A274406 Numbers m such that 9 divides m*(m + 1).

0, 8, 9, 17, 18, 26, 27, 35, 36, 44, 45, 53, 54, 62, 63, 71, 72, 80, 81, 89, 90, 98, 99, 107, 108, 116, 117, 125, 126, 134, 135, 143, 144, 152, 153, 161, 162, 170, 171, 179, 180, 188, 189, 197, 198, 206, 207, 215, 216, 224, 225, 233, 234, 242, 243, 251, 252, 260, 261, 269

Offset

1,2

Comments

Equivalently, numbers congruent to 0 or 8 mod 9.

Terms of A007494 with indices in A047264. Also, terms of A060464 with indices in A047335.

Links

Bruno Berselli, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Formula

G.f.: x^2*(8 + x)/((1 + x)*(1 - x)^2).

a(n) = (18*n + 7*(-1)^n - 11)/4. Therefore: a(2*m) = 9*m-1, a(2*m+1) = 9*m. It follows that a(j)+a(k) and a(j)*a(k) belong to the sequence if j and k are not both even.

a(n) = -A090570(-n+2).

a(n) = a(n-1) + a(n-2) - a(n-3).

a(2*r+1) + a(2*r+s+1) = a(4*r+s+1) and a(2*r) + a(2*r+2*s+1) = a(4*r+2*s). A particular case provided by these identities: a(n) = a(n - 2*floor(n/6)) + a(2*floor(n/6) + 1).

E.g.f.: 1 + ((9*x - 2)*cosh(x) + 9*(x - 1)*sinh(x))/2. - Stefano Spezia, Apr 24 2021

Mathematica

Select[Range[0, 300], Divisible[# (# + 1), 9] &]

Prog

(PARI) for(n=0, 300, if(n*(n+1)%9==0, print1(n", ")))
(Sage) [n for n in range(300) if 9.divides(n*(n+1))]
(Magma) [n: n in [0..300] | IsDivisibleBy(n*(n+1), 9)];

Crossrefs

Cf. A008591 (first bisection), A010689 (first differences), A017257 (second bisection).

Cf. similar sequences in which m*(m+1) is divisible by k: A014601 (k=4), A047208 (k=5), A007494 (k=3 and 6), A047335 (k=7), A047521 (k=8), this sequence (k=9).

Cf. A301451: numbers congruent to {1, 7} mod 9; A193910: numbers congruent to {2, 6} mod 9.

Sequence in context: A175053 A350075 A353057 * A261454 A022098 A129659

Adjacent sequences: A274403 A274404 A274405 * A274407 A274408 A274409

Keyword

nonn,easy

Author

Bruno Berselli, Jun 20 2016

Status

approved