A181865 -id:A181865 - cp's OEIS frontend

A181754 a(1) = 1, a(2) = 2. For n >= 3, a(n) is found by concatenating the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).

1, 2, 6, 21, 601, 21001, 60100001, 2100100000001, 601000010000000000001, 2100100000001000000000000000000001, 6010000100000000000010000000000000000000000000000000001

Offset: 1

Peter Bala, Nov 09 2010

The calculations for the first few values of the sequence are
... a(3) = 12/2 = 6
... a(4) = 126/6 = 21
... a(5) = 12621/21 = 601
... a(6) = 12621601/601 = 21001.
Similar sequences may be formed by
1) starting with different initial values. See A181755 and A181756.
2) concatenating the k-th powers of the first n-1 terms of the sequence before dividing by a(n-1). See A181864, A181865 and A181866.
3) concatenating the k-th powers of the first n-1 terms of the sequence in reverse order before dividing by a(n-1). See A181867, A181868, A181869 and A181870.

Cf. A000045, A181755, A181756, A181864, A181865, A181866, A181867, A181868, A181869, A181870

#A181754
M:=11:
a:=array(1..M):s:=array(1..M):
a[1]:=1:a[2]:=2:
s[1]:=convert(a[1],string):
s[2]:=cat(s[1],convert(a[2],string)):
for n from 3 to M do
a[n] := parse(s[n-1])/a[n-1];
s[n]:= cat(s[n-1],convert(a[n],string));
end do:
seq(a[n],n = 1..M);

DEFINITION
a(1) = 1, a(2) = 2, and for n >= 3
(1)... a(n) = concatenate(a(1),a(2),...,a(n-1))/a(n-1).
RECURRENCE RELATION
For n >= 2
(2)... a(n+2) = 10^F(n)*a(n)+1,
where F(n) = A000045(n) are the Fibonacci numbers.
For n >= 2, a(n) has F(n-1) digits.

A181756 a(1) = 1, a(2) = 10. For n >= 3, a(n) is found by concatenating the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).

Original entry on oeis.org

1, 10, 11, 1001, 110001, 1001000001, 1100010000000001, 10010000010000000000000001, 110001000000000100000000000000000000000001, 10010000010000000000000001000000000000000000000000000000000000000001

Offset: 1

Peter Bala, Nov 09 2010

The calculations for the first few values of the sequence are
... a(3) = 110/10 = 11
... a(4) = 11011/11 = 1001
... a(5) = 110111001/1001 = 110001.
The above calculations are in base 10, but we get exactly the same results when working in an arbitrary base b. For example, with starting values a(1) = 1, a(2) = b (= 10 in base b), we get
... a(3) = (b^2+b)/b = b+1 which equals 11 in base b,
... a(4) = (b^4+b^3+b+1)/(b+1) = b^3+1 = 1001 in base b,
and so on.
For similarly defined sequences see A181754, A181755 and A181864 through A181870.

Cf. A000045, A181754, A181755, A181864, A181865, A181866, A181867, A181868, A181869, A181870

#A181756
M:=10:
a:=array(1..M):s:=array(1..M):
a[1]:=1:a[2]:=10:
s[1]:=convert(a[1],string):
s[2]:=cat(s[1],convert(a[2],string)):
for n from 3 to M do
a[n] := parse(s[n-1])/a[n-1];
s[n]:= cat(s[n-1],convert(a[n],string));
end do:
seq(a[n],n = 1..M);

DEFINITION
a(1) = 1, a(2) = 10, and for n >= 3
(1)... a(n) = concatenate(a(1),a(2),...,a(n-1))/a(n-1).
RECURRENCE RELATION
For n >= 2
(2)... a(n+2) = 100^F(n)*a(n)+1,
where F(n) = A000045(n) are the Fibonacci numbers.
For n >= 2, a(n) has 2*F(n-1) digits.

A181864 a(1) = 1, a(2) = 2. For n >= 3, a(n) is found by concatenating the squares of the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).

Original entry on oeis.org

1, 2, 7, 207, 700207, 207000000700207, 70020700000000000000207000000700207, 2070000007002070000000000000000000000000000000000070020700000000000000207000000700207

Offset: 1

Peter Bala, Nov 28 2010

The calculations for the first few values of the sequence are
... 2^2 = 4 so a(3) = 14/2 = 7
... 7^2 = 49 so a(4) = 1449/7 = 207
... 207^2 = 42849 so a(5) = 144942849/207 = 700207.
For similarly defined sequences see A181754 through A181756 and A181865 through
 A181870.

A000129, A113225, A181754, A181755, A181756, A181865, A181866,A181867, A181868, A181869, A181870

#A181864
M:=8: a:=array(1..M):s:=array(1..M):
a[1]:=1:a[2]:=2:
s[1]:=convert(a[1]^2,string):
s[2]:=cat(s[1],convert(a[2]^2,string)):
for n from 3 to M do
a[n] := parse(s[n-1])/a[n-1];
s[n]:= cat(s[n-1],convert(a[n]^2,string));
end do:
seq(a[n],n = 1..M);

DEFINITION
a(1) = 1, a(2) = 2, and for n >= 3
(1)... a(n) = concatenate(a(1)^2,a(2)^2,...,a(n-1)^2)/a(n-1).
RECURRENCE RELATION
For n >= 2
(2)...a(n+2) = a(n+1) + 10^F(n,2)*a(n) = a(n+1) + 10^Pell(n)*a(n),
where F(n,2) is the Fibonacci polynomial F(n,x) evaluated at x = 2
and where Pell(n) = A000129(n).
RELATION WITH OTHER SEQUENCES
a(n) has A113225(n-2) digits.
a(n)^2 has Pell(n-1) digits.

A181870 a(1) = 2, a(2) = 1. For n >= 3, a(n) is found by concatenating the cubes of the first n-1 terms of the sequence in reverse order and then dividing the resulting number by a(n-1).

Original entry on oeis.org

2, 1, 18, 32401, 1049824801000018, 110213211279472739469283600032400000000000000032401

Offset: 1

Peter Bala, Dec 09 2010

The calculations for the first few values of the sequence are
... 2^3 = 8 so a(3) = 18/1 = 18
... 18^3 = 5832 so a(4) = 583218/18 = 32401
... 32401^3 = 34015373377201 so a(5) = 34015373377201583218/32401 = 1049824801000018.
The value of a(7) is given in the Example section below. For similarly defined sequences see A181754 through A181756 and A181864 through A181869.

			a(7) = 12 14695 19405 33697 08585 73749 64887 08343 06977 30753 71161 37983 82152 46308 85158 49299 58480 00000 00001 04982 48010 00000 00000 00000 00000 00000 00000 00000 00000 00000 00001 04982 48010 00018 has 167 digits.

Cf. A006190, A052991, A181754, A181755, A181756, A181864, A181865, A181866, A181867, A181868, A181869

#A181870
M:=6: a:=array(1..M):s:=array(1..M):
a[1]:=2: a[2]:=1:
s[1]:=convert(a[1]^3,string): s[2]:=cat(convert(a[2]^3,string),s[1]):
for n from 3 to M do
a[n] := parse(s[n-1])/a[n-1];
s[n]:= cat(convert(a[n]^3,string),s[n-1]);
end do:
seq(a[n],n = 1..M);

DEFINITION
a(1) = 2, a(2) = 1, and for n >= 3,
(1)... a(n) = concatenate (a(n-1)^3,a(n-2)^3,...,a(1)^3)/a(n-1).
RECURRENCE RELATION
It appears that for n >= 2,
(2)... a(n+2) = 100^F(n-1,3)*a(n+1)^2 + a(n)
= 100^A006190(n-1)*a(n+1)^2 + a(n)
= 10^A052991(n-1)*a(n+1)^2 + a(n),
where F(n,3) is the n-th Fibonacci polynomial F(n,x) evaluated at x = 3.

A181755 a(1) = 1, a(2) = 5. For n >= 3, a(n) is found by concatenating the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).

Original entry on oeis.org

1, 5, 3, 51, 301, 51001, 30100001, 5100100000001, 301000010000000000001, 5100100000001000000000000000000001, 3010000100000000000010000000000000000000000000000000001

Offset: 1

Peter Bala, Nov 09 2010

The calculations for the first few values of the sequence are
... a(3) = 15/5 = 3
... a(4) = 153/3 = 51
... a(5) = 15351/51 = 301
... a(6) = 15351301/301 = 51001.
For similarly defined sequences see A181754, A184756 and A181864 through A181870.

Cf. A000045, A181754, A181756, A181864, A181865, A181866, A181867, A181868, A181869, A181870

#A181755
M:=11:
a:=array(1..M):s:=array(1..M):
a[1]:=1:a[2]:=5:
s[1]:=convert(a[1],string):
s[2]:=cat(s[1],convert(a[2],string)):
for n from 3 to M do
a[n] := parse(s[n-1])/a[n-1];
s[n]:= cat(s[n-1],convert(a[n],string));
end do:
seq(a[n],n = 1..M);

nxt[lst_]:=Module[{nt=FromDigits[Flatten[IntegerDigits/@lst]]/Last[ lst]},Flatten[{lst,nt}]]; Nest[nxt[#]&,{1,5},10] (* Harvey P. Dale, Aug 08 2014 *)

DEFINITION
a(1) = 1, a(2) = 5, and for n >= 3
(1)... a(n) = concatenate(a(1),a(2),...,a(n-1))/a(n-1).
RECURRENCE RELATION
For n >= 2
(2)... a(n+2) = 10^F(n)*a(n)+1,
where F(n) = A000045(n) are the Fibonacci numbers.
For n >= 2, a(n) has F(n-1) digits.

A181866 a(1) = 1, a(2) = 2. For n >= 3, a(n) is found by concatenating the fourth powers of the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).

Original entry on oeis.org

1, 2, 58, 200195112, 580000000008023436288643185139644928

Offset: 1

Peter Bala, Nov 29 2010

The calculations for the first few values of the sequence are
... 2^4 = 16 so a(3) = 116/2 = 58
... 58^4 = 11316496 so a(4) = 11611316496/58 = 200195112.
The value of a(6) is calculated in the Example section below.
For similarly defined sequences see A181754 through A181756 and
A181864 through A181870.

			The recurrence relation (2) above gives
a(6) = a(5)^3+10^144*a(4)
= 200 19511 20000 00000 00000 00000 00000 00000 00000 00195 11200
00080 97251 90261 07158 64917 75226 28886 69453 43420 83613 55167
37330 42401 44438 01550 47183 94579 01959 53586 66752.
a(6) has 153 digits.

Cf. A000045, A001076 (F(n,4)), A181754, A181755, A181756, A181864, A181865,

A181867, A181868, A181869, A181870

#A181866
M:=5:
a:=array(1..M):s:=array(1..M):
a[1]:=1:a[2]:=2:
s[1]:=convert(a[1]^4,string):
s[2]:=cat(s[1],convert(a[2]^4,string)):
for n from 3 to M do
a[n] := parse(s[n-1])/a[n-1];
s[n]:= cat(s[n-1],convert(a[n]^4,string));
end do:
seq(a[n],n = 1..M);

DEFINITION
a(1) = 1, a(2) = 2, and for n >= 3
(1)... a(n) = concatenate(a(1)^4,a(2)^4,...,a(n-1)^4)/a(n-1).
RECURRENCE RELATION
For n >= 2
(2)... a(n+2) = a(n+1)^3 + (100^F(n,4))*a(n)
= a(n+1)^3 + (10^F(3*n))*a(n),
where F(n,4) is the Fibonacci polynomial F(n,x) evaluated at x = 4
and where F(n) denotes the n-th Fibonacci number A000045(n).
F(n,4) = A001076(n).

A181867 a(1) = 2, a(2) = 1. For n >= 3, a(n) is found by concatenating the first n-1 terms of the sequence in reverse order and then dividing the resulting number by a(n-1).

Original entry on oeis.org

2, 1, 12, 101, 10012, 10000101, 1000000010012, 100000000000010000101, 1000000000000000000001000000010012, 1000000000000000000000000000000000100000000000010000101

Offset: 1

Peter Bala, Nov 28 2010

Compare with A181754. Here we concatenate the terms of the sequence in reverse order before dividing by a(n-1).
The calculations for the first few values of the sequence are
... a(3) = 12/1 = 12
... a(4) = 1212/12 = 101
... a(5) = 1011212/101 = 10012
... a(6) = 100121011212/10012 = 10000101.
For similarly defined sequences see A181864 through A181870.

A000045, A181754, A181755, A181756, A181864, A181865, A181866, A181868, A181869, A181870

#A181867
M:=10:
a:=array(1..M):s:=array(1..M):
a[1]:=2:a[2]:=1:
s[1]:=convert(a[1],string):
s[2]:=cat(convert(a[2],string),s[1]):
for n from 3 to M do
a[n] := parse(s[n-1])/a[n-1];
s[n]:= cat(convert(a[n],string),s[n-1]);
end do:
seq(a[n],n = 1..M);

DEFINITION
a(1) = 2, a(2) = 1, and for n >= 3
(1)... a(n) = concatenate(a(n-1),a(n-2),...,a(1))/a(n-1).
RECURRENCE RELATION
For n >= 2
(2)... a(n+2) = a(n) + 10^(F(n)-1),
where F(n) = A000045(n) are the Fibonacci numbers.
a(n) has F(n) digits.

A181869 a(1) = 2, a(2) = 1. For n >= 3, a(n) is found by concatenating the squares of the first n-1 terms of the sequence in reverse order and then dividing the resulting number by a(n-1).

Original entry on oeis.org

2, 1, 14, 1401, 140100014, 140100014000000001401, 14010001400000000140100000000000000000000140100014

Offset: 1

Peter Bala, Nov 29 2010

The calculations for the first few values of the sequence are
... 2^2 = 4 so a(3) = 14/1 = 14
... 14^2 = 196 so a(4) = 19614/14 = 1401
... 1401^2 = 1962801 so a(5) = 196280119614/1401 = 140100014
For similarly defined sequences see A181754 through A181756 and A181864 through A181870.

A000129, A024537, A181754, A181755, A181756, A181864, A181865, A181866, A181867, A181868, A181870

#A181869
M:=7:
a:=array(1..M):s:=array(1..M):
a[1]:=2:a[2]:=1:
s[1]:=convert(a[1]^2,string):
s[2]:=cat(convert(a[2]^2,string),s[1]):
for n from 3 to M do
a[n] := parse(s[n-1])/a[n-1];
s[n]:= cat(convert(a[n]^2,string),s[n-1]);
end do:
seq(a[n],n = 1..M);

DEFINITION
a(1) = 2, a(2) = 1, and for n >= 3
(1)... a(n) = concatenate (a(n-1)^2,a(n-2)^2,...,a(1)^2)/a(n-1).
RECURRENCE RELATION
For n >= 2,
(2)... a(n+2) = 10^F(n,2)*a(n+1) + a(n) = 10^Pell(n)*a(n+1) + a(n),
where F(n,2) is the n-th Fibonacci polynomial F(n,x) evaluated at
x = 2, and Pell(n) = A000129(n).
a(n) has A024537(n-2) digits.

cp's OEIS Frontend

A181754 a(1) = 1, a(2) = 2. For n >= 3, a(n) is found by concatenating the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).

Views

Author

Keywords

Comments

Crossrefs

Programs

Maple

Formula

A181756 a(1) = 1, a(2) = 10. For n >= 3, a(n) is found by concatenating the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).

Views

Author

Keywords

Comments

Crossrefs

Programs

Maple

Formula

A181864 a(1) = 1, a(2) = 2. For n >= 3, a(n) is found by concatenating the squares of the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).

Views

Author

Keywords

Comments

Crossrefs

Programs

Maple

Formula

A181870 a(1) = 2, a(2) = 1. For n >= 3, a(n) is found by concatenating the cubes of the first n-1 terms of the sequence in reverse order and then dividing the resulting number by a(n-1).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Formula

A181755 a(1) = 1, a(2) = 5. For n >= 3, a(n) is found by concatenating the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).

Views

Author

Keywords

Comments

Crossrefs

Programs

Maple

Mathematica

Formula

A181866 a(1) = 1, a(2) = 2. For n >= 3, a(n) is found by concatenating the fourth powers of the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Formula

A181867 a(1) = 2, a(2) = 1. For n >= 3, a(n) is found by concatenating the first n-1 terms of the sequence in reverse order and then dividing the resulting number by a(n-1).

Views

Author

Keywords

Comments

Crossrefs

Programs

Maple

Formula

A181869 a(1) = 2, a(2) = 1. For n >= 3, a(n) is found by concatenating the squares of the first n-1 terms of the sequence in reverse order and then dividing the resulting number by a(n-1).

Views

Author

Keywords

Comments

Crossrefs

Programs

Maple

Formula