cp's OEIS Frontend

A269025 a(n) = Sum_{k = 0..n} 60^k.

%I A269025 #28 Nov 16 2024 22:15:34
%S A269025 1,61,3661,219661,13179661,790779661,47446779661,2846806779661,
%T A269025 170808406779661,10248504406779661,614910264406779661,
%U A269025 36894615864406779661,2213676951864406779661,132820617111864406779661,7969237026711864406779661,478154221602711864406779661
%N A269025 a(n) = Sum_{k = 0..n} 60^k.
%C A269025 Partial sums of powers of 60 (A159991).
%C A269025 Converges in a 10-adic sense to ...762711864406779661.
%C A269025 More generally, the ordinary generating function for the Sum_{k = 0..n} m^k is 1/((1 - m*x)*(1 - x)). Also, Sum_{k = 0..n} m^k = (m^(n + 1) - 1)/(m - 1).
%H A269025 <a href="/index/Par#partial">Index entries related to partial sums</a>
%H A269025 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (61,-60).
%F A269025 G.f.: 1/((1 - 60*x)*(1 - x)).
%F A269025 a(n) = (60^(n + 1) - 1)/59 = 60^n + floor(60^n/59).
%F A269025 a(n+1) = 60*a(n) + 1, a(0)=1.
%F A269025 a(n) = Sum_{k = 0..n} A159991(k).
%F A269025 Sum_{n>=0} 1/a(n) = 1.016671221665660580331...
%F A269025 E.g.f.: exp(x)*(60*exp(59*x) - 1)/59. - _Stefano Spezia_, Mar 23 2023
%t A269025 Table[Sum[60^k, {k, 0, n}], {n, 0, 15}]
%t A269025 Table[(60^(n + 1) - 1)/59, {n, 0, 15}]
%t A269025 LinearRecurrence[{61, -60}, {1, 61}, 15]
%o A269025 (PARI) a(n)=60^n + 60^n\59 \\ _Charles R Greathouse IV_, Jul 26 2016
%Y A269025 Cf. A159991.
%Y A269025 Cf. similar sequences of the form (k^n-1)/(k-1): A000225 (k=2), A003462 (k=3), A002450 (k=4), A003463 (k=5), A003464 (k=6), A023000 (k=7), A023001 (k=8), A002452 (k=9), A002275 (k=10), A016123 (k=11), A016125 (k=12), A091030 (k=13), A135519 (k=14), A135518 (k=15), A131865 (k=16), A091045 (k=17), A218721 (k=18), A218722 (k=19), A064108 (k=20), A218724-A218734 (k=21..31), A132469 (k=32), A218736-A218753 (k=33..50), this sequence (k=60), A133853 (k=64), A094028 (k=100), A218723 (k=256), A261544 (k=1000).
%K A269025 nonn,easy
%O A269025 0,2
%A A269025 _Ilya Gutkovskiy_, Feb 18 2016