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0, 1, 13, 136, 1370, 13715, 137171, 1371738, 13717416, 137174205, 1371742105, 13717421116, 137174211238, 1371742112471, 13717421124815, 137174211248270, 1371742112482836, 13717421124828513, 137174211248285301, 1371742112482853200, 13717421124828532210
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OFFSET
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0,3
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COMMENTS
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This sequence may be viewed as a generalization of A014824 and appears to share similar properties. See also A262183.
It follows from the explicit formula for a(n) given below that the decimal expansion of 1/a(6*n+4) for n >= 1 begins with long strings of 0's interlaced successively with the digits of the numbers 3^6*(1458*n^2 + 2727*n + 1270)^k for k = 0,1,2,.... The strings of 0's gradually shorten in length so this pattern eventually breaks down. For example, for n = 3 we have 1/a(22) = 0.000...000729000...00016455717000...000371454899841000...00083848514541108930001... with 729 = 3^6, 16455717 = (3^6)*22573, 371454899841 = (3^6)*(22573^2) and 8384851454110893 = (3^6)*(22573^3). Similar results can be found for the decimal expansions of the reciprocal of a(6*n+r) for r = 0,1,2,3 and 5.
These results should be helpful in explaining the following empirical observations concerning the decimal expansions of certain roots of the sequence terms and of their reciprocals. The decimal expansion of a(6*n+4)^(1/6) begins with strings of repeated digits (giving the appearance of rationality) alternating with strings of apparently random digits. The strings of repeating digits gradually shorten in length until they disappear from the expansion and the pattern breaks down. Brown calls numbers with these properties schizophrenic numbers or mock-rational numbers. The powers (a(6*n+4)^1/6 )^k, k = 2,3,... may also be examples of schizophrenic numbers. The decimal expansions of the numbers a(6*n+4)^(2/3) are particularly interesting as they seem to begin with repeating strings of the digits 123456790. See example (1) below.
The decimal expansion of 1/a(3*n+1)^(1/3) for n >= 5 starts with 4 long strings of 0's interlaced with 3 blocks of digits. If we read these blocks of digits as ordinary integers and factorize them, we find the numbers are related in a surprising manner. See example (2) below. Next in the expansion we find blocks of apparently random digits interlaced with strings of repeated digits. As n increases the number of strings of repeating digits increases. The repeating digits appear to always be an initial subsequence of [5, 851, 975308641, ....] independent of the value of n. Cf. A014824, A060011 and A262183.
The decimal expansions of the numbers 1/a(6*n+4)^(1/6) also show the mock-irrational behavior described by Brown.
A theorem of Kuzmin in the measure theory of continued fractions says that for a random real number alpha, the probability that some given partial quotient of alpha is equal to a positive integer k is given by 1/log(2)*( log(1 + 1/k) - log(1 + 1/(k+1)) ). Thus large partial quotients are the exception in continued fraction expansions. Empirically, we observe the presence of unexpectedly large partial quotients early in the continued fraction expansions of the numbers (10*a(6*n))^(1/3), (a(6*n+1))^(1/3), (100*a(6*n+2))^(1/3), (10*a(6*n+3))^(1/6), (a(6*n+4))^(1/6), (100*a(6*n+5))^(1/3), and their powers.
(End)
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REFERENCES
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A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
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LINKS
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FORMULA
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a(n) = 100*(10^n-1)/729 - 10*n/81 - binomial(n+1, 2)/9.
a(n) = 10*a(n-1) + binomial(n+1,2).
G.f. : x/((1-10x)(1-x)^3); a(n) = Sum_{k=0..n-1} binomial(n+1, k+2)9^k. - Paul Barry, Aug 24 2004
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EXAMPLE
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(1) Repeating digits in the decimal expansion of a(64)^(2/3) = 1.23456790 123456790 123456790 123456790 123456790 123456790 123456775774... * 10^42.
(2) 1/a(121)^(1/3) = 9.000 ... 0001826553000 ... 000741399080402000 ... 000(351 ... 301)555 ... 555(733 ... 556)851851851 ... 851851851(945 ... 936)308641975308641975308641975 ... 308641975308641975308641975(30864202 ...).
The three blocks of digits [9, 1826553, 741399080402] at the start of the decimal expansion shown above factorize as [3^2, 3*608851, 3*608851^2] showing they are related. These 3 blocks of digits are interlaced with long strings of zeros. There then follows 4 blocks of apparently random digits (enclosed in parentheses above for clarity) interlaced with 3 blocks of repeating digits. The repeating digits are 5, 851 and 308641975.
(End)
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MAPLE
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100*(10^n-1)/729 - 10*n/81 - binomial(n+1, 2)/9 ;
end proc:
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MATHEMATICA
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Table[(1/1458) (2 10^(n + 2) - 81 n^2 - 261 n - 200), {n, 0, 30}] (* Vincenzo Librandi, Sep 20 2015 *)
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PROG
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(PARI) a(n) = (100*((10^n)-1)/729) - (10*n/81) - binomial(n+1, 2)/9;
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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