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a(n) is the number of terms m such that d(m!) divides d((m!)^(2n+1)), where d is A000005.
2

%I #23 Aug 07 2017 21:25:54

%S 5,8,15,6,29,27,5,54,60,6,63,7,6,54,75,6,12,52,7,76,69,5,74,27,6,78,

%T 12,6,97,33,6,15,85,5,99,46,5,15,95,6,56,13,6,82,20,5,7,81,6,126,141,

%U 5,130,67,6,52,13,5,17,38,5,8,55,6,85,15,5,106,143,5,22,12,6,95,94,6

%N a(n) is the number of terms m such that d(m!) divides d((m!)^(2n+1)), where d is A000005.

%C Inspired by A069784.

%C Bisection of A286872.

%C d(m!) divides d((m!)^(2n)) only when m = 1.

%C The largest m for a(n) is: 5, 15, 91, 9, 275, 488, 5, 655, 1205, 21, 1687, 14, 9, 1462, 2313, 21, 35, 3436, 21, 7447, 4687, 5, 2555, 220, 9, 4627, 38, 9, 5114, 2606, 21, 65, 6071, 5, 4935, 5509, 5, 77, 10173, 9, 1646, 39, 9, 6715, 95, 5, 65, 2321, 9, 3786, 7059, 5, 7014, 1264, 9, 6272, 35, 5, 33, 215, 5, 27, 4283, 9, 2471, ..., ; and their records: 5, 15, 91, 275, 488, 655, 1205, 1687, 2313, 3436, 7447, 10173, 15464, 20004, 38539, 40605, 49143, ..., .

%H Robert G. Wilson v, <a href="/A286835/b286835.txt">Table of n, a(n) for n = 1..625</a>

%e a(1) = 5 since d(m!) divides d(m!^3) only for m = {1, 2, 3, 4, 5};

%e a(2) = 8 since d(m!) divides d(m!^5) only for m = {1, 2, 3, 4, 5, 12, 13, 15};

%e a(3) = 15 since d(m!) divides d(m!^7) only for m = {1, 2, 3, 4, 5, 6, 7, 8, 9, 14, 32, 33, 34, 35, 91};

%e a(4) = 6 since d(m!) divides d(m!^9) only for m = {1, 2, 3, 4, 5, 9};

%e a(5) = 29 since d(m!) divides d(m!^11) only for m = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ..., 274, 275}; etc.

%t factExpLst[nbr_] := factExpLst[nbr] = Table[Plus @@ Rest@ NestWhileList[ Floor[#/prm] &, nbr, # > 0 &], {prm, Prime@ Range@ PrimePi@ nbr}] (* which is the same as Transpose[ FactorInteger[ nbr!]][[2]] *);

%t ds0[nbr_, exp_] := Times @@ (1 + exp*factExpLst[ nbr]);

%t fQ[nbr_, exp_] := Mod[ds0[nbr, exp], ds0[nbr, 1]] == 0;

%t f[n_] := f[n] = If[EvenQ@ n, {1}, Select[Range@ 100000, fQ[#, n] &]];

%t f[1] = {};

%t Array[ Length@ f[2# +1] &, 60]

%o (PARI) valp(n,p)=my(s); while(n\=p, s+=n); s

%o f(m,e)=my(s=1); forprime(p=2,m\2, s*=e*valp(m,p)+1); s*(e+1)^(primepi(m)-primepi(m\2))

%o search(n,lim=100*n^2)=my(v=List(),e=2*n+1); for(m=1,lim, if(f(m,e)%f(m,1)==0, listput(v,m))); Vec(v) \\ N.B., empirical upper bound

%o a(n)=#search(n) \\ _Charles R Greathouse IV_, Aug 01 2017

%Y Cf. A000005, A027423, A069784, A286872.

%K nonn

%O 1,1

%A _Robert G. Wilson v_, Jul 31 2017