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

1,1

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.

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).

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)

Example

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

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 *)

Prog

(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

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.

Sequence in context: A255850 A108464 A128815 * A147991 A033160 A350147

Adjacent sequences: A056524 A056525 A056526 * A056528 A056529 A056530

Keyword

base,easy,nonn

Author

Henry Bottomley, Jun 19 2000

Status

approved