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Array[5 (10^(2 # + 1)-1)/9 - 2*10^# &, 15, 0]

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allocated for M. F. Hasler

a(n) = 5*(10^(2*n+1)-1)/9 - 2*10^n.

3, 535, 55355, 5553555, 555535555, 55555355555, 5555553555555, 555555535555555, 55555555355555555, 5555555553555555555, 555555555535555555555, 55555555555355555555555, 5555555555553555555555555, 555555555555535555555555555, 55555555555555355555555555555, 5555555555555553555555555555555

0,1

<a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (111,-1110,1000).

a(n) = 5*A138148(n) + 3*10^n = A002279(2n+1) - 2*10^n.

G.f.: (3 + 202*x - 700*x^2)/((1 - x)(1 - 10*x)(1 - 100*x)).

a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 2.

A332153 := n -> 5*(10^(2*n+1)-1)/9-2*10^n;

Array[5 (10^(2 # + 1)-1)/9 - 2*10^# &, 15]

(PARI) apply( {A332153(n)=10^(n*2+1)\9*5-2*10^n}, [0..15])

(Python) def A332153(n): return 10**(n*2+1)//9*5-2*10**n

Cf. A002275 (repunits R_n = (10^n-1)/9), A002279 (5*R_n), A011557 (10^n).

Cf. A138148 (cyclops numbers with binary digits), A002113 (palindromes).

Cf. A332113 .. A332193 (variants with different repeated digit 1, ..., 9).

Cf. A332150 .. A332159 (variants with different middle digit 0, ..., 9).

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nonn,base,easy

M. F. Hasler, Feb 09 2020

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