2] + a[j - 3] + a[j - 4], Print[j]]]

/* find a(5) */ v=[0,0,0,0,0]; s=0;for(i=1,5,v[i]=omega(i);s+=v[I])
for(i=6,10^10,o=omega(i);if(o==s,print(i);break);s-=v[i%5+1];s+=o;v[i%5+1]=o) \\ Lambert Klasen (lambert.klasen(AT)gmx.net), Nov 05 2005

a(5) from Lambert Klasen (lambert.klasen(AT)gmx.net), Nov 05 2005

a(6)-a(7) from Donovan Johnson, Feb 07 2009

cp's OEIS Frontend

A076251 Numbers m such that omega(m) = omega(m-1) + omega(m-2), where omega(m) is the number of distinct prime factors of m.

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