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%I #13 Sep 20 2017 19:26:16
%S 1,15,129,547,1593,3711,7465,13539,22737,35983,54321,78915,111049,
%T 152127,203673,267331,344865,438159,549217,680163,833241,1010815,
%U 1215369,1449507,1715953,2017551,2357265,2738179,3163497,3636543
%N 1+2n+3n^2+4n^3+5n^4.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F a(n) =(A053700(n+1)-A053700(n-1))/2-10n^2-4n-2.
%F G.f.: -(3*x^4+42*x^3+64*x^2+10*x+1) / (x-1)^5. - _Colin Barker_, May 04 2013
%e For n>5 this is 54321 translated from base n to base 10
%t Join[{1},Table[Total[Table[i n^(i-1),{i,5}]],{n,30}]] (* or *) LinearRecurrence[{5,-10,10,-5,1},{1,15,129,547,1593},30] (* _Harvey P. Dale_, Sep 20 2017 *)
%o (PARI) a(n)=1+2*n+3*n^2+4*n^3+5*n^4 \\ _Charles R Greathouse IV_, Oct 07 2015
%Y Note: 1+2x+3x^2+4x^3+5x^4 is derivative of 1+x+x^2+x^3+x^4 +x^5, i.e. A053700. Cf. A000012, A005408, A056109, A056578.
%K easy,nonn
%O 0,2
%A _Henry Bottomley_, Jun 29 2000
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