A262183 a(0) = 0, a(n) = 10*a(n-1) + n*(n+1)*(n+2)/6.

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

Offset: 0

Author

Peter Bala, Sep 14 2015

Keywords

• nonn
• easy

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 mock-rational 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.

Examples

			(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 ].
		

Links

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

Crossrefs

Cf. A060011, A014824, A030512, A052262.

Programs

Magma

[0] cat [n eq 1 select 1  else 10*Self(n-1) + n*(n+1)*(n+2)/6: n in [1..30]]; // Vincenzo Librandi, Sep 20 2015

Maple

#A262183
seq((1/13122)*(2*10^(n+3)-243*n^3-1539*n^2-3096*n-2000), 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 *)
nxt[{n_,a_}]:={n+1,10a+((n+1)(n+2)(n+3))/6}; NestList[nxt,{0,0},20][[All,2]] (* or *) LinearRecurrence[ {14,-46,64,-41,10},{0,1,14,150,1520},30] (* Harvey P. Dale, Feb 29 2020 *)

PARI

concat(0, Vec(-x/((x-1)^4*(10*x-1)) + O(x^40))) \\ Colin Barker, Sep 20 2015

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(n-1)-46*a(n-2)+64*a(n-3)-41*a(n-4)+10*a(n-5) for n>4. - Colin Barker, Sep 20 2015