OFFSET
0,2
COMMENTS
Quasipolynomial of order 12. - Charles R Greathouse IV, Dec 03 2014
LINKS
Index entries for linear recurrences with constant coefficients, signature (1,0,1,0,-1,0,-1,1,0,0,0,2,-2,0,-2,0,2,0,2,-2,0,0,0,-1,1,0,1,0,-1,0,-1,1).
FORMULA
a(n) = binomial(n+1,4) - presumably quadratic (PORC) correction term which depends on n mod 24.
From Charles R Greathouse IV, Dec 03 2014: (Start)
n == 0 (mod 12): a(n) = (n^4 + 4*n^3 + 12*n^2 + 24*n + 24)/24
n == 1 (mod 12): a(n) = (n^4 + 4*n^3 + 12*n^2 + 20*n + 11)/24
n == 2 (mod 12): a(n) = (n^4 + 4*n^3 + 10*n^2 + 12*n + 8)/24
n == 3 (mod 12): a(n) = (n^4 + 4*n^3 + 8*n^2 + 8*n + 3)/24
n == 4 (mod 12): a(n) = (n^4 + 4*n^3 + 12*n^2 + 20*n + 8)/24
n == 5 (mod 12): a(n) = (n^4 + 4*n^3 + 10*n^2 + 12*n + 5)/24
n == 6 (mod 12): a(n) = (n^4 + 4*n^3 + 12*n^2 + 12*n )/24
n == 7 (mod 12): a(n) = (n^4 + 4*n^3 + 8*n^2 + 8*n + 3)/24
n == 8 (mod 12): a(n) = (n^4 + 4*n^3 + 10*n^2 + 12*n + 8)/24
n == 9 (mod 12): a(n) = (n^4 + 4*n^3 + 12*n^2 + 12*n + 3)/24
n == 10 (mod 12): a(n) = (n^4 + 4*n^3 + 12*n^2 + 8*n + 8)/24
n == 11 (mod 12): a(n) = (n^4 + 4*n^3 + 6*n^2 + 4*n + 1)/24
(End)
a(n) = (19958400*(n^4+4*n^3+12*n^2+24*n+24) - (1235*n^2+2*1127*n+215)*m^11 +(74987*n^2+2*69047*n+13541)*m^10 -(1983300*n^2+2*1844700*n+377520)*m^9 +(29983800*n^2+2*28201800*n+6115890)*m^8 - (285731655*n^2+2*272034411*n+63415275)*m^7 +(1784142591*n^2+2*1720539051*n+436295013)*m^6 -(7344548530*n^2+2*7175131810*n+1995595030)*m^5 +(19515989350*n^2+2*19301456350*n+5911801060)*m^4 -(31672473360*n^2+2*31658103312*n+10685562360)*m^3 +(27907182072*n^2+2*28127231352*n+10490664096)*m^2 -(9932634720*n^2+2*10110299040*n+4359398400)*m)/479001600 where m=n-12*floor(n/12). - Luce ETIENNE, Sep 27 2017
PROG
(PARI) a(n)=my(s); for(i=0, n, forstep(j=i%2, n, 2, forstep(k=j%3, n, 3, s+=(n-(k%4))\4+1))); s \\ naive; Charles R Greathouse IV, Dec 03 2014
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane and Vinay Vaishampayan, Mar 09 2006
STATUS
approved