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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
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.
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 A373769
KEYWORD
base,easy,nonn
AUTHOR
Henry Bottomley, Jun 19 2000
STATUS
approved