%I #28 May 14 2022 22:22:38

%S 1,4,8,44,216,1474,11130,92489

%N Number of integers x in range A002110(n) .. A002110(1+n)-1 such that the k-th arithmetic derivative of A276086(x) is zero for some k, where A002110(n) is the n-th primorial.

%C a(n) is the number of terms of A328116 in range A002110(n) .. A002110(1+n)-1.

%C a(n) is the number of terms in A351255 (and in A099308) whose largest prime factor (A006530) is A000040(1+n).

%C Ratio a(n) / A061720(n) develops as:

%C 0: 1 / 1 = 1.0

%C 1: 4 / 4 = 1.0

%C 2: 8 / 24 = 0.333...

%C 3: 44 / 180 = 0.244...

%C 4: 216 / 2100 = 0.1029...

%C 5: 1474 / 27720 = 0.05317...

%C 6: 11130 / 480480 = 0.02316...

%C 7: 92489 / 9189180 = 0.01006...

%C Computing term a(8) would need processing over 213393180 integers whose greatest prime factor is 23, from single A351255(105368) = 23 at start to product (2^1)*(3^2)*(5*4)*(7^6)*(11^10)*(13^12)*(17^16)*(19^18)*(23^22) at the end of the batch [number whose size in binary is 346 bits], and would required factoring integers of comparable size and more (see A351261), that might not all be easily factorable.

%F a(n) = Sum_{k=A002110(n) .. A002110(1+n)-1} A328306(k).

%F a(n) = A328307(A002110(1+n)) - A328307(A002110(n)).

%e There are eight terms [6, 7, 9, 12, 15, 20, 21, 28] that are >= A002110(2) and < A002110(3) in A328116 for which the corresponding terms [5, 10, 30, 25, 150, 375, 750, 5625] in A276086 (and A351255) are all in A099308, therefore a(2) = 8.

%o (PARI)

%o \\ Memoization would work quite badly here. (See comments in A351255. In practice sequence A328306 was computed first, up to its term a(9699690). Same data is available in A328116.)

%o A002110(n) = prod(i=1,n,prime(i));

%o A003415checked(n) = if(n<=1, 0, my(f=factor(n), s=0); for(i=1, #f~, if(f[i, 2]>=f[i, 1], return(0), s += f[i, 2]/f[i, 1])); (n*s));

%o A328308(n) = if(!n, 1, while(n>1, n = A003415checked(n)); (n));

%o A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };

%o A328306(n) = A328308(A276086(n));

%o A351071(n) = sum(k=A002110(n),A002110(1+n)-1,A328306(k));

%Y Cf. A002110, A003415, A061720, A099308, A328306, A328307, A328308, A328116, A351067, A351069, A351255, A351261, A351072 (partial sums).

%Y Cf. also A327969.

%K nonn,hard,more

%O 0,2

%A _Antti Karttunen_, Feb 02 2022