A061588 -id:A061588 - cp's OEIS frontend

A082026 a(1)=5; a(n) is concatenation of squares of digits in a(n-1) (in base 10).

5, 25, 425, 16425, 13616425, 193613616425, 181936193613616425, 164181936181936193613616425, 13616164181936164181936181936193613616425, 19361361361616418193613616164181936164181936181936193613616425

Offset: 1

Matthew Vandermast, Apr 01 2003

If m is a member of the sequence represented by k digits in base 10, all members greater than m are congruent to m mod 10^k.

			The squares of the digits in the third term, 425, are 16, 4 and 25; hence a(4)=16425.

Paolo Xausa, Table of n, a(n) for n = 1..16

Cf. A061588.

NestList[FromDigits[Flatten[IntegerDigits[IntegerDigits[#]^2]]] &, 5, 10] (* Paolo Xausa, Jan 10 2025 *)

A308170 Start with the number 7, repeatedly square every digit in place to get a new number; in the limit this process converges (reading from right to left) to the string shown here.

Original entry on oeis.org

1, 6, 3, 9, 1, 8, 1, 4, 6, 1, 6, 1, 6, 3, 1, 6, 3, 1, 6, 3, 9, 1, 6, 3, 9, 1, 6, 3, 9, 1, 8, 1, 6, 3, 9, 1, 8, 1, 6, 3, 9, 1, 8, 1, 4, 6, 1, 6, 3, 9, 1, 8, 1, 4, 6, 1, 6, 3, 9, 1, 8, 1, 4, 6, 1, 6, 1, 6, 3, 1, 6, 3, 9, 1, 8, 1, 4, 6, 1, 6, 1, 6, 3, 1, 6, 3, 9

Offset: 1

N. J. A. Sloane, May 15 2019, following a suggestion from Jeremy Gardiner

If we start with 3 or 9, we get the same sequence. If instead we start with 2, 4, 6, or 8 we get the same sequence but without the initial 1. If we start with 5 we get A308171. [Corrected by M. F. Hasler, May 15 2019]
Comment from Jean-Paul Allouche, May 15 2019: These sequences can be obtained in the order shown (that is, in the right-to-left order) by starting with 7 (say), and repeatedly applying the morphism on the alphabet {1,2,3,...,9} defined by 1 -> 1, 3 -> 9, 4 -> 61, 5 -> 52, 6 -> 63, 7 -> 94, 8 -> 46, 9 -> 18.
If we start with 1, application of the morphism will never change that initial sequence of length 1: this is the third fixed point of the morphism. - M. F. Hasler, May 15 2019

			The successive numbers that arise are
7
49
1681
136641
193636161
1819369361361
1641819368193619361
...
and reading from the right we see 1,6,3,9, ...

Rémy Sigrist, Table of n, a(n) for n = 1..25000

Cf. A048385, A061588, A082027, A308171.

{ wanted = 87; a = [7]; while (1, b = concat(apply(d -> if (d,digits(d^2),[0]),a)); if (#b > wanted, b = b[#b-wanted+1..#b]); if (a==b, break, a = b)); print (Vecrev(a)) } \\ Rémy Sigrist, May 15 2019

A308170_vec(N,a=[9])={while(a!=a=concat(apply(t->digits(t^2),if(#a>N,a[-N..-1],a))),); Vecrev(a[-N..-1])} \\ M. F. Hasler, May 15 2019

More terms from Rémy Sigrist, May 15 2019

A082027 a(1)=6; a(n) is concatenation of the squares of each digit of a(n-1), in order (in base 10).

Original entry on oeis.org

6, 36, 936, 81936, 64181936, 3616164181936, 93613613616164181936, 819361936193613613616164181936, 641819361819361819361936193613613616164181936

Offset: 1

Matthew Vandermast, Apr 01 2003

If m is a member of the sequence consisting of k base-10 digits, every member of the sequence greater than m is congruent to m modulo 10^k.

			a(4)=81936 because the squares of the digits of a(3)--namely, 9, 3 and 6--are 81, 9 and 36 respectively.

Cf. A061588, A082026.

NestList[FromDigits[Flatten[IntegerDigits/@(IntegerDigits[#]^2)]]&,6,10] (* Harvey P. Dale, Feb 28 2013 *)

cp's OEIS Frontend

A082026 a(1)=5; a(n) is concatenation of squares of digits in a(n-1) (in base 10).

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A308170 Start with the number 7, repeatedly square every digit in place to get a new number; in the limit this process converges (reading from right to left) to the string shown here.

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A082027 a(1)=6; a(n) is concatenation of the squares of each digit of a(n-1), in order (in base 10).

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