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

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

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

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

Cf. A000217, A060544.

Sequence in context: A046104 A116984 A229055 * A186429 A165393 A107651

Adjacent sequences: A061790 A061791 A061792 * A061794 A061795 A061796

Keyword

nonn,easy

Author

Jason Earls, Jun 22 2001

Extensions

Corrected by T. D. Noe, Oct 25 2006

Status

approved