OFFSET
1,2
COMMENTS
Form the infinite matrix:
1 2 4 7 11 ...
3 5 8 12 17 ...
6 9 13 18 24 ...
10 14 19 25 32 ...
15 20 26 33 41 ...
...
The diagonal elements are b(n) = 1, 5, 13, 25, 41, ... = 2*n*(n-1)+1 = A001844(n-1).
M(n,m) = ((n+m)^2-n-3*m+2)/2.
a(n) = M(n,b(n)) = M(1,1), M(2,5), M(3,13), M(4,25), M(5,41), ...
Let us define the PHI algebra as follows:
The basis of the PHI algebra is the PHI(1), PHI(2), PHI(3), ... elements, and the production rules are:
PHI(M(n,m))*PHI(n) = PHI(m) and every other production is zero.
An element of the PHI algebra is X = Sum_{n>=1} c(n)*PHI(n), where c(n) are real or complex constants.
UNIT = Sum_{n>=1} PHI(b(n)) = PHI(1) + PHI(5) + PHI(13) + PHI(25)+ ...
For every X elements: UNIT*X = X.
OMEGA = Sum_{n>=1} PHI(n) = PHI(1) + PHI(2) + PHI(3) + ...
ULTRA = Sum_{n>=1} PHI(a(n)) = PHI(1) + PHI(17) + PHI(108) + PHI(382) + ...
ULTRA*OMEGA = UNIT.
The PHI algebra is a nonassociative, but universal algebra; every finite or countable algebra can be modeled in the PHI algebra.
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
FORMULA
G.f.: (2*x^4 + 33*x^3 + 12*x^2 + x)/(1-x)^5.
E.g.f.: (1/2)*(4*x^4 + 20*x^3 + 15*x^2 + 2*x)*exp(x).
a(1)=1, a(2)=17, a(3)=108, a(4)=382, a(5)=995, a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - _Harvey P. Dale_, May 25 2012
MAPLE
seq(2*n^4-2*n^3-1/2*n^2+3/2*n, n=1..30); for n from 1 to 30 do b[n]:=2*n*(n-1)+1 od: seq(((n+b[n])^2-n-3*b[n]+2)/2, n=1..30);
MATHEMATICA
Table[2n^4-2n^3-n^2/2+(3n)/2, {n, 30}] (* or *) LinearRecurrence[ {5, -10, 10, -5, 1}, {1, 17, 108, 382, 995}, 30] (* _Harvey P. Dale_, May 25 2012 *)
PROG
(PARI) a(n)=n*(4*n^3-4*n^2-n+3)/2 \\ _Charles R Greathouse IV_, Oct 12 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
_Miklos Kristof_, Dec 11 2007
STATUS
approved