<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD>
<META http-equiv=Content-Type content="text/html; charset=iso-8859-1">
<META content="MSHTML 6.00.2722.900" name=GENERATOR>
<STYLE></STYLE>
</HEAD>
<BODY bgColor=#ffffff>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>Copy-paste of the sequence:</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><STRONG>ID Number: </STRONG>A064161<BR><STRONG>Sequence:
</STRONG>12,12,20,42,66,78,102,114,138,174,186,222,246,258,282,318,<BR><STRONG>
</STRONG>354,366,402,426,438,474,498,534,582,606,618,642,654,678,762,<BR><STRONG>
</STRONG>786,822,834,894,906,942,978,1002,1038,1074,1086,1146,1158,<BR><STRONG>
</STRONG>1182,1194,1266,1338,1362,1374<BR><STRONG>Name:
</STRONG>Least abandant number divisible by the n_th prime
number.<BR><STRONG>Comments: </STRONG>I know of no reason why this
sequence should be monotonic, but this
is<BR><STRONG>
</STRONG> true up to the 10^3-rd
prime.<BR><STRONG>Example: </STRONG>The first eight abundant numbers
are 12, 18, 20, 24, 30, 36, 40 and 42.
But<BR><STRONG>
</STRONG> only the last one is divisible by 7, the fourth prime
number.
Therefore<BR><STRONG>
</STRONG> a(4) = 42.<BR></DIV>
<DIV><FONT face=Arial
size=2>-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>I've just realized that this sequence is also
generated with another formula since the fourth term:</FONT></DIV>
<DIV><FONT face=Arial size=2>1) The odd primes: 3 5 7 11 13 17
19...</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>2) Their first order differences: <FONT
face="Times New Roman" size=3>2 2 4 2 4 2 4 6 ...<BR></FONT></FONT></DIV>
<DIV><FONT face=Arial size=2><FONT face="Times New Roman" size=3>Multiplying
by 6: 12 12 24 12 24 12 24 36 12 36 24 12... = = V.</FONT></FONT></DIV>
<DIV><FONT face=Arial size=2><FONT face=Arial size=2></FONT></FONT> </DIV>
<DIV><FONT face=Arial size=2><FONT face="Times New Roman" size=3>If we start at
a0 = 42 and then do a(n) = a(n-1)+V(n) we obtain the sequence
A064161:</FONT></FONT></DIV>
<DIV><FONT face=Arial size=2><FONT face=Arial size=2></FONT></FONT> </DIV>
<DIV><FONT face=Arial
size=2>66,78,102,114,138,174,186,222,246,258,282,318,354,366,402,426,438,474,498,534,58<BR>2,606,618,642,654,678,762,786,822,834,894,906,942,978,1002,1038,1074,1086,1146,1<BR>158,1182,1194,1266,1338,1362,1374,1398,1434,1446,1506,1542,1578,1614,1626,1662,1<BR>686,1698,1758,1842,1866,1878,1902,1986,2022,2082,2094,2118,2154,2202,2238,2274,2<BR>298,2334,2382,2406,2454,2514,2526,2586,2598,2634,2658,2694,2742,2766,2778,2802,2<BR>874,2922,2946,2994
(PARI-GP)</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>An example: a80 = 2454 --> 2454 is abundant
(sigma(2454) = 2465) and prime(80) = 409 = a80/6.</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>This implies that aN = 6*prime(N) if N>=4
(because these are the differences of the primes * 6 and in a4 the sequence fall
in the required 6*prime: the sequence would run over the 6*primes
forever if my conjecture is true).</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>This is easy to prove:</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>Necessary condition: if </FONT><FONT
face=Arial size=2><FONT face="Times New Roman" size=3>N is an abundant
number multiple of k prime, N is, at least, = 6k.<BR><BR>1) N=k -->
sigma(N) = 1+k < 2k. It isn't abundant.<BR><BR>2) N=2k --> sigma(N) =
1+2+k+2k<= 4k if k>=3 --> It isn't abundant if k>2.<BR><BR>3)
N=3k -->sigma(N) = 1+3+k+3k <= 6k if k>=2 --> It isn't
abundant.<BR><BR>4) N= 4k --> sigma(N) = 1+2+4+k+2k+4k <=
8k if k>=7 --> It isn't abundant if k>5 (and it is abundant if
k<=5).<BR><BR>Sufficient Condition: If k is prime, 6k is always
abundant.<BR><BR>N = 6k --> sigma(6k) = 1+2+3+6+k+2k+3k+6k = 12(k+1) > 12k
= 2*6k = 2*N --> 6k is always abundant.</FONT></FONT></DIV>
<DIV><FONT face=Arial size=2><FONT face=Arial size=2></FONT></FONT> </DIV>
<DIV><FONT face=Arial size=2><FONT face="Times New Roman" size=3>Conclusion: ak
= 6*prime(k) for every k>3 (prime(3) = 5) and ak = 4*prime(k) if
k<=3.</FONT></DIV>
<DIV><BR></DIV></FONT>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>Would we add this as a comment? It has become a
trivial sequence, hasn't it?</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>Regards. Jose Brox.</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>PD
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>And, of course, the sequence is monotonic, since
6*prime(k) < 6*prime(k+1), 4*prime(k) < 4*prime(k+1) and 4*prime(3) <
6*prime(4).</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV></FONT></DIV>
<DIV><FONT face=Arial size=2> </DIV>
<DIV><BR></DIV>
<DIV><BR></DIV></FONT>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV></BODY></HTML>