A280641 - OEIS

%I #13 Dec 23 2024 14:53:45

%S 1,6,8,23,44,45,102,106,110,114,117,121,137,148,152,153,162,168,176,

%T 185,189,194,206,210,478,488,512,533,553,560,574,580,626,639,655,662,

%U 669,671,676,682,683,684,685,693,704,710,730,731,737,742,758,761,767,771

%N Numbers k such that k^3 has an odd number of digits and the middle digit is 1.

%C The sequence of cubes starts: 1, 216, 512, 12167, 85184, 91125, 1061208, 1191016, ...

%H Lars Blomberg, <a href="/A280641/b280641.txt">Table of n, a(n) for n = 1..10000</a>

%H Jeremy Gardiner, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2016-December/017135.html">Middle digit in cube numbers</a>, Seqfan Mailing list, Dec 12 2016.

%e 1^3 = (1), 114^3 = 148(1)544, 560^3 = 1756(1)6000

%t a[n_]:=Part[IntegerDigits[n], (Length[IntegerDigits[n]] + 1)/2];

%t Select[Range[0, 771], OddQ[Length[IntegerDigits[#^3]]] && a[#^3]==1 &] (* _Indranil Ghosh_, Mar 06 2017 *)

%o (PARI)

%o isok(k) = my(d=digits(k^3)); (#d%2 == 1) && (d[#d\2 +1] == 1);

%o for(k=0, 771, if(isok(k)==1, print1(k, ", "))); \\ _Indranil Ghosh_, Mar 06 2017

%o (Python)

%o i=0

%o j=1

%o while i<=771:

%o     n=str(i**3)

%o     l=len(n)

%o     if l%2 and n[(l-1)//2]=="1":

%o         print(str(i), end=', ')

%o         j+=1

%o     i+=1 # _Indranil Ghosh_, Mar 06 2017

%Y Cf. A280640, A280642-A280649, A181354.

%Y See A279420-A279429 for a k^2 version.

%Y See A279430-A279431 for a k^2 version in base 2.

%K nonn,base,easy

%O 1,2

%A _Lars Blomberg_, Jan 07 2017