Tautócrona tautocrona at terra.es
Mon Mar 24 02:36:58 CET 2003

```Copy-paste of the sequence:

ID Number: A064161
Sequence:  12,12,20,42,66,78,102,114,138,174,186,222,246,258,282,318,
354,366,402,426,438,474,498,534,582,606,618,642,654,678,762,
786,822,834,894,906,942,978,1002,1038,1074,1086,1146,1158,
1182,1194,1266,1338,1362,1374
Name:      Least abandant number divisible by the n_th prime number.
Comments:  I know of no reason why this sequence should be monotonic, but this is
true up to the 10^3-rd prime.
Example:   The first eight abundant numbers are 12, 18, 20, 24, 30, 36, 40 and 42. But
only the last one is divisible by 7, the fourth prime number. Therefore
a(4) = 42.

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I've just realized that this sequence is also generated with another formula since the fourth term:
1) The odd primes:  3 5 7 11 13 17 19...

2) Their first order differences: 2 2 4 2 4 2 4 6 ...

Multiplying by 6: 12 12 24 12 24 12 24 36 12 36 24 12... = = V.

If we start at a0 = 42 and then do a(n) = a(n-1)+V(n) we obtain the sequence A064161:

66,78,102,114,138,174,186,222,246,258,282,318,354,366,402,426,438,474,498,534,58
2,606,618,642,654,678,762,786,822,834,894,906,942,978,1002,1038,1074,1086,1146,1
158,1182,1194,1266,1338,1362,1374,1398,1434,1446,1506,1542,1578,1614,1626,1662,1
686,1698,1758,1842,1866,1878,1902,1986,2022,2082,2094,2118,2154,2202,2238,2274,2
298,2334,2382,2406,2454,2514,2526,2586,2598,2634,2658,2694,2742,2766,2778,2802,2
874,2922,2946,2994 (PARI-GP)

An example: a80 = 2454 --> 2454 is abundant (sigma(2454) = 2465) and prime(80) = 409 = a80/6.

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).

This is easy to prove:

Necessary condition: if N is an abundant number multiple of k prime, N is, at least, = 6k.

1) N=k --> sigma(N) = 1+k < 2k. It isn't abundant.

2) N=2k --> sigma(N) = 1+2+k+2k<= 4k if k>=3 --> It isn't abundant if k>2.

3) N=3k -->sigma(N) = 1+3+k+3k <= 6k if k>=2 --> It isn't abundant.

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).

Sufficient Condition: If k is prime, 6k is always abundant.

N = 6k --> sigma(6k) = 1+2+3+6+k+2k+3k+6k = 12(k+1) > 12k = 2*6k = 2*N --> 6k is always abundant.

Conclusion: ak = 6*prime(k) for every k>3 (prime(3) = 5) and ak = 4*prime(k) if k<=3.

Would we add this as a comment? It has become a trivial sequence, hasn't it?

Regards. Jose Brox.

PD

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).

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