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%I #39 Apr 16 2018 07:26:14
%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
%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.
%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) v=[]; for(n=0,100,v=concat(v,25*(n*(n+1)/2)+3)); v
%o (PARI) { for (n=0, 1000, write("b061793.txt", n, " ", 25*n*(n + 1)/2 + 3) ) } \\ _Harry J. Smith_, Jul 28 2009
%o (GAP) List([0..40],n->25*(n*(n+1)/2)+3); # _Muniru A Asiru_, Mar 30 2018
%Y Cf. A000217, A060544.
%K nonn,easy
%O 0,1
%A _Jason Earls_, Jun 22 2001
%E Corrected by _T. D. Noe_, Oct 25 2006
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