A066825 -id:A066825 - cp's OEIS frontend

A014563 a(n+1) is the smallest number > a(n) such that the digits of a(n)^2 are all (with multiplicity) contained in the digits of a(n+1)^2, with a(0)=1.

1, 4, 13, 14, 31, 36, 54, 96, 113, 311, 487, 854, 1036, 1277, 1646, 3214, 8351, 10456, 11414, 11536, 11563, 17606, 17813, 30287, 36786, 41544, 54927, 56547, 56586, 57363, 62469, 62634, 72813, 72897, 76944, 78345, 95061, 97944, 100963, 101944

Offset: 0

Marc Paulhus, Jan 29 2002

Probably infinite. - David W. Wilson, Jan 29 2002

			13^2 = 169 and 14 is the next smallest number whose square (in this case 196) contains the digits 1,6,9.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

If "contained in" is replaced by "properly contained in" we get A065297.
Cf. A066825, A067633, A067634, A067635, A065297.
Cf. A000290, A000196.

import Data.List ((\\))
a014563 n = a014563_list !! n
a014563_list = 1 : f 1 (drop 2 a000290_list) where
   f x (q:qs) | null $ xs \\ (show q) = y : f y qs
              | otherwise             = f x qs
              where y = a000196 q; xs = show (x * x)
-- Reinhard Zumkeller, Nov 22 2012

snd[n_]:=Module[{k=n+1},While[!AllTrue[Select[Transpose[{DigitCount[n^2],
DigitCount[k^2]}],#[[1]]>0&],#[[1]]<=#[[2]]&],k++];k]; NestList[ snd,1,40] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Sep 21 2016 *)

A065297 a(n+1) is the smallest number > a(n) such that the digits of a(n)^2 are all (with multiplicity) properly contained in the digits of a(n+1)^2, with a(0)=1.

Original entry on oeis.org

1, 4, 13, 36, 113, 487, 1036, 3214, 10456, 36786, 100963, 319656, 1001964, 3165969, 10001786, 31626854, 100013919, 316256807, 1000029656, 3162322481, 10000115537

Offset: 0

Floor van Lamoen, Oct 29 2001

Probably infinite and at least O(10^(n/2)). - David W. Wilson

			13^2 = 169 and 36 is the next smallest number whose square (in this case 1296) properly contains the digits 1,6,9.

Cf. A050630, A050631, A048559, A050636, A065298, A014563, A066825.

Cf. A000290, A000196.

import Data.List ((\\), sort)
a065297 n = a065297_list !! n
a065297_list = 1 : f 1 (drop 2 a000290_list) where
   f x (q:qs) | null (xs \\ sq) && sort xs /= sort sq = y : f y qs
              | otherwise                             = f x qs
              where y = a000196 q; sq = show q; xs = show (x * x)
-- Reinhard Zumkeller, Nov 22 2012

More terms from Marc Paulhus, Jan 29 2002
More terms from David W. Wilson and Marc Paulhus, Feb 05 2002
a(19)-a(20) from Sean A. Irvine, Aug 26 2023

A067633 a(1) = 1; sequence of digits of a(n)^2 is a subsequence of the sequence of digits of a(n+1)^2.

Original entry on oeis.org

1, 4, 13, 37, 117, 367, 3667, 36667, 366667, 3666667, 36666667, 366666667, 3666666667, 36666666667, 366666666667, 3666666666667, 36666666666667, 366666666666667, 3666666666666667, 36666666666666667, 366666666666666667, 3666666666666666667, 36666666666666666667

Offset: 1

David W. Wilson, Feb 05 2002

Probably infinite. Does the obvious pattern continue?

The pattern continues until at least n=100 and is unlikely to change because insertion of 2 digits into a(n)^2 is sufficient to maintain the pattern. - Sean A. Irvine, Dec 28 2023

Sean A. Irvine, Table of n, a(n) for n = 1..100
Sean A. Irvine, Java program (github)

Cf. A014563, A066825, A068175.

a(n) = sqrt(A068175(n)). - Sean A. Irvine, Dec 28 2023

Revised with data from A091874 by Sean A. Irvine, Dec 28 2023

A067634 a(1) = 1; string of digits of a(n)^2 is a substring of the string of digits of a(n+1)^2.

Original entry on oeis.org

1, 4, 13, 130, 1300, 13000, 130000, 1300000, 13000000, 130000000, 1300000000, 13000000000, 130000000000, 1300000000000, 13000000000000, 130000000000000, 1300000000000000, 13000000000000000, 130000000000000000

Offset: 1

David W. Wilson, Feb 05 2002

Provably infinite.

The obvious pattern continues. Proof: By induction, assume that 13*10^k is the (k+2)nd element in the sequence for some k >= 1. Clearly 13*10^{k+1} satisfies the required condition; we need to show that no other number works. Equivalently, we need to show that 169*10^{2k+2} is the smallest square of one of the forms: 169*10^{2k+1}+a, a*10^{2k+3}+169*10^{2k}, 169*10^{2k+2}+a*10+b, a*10^{2k+4}+169*10^{2k+1}+b, a*10^{2k+4}+b*10^{2k+3}+169*10^{2k}, where 0 <= a,b <= 9. Insisting that the number be less than 169*10^{2k+2} and checking that it is a 2-adic, 3-adic and 5-adic square eliminates all but 169*10^{2k+1}+9 and 1169*10^{2k+1}+1. To eliminate these, reduce modulo the primes 101, 137=(10^4+1)/173 and 5882353=(10^8+1)/17; these all divide 10^16+1, so it suffices to check k=0,1,2,3,4,5,6,7. QED. - Eric Rains, Jan 29 2002

Cf. A014563, A066825.

Join[{1,4},NestList[10#&,13,20]] (* Harvey P. Dale, Jul 25 2024 *)

More terms from David W. Wilson, Feb 05 2002

A067635 a(1) = 1; set of digits of a(n)^2 is a proper subset of the set of digits of a(n+1)^2.

Original entry on oeis.org

1, 4, 13, 36, 113, 487, 1036, 3267, 10136, 32043

Offset: 1

David W. Wilson, Feb 05 2002

Cf. A014563, A066825.

A217368 Smallest number having a power that in decimal has exactly n copies of all ten digits.

Original entry on oeis.org

32043, 69636, 643905, 421359, 320127, 3976581, 47745831, 15763347, 31064268, 44626422, 248967789, 85810806, 458764971, 500282265, 2068553967, 711974055, 2652652791, 901992825, 175536645, 3048377607, 3322858521, 1427472867, 3730866429, 9793730157

Offset: 1

James G. Merickel, Oct 01 2012

The exponents that produce the number with a fixed number of copies of each digit are listed in sequence A217378. See there for further comments.
Since we allow A217378(n)=1, the sequence is well defined, with the upper bound a(n) <= 100...99 ~ 10^(10n-1) (n copies of each digit, sorted in increasing order, except for one "1" permuted to the first position). - M. F. Hasler, Oct 05 2012
What is the minimum value of a(n)? Can it be proved that a(n) > 2 for all n? - Charles R Greathouse IV, Oct 16 2012

			The third term raised to the fifth power (A217378(3)=5), 643905^5 = 110690152879433875483274690625, has three copies of each digit (in its decimal representation), and no number smaller than 643905 has a power with this feature.

Cf. A020666, A054038, A054039, A066825, A067635, A074205, A156977, A217378.

f[n_] := Block[{k = 2, t = Table[n, {10}], r = Range[0, 9]}, While[c = Count[ IntegerDigits[k^Floor[ Log[k, 10^(10 n)]]], #] & /@ r; c != t, k++]; k] (* Robert G. Wilson v, Nov 28 2012 *)

is(n,k)=my(v);for(e=ceil((10*n-1)*log(10)/log(k)), 10*n*log(10)/log(k), v=vecsort(digits(k^e)); for(i=1,9,if(v[i*n]!=i-1 || v[i*n+1]!=i, return(0))); return(1)); 0
a(n)=my(k=2); while(!is(n,k),k++); k \\ Charles R Greathouse IV, Oct 16 2012

a(13)-a(14) from James G. Merickel, Oct 06 2012 and Oct 08 2012
a(15)-a(16) from Charles R Greathouse IV, Oct 17 2012
a(17)-a(19) from Charles R Greathouse IV, Oct 18 2012
a(20) from Charles R Greathouse IV, Oct 22 2012
a(21)-a(24) from Giovanni Resta, May 05 2017

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A014563 a(n+1) is the smallest number > a(n) such that the digits of a(n)^2 are all (with multiplicity) contained in the digits of a(n+1)^2, with a(0)=1.

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A065297 a(n+1) is the smallest number > a(n) such that the digits of a(n)^2 are all (with multiplicity) properly contained in the digits of a(n+1)^2, with a(0)=1.

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A067633 a(1) = 1; sequence of digits of a(n)^2 is a subsequence of the sequence of digits of a(n+1)^2.

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A067634 a(1) = 1; string of digits of a(n)^2 is a substring of the string of digits of a(n+1)^2.

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A067635 a(1) = 1; set of digits of a(n)^2 is a proper subset of the set of digits of a(n+1)^2.

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A217368 Smallest number having a power that in decimal has exactly n copies of all ten digits.

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