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A074483 Consider the recursion b(1,n) = 1, b(k+1,n) = b(k,n) + (b(k,n) reduced mod(k+n)); then there is a number y such that b(k,n)-b(k-1,n) is a constant (= A074482(n)) for k > y. Sequence gives values of y. 3

%I #11 Mar 30 2012 18:50:25

%S 397,396,395,4,11,10,25,24,29,14,5,26,25,10,7,16,68265,14,13,12,17,

%T 1220,67,136,93,6,133,132,9,272,129,14,1209,126,125,124,48605,48604,

%U 269393,269392,292695,180,77,178,177,269386,24017,72,24015,172,67,44,11,16,65

%N Consider the recursion b(1,n) = 1, b(k+1,n) = b(k,n) + (b(k,n) reduced mod(k+n)); then there is a number y such that b(k,n)-b(k-1,n) is a constant (= A074482(n)) for k > y. Sequence gives values of y.

%C Conjecture: a(n) is defined for all n (as well as A074482);

%C A074484(n) = A074482(n)*(a(n)+ n + 1);

%C b(k, n) = A074482(n)*(k + n + 1) for k > a(n).

%H David W. Wilson, <a href="/A074483/b074483.txt">Table of n, a(n) for n = 0..10000</a>

%e A074482(0) = A073117(a(0)) mod a(0) = A073117(397) mod 397 = 38606 mod 397 = 97.

%Y Cf. A073117.

%K nonn

%O 0,1

%A _Reinhard Zumkeller_ and _Benoit Cloitre_, Aug 23 2002

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