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A034587
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Fibonacci iteration starting with (1, a(n)) leads to a "nine digits anagram".
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4
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718, 1790, 1993, 2061, 2259, 3888, 3960, 4004, 4396, 5093, 5832, 7031, 7310, 7712, 8039, 8955, 9236, 11598, 11742, 12312, 13295, 15095, 15432, 16044, 16355, 16472, 18109, 18559, 19144, 19950, 19968, 20116, 20180, 20494, 21170, 21376, 21998
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OFFSET
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1,1
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COMMENTS
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By "nine digits anagram" the author means a number whose digits are a permutation of {1, ..., 9}. These are more commonly known as restricted zeroless pandigital numbers and form the first 9! terms of A050289.
The largest term is a(750767) = 987654320.
More generally, the last N = 9! - 158323 = 204557 (> 56% of 9!) terms are given as A050289(k)-1 with indices k = 9!-N+1, ..., 9!. Indeed, a number > (987654321-1)/2 = 493827160 is a term if and only if it equals a "9-digit anagram" minus 1, since all results beyond the first iteration (1 + n = n+1) will be too large. Since 493827165 = A050289(158324) > 493827160, starting with a(546211) = 493827164 the terms are given by A050289(158324 .. 9!) - 1, for a total of 546211 + N - 1 = 750767 terms. (The term 493827164 is preceded by 493827160 (which yields 987654321 but is not in A050289 - 1) and 493827155 = A050289(158323) - 1.) - M. F. Hasler, Jan 07 2020
The ratio between consecutive terms in a Fibonacci sequence x(n+1) = x(n) + x(n-1) tends quickly to the golden ratio Phi = (sqrt(5)+1)/2 = A001622. We can tell whether a starting value N is in this sequence or not from the terms between 123456789 and 987654321 ~ 1e9. From N*Phi^k = 1e9 we get k = log(1e9/N)/log(Phi) ~ 43 - 2*log(N) for the maximum (and 3 less for the minimum) number of required iterations. - M. F. Hasler, Jan 06 2020
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LINKS
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FORMULA
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a(n) = A050289(m) with n = 387887 + m for 158324 <= m <= 9! or 546211 <= n <= 750767 = total number of terms in this sequence. - M. F. Hasler, Jan 07 2020
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EXAMPLE
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Denote by F(a,b) the Fibonacci-type sequence x(n+1) = x(n) + x(n-1) starting with x(0) = a, x(1) = b.
Then F(1,21998) = (1, 21998, 21999, 43997, 65996, 109993, 175989, 285982, 461971, 747953, 1209924, 1957877, 3167801, 5125678, 8293479, 13419157, 21712636, 35131793, 56844429, 91976222, 148820651, 240796873, 389617524, ...) where a nine-digits anagram has been reached.
The growth is roughly linear in three parts, with a slope of 700 up to a(292967) = 206993812, then an average slope of 1130 before it rises to (9.87e8 - 4.94e8)/2.05e5 ~ 2400 for 546211 <= n <= 750767 (cf. formula & comments): a(100) = 71960, a(200) = 149540, a(500) = 351868, a(1000) = 649921, a(2000) = 1400539, a(5000) = 3209798, a(10^4) = 6595301, a(2e4) = 13351498, a(5e4) = 32441506, a(10^5) = 67090523, a(2e5) = 134759627, a(3e5) = 214973567, a(4e5) = 327136594, a(5e5) = 439256717. - M. F. Hasler, Jan 07 2020
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PROG
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(PARI)
A034587=select( {is_A034587(n, s=1, L=[1..9])=while( 123456789 > n=s+s=n, ); n<1e9 && until( 987654321 < n=s+s=n, Set(digits(n))==L&&return(n))}, [1..22222]) \\ Function is_A034587 returns the 9-digit anagram if one is reached; null == false == 0 else.
nxt_A034587(n)={until(is_A034587(n+=1), ); n} \\ Returns next larger term
(Python)
def ok(n):
f, g = n, n+1
while g < 10**9:
if g > 123456788 and "".join(sorted(str(g))) == "123456789":
return True
f, g = g, f+g
return False
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CROSSREFS
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KEYWORD
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nonn,base,fini
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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