

A262183


a(0) = 0, a(n) = 10*a(n1) + n*(n+1)*(n+2)/6.


2



0, 1, 14, 150, 1520, 15235, 152406, 1524144, 15241560, 152415765, 1524157870, 15241578986, 152415790224, 1524157902695, 15241579027510, 152415790275780, 1524157902758616, 15241579027587129, 152415790275872430, 1524157902758725630, 15241579027587257840
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OFFSET

0,3


COMMENTS

This sequence may be viewed as a generalization of A014824 and appears to share similar properties. See also A052262. We make the following empirical observations.
The decimal expansion of a(4*n+1)^(1/4) begins with strings of repeated digits (giving the appearance of rationality) alternating with strings of apparently random digits. The strings of repeated digits gradually shorten in length until they disappear and this pattern breaks down. See example (1) below. Brown calls numbers with these properties schizophrenic or mockrational numbers. By increasing n we can prevent the disappearance of the repeating strings as long as we like. The repeating digits appear to always be an initial subsequence of [1, 3, 6, 7, 6, 3, 1, 9, 9, 3, 9, 9, 3, ....]. Cf. A060011.
It appears that the numbers (a(8*n+5))^(1/8) are also examples of Brown's schizophrenic numbers.
The decimal expansion of 1/(a(4*n+1))^(k/4) for k = 1,2,3,... has long strings of 0's (gradually shortening in length until they disappear) interlaced with blocks of digits. If we read these blocks of digits (for a fixed n and k) as ordinary integers and factorize them, we find the numbers are related in a surprising manner. See examples (2) through (5) below. Similar remarks apply to the decimal expansion of the numbers 1/(10*a(8*n))^(1/4), 1/(10*a(8*n+2))^(1/2), 1/(a(8*n+3))^(1/2), 1/(10*a(8*n+4))^(1/8), 1/(a(8*n+5))^(1/8), 1/(10*a(8*n+6))^(1/2), 1/(a(8*n+7))^(1/2) and their powers.
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 (a(8*n+1))^(1/4), (a(8*n+3))^(1/2), (a(8*n+5))^(1/8), (a(8*n+7))^(1/2),
(10*a(8*n))^(1/4), (10*a(8*n+2))^(1/2), (10*a(8*n+4))^(1/8), (10*a(8*n+6))^(1/2) and their powers.


LINKS

Colin Barker, Table of n, a(n) for n = 0..1000
K. S. Brown, Mockrational numbers.
Wikipedia, Schizophrenic Number
Index entries for linear recurrences with constant coefficients, signature (14,46,64,41,10).


FORMULA

a(n) = (1/9^4)*10^(n+3)  (243*n^3 + 1539*n^2 + 3096*n + 2000)/13122.
O.g.f. x/((1  10*x)*(1  x)^4).
a(n) = 14*a(n1)46*a(n2)+64*a(n3)41*a(n4)+10*a(n5) for n>4.  Colin Barker, Sep 20 2015


EXAMPLE

(1) The decimal expansion of a(61)^(1/4) (with the blocks of 'random' digits enclosed in parentheses to aid readability) begins
1.111...111(026286308)333...333(2361974965884332291)666...666(4936365745813146737399105902)777...777(414516002742700195101894168058610026041)666...666(5834699239217156417791785081497321498627522786458)333..333(1... * 10^15.
The repeating digits are 1, 3, 6, 7, 6 and 3, an initial subsequence of A060011.
(2) The decimal expansion of 1/a(61)^(1/4) (with now the strings of 0's enclosed in parentheses) begins
9.(000..000)6870809025(000...000)131133379605615140625(000...000)300330802691003816294298046875(000...000)74515840736091563874877683318366943359375(000...000)193416219724333545001418899430083738351541748046875(000...000)5... * 10^(16)
The long strings of 0's gradually shorten in length until they disappear and are interlaced with 5 strings of digits [6870809025, 131133379605615140625, 300330802691003816294298046875, 74515840736091563874877683318366943359375, 193416219724333545001418899430083738351541748046875]. Reading these strings as ordinary integers and factorizing we obtain [ (3^2)*(5^2)*30536929, (3^2)*(5^6)*(30536929)^2, (3^3)*(5^8)*(30536929)^3, (3^3)*(5^12)*13*(30536929)^4, (3^3)*(5^13)*13*17*(30536929)^3 ] showing how the numbers are related.
(3) The decimal expansion of 1/a(61)^(2/4) begins
8.1(000...000)12367456245(000...000)28324809994812870375(000...000)7207939264584091591063153125(000...000)192594788364052042015068473807471484375(000...000)52931280402387750233872466233174047490955859375(000...000)1... * 10^(31).
The long strings of 0's gradually shorten in length and are interlaced with 5 strings of digits
[12367456245, 28324809994812870375, 7207939264584091591063153125, 192594788364052042015068473807471484375, 52931280402387750233872466233174047490955859375].
Reading these strings as ordinary integers and factorizing we obtain [ (3^4)*5*30536929, (3^5)*(5^3)*(30536929)^2, (3^4)*(5^5)*(30536929)^3, (3^4)*(5^8)*7*(30536929)^4, (3^6)*(5^8)*7*(30536929)^3 ].
(4) The decimal expansion of 1/a(61)^(3/4) begins
7.29(000...000)1669606593075(000...000)44611575741830270840625(000...000)124877547758919386815169127890625(000...000)35750407590077160299047085450511894287109375(000...000)1... * 10^(46).
The long strings of 0's gradually shorten in length and are interlaced with 4 strings of digits [1669606593075, 44611575741830270840625, 124877547758919386815169127890625, 35750407590077160299047085450511894287109375]. Reading these strings as ordinary integers and factorizing we obtain [ (3^7)*(5^2)*30536929, 3^7)*(5^5)*7*(30536929)^2, (3^6)*(5^7)*7*11*(30536929)^3, (3^7)*(5^12)*7*11**(30536929)^4 ].
(5) The decimal expansion of 1/a(61) begins
6.561(000...000)200352791169(000...000)6118158958879580001(000...000)186829785738019654040356929(000...000)5705207902167118776034942675531041(000...000)174219528638716252198345946001761436313089(000...000)5320129376453944844526984070493622855630820563681(000...000)1... * 10^(61).
The long strings of 0's gradually shorten in length and are interlaced with 6 strings of digits [200352791169, 6118158958879580001, 186829785738019654040356929, 5705207902167118776034942675531041, 174219528638716252198345946001761436313089, 5320129376453944844526984070493622855630820563681]. Reading these strings as ordinary integers and factorizing we obtain [ (3^8)*30536929, (3^8)*(30536929)^2, (3^8)*(30536929)^3, (3^8)*(30536929)^4, (3^8)*(30536929)^5, (3^8)*(30536929)^6 ].


MAPLE

#A262183
seq((1/13122)*(2*10^(n+3)243*n^31539*n^23096*n2000), n = 0..22);


MATHEMATICA

Table[(1/9^4) 10^(n + 3)  (243 n^3 + 1539 n^2 + 3096 n + 2000)/13122, {n, 0, 30}] (* Vincenzo Librandi, Sep 20 2015 *)


PROG

(MAGMA) [0] cat [n eq 1 select 1 else 10*Self(n1) + n*(n+1)*(n+2)/6: n in [1..30]]; // Vincenzo Librandi, Sep 20 2015
(PARI) concat(0, Vec(x/((x1)^4*(10*x1)) + O(x^40))) \\ Colin Barker, Sep 20 2015


CROSSREFS

Cf. A060011, A014824, A030512, A052262.
Sequence in context: A009802 A153598 A180347 * A037960 A222677 A016163
Adjacent sequences: A262180 A262181 A262182 * A262184 A262185 A262186


KEYWORD

nonn,easy


AUTHOR

Peter Bala, Sep 14 2015


STATUS

approved



