OFFSET
0,2
COMMENTS
Question: Will a(n)/(2^n) converge towards zero? After local maximum a(7)/(2^7) = 32/128 = 1/4, the ratio seems to be decreasing, and is already only 0.02978... (< 1/32) at a(27) = 3997263.
MATHEMATICA
Block[{a, b, c, f, nn = 12}, b[0] = c[1] = 1; f[n_] := DivisorSigma[1, n]/(n Log[Log[n]]); Do[b[i] = Prime[1 + BitLength[i] - DigitCount[i, 2, 1]]*b[i - 2^Floor@ Log2@ i]; c[i + 1] = Apply[Times, Flatten@ MapIndexed[ConstantArray[Prime[First[#2]], #1] &, Table[LengthWhile[#1, # >= j &], {j, #2}] & @@ {#, Max[#]} &@ Sort[Flatten[ConstantArray[PrimePi@ #1, #2] & @@@ FactorInteger[b[i]]], Greater]]]; a[i - 1] = Boole[f[c[i]] >= f[c[Floor[(i + 1)/2]]]], {i, 2^nn}]; Total /@ Table[a[2^n + k], {n, 0, nn - 2}, {k, 0, 2^n - 1}]] (* Michael De Vlieger, Mar 07 2021 *)
PROG
(PARI)
default(parisizemax, 2^31);
default(realprecision, 121);
A283980(n) = {my(f=factor(n)); prod(i=1, #f~, my(p=f[i, 1], e=f[i, 2]); if(p==2, 6, nextprime(p+1))^e)}; \\ From A283980
Gi(n) = (log(n)^(n/sigma(n)));
A342000(n) = if(n<=3, !!(n-1), my(p=A329886(n\2)); if(n%2, Gi(2*p)<=Gi(p), Gi(A283980(p))<=Gi(p))); \\ Program less vulnerable to the loss of precision.
s=0; k=0; for(n=2, 2^31, if(!bitand(n, n-1), write("b342020.txt", k, " ", s); print(s, "/2^", k, " = ", (1.0)*(s/(2^k))); k++; s = 0); s += A342000(n));
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Antti Karttunen and Michael De Vlieger, Mar 07 2021
STATUS
approved