A147875 - OEIS

A147875

Second heptagonal numbers: a(n) = n*(5*n+3)/2.

0, 4, 13, 27, 46, 70, 99, 133, 172, 216, 265, 319, 378, 442, 511, 585, 664, 748, 837, 931, 1030, 1134, 1243, 1357, 1476, 1600, 1729, 1863, 2002, 2146, 2295, 2449, 2608, 2772, 2941, 3115, 3294, 3478, 3667, 3861, 4060, 4264, 4473, 4687, 4906, 5130, 5359, 5593

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OFFSET

0,2

COMMENTS

Zero followed by partial sums of A016897.

Apparently = every 2nd term of A111710 and A085787.

Bisection of A085787. Sequence found by reading the line from 0, in the direction 0, 13, ... and the line from 4, in the direction 4, 27, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. - Omar E. Pol, Jul 18 2012

Numbers of the form m^2 + k*m*(m+1)/2: in this case is k=3. See also A254963. - Bruno Berselli, Feb 11 2015

LINKS

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Leo Tavares, Illustration: Sliced Hexagons

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

FORMULA

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

a(n) = Sum_{k=0..n-1} A016897(k).

a(n) - a(n-1) = 5*n -1. - Vincenzo Librandi, Nov 26 2010

G.f.: U(0) where U(k) = 1 + 2*(2*k+3)/(k + 2 - x*(k+2)^2*(k+3)/(x*(k+2)*(k+3) + (2*k+2)*(2*k+3)/U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 14 2012

E.g.f.: U(0) where U(k) = 1 + 2*(2*k+3)/(k + 2 - 2*x*(k+2)^2*(k+3)/(2*x*(k+2)*(k+3) + (2*k+2)^2*(2*k+3)/U(k+1))); (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Nov 14 2012

a(n) = A130520(5n+3). - Philippe Deléham, Mar 26 2013

a(n) = A131242(10n+7)/2. - Philippe Deléham, Mar 27 2013

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=4, a(2)=13. - Harvey P. Dale, May 15 2013

Sum_{n>=1} 1/a(n) = 10/9 + sqrt(1 - 2/sqrt(5))*Pi/3 - 5*log(5)/6 + sqrt(5)*log((1 + sqrt(5))/2)/3 = 0.4688420784500060750083432... . - Vaclav Kotesovec, Apr 27 2016

a(n) = A000217(n) + A000217(2*n). - Bruno Berselli, Jul 01 2016

From Ilya Gutkovskiy, Jul 01 2016: (Start)

E.g.f.: x*(8 + 5*x)*exp(x)/2.

Dirichlet g.f.: (5*zeta(s-2) + 3*zeta(s-1))/2. (End)

a(n) = A000566(-n) for all n in Z. - Michael Somos, Jan 25 2019

From Leo Tavares, Feb 14 2022: (Start)

a(n) = A003215(n) - A000217(n+1). See Sliced Hexagons illustration in links.

a(n) = A000096(n) + 2*A000290(n). (End)

EXAMPLE

G.f. = 4*x + 13*x^2 + 27*x^3 + 46*x^4 + 70*x^5 + 99*x^6 + 133*x^7 + ... - Michael Somos, Jan 25 2019

MATHEMATICA

Table[(n(5n+3))/2, {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 4, 13}, 50] (* Harvey P. Dale, May 15 2013 *)