A102807 -id:A102807 - cp's OEIS frontend

A348589 a(n) = (10^n+2)^2 / 6.

24, 1734, 167334, 16673334, 1666733334, 166667333334, 16666673333334, 1666666733333334, 166666667333333334, 16666666673333333334, 1666666666733333333334, 166666666667333333333334, 16666666666673333333333334, 1666666666666733333333333334

Offset: 1

Bernard Schott, Oct 24 2021

Numbers q.r such that q.r = 3*q*r, when q and r have the same number of digits, "." means concatenation, r = 2q and r may not begin with 0.
We must solve the Diophantine equation q.r = q*10^m+r = 3 * q*r where m = length(q) = length(r).
The number of solutions is infinite with (r, q) = ((10^n+2)/3, (10^n+2)/6) and n >= 1.
Note that 15 satisfies also q.r = 3*q*r, 15 = 3*1*5 with here r = 5*q.
For further information about the general equation q.r = k * q*r, see A347541.
Problem proposed on the French website Diophante (see link).

			a(1) = 12^2 / 6 = 24 and 2.4 = 3 * 2*4.
a(2) = 102^2 / 6 = 1734 and 17.34 = 3 * 17*34.

Diophante, A1945 - Concaténations en tous genres (in French).
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A102807, A133384, A147553.

Subsequence of A347541.

Maple
```
seq((10^n+2)^2 / 6, n=1..14);
```

Table[(10^n + 2)^2/6, {n, 1, 14}] (* Amiram Eldar, Oct 24 2021 *)
LinearRecurrence[{111,-1110,1000},{24,1734,167334},20] (* Harvey P. Dale, Sep 05 2025 *)

def a(n): return (10**n+2)**2//6
print([a(n) for n in range(1, 15)]) # Michael S. Branicky, Oct 24 2021

a(n) = (10^n+2)^2 / 6.
a(n) = A133384(n-1)^2/6.
G.f.: 6*x*(4-155*x+250*x^2)/((1-x)*(1-10*x)*(1-100*x)). - Stefano Spezia, Oct 25 2021
a(n) = 3*A102807(n)/2. - Hugo Pfoertner, Oct 30 2021

A102832 Number of n-digit squares which contain the string "666" but not "6666".

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 3, 21, 78, 302, 1139, 4156, 14791, 52529, 183565, 635691, 2183533, 7314869, 25303217

Offset: 1

James R. Buddenhagen, Feb 27 2005

Cf. A052041, A075415, A102807 A032746.

fQ[n_] := StringPosition[ ToString[n^2], "666"] != {} && StringPosition[ ToString[n^2], "6666"] == {}; t = Table[0, {20}]; Do[ If[ fQ[n], t[[ Ceiling[ Log[10, n^2]] + 1]]++ ], {n, 3162277661}]; t (* Robert G. Wilson v, Mar 10 2005 *)
t6[digs_]:=Module[{a=Floor[Sqrt[10^(digs-1)]],b=Floor[ Sqrt[ 10^digs]]},Count[ Range[a,b]^2,?(SequenceCount[IntegerDigits[#],{6,6,6}]> 0&&SequenceCount[IntegerDigits[#],{6,6,6,6}]==0&)]]; Table[ t6[n],{n,20}] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale, Mar 26 2021 *)

a(13)-a(19) from Robert G. Wilson v, Mar 03 2005

cp's OEIS Frontend

A348589 a(n) = (10^n+2)^2 / 6.

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