

A056527


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


4



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
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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(n1) + a(n4)  a(n5) 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)  (33*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 A110924
Adjacent sequences: A056524 A056525 A056526 * A056528 A056529 A056530


KEYWORD

base,easy,nonn


AUTHOR

Henry Bottomley, Jun 19 2000


STATUS

approved



