OFFSET
0,1
COMMENTS
"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.
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
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
REFERENCES
D. M. Burton, Elementary Number Theory, Allyn and Bacon, Inc. Boston, MA, 1976, p. 17.
LINKS
Harry J. Smith, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
From Elmo R. Oliveira, Oct 23 2024: (Start)
G.f.: (3 + 19*x + 3*x^2)/(1 - x)^3.
E.g.f.: (3 + 25*x + 25*x^2/2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. (End)
MAPLE
[seq(25*(n*(n+1)/2)+3, n=0..40)]; # Muniru A Asiru, Mar 30 2018
MATHEMATICA
25*Accumulate[Range[0, 40]]+3 (* Harvey P. Dale, Aug 26 2013 *)
PROG
(PARI) v=[]; for(n=0, 100, v=concat(v, 25*(n*(n+1)/2)+3)); v
(PARI) { for (n=0, 1000, write("b061793.txt", n, " ", 25*n*(n + 1)/2 + 3) ) } \\ Harry J. Smith, Jul 28 2009
(GAP) List([0..40], n->25*(n*(n+1)/2)+3); # Muniru A Asiru, Mar 30 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Jason Earls, Jun 22 2001
EXTENSIONS
Corrected by T. D. Noe, Oct 25 2006
STATUS
approved