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A384819 Nonnegative numbers a(n) < n for n >= 1 such that exp( Sum_{n>=1} (n^2 - a(n))*x^n/n ) is a power series with integral coefficients.

%I A384819 #14 Jun 22 2025 00:51:48
%S A384819 0,1,2,1,4,3,6,1,2,7,10,3,12,11,3,1,16,3,18,15,14,19,22,3,4,23,2,27,
%T A384819 28,12,30,1,15,31,7,3,36,35,32,7,40,37,42,7,21,43,46,3,6,7,27,11,52,3,
%U A384819 34,35,50,55,58,44,60,59,5,1,18,0,66,19,39,40,70,3,72,71,3,23,1,19,78,55,2,79,82,41,47,83,51,47,88,84,74,31,86,91,17,3,96,11,42,15,100
%N A384819 Nonnegative numbers a(n) < n for n >= 1 such that exp( Sum_{n>=1} (n^2 - a(n))*x^n/n ) is a power series with integral coefficients.
%C A384819 Conjecture: a(p^n) = p - 1 when p is prime for n >= 1.
%H A384819 Paul D. Hanna, <a href="/A384819/b384819.txt">Table of n, a(n) for n = 1..520</a>
%e A384819 L.g.f.: A(x) = 0*x + 1*x^2/2 + 2*x^3/3 + 1*x^4/4 + 4*x^5/5 + 3*x^6/6 + 6*x^7/7 + 1*x^8/8 + 2*x^9/9 + 7*x^10/10 + 10*x^11/11 + 3*x^12/12 + 12*x^13/13 + 11*x^14/14 + 3*x^15/15 + 1*x^16/16 + ...
%e A384819 where the following is a power series with integral coefficients
%e A384819 exp( x/(1-x)^2 - A(x) ) = 1 + x + 2*x^2 + 4*x^3 + 8*x^4 + 14*x^5 + 25*x^6 + 43*x^7 + 74*x^8 + 124*x^9 + 205*x^10 + 335*x^11 + 543*x^12 + 869*x^13 + 1379*x^14 + 2170*x^15 + 3388*x^16 + ... + A384820(n)*x^n + ...
%e A384819 which is equivalent to
%e A384819 exp( Sum_{n>=1} (n^2 - a(n))*x^n/n ) = exp(x + 3*x^2/2 + 7*x^3/3 + 15*x^4/4 + 21*x^5/5 + 33*x^6/6 + 43*x^7/7 + 63*x^8/8 + 79*x^9/9 + 93*x^10/10 + 111*x^11/11 + 141*x^12/12 + 157*x^13/13 + 185*x^14/14 + 222*x^15/15 + 255*x^16/16 + ...).
%o A384819 (PARI) {a(n) = my(A=[1]); for(i=2,n, A = concat(A,t);
%o A384819 for(t=1,(#A)^2+1, if( denominator( eval(polcoef( exp( intformal(Ser(A)) ),#A)) )==1, A[#A] = t + (#A)*(#A-1); break)) ); n^2 - A[n]}
%o A384819 for(n=1,101, print1(a(n),", "))
%Y A384819 Cf. A384820, A082579 (exp(x/(1-x)^2)).
%K A384819 nonn
%O A384819 1,3
%A A384819 _Paul D. Hanna_, Jun 18 2025