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A254618 a(n) = k-tuple deficiency of n-th deficient number.

%I A254618 #19 Mar 28 2015 14:48:24
%S A254618 1,2,2,3,2,2,3,4,4,2,2,5,5,6,2,2,3,6,2,1,7,3,2,2,3,6,1,3,2,7,3,2,2,1,
%T A254618 7,8,2,4,3,4,9,2,3,3,4,2,2,2,3,4,3,2,5,4,2,2,1,5,5,3,2,1,2,2,3,9,7,2,
%U A254618 4,6,4,4,2,2,3,4,2,2,8,1,2,2,2,3,2,3,5
%N A254618 a(n) = k-tuple deficiency of n-th deficient number.
%C A254618 For any deficient number x iterate the process f(x)=sigma(x)-x. Sequence lists how many times f(x) keeps deficient until it reaches zero.
%C A254618 Non-deficient numbers are excluded from this sequence.
%C A254618 k-tuple deficiency records is A000027.
%C A254618 k-tuple deficiency record-holders is A234899.
%H A254618 Paolo P. Lava, <a href="/A254618/b254618.txt">Table of n, a(n) for n = 1..1000</a>
%e A254618 a(20) = 1 because the 20th deficient number is 25 and:
%e A254618 1) f(25) = sigma(25) - 25 = 6 < 25.
%e A254618 We must stop here because 6 is abundant.
%e A254618 a(21) = 7 because the 21st deficient number is 26 and:
%e A254618 1) f(26) = sigma(26) - 26 = 16 < 26;
%e A254618 2) f(16) = sigma(16) - 16 = 15 < 16;
%e A254618 3) f(15) = sigma(15) - 15 = 9 < 15;
%e A254618 4) f(9) = sigma(9) - 9 = 4 < 9;
%e A254618 5) f(4) = sigma(4) - 4 = 3 < 4;
%e A254618 6) f(3) = sigma(3) - 3 = 2 < 1;
%e A254618 7) f(1) = sigma(1) - 1 = 0 < 1.
%e A254618 We must stop here because sigma(0) is not defined.
%p A254618 with(numtheory): P:=proc(q) local a,b,n,t;
%p A254618 for n from 1 to q do t:=0; b:=sigma(n)-n; a:=n;
%p A254618 if b<a then while b<a do t:=t+1; a:=b; b:=sigma(b)-b; od;
%p A254618 print(t); fi; od; end: P(10^3);
%Y A254618 Cf. A000027, A005100, A081705, A098007, A098008, A234899.
%K A254618 nonn
%O A254618 1,2
%A A254618 _Paolo P. Lava_, Feb 03 2015