A277959 Numbers k such that 2 is the largest decimal digit of k^2.

11, 101, 110, 149, 1001, 1010, 1011, 1100, 1101, 1490, 10001, 10010, 10011, 10100, 10110, 11000, 11001, 11010, 14499, 14900, 100001, 100010, 100011, 100100, 100101, 100110, 101000, 101001, 101100, 110000, 110001, 110010, 110100, 144990, 149000, 316261

Offset: 1

Author

Colin Barker, Nov 06 2016

Keywords

• nonn
• base

Comments

The terms > 1 of A058411 can be considered as primitive elements of this sequence, obtained by multiplying those by powers of 10 (cf. formula). These terms of A058411 have at least 2 nonzero digits, and therefore their square has at least one digit 2. - M. F. Hasler, Nov 15 2017

Links

• M. F. Hasler, Table of n, a(n) for n = 1..1380 (first 50 terms from Colin Barker)
• OEIS Index to sequences related to squares.

Crossrefs

Cf. A277946 (the squares); A277960, A277961, A295005, ..., A295009 (analog for largest digit 3, 4, 5, ..., 9).

Cf. A058411, A058412 and A058413, ..., A058474. (Similar but no trailing 0's allowed.)

Cf. A136808 and A136809, ..., A137147 for other digit combinations. (Numbers must satisfy the same restriction as their squares.)

Programs

Mathematica

Select[Range[4*10^5], And[#[[2]] > 0, Union@ Take[RotateLeft[#, 2], 7] == {0}] &@ DigitCount[#^2] &] (* Michael De Vlieger, Nov 16 2017 *)

PARI

L=List(); for(n=1, 10000, if(vecmax(digits(n^2))==2, listput(L, n))); Vec(L)

PARI

A277959(LIM=1e15, L=List(), N=1)={while(LIM>N=next_A058411(N),my(t=N); until(LIMM. F. Hasler, Nov 15 2017

Formula

Equals (A058411 \ {1})*A011557, where A011557 = { 10^k; k >= 0 }. - M. F. Hasler, Nov 16 2017

Extensions

Edited by M. F. Hasler, Nov 16 2017