A101706 -id:A101706 - cp's OEIS frontend

A101704 Numbers n such that reversal(n)=2n/3.

0, 6534, 65934, 659934, 6599934, 65346534, 65999934, 653406534, 659999934, 6534006534, 6593465934, 6599999934, 65340006534, 65934065934, 65999999934, 653400006534, 653465346534, 659340065934, 659934659934, 659999999934, 6534000006534, 6534659346534, 6593400065934, 6599340659934, 6599999999934

Offset: 1

Farideh Firoozbakht, Dec 31 2004

If n=0 or n>1 then 66*(10^n-1) is in the sequence (the first five terms of this sequence are of this form) so this sequence is infinite. Let g(s,t,r) be (s.(0)(t))(r).s where dot between numbers means concatenation and "(m)(n)" means number of m's is n, for example g(2005,1,2)=20050200502005. It is interesting that, if n is in the sequence then all numbers of the form g(n,t,r) for nonnegative integers t and r are in the sequence, for example since 6534 is in the sequence so g(6534,1,2)=(6534.(0)(1))(2).6534=65340653406534 is in the sequence.
It seems that all similar sequences (sequences with the definition "numbers n such that reversal(n) =r*n for a fixed rational number r" ) have the same property (see A101705 and A101706). All sequences of the form 10^s*A002113 are in this category.
There are Fibonacci(floor((n-2)/2)) terms with n digits, n>1 (this is essentially A103609). - Ray Chandler, Oct 12 2017

			g(65934,3,4)=6593400065934000659340006593400065934 is in the sequence
because reversal(6593400065934000659340006593400065934)
= 4395600043956000439560004395600043956
=2/3*6593400065934000659340006593400065934.

Ray Chandler, Table of n, a(n) for n = 1..10000

Cf. A101703, A101705.

Do[If[FromDigits[Reverse[IntegerDigits[n]]] == 2/3*n, Print[n]], {n, 150000000}]

a(8)-a(25) from Max Alekseyev, Aug 18 2013

A101705 Numbers n such that n = 12*reversal(n).

Original entry on oeis.org

0, 540, 5940, 54540, 59940, 540540, 599940, 5400540, 5454540, 5945940, 5999940, 54000540, 54594540, 59405940, 59999940, 540000540, 540540540, 545454540, 545994540, 594005940, 594545940, 599459940, 599999940, 5400000540, 5405940540, 5454054540, 5459994540, 5940005940, 5945945940, 5994059940, 5999999940

Offset: 1

Farideh Firoozbakht, Jan 02 2005

60 divides all terms of the sequence. For all nonnegative integers m and n all numbers of the form f(m,n) = (100*(6*10^m - 1)+ 40)*(10^((m + 2)*n) - 1)/(10^(m + 2) - 1) are in the sequence, in fact f(m,n) = (5.(9)(m))(n).0 where dot between numbers means concatenation and "(r)(t)" means number of r's is t. f(m,1) = 100*(6*10^m - 1)+ 40 = 5.(9)(m).40; f(0,1) = 540, f(1,1) = 5940, f(2,1)=59940, etc. f(m,2) = 5.(9)(m).50(9)(m).40; f(0,2) = 54540, f(1,2) = 5945940, etc. Let g(s,t,r) = s*(10^((L+t)(1+r))-1)/(10^(L+t)-1) where L = number of digits of s. If s is in the sequence then all numbers of the form g(s,t,r) for nonnegative integers t and r are in the sequence (the function g is the same function that has been defined in the sequence A101704). If n and m are nonnegative integers then g(n,0,m) = (n)(m+1) for example g(13,0,3) = (13)(4) = 13131313.

			g(540,0,5)= (540)(6) = 540540540540540540 is in the sequence because reversal(540540540540540540) = 45045045045045045 and 12*45045045045045045 = 540540540540540540.

Ray Chandler, Table of n, a(n) for n = 1..10000

Cf. A101704, A101706, A001232, A008918.

Do[If[n == 12*FromDigits[Reverse[IntegerDigits[n]]], Print[n]], {n, 0, 6000000000, 60}]
Select[Range[0,6*10^9,60],#==12IntegerReverse[#]&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Apr 12 2017 *)

def A101705(n):
    if n == 1: return 0
    a = 1<Chai Wah Wu, Jul 23 2024

a(n) = 540*A057148. - Ray Chandler, Oct 09 2017

a(25)-a(31) from Max Alekseyev, Aug 18 2013

A102277 Numbers n such that n = 15*reversal(n).

Original entry on oeis.org

0, 65340, 659340, 6599340, 65999340, 653465340, 659999340, 6534065340, 6599999340, 65340065340, 65934659340, 65999999340, 653400065340, 659340659340, 659999999340, 6534000065340, 6534653465340, 6593400659340, 6599346599340, 6599999999340

Offset: 1

Farideh Firoozbakht, Jan 04 2005

30 divides all terms of the sequence. For all nonnegative integers m and n all numbers of the form f1(m,n) = 660(10^(m + 2) - 1)*(10^((m + 4)*n) - 1)/(10^(m + 4) - 1) are in the sequence, in fact f1(m,n) = (65.(9)(m).34)(n).0 where dot between numbers means concatenation and "(r)(t)" means number of r's is t. With this definition a(1) = 0 = f1(0,0), a(2) = 65340 = f1(0,1), a(3) = 659340 = f1(1,1), a(4) = 6599340 = f1(2,1), a(5) = 65999340 = f1(3,1), a(6) = 653465340 = f1(0,2), a(7) = 659999340 = f1(4,1), a(9) = 6599999340 = f1(5,1), etc. f1(m,1) = 660(10^(m + 2) - 1) = 65.(9)(m).340, f1(m,2) = 65.(9)(m).34.65.(9)(m).340, etc. Let g(s,t,r) = s*(10^((L+t)*(1+r))-1)/(10^(L+t)-1) where L = number of digits of s, in fact g(s,t,r) = (s.(0)(t))(r).s so the function g is the same function that has been defined in the sequence A101704. If s is in the sequence then all numbers of the form g(s,t,r) for nonnegative integers t and r are in the sequence. Next term is greater than 11*10^9. It seems that the eleven next terms are 65340065340, 65934659340, 65999999340, 653400065340, 659340659340 659999999340, 6534000065340, 6534653465340, 6593400659340, 6599346599340 and 6599999999340. Is it true that, all terms of this sequence are of the form g(f1(m,n),r,t)?

			g(65340,0,2)= (65340)(3) = 653406534065340 is in the sequence because reversal(653406534065340) = 43560435604356 = (1/15)*653406534065340.

Ray Chandler, Table of n, a(n) for n = 1..10000

Cf. A001232, A008918, A101704, A101705, A101706.

Do[If[n == 15*FromDigits[Reverse[IntegerDigits[n]]], Print[n]], {n, 0, 11000000000, 30}]

a(n) = 10*A101704(n) = 20*A101706(n). - Ray Chandler, Oct 09 2017

More terms from Ray Chandler, Oct 09 2017

A285040 Numbers n such that three-halves of n equals the reverse of n.

Original entry on oeis.org

4356, 43956, 439956, 4399956, 43564356, 43999956, 435604356, 439999956, 4356004356, 4395643956, 4399999956, 43560004356, 43956043956, 43999999956, 435600004356, 435643564356, 439560043956, 439956439956, 439999999956

Offset: 1

Harvey P. Dale, Apr 08 2017

There are Fibonacci(floor((n-2)/2)) terms with n digits (this is essentially A103609). - Ray Chandler, Oct 12 2017

			439956 times 3/2 equals 659934 which is the reverse of 439956.

David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Rev. ed. 1997), p. 158.

Ray Chandler, Table of n, a(n) for n = 1..10000

Cf. A101704, A101705, A101706, A001232, A008918.

Select[2 Range[10^7], 3(#/2) == FromDigits@ Reverse@ IntegerDigits@ # &] (* Giovanni Resta, Apr 08 2017 *)

isok(n) = 3*n/2 == fromdigits(Vecrev(digits(n))); \\ Michel Marcus, Apr 09 2017

Data corrected by Giovanni Resta, Apr 08 2017

cp's OEIS Frontend

A101704 Numbers n such that reversal(n)=2n/3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Extensions

A101705 Numbers n such that n = 12*reversal(n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Python

Formula

Extensions

A102277 Numbers n such that n = 15*reversal(n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A285040 Numbers n such that three-halves of n equals the reverse of n.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

Extensions