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#23 by N. J. A. Sloane at Mon Aug 30 22:24:19 EDT 2021
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#22 by Michel Marcus at Fri Aug 20 12:25:24 EDT 2021
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#21 by Michel Marcus at Fri Aug 20 12:25:21 EDT 2021
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| COMMENTS
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a(9*k+0 or 8) is congruent to 0 (mod 9), a(9*k+1 or 4 or 7) is congruent to 1 (mod 9), a(9*k+2 or 6) is congruent to 3 (mod 9), a(9*k+3 or 5) is congruent to 6 (mod 9).
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| FORMULA
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a(9*k+0 or 8) == 0 (mod 9);
a(9*k+1 or 4 or 7) == 1 (mod 9);
a(9*k+2 or 6) == 3 (mod 9);
a(9*k+3 or 5) == 6 (mod 9).
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proposed
editing
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#20 by Robert P. P. McKone at Mon Aug 16 05:13:54 EDT 2021
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Discussion
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| Mon Aug 16
| 08:39
| Ctibor O. Zizka: Based on numerical experiments I assume the linear non-homogenous recurrences with constant coefficients of the form a(n) = A004185(c_1*a(n-1) + f(n)) are not reaching a cycle for f(n) polynomial of degree >= 1, but allways reaching a cycle for f(n) = const.
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#19 by Robert P. P. McKone at Mon Aug 16 05:12:48 EDT 2021
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| LINKS
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Robert P. P. McKone, <a href="/A345681/b345681.txt">Table of n, a(n) for n = 0..2999</a>
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proposed
editing
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#18 by Robert P. P. McKone at Mon Aug 16 05:05:12 EDT 2021
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#17 by Robert P. P. McKone at Mon Aug 16 05:04:48 EDT 2021
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| MATHEMATICA
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a[0] := 0;
a[n_] := FromDigits[Sort[DeleteCases[IntegerDigits[a[n - 1] + n], 0]]];
Table[a[n], {n, 0, 59}] (* Robert P. P. McKone, Aug 16 2021 *)
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proposed
editing
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#16 by Ctibor O. Zizka at Sun Aug 15 06:40:20 EDT 2021
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#15 by Ctibor O. Zizka at Sun Aug 15 06:37:40 EDT 2021
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| COMMENTS
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a(9*k+0 or 8) is congruent to 0 (mod 9), a(9*k+1 or 4 or 7) is congruent to 1 (mod 9), a(9*k+2 or 6) is congruent to 3 (mod 9), a(9*k+3 or 5) is congruent to 6 (mod 9).
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approved
editing
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#14 by Bruno Berselli at Mon Jun 28 10:49:11 EDT 2021
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