A061793 a(n) = 25*n*(n + 1)/2 + 3.

3, 28, 78, 153, 253, 378, 528, 703, 903, 1128, 1378, 1653, 1953, 2278, 2628, 3003, 3403, 3828, 4278, 4753, 5253, 5778, 6328, 6903, 7503, 8128, 8778, 9453, 10153, 10878, 11628, 12403, 13203, 14028, 14878, 15753, 16653, 17578, 18528, 19503, 20503, 21528, 22578, 23653

Offset: 0

Author

Jason Earls, Jun 22 2001

Keywords

• nonn
• easy

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).

Crossrefs

Cf. A000217, A000290, A060544, A123296.

Programs

GAP

List([0..40],n->25*(n*(n+1)/2)+3); # Muniru A Asiru, Mar 30 2018

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 *)

PARI

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

Formula

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

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)

Extensions

Corrected by T. D. Noe, Oct 25 2006