[seqfan] A113571

[Don Reble]

> %I A113571
> %S 1,2,6,4,10,7,14,8,9,61,22,13,26,1049,41233,16,34,19,38
> %N Last appearance of n or zero if no such appearance exists.
> %Y Cf. A110545.
> %E Sequence continues: a(20)=?, a(21)=?, 23, 46, 138, 50, a(26)=?, 27,
>    841, 58, 31, 62, 32, 269, 3583, 397, 1297, 74, 199, a(39)=?, 41,
>    82, 301, 86, 407, 1553, 47, 94, a(48)=?, 98, a(50)=?, 76943, 809,
>    106, 5953, a(55)=?, a(56)=?, 26693, 59, 118, 2207, ...

    Actually, it's the _first_ appearance of n in A110545.

    One can define this without referring to A110545:

        Let H(n) be the reduced fraction Sum_{i=1..n} 1/i.
        a(n) is the least factor of H(n)'s numerator or denominator
        which doesn't divide either part of any earlier H(m).

    so there's a far better way of calculating the sequence, than
    scanning A110545.

    More terms:

%S A113571 1,2,6,4,10,7,14,8,9,61,22,13,26,1049,41233,16,34,19,38,11167027,
%T A113571 18858053,23,46,138,50,34395742267,27,841,58,31,62,32,269,3583,397,
%U A113571 1297,74,199,737281,41,82,301,86,407,1553,47,94,2323031,98,587948341,
           76943,2809,106,5953,51862596437,252476961434436524654789,26693,59,
           118,2207,122,928551009361054917576341971,347,64,2473,67,134

    (a(52)=2809, not 809.)


> %C Conjectured last occurrence of n:
>    1,3,11,25,137,49,363,761,7129,7381,83711, [sic] 6617,72072,1117,

    This might be A001008.
    (That sequence differs after 83711: ...,83711,86021,1145993,1171733,...
     Maybe the conjecturer scanned A113571, and stopped around a(85000).)
    But if one of those numerators divides a previous numerator or a
    previous denominator (A002805), then they're different sequences.
    Does anyone know whether that happens?

-- 
Don Reble  djr at nk.ca


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