cp's OEIS Frontend

A099150 Positive integers k such that f(k)+f(k)=concatenation of k and k, where f(k)=k(k+3)/2 (A000096).

%I A099150 #38 May 03 2025 09:40:21
%S A099150 8,98,998,9998,99998,999998,9999998,99999998,999999998,9999999998,
%T A099150 99999999998,999999999998,9999999999998,99999999999998,
%U A099150 999999999999998,9999999999999998,99999999999999998,999999999999999998,9999999999999999998,99999999999999999998
%N A099150 Positive integers k such that f(k)+f(k)=concatenation of k and k, where f(k)=k(k+3)/2 (A000096).
%C A099150 By the definition, k*(k+3) = k*10^m+k. So k+3 = 10^m+1, that is k = 10^m-2. - _Seiichi Manyama_, Aug 31 2019
%H A099150 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11,-10).
%F A099150 a(n) = A002283(n) - 1 = 10^n - 2. - _Seiichi Manyama_, Aug 31 2019
%F A099150 From _Chai Wah Wu_, Jun 15 2020: (Start)
%F A099150 a(n) = 11*a(n-1) - 10*a(n-2) for n > 2.
%F A099150 G.f.: x*(10*x + 8)/((x - 1)*(10*x - 1)). (End)
%F A099150 E.g.f.: 1 - 2*exp(x) + exp(10*x). - _Stefano Spezia_, May 02 2025
%F A099150 a(n) = 2*A198971(n-1) = A177096(n)/5. - _Elmo R. Oliveira_, May 02 2025
%e A099150 99998*(99998+3) = 9999899998 (concatenation of 99998 and 99998).
%o A099150 (PARI) for(k=1, 1e9, if(k*(k+3)==eval(Str(k, k)), print1(k", "))) \\ _Seiichi Manyama_, Aug 31 2019
%o A099150 (PARI) {a(n) = 10^n-2} \\ _Seiichi Manyama_, Aug 31 2019
%Y A099150 Cf. A000096, A002283, A096032, A099148, A099149, A177096, A198971.
%K A099150 nonn,base,easy
%O A099150 1,1
%A A099150 _John W. Layman_, Sep 30 2004
%E A099150 a(9)-a(20) from _Seiichi Manyama_, Aug 31 2019