A317313 Multiples of 13 and odd numbers interleaved.

0, 1, 13, 3, 26, 5, 39, 7, 52, 9, 65, 11, 78, 13, 91, 15, 104, 17, 117, 19, 130, 21, 143, 23, 156, 25, 169, 27, 182, 29, 195, 31, 208, 33, 221, 35, 234, 37, 247, 39, 260, 41, 273, 43, 286, 45, 299, 47, 312, 49, 325, 51, 338, 53, 351, 55, 364, 57, 377, 59, 390, 61, 403, 63, 416, 65, 429, 67, 442, 69

Offset: 0

Author

Omar E. Pol, Jul 25 2018

Keywords

• nonn
• easy
• mult

Comments

Partial sums give the generalized 17-gonal numbers (A303305).

More generally, the partial sums of the sequence formed by the multiples of m and the odd numbers interleaved, give the generalized k-gonal numbers, with m >= 1 and k = m + 4.

a(n) is also the length of the n-th line segment of the rectangular spiral whose vertices are the generalized 17-gonal numbers.

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Crossrefs

Cf. A008595 and A005408 interleaved.

Column 13 of A195151.

Sequences whose partial sums give the generalized k-gonal numbers: A026741 (k=5), A001477 (k=6), zero together with A080512 (k=7), A022998 (k=8), A195140 (k=9), zero together with A165998 (k=10), A195159 (k=11), A195161 (k=12), A195312 (k=13), A195817 (k=14), A317311 (k=15), A317312 (k=16).

Cf. A303305.

Programs

Mathematica

Table[{13n, 2n + 1}, {n, 0, 35}] // Flatten (* or *)
CoefficientList[Series[(x^3 + 13 x^2 + x)/(x^2 - 1)^2, {x, 0, 69}], x] (* or *)
LinearRecurrence[{0, 2, 0, -1}, {0, 1, 13, 3}, 70] (* Robert G. Wilson v, Jul 26 2018 *)

PARI

a(n) = if(n%2==0, return((n/2)*13), return(n)) \\ Felix Fröhlich, Jul 26 2018

PARI

concat(0, Vec(x*(1 + 13*x + x^2) / ((1 - x)^2*(1 + x)^2) + O(x^60))) \\ Colin Barker, Jul 29 2018

Formula

a(2n) = 13*n, a(2n+1) = 2*n + 1.

From Colin Barker, Jul 29 2018: (Start)

G.f.: x*(1 + 13*x + x^2) / ((1 - x)^2*(1 + x)^2).

a(n) = 2*a(n-2) - a(n-4) for n>3. (End)

Multiplicative with a(2^e) = 13*2^(e-1), and a(p^e) = p^e for an odd prime p. - Amiram Eldar, Oct 14 2023

Dirichlet g.f.: zeta(s-1) * (1 + 11/2^s). - Amiram Eldar, Oct 25 2023

a(n) = (15 + 11*(-1)^n)*n/4. - Aaron J Grech, Aug 20 2024