|
Search: id:A115386
|
|
|
| A115386 |
|
a(1) = 1. For n >= 2, a(n) = sum of the two (not necessarily distinct) earlier terms, a(j) and a(k), which maximizes d(a(j)+a(k)), where d(m) is the number of positive divisors of m. a(n) = the maximum (a(j)+a(k)) if more than one such sum has the maximum number of divisors. |
|
+0
2
|
|
| 1, 2, 4, 8, 12, 24, 48, 96, 120, 240, 480, 720, 1440, 2880, 5760, 8640, 10080, 20160, 40320, 60480, 120960, 241920, 302400, 604800, 665280, 1330560, 2661120, 3326400, 6652800, 13305600, 26611200, 39916800, 43243200, 86486400, 172972800
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
In A115387 the minimum (a(j)+a(k)) is taken in case of a tie.
|
|
EXAMPLE
|
Sequence begins 1, 1, 2, 4. Now d(1+1) = 2, d(1+2) = 2, d(1+4) = 2, d(2+2) = 3, d(2+4) = 4, d(4+4)=4. So d(2+4) and d(4+4) are tied for the maximum number of divisors of a sum of two earlier terms of the sequence. But we want the maximum sum among these two values. So a(5) = 4+4 = 8.
|
|
PROGRAM
|
(PARI) {print1(a=1, ", "); v=[a]; for(n=2, 35, dsmax=0; smax=0; for(j=1, #v, for(k=j, #v, s=v[j]+v[k]; d=numdiv(s); if(dsmax==d, smax=max(smax, s), if(dsmax<(d), dsmax=d; smax=s)))); print1(smax, ", "); v=concat(v, smax))}
|
|
CROSSREFS
|
Cf. A115387.
Adjacent sequences: A115383 A115384 A115385 this_sequence A115387 A115388 A115389
Sequence in context: A089821 A097942 A004653 this_sequence A058771 A036493 A082906
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), following a suggestion of Leroy Quet (qq-quet(AT)mindspring.com), Jan 22 2006
|
|
|
Search completed in 0.002 seconds
|
