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A194641
Digital roots of the nonzero heptagonal numbers.
0
1, 7, 9, 7, 1, 9, 4, 4, 9, 1, 7, 9, 7, 1, 9, 4, 4, 9, 1, 7, 9, 7, 1, 9, 4, 4, 9, 1, 7, 9, 7, 1, 9, 4, 4, 9, 1, 7, 9, 7, 1, 9, 4, 4, 9, 1, 7, 9, 7, 1, 9, 4, 4, 9, 1, 7, 9, 7, 1, 9, 4, 4, 9, 1, 7, 9, 7, 1, 9, 4, 4, 9, 1, 7, 9, 7, 1, 9, 4, 4, 9, 1, 7, 9, 7, 1
OFFSET
1,2
COMMENTS
This is a periodic sequence with period 9 and cycle 1, 7, 9, 7, 1, 9, 4, 4, 9, which are also the coefficients of x in the numerator of the generating function.
FORMULA
a(n) = a(n-9), and as the sum of the terms contained in each cycle is 51 they also satisfy the eighth-order inhomogeneous recurrence a(n) = 51 -a(n-1) -a(n-2) -a(n-3) -a(n-4) -a(n-5) -a(n-6) -a(n-7) -a(n-8).
a(n) = 3*(1+cos(2*n*Pi/3) +cos(4*n*Pi/3)) +(n^3 +8*n^4 +7*n^5 +5*n^6 +4*n^7 +3*n^8) mod 9.
G.f.: x*(1 +7*x +9*x^2 +7*x^3 +x^4 +9*x^5 +4*x^6 +4*x^7 +9*x^8)/((1-x)*(1 +x +x^2)*(1 +x^3 +x^6)).
a(n) = A010888(A000566(n)). - Michel Marcus, Aug 10 2015
EXAMPLE
The sixth nonzero heptagonal number is A000566(6)=81. As 8+1=9, this has digital root 9 and so a(6)=9.
MATHEMATICA
DigitalRoot[n_]:=FixedPoint[Plus@@IntegerDigits[#]&, n]; Table[DigitalRoot[ n*(5*n - 3)/2], {n, 100}]
PadRight[{}, 120, {1, 7, 9, 7, 1, 9, 4, 4, 9}] (* Harvey P. Dale, Dec 31 2023 *)
CROSSREFS
Sequence in context: A154168 A019644 A100639 * A248674 A108743 A177271
KEYWORD
nonn,easy,base
AUTHOR
Ant King, Aug 31 2011
STATUS
approved