login
a(n) = 25*n*(n + 1)/2 + 3.
2

%I #47 Dec 25 2024 21:00:19

%S 3,28,78,153,253,378,528,703,903,1128,1378,1653,1953,2278,2628,3003,

%T 3403,3828,4278,4753,5253,5778,6328,6903,7503,8128,8778,9453,10153,

%U 10878,11628,12403,13203,14028,14878,15753,16653,17578,18528,19503,20503,21528,22578,23653

%N a(n) = 25*n*(n + 1)/2 + 3.

%C "If m is a triangular number, then so are 9*m+1, 25*m+3 and 49*m+6. (Euler, 1775)", see Burton in References. Note that A060544 is the same as 9*m+1 when m is triangular and that 9*(m*(m+1)/2)+1 is another formula for it.

%C 9*m+1, 25*m+3 and 49*m+6 are special cases of the identity A000290(2*r + 1)*A000217(s) + A000217(r) = A000217((2*r + 1)*s + r). - _Bruno Berselli_, Mar 01 2018

%C Complementing the previous comment, with T(n) = A000217(n), 4*T(s)+1+s = T(2*s+1), 16*T(s)+3+2s = T(4*s+2) and 36*T(s)+6+3s = T(6*s+3) are special cases of the identity A000290(2*r)*T(s) + T(r) + r*s = T(2*r*s + r). - _Charlie Marion_, Mar 28 2018

%D D. M. Burton, Elementary Number Theory, Allyn and Bacon, Inc. Boston, MA, 1976, p. 17.

%H Harry J. Smith, <a href="/A061793/b061793.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F a(n) = 25*A000217(n) + 3 = A123296(n) + 3.

%F From _Elmo R. Oliveira_, Oct 23 2024: (Start)

%F G.f.: (3 + 19*x + 3*x^2)/(1 - x)^3.

%F E.g.f.: (3 + 25*x + 25*x^2/2)*exp(x).

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. (End)

%p [seq(25*(n*(n+1)/2)+3,n=0..40)]; # _Muniru A Asiru_, Mar 30 2018

%t 25*Accumulate[Range[0,40]]+3 (* _Harvey P. Dale_, Aug 26 2013 *)

%o (PARI) a(n) = 25*n*(n + 1)/2 + 3

%o (GAP) List([0..40],n->25*(n*(n+1)/2)+3); # _Muniru A Asiru_, Mar 30 2018

%Y Cf. A000217, A000290, A060544, A123296.

%K nonn,easy

%O 0,1

%A _Jason Earls_, Jun 22 2001

%E Corrected by _T. D. Noe_, Oct 25 2006