A210853 a(n) = (A210852(n)^3 + 1)/7^n, n >= 0.

1, 4, 608, 100082, 1033865, 147695, 363432817, 493771113103, 2362056468993, 408352474516087, 11132773648769182, 1051698129414636470, 55996715400581424222, 4972138747809482684591, 29726859239716779753649, 180817068189496094994710, 34294232575354274959952776, 358207669631705219617812791

Offset: 0

Author

Wolfdieter Lang, May 02 2012

Keywords

• nonn
• easy

Comments

a(n) is an integer because A210852(n) is one of the three solutions of X(n)^3 + 1 == 0 (mod 7^n), namely the one satisfying also X(n) == 3 (mod 7).

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

Examples

			a(0) = 1/1 = 1.
a(3) = (325^3 + 1)/7^3 = 34328126/343  = 100082, (b(3) = 31^7 (mod 7^3) = 325).
		

Links

• Paolo Xausa, Table of n, a(n) for n = 0..500

Crossrefs

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

Programs

Mathematica

Join[{1}, MapIndexed[(#^3 + 1)/7^#2[[1]] &, FoldList[PowerMod[#, 7, 7^#2] &, 3, Range[2, 20]]]] (* Paolo Xausa, Jan 14 2025 *)

Formula

a(n) = (b(n)^3 + 1)/7^n, n>=0, with b(n):=A210852(n) given by a recurrence. See also a Maple program for b(n) there.