A212156 a(n) = ((6*A023000(n))^3 + 1)/7^n, n >= 0.

1, 31, 2257, 116623, 5757601, 282424831, 13840934257, 678220602223, 33232913275201, 1628413476849631, 79792265450186257, 3909821042651007823, 191581231339042552801, 9387480337357087274431, 459986536542705291758257

Offset: 0

Author

Wolfdieter Lang, May 02 2012

Keywords

• nonn
• easy

Comments

a(n) is an integer because 6*A023000(n) is one of three solution of X(n)^3+1 == 0 (mod 7^n), namely the one satisfying also X(n) == 6 (mod 7) == -1 (mod 7).

See the comments on A210852, and the Nagell reference given in A210848.

Examples

			a(0) = 1/1 = 1.
a(3) = ((6*57)^3 + 1)/7^3 = 40001689/343  = 116623,  (b(3) = 48^7 (mod 7^3) = 342 = 6*57).
		

Crossrefs

Cf. A210848, A210849 (the p=5 case). A210853, A212154.

Formula

a(n) = (b(n)^3+1)/7^n, n>=0, with b(n):=6*A023000(n) given by a recurrence obtained from the one of A023000. There also programs for b(n)/6 are given.