A122061 First pentagonal number, 2nd hexagonal number, 3rd heptagonal number, 4th octagonal number and then repeat the same pattern: 5th pentagonal, 6th hexagonal, 7th heptagonal, 8th octagonal, etc.
1, 6, 18, 40, 35, 66, 112, 176, 117, 190, 286, 408, 247, 378, 540, 736, 425, 630, 874, 1160, 651, 946, 1288, 1680, 925, 1326, 1782, 2296, 1247, 1770, 2356, 3008, 1617, 2278, 3010, 3816, 2035, 2850, 3744, 4720, 2501, 3486, 4558, 5720, 3015, 4186, 5452
Offset: 1
Keywords
References
- A. Wareham, Test Your Brain Power, Ward Lock Ltd (1995).
Links
- Harvey P. Dale, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 3, 0, 0, 0, -3, 0, 0, 0, 1).
Crossrefs
Cf. A060354.
Programs
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Mathematica
fn[n_]:=Module[{r=Mod[n,4]},Which[r==1,(n(3n-1))/2,r==2,(n(4n-2))/2,r==3,(n(5n-3))/2,r==0,(n(6n-4))/2]]; Array[fn,50] (* or *) LinearRecurrence[ {0,0,0,3,0,0,0,-3,0,0,0,1},{1,6,18,40,35,66,112,176,117,190,286,408},50] (* Harvey P. Dale, Mar 01 2015 *) -
PARI
for(n=1,60,m=(n+3)%4;print1(n*((m+3)*n-m-1)/2,","))
Formula
a(n) = n*(3*n-1)/2 if n=1 mod 4 or n*(4*n-2)/2 if n=2 mod 4 or n*(5*n-3)/2 if n=3 mod 4 or n*(6*n-4)/2 if n=0 mod 4
a(n)=3*a(n-4)-3*a(n-8)+a(n-12) for n>11. - Harvey P. Dale, Mar 01 2015
Comments