A269132 a(n) = n + floor(n*(2*n+1)/5).

0, 1, 4, 7, 11, 16, 21, 28, 35, 43, 52, 61, 72, 83, 95, 108, 121, 136, 151, 167, 184, 201, 220, 239, 259, 280, 301, 324, 347, 371, 396, 421, 448, 475, 503, 532, 561, 592, 623, 655, 688, 721, 756, 791, 827, 864, 901, 940, 979, 1019, 1060, 1101, 1144, 1187, 1231

Offset: 0

Author

Mikk Heidemaa, Feb 19 2016

Keywords

• nonn
• easy

Comments

Differences between the two adjacent prime terms (i.e. between two primes in the blocks of length two) are divisible by 4 (checked up to n=10^8).

Links

• Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,1,-2,1).

Crossrefs

Cf. A139570.

Programs

Magma

[n + n*(2*n+1) div 5: n in [0..60]]; // Bruno Berselli, Mar 03 2016

Mathematica

Table[Floor[2 n (n + 3)/5], {n, 0, 1000}]
LinearRecurrence[{2, -1, 0, 0, 1, -2, 1}, {0, 1, 4, 7, 11, 16, 21}, 100]
CoefficientList[Series[(x^6 - x^5 - x^4 - 2 x^2-x)/((x - 1)^3 (x^4 + x^3 + x^2 + x + 1)), {x, 0, 100}], x]

PARI

a(n) = 2*n*(n+3)\5; \\ Michel Marcus, Mar 03 2016

Formula

G.f.: (x^6-x^5-x^4-2x^2-x)/((x-1)^3*(x^4+x^3+x^2+x+1)).

a(n) = ceiling(((n*(n+2)+(n+1)*ceiling(n))*(n-1))/ (2*ceiling(n)+ceiling(ceiling(n)+n)+n)+n) for n>0.

a(n) = floor(A139570(n)/5). - Michel Marcus, Mar 03 2016