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a(n) = n^n - a(n-1), with a(1) = 1.
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%I #22 Jun 17 2022 16:34:25

%S 1,3,24,232,2893,43763,779780,15997436,371423053,9628576947,

%T 275683093664,8640417354592,294234689237661,10817772136320355,

%U 427076118244539020,18019667955465012596,809220593930871751581,38537187481365665823843,1939882468178947923300136

%N a(n) = n^n - a(n-1), with a(1) = 1.

%H T. D. Noe, <a href="/A001099/b001099.txt">Table of n, a(n) for n = 1..100</a>

%F Absolute value of Sum_{k=1..n} k^k*(-1)^(k+1). a(n) = n^n - (n-1)^(n-1) + (n-2)^(n-2) - ... - (-1)^n*1^1. - _Alexander Adamchuk_, Jun 30 2006

%t Abs[Table[Sum[k^k*(-1)^(k+1),{k,1,n}],{n,1,30}]] (* _Alexander Adamchuk_, Jun 30 2006 *)

%t RecurrenceTable[{a[1]==1,a[n]==n^n-a[n-1]},a,{n,20}] (* _Harvey P. Dale_, Jan 21 2015 *)

%o (Python)

%o from itertools import accumulate, count, islice

%o def A001099_gen(): # generator of terms

%o yield from accumulate((k**k for k in count(1)),func=lambda x,y:y-x)

%o A001099_list = list(islice(A001099_gen(),20)) # _Chai Wah Wu_, Jun 17 2022

%Y Cf. A001923.

%K nonn

%O 1,2

%A _Ed Smiley_