A397140 - OEIS

A397140

a(n) = number of positive g < prime(n) such that g and g+1 are both primitive roots modulo prime(n).

0, 0, 1, 0, 2, 1, 4, 3, 4, 5, 3, 3, 6, 5, 10, 11, 14, 5, 4, 7, 8, 8, 20, 18, 12, 15, 13, 26, 12, 20, 10, 17, 30, 12, 35, 11, 16, 19, 40, 41, 44, 15, 31, 20, 37, 15, 10, 22, 56, 22, 54, 38, 16, 40, 64, 64, 65, 19, 28, 38, 32, 71, 30, 42, 28, 77, 24, 22, 86, 37, 74, 88, 39, 39, 34, 94

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OFFSET

1,5

COMMENTS

a(n) = 0 only for prime(n) = 2, 3 and 7. Cohen proved that every finite field of order other than 2, 3 or 7 contains a pair of consecutive primitive roots.

a(n) is not determined by A008330(n) = phi(prime(n)-1): the deviation from the asymptotic main term is of order sqrt(prime(n)).

REFERENCES

S. D. Cohen, Pairs of primitive roots, Mathematika 32 (1985), 276-285.

LINKS

Daniel Okwor, Table of n, a(n) for n = 1..9592

S. D. Cohen, T. Oliveira e Silva and T. Trudgian, On consecutive primitive elements in a finite field, arXiv:1410.6210 [math.NT], 2014; Bull. London Math. Soc. 47 (2015), 418-426.

FORMULA

a(n) ~ phi(p-1)^2/(p-1), where p = prime(n), the count predicted if being a primitive root were independent at g and g+1.

EXAMPLE

prime(5) = 11 has primitive roots 2, 6, 7 and 8. The consecutive pairs among them are (6,7) and (7,8), so a(5) = 2.

PROG

(PARI) a(n) = my(p=prime(n), c=0); for(x=1, p-2, if(znorder(Mod(x, p))==p-1 && znorder(Mod(x+1, p))==p-1, c++)); c

CROSSREFS

Cf. A229899, A261388, A261438, A268397, A060749, A008330, A249412, A229910.

Sequence in context: A276638 A116449 A316433 * A071046 A324757 A144334

Adjacent sequences: A397137 A397138 A397139 * A397141 A397142 A397144

KEYWORD

nonn

AUTHOR

Daniel Okwor, Jul 20 2026

STATUS

approved