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Dirichlet inverse of A344005, the smallest positive m such that n divides the oblong number m*(m+1).
4

%I #11 Jun 12 2022 15:38:57

%S 1,-1,-2,-2,-4,2,-6,-2,-4,4,-10,7,-12,6,11,0,-16,4,-18,16,18,10,-22,4,

%T -8,12,-2,23,-28,-11,-30,4,29,16,34,12,-36,18,36,5,-40,-18,-42,39,27,

%U 22,-46,-6,-12,8,47,48,-52,2,70,25,54,28,-58,-78,-60,30,21,8,71,-29,-66,64,65,-34,-70,24,-72,36,16,71,99

%N Dirichlet inverse of A344005, the smallest positive m such that n divides the oblong number m*(m+1).

%H Antti Karttunen, <a href="/A354875/b354875.txt">Table of n, a(n) for n = 1..20000</a>

%F a(1) = 1, and for n > 1, a(n) = -Sum_{d|n, d<n} A344005(n/d) * a(d).

%F a(n) = A354876(n) - A344005(n).

%t f[n_] := Module[{m = 1}, While[! Divisible[m*(m + 1), n], m++]; m]; a[1] = 1; a[n_] := a[n] = -DivisorSum[n, a[#]*f[n/#] &, # < n &]; Array[a, 100] (* _Amiram Eldar_, Jun 12 2022 *)

%o (PARI)

%o A344005(n) = for(m=1, oo, if((m*(m+1))%n==0, return(m))); \\ From A344005

%o memoA354875 = Map();

%o A354875(n) = if(1==n,1,my(v); if(mapisdefined(memoA354875,n,&v), v, v = -sumdiv(n,d,if(d<n,A344005(n/d)*A354875(d),0)); mapput(memoA354875,n,v); (v)));

%Y Cf. A002378, A344005, A354876, A354877 (positions of 0's).

%Y Cf. also A345055.

%K sign

%O 1,3

%A _Antti Karttunen_, Jun 12 2022