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A257850 a(n) = floor(n/10) * (n mod 10).

%I A257850 #30 Sep 08 2022 08:46:12
%S A257850 0,0,0,0,0,0,0,0,0,0,0,1,2,3,4,5,6,7,8,9,0,2,4,6,8,10,12,14,16,18,0,3,
%T A257850 6,9,12,15,18,21,24,27,0,4,8,12,16,20,24,28,32,36,0,5,10,15,20,25,30,
%U A257850 35,40,45,0,6,12,18,24,30,36,42,48,54,0,7,14,21,28,35,42,49,56,63,0,8
%N A257850 a(n) = floor(n/10) * (n mod 10).
%C A257850 Equivalently, write n in base 10, multiply the last digit by the number with its last digit removed.
%C A257850 See A142150(n-1) for the base 2 analog and A257843 - A257849 for the base 3 - base 9 variants.
%C A257850 The first 100 terms coincide with those of A035930 (maximal product of any two numbers whose concatenation is n), A171765 (product of digits of n, or 0 for n<10), A257297 ((initial digit of n)*(n with initial digit removed)), but the sequence is of course different from each of these three.
%C A257850 The terms a(10) - a(100) also coincide with those of A007954 (product of decimal digits of n).
%H A257850 Colin Barker, <a href="/A257850/b257850.txt">Table of n, a(n) for n = 0..1000</a>
%H A257850 <a href="/index/Rec#order_20">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,-1).
%F A257850 a(n) = 2*a(n-10)-a(n-20). - _Colin Barker_, May 11 2015
%F A257850 G.f.: x^11*(9*x^8+8*x^7+7*x^6+6*x^5+5*x^4+4*x^3+3*x^2+2*x+1) / ((x-1)^2*(x+1)^2*(x^4-x^3+x^2-x+1)^2*(x^4+x^3+x^2+x+1)^2). - _Colin Barker_, May 11 2015
%t A257850 Table[Floor[n/10] Mod[n, 10], {n, 100}] (* _Vincenzo Librandi_, May 11 2015 *)
%o A257850 (PARI) a(n,b=10)=(n=divrem(n,b))[1]*n[2]
%o A257850 (Magma) [Floor(n/10)*(n mod 10): n in [0..100]]; // _Vincenzo Librandi_, May 11 2015
%o A257850 (Python) def A257850(n): return n//10*(n%10) # _M. F. Hasler_, Sep 01 2021
%Y A257850 Cf. A142150 (the base 2 analog), A115273, A257844 - A257849.
%Y A257850 Cf. also A007954, A035930, A171765, A257297.
%K A257850 nonn,base,easy
%O A257850 0,13
%A A257850 _M. F. Hasler_, May 10 2015