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A301451
Numbers congruent to {1, 7} mod 9.
4
1, 7, 10, 16, 19, 25, 28, 34, 37, 43, 46, 52, 55, 61, 64, 70, 73, 79, 82, 88, 91, 97, 100, 106, 109, 115, 118, 124, 127, 133, 136, 142, 145, 151, 154, 160, 163, 169, 172, 178, 181, 187, 190, 196, 199, 205, 208, 214, 217, 223, 226, 232, 235, 241, 244, 250, 253, 259, 262, 268
OFFSET
1,2
COMMENTS
First bisection of A056991, second bisection of A242660.
The squares of the terms of A174396 are the squares of this sequence.
FORMULA
O.g.f.: x*(1 + 6*x + 2*x^2)/((1 + x)*(1 - x)^2).
E.g.f.: (3 + 8*exp(x) - 11*exp(2*x) + 18*x*exp(2*x))*exp(-x)/4.
a(n) = a(n-1) + a(n-2) - a(n-3).
a(n) = 2*(2*n - 1) + (2*n - 3*(1 - (-1)^n))/4. Therefore, for n even a(n) = (9*n - 4)/2, otherwise a(n) = (9*n - 7)/2.
a(2n+1) = A017173(n). a(2n) = A017245(n-1). - R. J. Mathar, Feb 28 2019
MATHEMATICA
Table[2 (2 n - 1) + (2 n - 3 (1 - (-1)^n))/4, {n, 1, 60}]
{#+1, #+7}&/@(9*Range[0, 30])//Flatten (* or *) LinearRecurrence[{1, 1, -1}, {1, 7, 10}, 60] (* Harvey P. Dale, Nov 08 2020 *)
PROG
(GAP) a := [1, 7, 10];; for n in [4..60] do a[n] := a[n-1] + a[n-2] - a[n-3]; od; a;
(Python) [2*(2*n-1)+(2*n-3*(1-(-1)**n))/4 for n in range(1, 70)]
(Sage) [n for n in (1..300) if n % 9 in (1, 7)]
(Magma) &cat [[9*n+1, 9*n+7]: n in [0..40]];
(PARI) Vec(x*(1 + 6*x + 2*x^2) / ((1 - x)^2*(1 + x)) + O(x^60)) \\ Colin Barker, Mar 22 2018
CROSSREFS
Cf. A274406: numbers congruent to {0, 8} mod 9.
Cf. A193910: numbers congruent to {2, 6} mod 9.
Sequence in context: A083390 A234093 A287567 * A033817 A286873 A218128
KEYWORD
nonn,easy
AUTHOR
Bruno Berselli, Mar 21 2018
STATUS
approved