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%I #41 Aug 03 2023 18:36:07
%S 1,4,7,16,25,52,79,160,241,484,727,1456,2185,4372,6559,13120,19681,
%T 39364,59047,118096,177145,354292,531439,1062880,1594321,3188644,
%U 4782967,9565936,14348905,28697812,43046719,86093440,129140161,258280324,387420487,774840976,1162261465
%N Partial sums of A162436.
%C Interleaving of A058481 and A100774 without initial term 0.
%C Apparently a(n) = A062318(n+2) - 1.
%C The terms beginning with a(2) are the row numbers in Pascal's Triangle where every 3rd element in those rows is divisible by 3 and none of the other elements in those rows are divisible by 3. - _Thomas M. Green_, Apr 03 2013
%D Thomas M. Green, Prime Patterns in Pascal's Triangle, paper in review process, 2013.
%H Vincenzo Librandi, <a href="/A164123/b164123.txt">Table of n, a(n) for n = 1..1000</a>
%H Barry Brent, <a href="https://doi.org/10.20944/preprints202306.1164.v6">On the Constant Terms of Certain Laurent Series</a>, Preprints (2023) 2023061164.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-3).
%F a(n) = A038754(n+1) - 2.
%F a(n) = 3*a(n-2) + 4 for n > 2; a(1) = 1, a(2) = 4.
%F a(n) = (5 - (-1)^n)*3^(1/4*(2*n - 1 + (-1)^n))/2 - 2.
%F G.f.: x*(1 + 3*x)/((1 - x)*(1 - 3*x^2)).
%F E.g.f.: 2*(cosh(sqrt(3)*x) - cosh(x)) + sqrt(3)*sinh(sqrt(3)*x) - 2*sinh(x). - _Stefano Spezia_, Dec 31 2022
%e For n = 3, a(3) = 7. The binomial coefficients of the 7th row of Pascal's Triangle are 1 7 21 35 35 21 7 1 and every 3rd element is a multiple of 3. - _Thomas M. Green_, Apr 03 2013
%t Accumulate[Transpose[NestList[{Last[#],3*First[#]}&,{1,3},40]][[1]]] (* _Harvey P. Dale_, Feb 17 2012 *)
%o (Magma) T:=[ n le 2 select 2*n-1 else 3*Self(n-2): n in [1..33] ]; [ n eq 1 select T[1] else Self(n-1)+T[n]: n in [1..#T]];
%o (PARI) a(n) = (2+n%2)*3^(n\2)-2 \\ _Charles R Greathouse IV_, Jul 15 2011
%Y Cf. A162436, A058481 (3^n-2), A100774 (2*(3^n - 1)), A062318, A038754, A038754.
%K nonn,easy
%O 1,2
%A _Klaus Brockhaus_, Aug 10 2009
%E Incorrect formula removed by _Stefano Spezia_, Dec 31 2022
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