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A010073
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a(n) = sum of base-6 digits of a(n-1) + sum of base-6 digits of a(n-2); a(0)=0, a(1)=1.
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11
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0, 1, 1, 2, 3, 5, 8, 8, 6, 4, 5, 9, 9, 8, 7, 5, 7, 7, 4, 6, 5, 6, 6, 2, 3, 5, 8, 8, 6, 4, 5, 9, 9, 8, 7, 5, 7, 7, 4, 6, 5, 6, 6, 2, 3, 5, 8, 8, 6, 4, 5, 9, 9, 8, 7, 5, 7, 7, 4, 6, 5, 6, 6, 2, 3, 5, 8, 8, 6, 4, 5, 9, 9, 8, 7, 5, 7, 7, 4, 6, 5, 6, 6
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OFFSET
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0,4
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COMMENTS
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The digital sum analog (in base 6) of the Fibonacci recurrence. - Hieronymus Fischer, Jun 27 2007
For general bases p > 2, we have the inequality 2 <= a(n) <= 2p-3 (for n > 2). Actually, a(n) <= 9 = A131319(6) for the base p=6. - Hieronymus Fischer, Jun 27 2007
a(n) and Fibonacci(n)=A000045(n) are congruent modulo 5 which implies that (a(n) mod 5) is equal to (Fibonacci(n) mod 5) = A082116(n) (for n > 0). Thus (a(n) mod 6) is periodic with the Pisano period A001175(5)=20. - Hieronymus Fischer, Jun 27 2007
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LINKS
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FORMULA
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a(n) = a(n-1) + a(n-2) - 5*(floor(a(n-1)/6) + floor(a(n-2)/6)). - Hieronymus Fischer, Jun 27 2007
a(n) = floor(a(n-1)/6) + floor(a(n-2)/6) + (a(n-1) mod 6) + (a(n-2) mod 6). - Hieronymus Fischer, Jun 27 2007
a(n) = Fibonacci(n) - 5*Sum_{k=2..n-1} Fibonacci(n-k+1)*floor(a(k)/6). - Hieronymus Fischer, Jun 27 2007
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MATHEMATICA
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nxt[{a_, b_, c_}]:={b, c, Total[IntegerDigits[c, 6]]+Total[ IntegerDigits[ b, 6]]}; Transpose[NestList[nxt, {0, 1, 1}, 90]][[1]] (* Harvey P. Dale, Oct 09 2014 *)
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PROG
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(Magma) [0] cat [n le 2 select 1 else Self(n-1)+Self(n-2)-5*((Self(n-1) div 6)+(Self(n-2) div 6)): n in [1..100]]; // Vincenzo Librandi, Jul 11 2015
(PARI) lista(nn) = {va = vector(nn); va[2] = 1; for (n=3, nn, va[n] = sumdigits(va[n-1], 6) + sumdigits(va[n-2], 6); ); va; } \\ Michel Marcus, Apr 24 2018
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CROSSREFS
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Cf. A000045, A010074, A010075, A010076, A010077, A131294, A131295, A131296, A131297, A131318, A131319, A131320.
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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