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A152950
a(n) = 3 + n*(n-1)/2.
17
3, 4, 6, 9, 13, 18, 24, 31, 39, 48, 58, 69, 81, 94, 108, 123, 139, 156, 174, 193, 213, 234, 256, 279, 303, 328, 354, 381, 409, 438, 468, 499, 531, 564, 598, 633, 669, 706, 744, 783, 823, 864, 906, 949, 993, 1038, 1084, 1131, 1179, 1228, 1278, 1329, 1381, 1434, 1488
OFFSET
1,1
COMMENTS
a(1)=3; then add 1 to the first number, then 2, 3, 4, ... and so on.
Numbers m such that 8*m - 23 is a square. - Bruce J. Nicholson, Jul 25 2017
Alternating sums of 3 consecutive terms of A152948 give this sequence. The a(n) are never divisible by primes listed in A191065. - Klaus Purath, Jan 22 2026
LINKS
Ângela Mestre and José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
FORMULA
a(n) = A152949(n+1) = 3 + A000217(n-1). - R. J. Mathar, Jan 03 2009
a(n) = 3 + C(n,2), n >= 1. - Zerinvary Lajos, Mar 12 2009
a(n) = a(n-1) + n - 1 (with a(1)=3). - Vincenzo Librandi, Nov 27 2010
Sum_{n>=1} 1/a(n) = 2*Pi*tanh(sqrt(23)*Pi/2)/sqrt(23). - Amiram Eldar, Dec 13 2022
From Elmo R. Oliveira, Nov 18 2024: (Start)
G.f.: x*(3 - 5*x + 3*x^2)/(1-x)^3.
E.g.f.: exp(x)*(3 + x^2/2) - 3.
a(n) = A027691(n-1)/2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)
a(n) = 2*a(n-1) - a(n-2) + 1 for n >= 3. - Klaus Purath, Jan 22 2026
MAPLE
A152950:=n->3 + n*(n-1)/2; seq(A152950(n), n=1..100); # Wesley Ivan Hurt, Jan 28 2014
MATHEMATICA
s=3; lst={3}; Do[s+=n; AppendTo[lst, s], {n, 1, 5!}]; lst
Table[3 + n*(n-1)/2, {n, 100}] (* Wesley Ivan Hurt, Jan 28 2014 *)
PROG
(SageMath) [3+binomial(n, 2) for n in range(1, 55)] # Zerinvary Lajos, Mar 12 2009
(PARI) a(n)=3+n*(n-1)/2 \\ Charles R Greathouse IV, Oct 07 2015
(Magma) [3+n*(n-1)/2 : n in [1..50]]; // Wesley Ivan Hurt, Mar 25 2020
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved