A003226 -id:A003226 - cp's OEIS frontend

A054869 Digits of an idempotent 6-adic number.

3, 1, 2, 0, 5, 3, 1, 2, 2, 2, 5, 1, 5, 5, 1, 4, 1, 3, 1, 2, 5, 5, 5, 0, 5, 2, 5, 5, 5, 3, 1, 4, 3, 3, 0, 4, 2, 2, 4, 0, 1, 3, 3, 1, 4, 0, 2, 0, 1, 2, 5, 2, 4, 0, 2, 3, 3, 0, 3, 4, 5, 5, 2, 5, 5, 4, 3, 2, 3, 1, 5, 4, 5, 4, 0, 1, 1, 0, 4, 2, 0, 1, 3, 0, 1, 5, 0, 4, 3, 5, 0, 1, 0, 2, 4, 0, 3, 4, 2

Offset: 0

Paolo Dominici (pl.dm(AT)libero.it), May 23 2000

( a(0) + a(1)*6 + a(2)*6^2 + ... )^k = a(0) + a(1)*6 + a(2)*6^2 + ... for each k. Apart from 0 and 1, in base 6 there are only 2 numbers with this property. For the other see A055620.

V. deGuerre and R. A. Fairbairn, Automorphic numbers, J. Rec. Math., 1 (No. 3, 1968), 173-179.

Kenny Lau, Table of n, a(n) for n = 0..9999
V. deGuerre and R. A. Fairbairn, Automorphic numbers, Jnl. Rec. Math., 1 (No. 3, 1968), 173-179

The six examples given by deGuerre and Fairbairn are A055620, A054869, A018247, A018248, A259468, A259469.

n=10000;res=1-pow((3**n+1)//2,n,3**n)*2**n
for i in range(n):print(i,res%6);res//=6
# Kenny Lau, Jun 09 2018

a(n) == 3^(2^n) (mod 6^n). - Robert Dawson, Oct 28 2022

A201921 Automorphic numbers: n^2 ends with n in base 15 (written in base 10).

Original entry on oeis.org

0, 1, 6, 10, 100, 126, 1000, 2376, 4375, 46251, 156250, 603126, 3640626, 7750000, 19140625, 151718751, 835156251, 1727734375, 5960937501, 32482421875, 236621093751, 340029296875, 8413134765625, 60784912109376, 68961425781250, 709516601562501, 1236678466796875

Offset: 1

Martin Renner, Dec 06 2011

			a(3) = 6 = (6)_15 since 6^2 = 36 = (26)_15 ends with 6 in base 15.
a(4) = 10 = (A)_15 since 10^2 = 100 = (6A)_15 ends with A in base 15.
a(5) = 100 = (6A)_15 since 100^2 = 10000 = (2E6A)_15 ends with 6A in base 15.

Eric M. Schmidt, Table of n, a(n) for n = 1..1000

Cf. A003226, A201821, A201918, A201919, A201948, A237583.

# See A003226. - Eric M. Schmidt, Feb 09 2014

More terms from Eric M. Schmidt, Feb 09 2014

A201948 Automorphic numbers: n^2 ends with n in base 18 (written in base 10).

Original entry on oeis.org

0, 1, 9, 10, 81, 244, 729, 5104, 6561, 98416, 413344, 1476225, 9034497, 24977728, 263063296, 349156737, 2711943424, 8308017153, 96467701761, 101891588608, 1286623443969, 2283843782656, 30847581595648, 33420828483585, 352189631991808, 804641749434369

Offset: 1

Martin Renner, Dec 06 2011

			a(3) = 9 = (9)_18 since 9^2 = 81 = (49)_18 ends with 9 in base 18.
a(4) = 10 = (A)_18 since 10^2 = 100 = (5A)_18 ends with A in base 18.
a(5) = 81 = (49)_18 since 81^2 = 6561 = (1249)_18 ends with 49 in base 18.

Eric M. Schmidt, Table of n, a(n) for n = 1..1000

Cf. A003226, A201821, A201918, A201919, A201921, A237583.

# See A003226. - Eric M. Schmidt, Feb 09 2014

More terms from Eric M. Schmidt, Feb 09 2014

A306686 Values of n such that 9^n ends in n, or expomorphic numbers relative to "base" 9.

Original entry on oeis.org

9, 89, 289, 5289, 45289, 745289, 2745289, 92745289, 392745289, 7392745289, 97392745289, 597392745289, 7597392745289, 87597392745289, 8087597392745289, 48087597392745289, 748087597392745289, 10748087597392745289, 610748087597392745289, 5610748087597392745289

Offset: 1

Bernard Schott, Mar 05 2019

Definition: For positive integers b (as base) and n, the positive integer (allowing initial zeros) k(n) is expomorphic relative to base b (here 9) if k(n) has exactly n decimal digits and if b^k(n) == k(n) (mod 10^n) or, equivalently, b^k(n) ends in k(n). [See Crux Mathematicorum link.]
For sequences in the OEIS, no term is allowed to begin with a digit 0 (except for the 1-digit number 0 itself). However, in the problem as defined in the Crux Mathematicorum article, leading 0 digits are allowed; under that definition a(n) = k(n) until the first k(n) which begins with digit 0. When k(n) begins with 0, then, a(n) is the next term of the sequence k(n) which doesn't begin with digit 0.
Conjecture: if k(n) is expomorphic relative to "base" b, then, the next one in the sequence, k(n+1), consists of the last n+1 digits of b^k(n).
a(n) is the backward concatenation of A133619(0) through A133619(n-1). So, a(1) = 9, a(2) = 89 and so on, with recognition of the former comments about the OEIS and terms beginning with 0. - Davis Smith, Mar 07 2019

			9^9 = 387420489 ends in 9, so 9 is a term; 9^89 = .....289 ends in 89, so 89 is another term.

Davis Smith, Table of n, a(n) for n = 1..944
Charles W. Trigg, Problem 559, Crux Mathematicorum, page 192, Vol. 7, Jun. 81.
Emil Vaughan, Problem 226.8 - 999 nines, M500 Magazine of the Open University, number 226, February 2009, page 21; and Tony Forbes, Solution 226.8 - 999 nines, M500 Magazine of the Open University, number 232, February 2010, pages 8-9, calculating a(9) = 392745289.

Cf. A064541 (base 2), A183613 (base 3), A288845 (base 4), A306570 (base 5), A290788 (base 6), A321970 (base 7), A289138 (smallest expomorphic number in base n).
Cf. A003226 (automorphic numbers), A033819 (trimorphic numbers).
Cf. A133619 (leading digits).

tetrmod(b, n, m)=my(t=b); for(i=1, n, if(i>1, t=lift(Mod(b,m)^t), t)); t
for(n=1, 21,if(tetrmod(9,n,10^n)!=tetrmod(9,n-1,10^(n-1)),print1(tetrmod(9,n,10^(n-1)),", "))) \\ Davis Smith, Mar 09 2019

a(8)-a(20) from Davis Smith, Mar 07 2019

A035383 Automorphic numbers: n ends with square root of n.

Original entry on oeis.org

0, 1, 25, 36, 625, 5776, 141376, 390625, 87909376, 8212890625, 11963109376, 793212890625, 8355712890625, 50543227109376, 166168212890625, 7588043387109376, 45322418212890625, 619541169787109376

Offset: 1

Patrick De Geest, Nov 15 1998

Subsequence of A000290. - Georg Fischer, Sep 03 2020

R. A. Fairbairn, More on automorphic numbers, J. Rec. Math., 2 (No. 3, 1969), 170-174.
V. de Guerre and R. A. Fairbairn, Automorphic numbers, J. Rec. Math., 1 (No. 3, 1968), 173-179.

Eric Weisstein's World of Mathematics, Automorphic Numbers
Index entries for sequences related to automorphic numbers

Cf. A000290, A003226.

a(0)=0 added by Georg Fischer, Sep 03 2020

A062118 Numbers k such that k^2 has k as its middle digits.

Original entry on oeis.org

1, 50, 60, 250, 3792, 7600, 376000, 495475, 625000, 971582, 66952741, 93760000, 177656344, 3199268655, 9062500000, 10937600000, 788138178328, 860628177919, 890625000000, 2291665833333, 2780225311054, 2890625000000, 71093760000000, 128906250000000

Offset: 1

Brian Wallace (wallacebrianedward(AT)yahoo.co.uk), Jun 28 2001

Some of the terms are automorphic numbers (A003226) multiplied by an appropriate power of 10. a(25) > 10^15. - Giovanni Resta, Jul 29 2013

			a(5)=3792 because 3792^2 = 14379264 has 3792 as its middle digits.

Computed by Robert Israel.

k^2 is given in A062120.

Do[ If[ StringPosition[ ToString[n^2], ToString[n]] [[1, 1]] == (Ceiling[ Log[10, n^2] ] - Ceiling[ Log[10, n] ])/2 + 1, Print[n] ], {n, 1, 10^9} ]

Corrected and extended by Robert G. Wilson v, Aug 08 2001

a(15)-a(24) from Giovanni Resta, Jul 29 2013

A072495 Automorphic numbers: numbers k such that k^21 ends with k. Also m-morphic numbers for any m such that (m-1)/10 is an even integer not divisible by 10.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 16, 17, 19, 21, 23, 24, 25, 27, 28, 29, 31, 32, 33, 36, 37, 39, 41, 43, 44, 47, 48, 49, 51, 52, 53, 56, 57, 59, 61, 63, 64, 67, 68, 69, 71, 72, 73, 75, 76, 77, 79, 81, 83, 84, 87, 88, 89, 91, 92, 93, 96, 97, 99, 101, 107, 125, 143

Offset: 1

Benoit Cloitre, Oct 19 2002

Definition: k is an m-morphic number if k^m ends with k. For this sequence m can be 21, 41, 61, ...

3-morphic numbers = 7-morphic numbers, see A033819; 5-morphic numbers = 13-morphic numbers, see A068407.

Cf. A003226, A033819, A068407, A068408, A072496.

isok(n, m=21)={n == 0 || (n^m)%(10^(1+logint(n,10))) == n}

Missing terms inserted by Sean A. Irvine, Oct 05 2024

A113627 a(n) is the smallest number k such that k and 2^k have the same last n digits. Here k may have fewer than n digits and can be padded with leading zeros (cf. A121319).

Original entry on oeis.org

14, 36, 736, 8736, 48736, 948736, 2948736, 32948736, 432948736, 3432948736, 53432948736, 353432948736, 5353432948736, 75353432948736, 75353432948736, 5075353432948736, 15075353432948736, 615075353432948736, 8615075353432948736, 98615075353432948736, 98615075353432948736

Offset: 1

Jon E. Schoenfield, Apr 23 2007

			2^14 = 16384 and 14 end with the same single digit 4, thus a(1) = 14.

Jon E. Schoenfield, Table of n, a(n) for n = 1..81

See A121319, the main entry for this sequence, for further information.
Same as A109405 except for the initial term (14). - Max Alekseyev, May 11 2007
Cf. A007185, A016090, A003226, A035383, A064540, A064541, A109405, A121319, A206636.

A121319 a(n) is the smallest number k such that k and 2^k have the same last n digits. Here k must have at least n digits (cf. A113627).

Original entry on oeis.org

14, 36, 736, 8736, 48736, 948736, 2948736, 32948736, 432948736, 3432948736, 53432948736, 353432948736, 5353432948736, 75353432948736, 1075353432948736, 5075353432948736, 15075353432948736, 615075353432948736, 8615075353432948736, 98615075353432948736

Offset: 1

Tanya Khovanova, Aug 25 2006

			2^14 = 16384 and 14 end with the same single digit 4, thus a(1) = 14.

Jon E. Schoenfield, Table of n, a(n) for n = 1..80
Jon E. Schoenfield, Excel program

Cf. A007185, A016090, A003226, A035383, A064540, A064541, A109405.

f[n_] := Block[{k = If[n == 1, 2, 10], m = 10^n}, While[ PowerMod[2, k, m] != Mod[k, m], k += 2]; k]; Do[ Print@f@n, {n, 9}] (* Robert G. Wilson v *)

A121319(n) = { local(k,tn); tn=10^n ; forstep(k=2,1000000000,2, if ( k % tn == (2^k) % tn, return(k) ; ) ; ) ; return(0) ; } { for(n = 1,13, print( A121319(n)) ; ) ; } \\ R. J. Mathar, Aug 27 2006

If A109405(n) has n digits, a(n) = A109405(n), otherwise a(n) = A109405(n) + 10^n. - Max Alekseyev, May 05 2007

a(6)-a(9) from Robert G. Wilson v and Jon E. Schoenfield, Aug 26 2006
a(10) from Robert G. Wilson v, Sep 26 2006
a(11)-a(16) from Alexander Adamchuk, Jan 28 2007
a(16) corrected by Max Alekseyev, Apr 12 2007

A306570 Values of n such that 5^n ends in n, or expomorphic numbers relative to "base" 5.

Original entry on oeis.org

5, 25, 125, 3125, 203125, 8203125, 408203125, 8408203125, 18408203125, 618408203125, 2618408203125, 52618408203125, 152618408203125, 3152618408203125, 93152618408203125, 493152618408203125, 7493152618408203125, 17493152618408203125, 117493152618408203125, 7117493152618408203125, 87117493152618408203125

Offset: 1

Bernard Schott, Feb 24 2019

Definition: For positive integers b (as base) and n, the positive integer (allowing initial zeros) k(n) is expomorphic relative to base b (here 5) if k(n) has exactly n decimal digits and if b^k(n) == k(n) (mod 10^n) or, equivalently, b^k(n) ends in k(n). [See Crux Mathematicorum link.]
For sequences in the OEIS, no term is allowed to begin with a digit 0 (except for the 1-digit number 0 itself). However, in the problem as defined in the Crux Mathematicorum article, leading 0 digits are allowed; under that definition a(n) = k(n) until the first k(n) which begins with digit 0. When k(n) begins with 0, then, a(n) is the next term of the sequence k(n) which doesn't begin with digit 0.
Under that definition, the term after a(4) = 3125 is not "03125" but a(5) = 203125. [Comments from Jon E. Schoenfield in A288845 and discussion with Rémy Sigrist]
Conjecture: if k(n) is expomorphic relative to "base" b, then, the next one in the sequence, k(n+1), consists of the last n+1 digits of b^k(n).
a(n) is the backward concatenation of A133615(0) through A133615(n-1). So, a(1) is 5, a(2) is 25, and so on, with recognition of the comments about the OEIS and terms beginning with 0 (for example, when n = 5, A133615(n-1) = 0, so the next nonzero digit is concatenated as well, reducing the amount subtracted from n by 1). - Davis Smith Mar 07 2019

			5^5 = 25 ends in 5, so 5 is a term; 5^25 = ...125 ends in 25, so 25 is another term.

Davis Smith, Table of n, a(n) for n = 1..894
Charles W. Trigg, Problem 559, Crux Mathematicorum, page 192, Vol. 7, Jun. 81.

Cf. A064541 (base 2), A183613 (base 3), A288845 (base 4), A290788 (base 6), A321970 (base 7), A306686 (base 9), A289138 (smallest expomorphic number in base n).
Cf. A003226 (automorphic numbers), A033819 (trimorphic numbers).
Cf. A133615 (leading digits).

is(n) = my(t=#digits(n)); lift(Mod(5, 10^t)^n)==n
for(n=1, oo, my(x=n*5); if(lift(Mod(5, 10)^x)==x%10, if(is(x), print1(x, ", ")))) \\ Felix Fröhlich, Feb 24 2019

tetrmod(b,n,m)=my(t=b); for(i=1, n, if(i>1, t=lift(Mod(b,m)^t), t)); t
for(n=1, 21,if(tetrmod(5,n,10^n)!=tetrmod(5,n-1,10^(n-1)),print1(tetrmod(5,n,10^(n-1)),", "))) \\ Davis Smith, Mar 09 2019

a(5)-a(7) from Felix Fröhlich, Feb 24 2019
a(8) from Michel Marcus, Mar 02 2019
a(9)-a(21) from Davis Smith, Mar 07 2019

cp's OEIS Frontend

A054869 Digits of an idempotent 6-adic number.

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A201921 Automorphic numbers: n^2 ends with n in base 15 (written in base 10).

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A201948 Automorphic numbers: n^2 ends with n in base 18 (written in base 10).

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A306686 Values of n such that 9^n ends in n, or expomorphic numbers relative to "base" 9.

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A035383 Automorphic numbers: n ends with square root of n.

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A062118 Numbers k such that k^2 has k as its middle digits.

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A072495 Automorphic numbers: numbers k such that k^21 ends with k. Also m-morphic numbers for any m such that (m-1)/10 is an even integer not divisible by 10.

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A113627 a(n) is the smallest number k such that k and 2^k have the same last n digits. Here k may have fewer than n digits and can be padded with leading zeros (cf. A121319).

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A121319 a(n) is the smallest number k such that k and 2^k have the same last n digits. Here k must have at least n digits (cf. A113627).