A036498 Numbers of the form m*(6*m-1) and m*(6*m+1), where m is an integer.
0, 5, 7, 22, 26, 51, 57, 92, 100, 145, 155, 210, 222, 287, 301, 376, 392, 477, 495, 590, 610, 715, 737, 852, 876, 1001, 1027, 1162, 1190, 1335, 1365, 1520, 1552, 1717, 1751, 1926, 1962, 2147, 2185, 2380, 2420, 2625, 2667, 2882, 2926, 3151, 3197, 3432, 3480
Offset: 1
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).
Programs
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Magma
[1/8*(-1+(-1)^n+2*n)*(-3+(-1)^n+6*n): n in [1..50]]; // Vincenzo Librandi, Apr 24 2012
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Magma
/* By definition: */ A036498:=func
; [0] cat [A036498(n*m): m in [-1,1], n in [1..25]]; // Bruno Berselli, Nov 13 2012 -
Maple
p1 := n->n*(6*n-1): p2 := n->n*(6*n+1): S:={}: for n from 0 to 100 do S := S union {p1(n), p2(n)} od: S
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Mathematica
Table[ 1/8*(-1 + (-1)^k + 2*k)*(-3 + (-1)^k + 6*k), {k, 64} ] CoefficientList[Series[x*(5+2*x+5*x^2)/((1+x)^2*(1-x)^3),{x,0,50}],x] (* Vincenzo Librandi, Apr 24 2012 *) Rest[Flatten[{#(6#-1),#(6#+1)}&/@Range[0,30]]] (* or *) LinearRecurrence[ {1,2,-2,-1,1},{0,5,7,22,26},60] (* Harvey P. Dale, Aug 13 2012 *)
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PARI
\ps 5000; for(n=1,5000,if(polcoeff(eta(x),n,x)==1,print1(n,",")))
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PARI
concat(0, Vec(x^2*(5+2*x+5*x^2)/((1+x)^2*(1-x)^3) + O(x^100))) \\ Altug Alkan, Jan 19 2016
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Python
def A036498(n): return (n*(3*n-5)>>1)+1 if n&1 else n*(3*n-1)>>1 # Chai Wah Wu, Mar 25 2025
Formula
a(n) = n(n+1)/6 for n=0 or 5 (mod 6).
a(n) = 1/8*(-1+(-1)^n+2*n)*(-3+(-1)^n+6*n) (see MATHEMATICA code).
G.f.: x^2*(5+2*x+5*x^2)/((1+x)^2*(1-x)^3). - Colin Barker, Apr 02 2012
a(1)=0, a(2)=5, a(3)=7, a(4)=22, a(5)=26, a(n)=a(n-1)+2*a(n-2)- 2*a(n-3)- a(n-4)+a(n-5). - Harvey P. Dale, Aug 13 2012
Bisections: a(2*k+1) = A001318(4*k) = k*(1+6*k) = A049453(k), k >= 0; a(2*k) = A001318(4*k-1) = k*(-1+6*k) = A049452(k), k >= 1. - Wolfdieter Lang, Jan 18 2016
From Amiram Eldar, Feb 13 2024: (Start)
Sum_{n>=2} 1/a(n) = 6 - sqrt(3)*Pi.
Sum_{n>=2} (-1)^n/a(n) = 4*log(2) + 3*log(3) - 6. (End)
Extensions
Better description from Claude Lenormand (claude.lenormand(AT)free.fr), Feb 12 2001
Additional comments and more terms from James Sellers, Feb 14 2001
Comments