A383300 Numbers k such that primorial base expansion of k has the primorial base expansion of k' as its suffix, where k' stands for the arithmetic derivative of k (A003415).

0, 1, 3, 4, 5, 7, 11, 13, 17, 19, 23, 27, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331

Offset: 1

Author

Antti Karttunen, May 15 2025

Keywords

• nonn

Comments

a(n) = A348283(n) for n=1 and n=3..334432. a(334433) = 4784261, which is not present in A348283 (see examples). - R. J. Mathar and Antti Karttunen, May 16 2025

Examples

			0 is a term as A003415(0) = 0.
1 is a term as A003415(1) = 0, whose primorial base expansion is here understood as an empty sequence of digits, thus it is a suffix of A049345(1) = 1.
3, like all odd primes, is a term as A003415(3) = 1, with A049345(3) = 11 and A049345(1) = 1.
4 and 27 are terms as they are in A051674 (the nonzero fixed points of A003415).
4784261 is a term as A003415(4784261) = 189671, with A049345(4784261) = 96411121 and A049345(189671) = 6411121. 4784261 is the first term > 1 of this sequence that is not in A348283. See more examples in A383301.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..20000
• Index entries for sequences related to primorial base.

Crossrefs

Cf. A003415, A049345, A235224.

Disjoint union of {1}, A348283\{2} and A383301.

Cf. A006005, A051674 (other subsequences).

Subsequence of A383299.

Cf. also A383933.

Programs

PARI

A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
isA383300(n) = if(n<2, 1, my(p=2, k=A003415(n)); while(k, if((k%p)!=(n%p), return(0)); n = n\p; k = k\p; p = nextprime(1+p)); (1));

PARI

A002110(n) = prod(i=1,n,prime(i));
A235224(n) = { my(s=0, p=2); while(n, s++; n = n\p; p = nextprime(1+p)); (s); };
isA383300(n) = { my(ad=A003415(n)); (n%A002110(A235224(ad))==ad); };

Formula

{k such that A003415(k) is equal to k modulo A002110(A235224(A003415(k)))}.