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A306979
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a(1) = 0; a(n) is the sum of powers of 2 where the exponents are the digits of a(n-1).
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1
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0, 1, 2, 4, 16, 66, 128, 262, 72, 132, 14, 18, 258, 292, 520, 37, 136, 74, 144, 34, 24, 20, 5, 32, 12, 6, 64, 80, 257, 164, 82, 260, 69, 576, 224, 24, 20, 5, 32, 12, 6, 64, 80, 257, 164, 82, 260, 69, 576, 224, 24, 20, 5, 32, 12, 6, 64, 80, 257, 164, 82, 260, 69, 576, 224
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OFFSET
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1,3
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COMMENTS
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Let f(n) be the sum of powers of 2 where the exponents are the digits of n.
If n is a d-digit number, then f(n) <= 512*n while f(n) >= 10^(n-1). So if d >= 5, f(n) is strictly smaller than n. If d <= 4, then f(n) <= 4*512 = 2048, so if 2048 < n < 10000, f(n) is strictly smaller than n. If n <= 2048, then f(n) <= 3*512 = 1536. So if 1536 < n <= 2048, f(n) is also strictly smaller than n. As a result, the terms of the iteration sequence {n, f(n), f(f(n)), ...} must eventually become no more than 1536, so the iteration sequence itself must be eventually periodic. Use a program we can see that the terms eventually fall into one of the four cycles:
- (148, 274) (length = 2);
- (98, 768, 448, 288, 516) (length = 5);
- (70, 129, 518, 290, 517, 162) (length = 6);
- (5, 32, 12, 6, 64, 80, 257, 164, 82, 260, 69, 576, 224, 24, 20) (length = 15).
For all n <= 1536, the iteration f(f(...(f(n))...)) falls into one of the four cycles after no more than 28 steps (n = 127, 172, 217, 271, 712, 721, 1117, 1171, 1266 requires exactly 28).
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,1).
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FORMULA
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a(n) = a(n-15) for n >= 36.
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EXAMPLE
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a(2) = 2^0 = 1; a(3) = 2^1 = 2; a(4) = 2^2 = 4; a(5) = 2^4 = 16; a(6) = 2^1 + 2^6 = 66; a(7) = 2^6 + 2^6 = 128 ...
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PROG
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(PARI) f(n) = if(n, my(d=digits(n)); sum(i=1, #d, 2^d[i]), 1)
a(n) = my(v=vector(n)); v[1] = 0; for(i=2, n, v[i]=f(v[i-1])); v[n]
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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STATUS
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approved
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