OFFSET
1,3
COMMENTS
This sequence would be better defined as a(n) = (n^7-n)/7 with offset 0, which is an integer by Fermat's little theorem. - N. J. A. Sloane, Nov 13 2023
Row 7 of A208535.
Also, row 7 (with different offset) of A074650. - Eric M. Schmidt, Dec 08 2017
REFERENCES
J. Jeffries, Differentiating by prime numbers, Notices Amer. Math. Soc., 70:11 (2023), 1772-1779.
LINKS
R. H. Hardin, Table of n, a(n) for n = 1..210
Wikipedia, p-derivation.
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).
FORMULA
Empirical: a(n) = (1/7)*n^7 - 1*n^6 + 3*n^5 - 5*n^4 + 5*n^3 - 3*n^2 + (6/7)*n.
Empirical formula confirmed by Petros Hadjicostas, Nov 05 2017 (see A208535).
a(n+2) = delta(-n) = -delta(n) for n >= 0, where delta is the p-derivation over the integers with respect to prime p = 7. - Danny Rorabaugh, Nov 10 2017
From Colin Barker, Nov 11 2017: (Start)
G.f.: 6*x^3*(3 + 28*x + 58*x^2 + 28*x^3 + 3*x^4) / (1 - x)^8.
a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8) for n>8.
(End)
a(n) = ((n-1)^7 - (n-1))/7. (inspired by Hassler's formula in A208536) - Eric M. Schmidt, Dec 08 2017
EXAMPLE
All solutions for n=3:
..1...1...1...1...1...1...1...1...1...1...1...1...1...1...1...1...1...1
..2...2...2...3...2...2...2...2...2...3...3...3...2...2...2...2...2...2
..1...1...1...1...1...1...3...3...3...2...2...1...3...1...3...1...3...1
..2...3...2...3...3...3...2...1...2...3...1...3...2...2...1...3...1...2
..3...2...1...2...1...2...1...3...3...2...3...1...3...3...3...1...2...1
..1...3...3...3...2...1...3...1...2...3...2...3...1...2...2...3...3...2
..3...2...2...2...3...3...2...3...3...2...3...2...3...3...3...2...2...3
MATHEMATICA
PROG
(PARI) Vec(6*x^3*(3 + 28*x + 58*x^2 + 28*x^3 + 3*x^4) / (1 - x)^8 + O(x^40)) \\ Colin Barker, Nov 11 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
R. H. Hardin, Feb 27 2012
STATUS
approved