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%I #19 Apr 15 2023 04:21:07
%S 1,4,11,14,21,24,31,34,41,44,51,54,61,64,71,74,81,84,91,94,101,104,
%T 111,114,121,124,131,134,141,144,151,154,161,164,171,174,181,184,191,
%U 194,201,204,211,214,221,224,231,234,241,244,251,254,261,264,271,274
%N Numbers that are congruent to {1,4} mod 10.
%C Numbers ending in 1 or 4, Union of A017281 and A017317.
%C a(n+3) gives the sum of 5 consecutive terms of A004442 starting at A004442(n) for n>0. (i.e., a(4) = 14 = 0+3+2+5+4 = Sum_{i=0..4} A004442(n+i)).
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F G.f.: x*(1+3*x+6*x^2)/((-1+x)^2*(1+x)).
%F a(n) = a(n-1) + a(n-2) - a(n-3) for n>3.
%F a(n) = 5*n - 5 - (-1)^n.
%F a(n) = -n + 2*A047241(n).
%F a(n+1) = n + 1 + 2*A042948(n).
%F Shifted bisections: a(2n+2) = A017317(n), a(2n+1) = A017281(n).
%F E.g.f.: 5*(x-1)*exp(x) - exp(-x). - _G. C. Greubel_, Apr 08 2016
%F Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(1+2/sqrt(5))*Pi/10 + log(phi)/sqrt(5) + log(2)/5, where phi is the golden ratio (A001622). - _Amiram Eldar_, Apr 15 2023
%p A271508:=n->5*n-5-(-1)^n: seq(A271508(n), n=1..100);
%t Table[5 n - 5 - (-1)^n, {n, 60}] (* or *)
%t Select[Range[0, 300], MemberQ[{1, 4}, Mod[#, 10]] &]
%o (Magma) [5*n-5-(-1)^n : n in [1..100]];
%o (PARI) my(x='x+O('x^99)); Vec(x*(1+3*x+6*x^2)/((-1+x)^2*(1+x))) \\ _Altug Alkan_, Apr 09 2016
%Y Cf. A001622, A004442, A017281, A017317, A042948, A047241.
%K nonn,easy
%O 1,2
%A _Wesley Ivan Hurt_, Apr 08 2016
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