|
Search: id:A160691
|
|
| |
|
| 1, 2, 2, 2, 4, 2, 2, 4, 2, 2, 4, 2, 2, 4, 2, 4, 2, 4, 4, 4, 2, 4, 4, 4, 2, 4, 6, 4, 6, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2, 2, 4, 4, 4, 2, 4, 4, 2, 4, 2, 4, 4, 4, 2, 2, 4, 4, 2, 4, 4, 4, 2, 4, 2, 4, 4, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2, 2, 4, 2, 4, 4, 2, 4, 4, 4, 4, 4, 4, 4, 4, 2, 4, 2, 4, 4, 2, 4, 4, 2
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
Contribution from Farideh Firoozbakht (mymontain(AT)yahoo.com), May 28 2009: (Start)
For the first 200000 natural numbers n, a(n) is in the set {1,2,4,6,8,12}
and in fact we have:
For one number n, A160691(n)=1.
For 13 numbers n, A160691(n)=12 (see the sequence A158963).
For 4785 numbers n, A160691(n)=6.
For 6706 numbers n, A160691(n)=8.
For 26790 numbers n, A160691(n)=2.
For 161705 numbers n, A160691(n)=4.
Also n=2 is the only number n (less than 200000) that a(n)=a(n+1)=a(n+2)=2.
and for the 53 consecutive numbers 64833, 64834, ... , 64885 we have a(n)=4. (End)
|
|
MATHEMATICA
|
Contribution from Farideh Firoozbakht (mymontain(AT)yahoo.com), May 28 2009: (Start)
c[1]=1; c[n_]:=c[n]=(s=Sum[c[k], {k, n-1}]; For[m=1, DivisorSigma[0, m]!=
DivisorSigma[0, s+m], m++ ]; m); a[n_]:=a[n]=DivisorSigma[0, c[n]];
Table[a[n], {n, 105}] (End)
|
|
CROSSREFS
|
A160689, A160690
Cf. A158963, A158964. [From Farideh Firoozbakht (mymontain(AT)yahoo.com), May 28 2009]
Sequence in context: A075016 A102445 A027389 this_sequence A049716 A066671 A159802
Adjacent sequences: A160688 A160689 A160690 this_sequence A160692 A160693 A160694
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Leroy Quet (q1qq2qqq3qqqq(AT)yahoo.com), May 24 2009
|
|
EXTENSIONS
|
More terms from Farideh Firoozbakht (mymontain(AT)yahoo.com), May 28 2009
|
|
|
Search completed in 0.002 seconds
|
