A193090 Digital roots of the nonzero pentagonal numbers.

1, 5, 3, 4, 8, 6, 7, 2, 9, 1, 5, 3, 4, 8, 6, 7, 2, 9, 1, 5, 3, 4, 8, 6, 7, 2, 9, 1, 5, 3, 4, 8, 6, 7, 2, 9, 1, 5, 3, 4, 8, 6, 7, 2, 9, 1, 5, 3, 4, 8, 6, 7, 2, 9, 1, 5, 3, 4, 8, 6, 7, 2, 9, 1, 5, 3, 4, 8, 6, 7, 2, 9, 1, 5, 3, 4, 8, 6, 7, 2, 9, 1, 5, 3, 4, 8

Offset: 1

Author

Ant King, Aug 15 2011

Keywords

• nonn
• easy
• base

Comments

This is a periodic sequence with period 9 and cycle 1,5,3,4,8,6,7,2,9 - which are also the coefficients of x in the numerator of the generating function.

Note that the cycle 1,5,3,4,8,6,7,2,9 is a permutation of the first 9 natural numbers A000027. - Omar E. Pol, Aug 15 2011

This sequence is the same as A002450(n+1) mod 9, except with a value of 9 where that would return 0. - Joe Slater, Mar 04 2018

Examples

			The sixth nonzero pentagonal number is A000326(6) = 51, which has digital root 5 + 1 = 6. Hence a(6) = 6.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,1).

Crossrefs

Cf. A000326, A002450, A010888.

Programs

Mathematica

DigitalRoot[n_]:=FixedPoint[Plus@@IntegerDigits[#]&,n]; DigitalRoot[1/2 # (3#-1)]&/@Range[90]
PadRight[{},120,{1,5,3,4,8,6,7,2,9}] (* Harvey P. Dale, Sep 12 2017 *)

PARI

a(n)=[9, 1, 5, 3, 4, 8, 6, 7, 2][n%9+1] \\ Charles R Greathouse IV, Oct 04 2012

Formula

a(n) = a(n-9).

As the sum of the terms contained in each cycle is 45, they also satisfy the eighth-order inhomogeneous recurrence a(n)=45-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) = cos(8n Pi/9) (1 + 2 cos(2n Pi/9))(1 + 2 cos(2n Pi/3)) + (n + 7n^3 + 5n^4 + n^5 + 5n^6 + 4n^7 + 5n^8) mod 9.

G.f.: x(1 + 5x + 3x^2 + 4x^3 + 8x^4 + 6x^5 + 7x^6 + 2x^7 + 9x^8)/((1-x)(1 + x + x^2)(1 + x^3 + x^6)).

a(n) = A010888(A000326(n)). - Jonathan Vos Post, Aug 15 2011

a(n) = 9-((8*(4^n-1)/3) mod 9). - Joe Slater, Mar 04 2018