A056527 Numbers where iterated sum of digits of square settles down to a cyclic pattern (in fact 13, 16, 13, 16, ...).

2, 4, 5, 7, 11, 13, 14, 16, 20, 22, 23, 25, 29, 31, 32, 34, 38, 40, 41, 43, 47, 49, 50, 52, 56, 58, 59, 61, 65, 67, 68, 70, 74, 76, 77, 79, 83, 85, 86, 88, 92, 94, 95, 97, 101, 103, 104, 106, 110, 112, 113, 115, 119, 121, 122, 124, 128, 130, 131, 133, 137, 139, 140

Offset: 1

Author

Henry Bottomley, Jun 19 2000

Keywords

• base
• easy
• nonn

Comments

Numbers == 2, 4, 5 or 7 mod 9, i.e. such that n^4 is not congruent to n^2 mod 9.

Numbers congruent to {2, 4, 5, 7} mod 9.

Examples

			a(1)=2 because iteration starts 2, 4, 7, 13, 16, 13, 16, ....
		

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 1, -1).

Crossrefs

Cf. A004159 for sum of digits of square, A056020 where iteration settles to 1, A056020 where iteration settles to 9, also A056528, A056529. Unhappy numbers A031177 deal with iteration of square of sum of digits not settling to a single result.

Programs

Mathematica

Flatten[Table[9n+{2,4,5,7},{n,0,20}]] (* or *) LinearRecurrence[{1,0,0,1,-1},{2,4,5,7,11},100] (* Harvey P. Dale, Apr 05 2015 *)

PARI

Vec(x*(2 + 2*x + x^2 + 2*x^3 + 2*x^4) / ((1 - x)^2*(1 + x)*(1 + x^2)) + O(x^80)) \\ Colin Barker, Dec 19 2017

Formula

a(n) = a(n-1) + a(n-4) - a(n-5) for n>5. - Harvey P. Dale, Apr 05 2015

From Colin Barker, Dec 19 2017: (Start)

G.f.: x*(2 + 2*x + x^2 + 2*x^3 + 2*x^4) / ((1 - x)^2*(1 + x)*(1 + x^2)).

a(n) = (-9 + (-1)^(1+n) - (3-3*i)*(-i)^n - (3+3*i)*i^n + 18*n) / 8 where i=sqrt(-1).

(End)