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A392879
Let p = A129252(n), if p = 1, then a(n) = n, otherwise a(n) = n * p^((e mod p)-e) * q^floor(e/p), where e is p-adic valuation of n and q is the least prime number greater than p.
3
1, 2, 3, 3, 5, 6, 7, 6, 9, 10, 11, 9, 13, 14, 15, 9, 17, 18, 19, 15, 21, 22, 23, 18, 25, 26, 5, 21, 29, 30, 31, 18, 33, 34, 35, 27, 37, 38, 39, 30, 41, 42, 43, 33, 45, 46, 47, 27, 49, 50, 51, 39, 53, 10, 55, 42, 57, 58, 59, 45, 61, 62, 63, 27, 65, 66, 67, 51, 69, 70, 71, 54, 73, 74, 75, 57, 77, 78, 79, 45, 15, 82
OFFSET
1,2
FORMULA
A276085(a(n)) = A276085(n).
EXAMPLE
For n = 12 = 2^2 * 3, the least prime that "overflows" (has an exponent >= the prime itself) is 2, so we divide by 2^(2-(2 mod 2)) and multiply by 3^floor(2/2), i.e., divide by 4 and multiply by 3, to obtain a(12) = 9.
For n = 27 = 3^3, the least prime which overflows is 3, so we divide by 3^(3-(3 mod 3)) and multiply by 5^floor(3/3), i.e., divide by 3^3 and multiply by 5, to obtain a(27) = 5.
For n = 36 = 2^2 * 3^2, the least prime which overflows is 2, so we divide by 2^2 and multiply by 3, to obtain a(36) = 3^3 = 27.
For n = 54 = 2^1 * 3^3, the least prime which overflows is 3, so we divide by 3^(3-(3 mod 3)) and multiply by 5^floor(3/3), i.e., divide by 3^3 and multiply by 5, to obtain a(54) = 2*5 = 10.
For n = 72 = 2^3 * 3^2, the least prime which overflows is 2, so we divide by 2^(3-(3 mod 2)), i.e., by 2^2, and multiply by 3^floor(3/2), i.e., by 3, to obtain a(72) = 54 = 2 * 3^3.
PROG
(PARI)
A129252(n) = { my(pp); forprime(p=2, , pp = p^p; if(!(n%pp), return(p)); if(pp > n, return(1))); };
A392879(n) = { my(p=A129252(n), e); if(1==p, return(n), e = valuation(n, p); (n * p^((e%p)-e) * nextprime(1+p)^(e\p))); };
CROSSREFS
Cf. A048103 (fixed points), A129252, A276085, A276086, A392878, A392880 (number of iterations needed to reach A392878(n)).
Cf. also A277886 for somewhat analogous sequence for base-2.
Sequence in context: A384249 A392877 A392878 * A255563 A331288 A115350
KEYWORD
nonn,easy
AUTHOR
Antti Karttunen, Jan 31 2026
STATUS
approved